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Why work is obtained from ordered molecular motion? I am trying to understand how entropy is actually a measure of randomness. The definition for Entropy is --> Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system.
I don't understand:
*
*what is ordered molecular motion
*why work comes from ordered molecular motion
*what is randomness in molecular terms
*is entropy applicable to quantum particles
| Heat is the transfer of energy that makes use of random motion (thermal motion). It is due to collision between molecules of bodies with different temperatures. It is random because linear momenta of the molecules are oriented at random, there's no preferred direction.
Work in thermodynamics is defined as in classical physics. $dw=\mathbf{F}d\mathbf{x}$. If you perform work on a system say a gas in a container with a piston above, by putting a weight on top of the piston. This is considered an ordered transfer of energy because the atoms of the weight move all in the same direction (contrary to heat).
Entropy is defined in terms of a variation $dS = q_{rev}/T$, rather than in absolute terms as you're text seems to suggest. When you'll study statistical thermodynamics you will see that entropy can also be defined as
$$S = K_b \ln W$$
where W can be be taken to be a quantitative measure of the disorder of the system. And you will apply this definition to systems composed of quantum particles.
See also https://gioretikto.github.io/chemistry/thermodynamics/heat.html
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If I pull a metal bar for long enough with a constant small force, will it eventually break? Let's say I have a strong metal bar. I pull it apart with a very small constant force -- obviously it doesn't break. However, this would disturb the internal configuration. If I let go, then eventually the internal configuration would return to what it was before I started pulling on the bar. However, if I keep pulling on the bar long enough, would the bar eventually break, no matter how small the force is?
| If in time no changes occur (whatever they are) in the elemental structure of the bar, then the bar will not break. If changes do occur, then there is a chance that the bar will break, so it eventually will break. How the breaking proceeds (breaking is a non-instantaneous process) is another story.
Taking quantum mechanics into account, there always is an accumulated chance of a large enough change, on top of the changes induced by interaction of the bar with the environment (the bar can rust or whatever).
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Is there an intuitive way to view the concept of momentum? Ideas like distance, velocity and force are very intuitive to understand because you can "see" their real-world applications and so one can come to understand them without having any knowledge of their mathematical formulas.
Momentum as it is defined is the product of mass times the velocity. I can see how it came to be mathematically derived via combining Newton's second law and the equation for acceleration (change in velocity over change in time).
However, besides just remembering that p=mv, is there an intuitive way to view what momentum is? Am I supposed to have an intuitive understanding of it? or is that just how every physics student thinks of it?
| If an object collides with you plastically its momentum is what knocks you back.
| {
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$PdV =VdP$ is true for any reversible cycles? My Question:
Does the following equation hold, at least for reversible cycles? If not, are there any special conditions that must be met for this to be true?
$$PdV=VdP$$
The reversible cycle can be drawn as a loop on the PV diagram. In one cycle, the work done by the cycle is the area of the region enclosed by the loop. This seems to be expressed in many thermodynamic textbooks as the following integral In other words, this may be regarded as a function of P as a function of V.
$$W=\oint P dV \tag{eq.1}$$
However, we can also consider V as a function of P, as shown in the figure below, which also seems to mathematically represent the area of the region enclosed by the loop.
$$\oint V dP = (4)- ((1)+(2)+(3))\tag{eq.2}$$
Fig.1
If I understand correctly, it seems to be mathematically possible to interpret eq1 as the integral around the differential form PdV, and eq2 as the integral around the differential form VdP.
In both cases, the loop of the line integral would be mathematically the same curve.
In other words.
$$PdV= VdP\tag{eq.3}$$
must always be true to say this. However, I have never seen this equation in my textbook. So I'm confused.
| What is true is that
$$\oint P \, dV = - \oint V \, dP$$
because of the area argument you made, noting that the signs are opposite. From this, you can't remove the integral sign to conclude that $P \, dV = -V \, dP$, which is certainly not true.
The integral result is easy to show in general, as
$$\oint P \, dV + \oint V \, dP = \oint d (PV)$$
and the net change of $PV$ along a cycle is zero, because it begins and ends in the same place. Of course, this argument also applies for literally any state function along any loop where state functions are always well-defined. (That includes irreversible cycles, but not cycles where you radically depart from equilibrium, like those involving free expansion.) The reason it's not emphasized in textbooks is probably just that it's not often useful.
| {
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Can an LC circuit be used to produce and receive FM signal? By simply changing the frequency of oscillation of an LC circuit (which you can do with the capacitor alone) you could emit (and receive) FM signal.
Are LC circuits actually used to do that? If not, why not?
| Yes, indeed, these are the basis of radio and television. Although the actual generators, modulators, and receivers are usually more complicated than a simple LC circuits, the LC is the basis of it (see, e.g., heterodyne, which is the simplest mixing device, used for the amplitude modulation.)
The field of radio wave generation, transmission, and reception has been so well established that it has been firmly divided into Radiophysics and Radioengineering, with (Radio)physicists often having only the basic knowledge about the relevant technology. It si exacerbated by the fact that most radio components are nowdays packed in a single chip, and thsu the popular hobby of assembling radio transmitters/receivers on one's tabletop has gone out of practice.
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Are soundproof foam shapes reflective of audio frequencies? I came across a previous question regarding how soundproof foam absorbs sound: How does foam "absorb" sound? where the answer explains the properties of the foam itself rather than the shape.
However, in analyzing audio spectrogram images (and messing around with Chrome Music Lab), I noticed a similarity in structure between acoustic foam and 3-dimensional spectrograms. The first two images are spectrograms and the latter two are two different acoustic foam designs.
It appears that, depending on the kinds of soundwaves (music recording, conversation, etc.), the foam would be designed to replicate similar shapes and fit together almost like a puzzle piece.
My question is: Are soundfoam shapes designed to sort of "mirror" the incoming soundwaves?
| I did a bit of search engine searching with the following search string
shape of sound absorbing material physics
In that search I did not encounter information about the shape of the sound absorbing material. The informative pages I found had information only about the properties of the material itself, such as elasticity, size of internal bubbles and what have you.
Hypothesis:
the purpose of the wedge shape is to increase surface area. The same material would already perform well when applied as flat panels, but the walls of a room have only so much area. The surface area of the wedge shapes is larger than the backing tile.
More surface area => more absorption.
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Why does closing someone's eyes not give them near-sightedness? When we close our eyes aren't we technically looking a really close up piece of skin? That is, our eyelids? If it's so close to our eyes why doesn't it give us bad vision? We know if screens or books or other objects constantly being close to us causes our vision to get worse, why doesn't the eyelid do the same thing? What's the difference?
| I think to answer your question you have to think about how the eye works. Light reflects off of an object and strikes the back of the eye to trigger the individual sensors that detect light. When you move your eyelid over your eye there is no light striking the light sensor in the back of your eye and the brain doesnt make any adjustments.
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Does universal speed limit of information contradict the ability of a particle to pick a trajectory using Principle of Least Action? I'm doing some self reading on Lagrangian Mechanics and Special Relavivity. The following are two statements that seem to be taken as absolute fundamentals and yet I'm unable to reconcile one with the other.
*
*Principle of Least action states that the particle's trajectory under the influence of a potential is determined by minimising the action that is $\delta S =0$.
*The universal speed limit of information propagation is $c$.
Question: Consider a potential that exists throughout space (like gravity perhaps). Now, I give a particle $x,p$ and set it motion. As per principle of least action the motion is now fully deterministic since the Lagrangian $L$ is known. We arrive at this claim by assuming that the particle is able to "calculate" the trajectory such that $\delta S = 0$ using Principle of Least Action.
This is where my confusion is. The first principle is a global statement whereas the second principle is a local statement.
What is the modern day physics answer to this?
I hope I could get my questions across! Look forward to your valuable opinions!
| The key to this is that the Lagrangian cannot be just any old function. It has to be a function such that, when the action is stationary, its solution describes the kinematics of the system.
Thus, if we assert that there is a universal speed limit in the real world, then that says something about the Lagrangians which can be used. They must be chosen such that the motion of the particle perscribed by solving the Lagrangian Mechanics problem is such that the path of any particle is the same, regardless of the state of the universe outside of its light-cone.
These are not mutually exclusive. From one perspective, you have a collection of every particle, and every particle obeys some rules. From another perspective, you have a global system whose time evolution ensures said rules for each particle are obeyed.
Perhaps obvious: Physicists who use Lagrangian Mechanics to explore relativistic systems do indeed use Lagrangians that have this property.
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Where is the potential energy? I read this question What is potential energy truly? and I find the answers not really satisfying.
When I move an object upward in a gravitational field, I have to work against that gravity. There is not really any energy mysteriously stored inside the object. Conversely, if the object is pushed over the edge it will fall because of that gravitational field and on impact it will exert the energy it obtained because of the acceleration.
If I move such a system into a zero gravity zone in space, then the object would still be "above" (even though "above" may not make much sense in zero gravity), but suddenly the potential energy is gone?
So isn't potential energy just an effect of gravitation and not something "hidden" inside an object?
| "If I move such a system into a zero gravity zone in space, then the object would still be "above" (even though "above" may not make much sense in zero gravity),but suddenly the potential energy is gone?"
Nope. The potential energy is now stored in the object.
If the object is in gravity it will " fall back" and the potential energy is converted to the kinetic energy while "falling back".
But if there is no gravity then the energy is still there in the object and will be released whenever finds any chance.
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Why don't evanescent waves give rise to electromagnetic waves? I'm reading about evanescent waves for the first time. I understand that even thought no electromagnetic wave is transmitted across the boundary, an electric field is transmitted which decays exponentially into the material.
As far as I understand this is still an oscillating electric field. Can anyone explain, therefore, why it doesn't generate a magnetic field and give rise to an electromagnetic wave?
Or is there an electromagnetic wave moving parallel to the boundary but people just talk about the E field?
| Indeed, an evanescent wave only has a wavevector real part parallel to the interface.
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Can pure qubit states be represented in a unit circle in $\mathbb R^2$? How does such representation relate with the Bloch sphere? In this video they are regarding 2D circles denoting real-valued states of qubit, like
Teacher says it can be extended to 3D to Bloch sphere.
But Bloch sphere has |0> at the top and |1> at the bottom, while 2D circle has |0> also at the top, while it has |1> at the right. How these can be corresponded?
| I think what they are doing is representing the (real part of) the complex vector space in which a (pure) qubit is embedded.
Remember that the state of a $d$-dimensional system, which we usually denote via a complex unit vector $|\psi\rangle\in\mathbb C^n$, is more precisely defined as an equivalence class of such vectors, where different unit vectors are identified when they differ by a global phase.
Formally, we call the corresponding set of equivalence classes a complex projective space $\mathbb{CP}^{n-1}$.
The Bloch representation (the Bloch sphere, for a single qubit) is a way to represent such complex projective space as a subset of $\mathbb R^{n^2-1}$.
It is worth stressing that this means that there is a one-to-one relation between states and representative points in the Bloch representation.
What they seem to be using in the video is a bit different. Rather then using the Bloch representation, they are directly representing (pure) states via the corresponding point in the embedding vector space. In other words, they represent $|\psi\rangle=\sum_i \psi_i |i\rangle$ as the vector $(\psi_1,...,\psi_n)\in\mathbb C^n$.
Restricting to $n=2$ (a single qubit) and considering only states with real coefficients, we thus have the set of states of the form
$$a |0\rangle + \sqrt{1-a^2} |1\rangle \simeq \begin{pmatrix}a\\ \sqrt{1-a^2}\end{pmatrix}\in\mathbb R^2, \qquad a\in[-1,1].$$
These are the vectors shown in the screenshot. Not that in this representation, $|0\rangle\to (1,0)$ and $|1\rangle\to (0,1)$, compatibly with the screenshot.
However, this representation has the severe shortcoming that different vectors correspond to the same state. For example, you might notice how $(0,1)$ and $(0,-1)$, which both represent the state $|1\rangle$, are assigned different points in the picture.
You can imagine every point in the Bloch representation as a direction in their "unit circle state machine" representation.
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What does it mean for the gravitational force to be "between" two bodies? What is the meaning of the word "between" in the law that the force between two masses at separation $r$ is given by $\frac{GM_1M_2}{r^2}$? I am confused about how can a force be in-between, either it is on body A or on body B, or on both.
Suppose body A exerts force $F$ on Body B, so according to Newton's 3rd law of motion B should also exert a force on A.
Let's consider this case for gravitational force between two bodies. If body A exerts force $g$ on Body B, then B body should also exert a force $g$ on A, but B is also exerting the gravitational force $X$ on A, hence A will also exert force $X$ on B.
So, how are two forces acting?
I have given the representation in this diagram.
|
**If A body Exerts force G on Body B,then B body should also exert A force G on A,
but B is also exerting the gravitational force X on A,hence A will
Also exert Force X on B **
Like you said, A experiences a force G towards B , given by $\frac{GM_1M_2}{r^2}$.
B experiences the same force towards A.
These are the 2 forces in this scenario.
Where does X come into the picture. There is no other force X.
In your diagram " the reaction force of G " and X are not two distinct forces. They are the same thing.
Similarly, in your diagram " the reaction force of X " and G are the same thing
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What is the formula to determine the change in pressure when there is a change in flow? [Updated to help clarify my question]
I have a current water flow of 9 GPM (gallons per minute) at 50 PSI through a 1/2 inch diameter pipe pouring out at the end.
I understand that if I reduce the flow at the end (exit) of the pipe, it will increase the pressure of the water exiting the pipe. So, if I add a fixture at the end of the pipe which reduces the out-coming water flow to 2 GPM what will my new exiting water PSI be? What is the formula to determine this?
| At a flow rate of 9 GPM translates into 1.2 CFM = 1.25 lb_m/sec. Water viscosity of 0.01 Poise = 0.00067 $lb_m/(ft-sec)$. So the Reynolds number is $$Re=\frac{4m}{\pi D \mu}=\frac{(4)(1.25)}{(3.14159)(1/24)(0.00067)}=57000$$That would give a Fanning friction factor of about f=0.005. Dividing the volume flow rate by the cross sectional area of the pipe gives and water velocity of v = 14.7 ft/sec.
So the shear stress at the wall would be $$\tau=\rho \frac{v^2}{2g_c}f=(62.4)\frac{14.7^2}{(2)(32.2)}(0.005)=1.04\ psf=0.0072 psi$$That would make the pressure gradient in the pipe $$\frac{\Delta p}{L}=\frac{4}{D}\tau=(4)(24)(0.0072)=0.70\ \frac{psi}{ft}$$
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Velocity is relative, which means acceleration is relative, which further implies that forces are relative as well So how would we know whether a force truly exists or not. I can be accelerating a car my 5 meters per second squared but another car accelerating with the same acceleration would think that my car is at rest relative to them. So is there any force on the car? Or are forces just relative and their existence just depends on our reference frame?
|
Velocity is relative, which means acceleration is relative
This is not correct. Velocity is relative, but (proper) acceleration is not relative. It is an invariant. Real forces lead to proper acceleration so the existence of real forces does not depend on the reference frame. In contrast, fictitious or inertial forces do not cause proper acceleration so their existence does depend on the reference frame. Reference frames where fictitious forces exist are called non-inertial reference frames.
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Why $U(c,b)U(b,a)=U(c,a)$ instead of $\int_a^c db U(c,b)U(b,a)=U(c,a)$? It's supposed to be a hw&exercise but the proof was in the textbook, in the purpose that one study it. However, somehow the proof in the book was confusing and felt wrong.
Consider a time evolution operator or time ordered product,
$$U(b,a)=T\{\exp [-i \int_a^b dt H(t)]\}\tag{1}$$
where $T$ was the time ordering operator.
The statement was that, since
$$i\frac{\partial U}{\partial t}=H(t)U(t,t_0)\tag{2}$$
... using boundary condition $$U(t_0,t_0)=1\tag{3}$$ with
$$i\frac{\partial U(t,t_0)U(t_0,t_2)}{\partial t}=H(t)U(t,t_0)U(t_0,t_2)\tag{4}$$
and, thus, $$U(t,t_0)U(t_0,t_2)=U(t,t_2)\tag{5}$$
But $t_0$ here was assumed to be a fixed value, an assumption that was ignored during the computation, in the sense that not all the path in temporal domain were counted for $U(t,t_0)U(t_0,t_2).$
After doing the algebra for time ordered product directly,
$$\int_a^cdb (T\{\exp [-i \int_b^c dt H(t)]\} T\{\exp [-i \int_a^b dt H(t)]\})=T\{\exp [-i \int_a^c dt H(t)]\}\tag{6}$$
i.e.
$$\int_a^c db U(c,b)U(b,a)=U(c,a)\tag{7}$$
Why $U(c,b)U(b,a)=U(c,a)$ instead of $\int_a^c db U(c,b)U(b,a)=U(c,a)$?
| Essentially because the group property (5) not (7) meet the correct boundary condition (3).
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Is there an alternative to radio waves that can go through metallic objects? Radar can pass through materials such as paper, wood, glass, brick, and concrete, but it reflects off of metal. Is there an alternative to radar that can pass through metal substances? If not, is it likely that we will ever find such a wave?
| For imaging, one alternative is neutrons. Neutrons tend to pass through heavy elements and are absorbed by light elements, so they can be used to image light elements that are present inside or behind heavy elements. You can see an example image here.
One disadvantage is that neutron radiation is harmful to living things and requires a nuclear reactor or a particle accelerator to create, so it's not practical for all uses.
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Can a photon be detected by a "lateral" detector? If I direct a laser pointer north and I put a photodetector eastwards (i.e. at $90^\circ$ ), and I wait for a very very long time (in a perfect vacuum if necessary), will the detector ever be triggered by a photon?
| Photons are elementary particles of the standard model. They have mass zero and E=hnu, where nu is the frequency of light the come from, the laser in this case.
As point particles generated by the lasing phenomenon they travel in straight lines because of conservation of energy and momentum. The laser beam will not be perfectly collimated, so there would be a statistical dispersion around the classical ray direction, but 90degrees would mean a very bad laser construction.
(p.s. path integrals cannot be used for paths of real, described by a real four vector , particle. They are a mathematical tool for calculating the probability distributions and the mathematical curves cannot be cut for a calculation that is meant to be going from A to B. )
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Kinematic behavior of a flat, a closed, an open universe
according to An Introduction to Modern Astrophysics, 1263p, there is explanation about evolution of scale factor.
“ For the early universe($R<1$) there is little difference among the kinematic behaviors of a flat, a closed, an open universe because the early universe was essentially flat.”
what does it mean? i don’t know relationship between ‘kinematic behaviors’ and ‘early universe flat’
add)
Help me please
|
For the early universe(R<1) there is little difference among the kinematic behaviors of a flat, a closed, and open universe because the early universe was essentially flat.
Indeed somewhat confusing, because if the early universe were essentially flat, how can it develop in a closed, flat, or open universe? Won't it stay flat forever?
f you imagine the universe to be a balloon (reducing the number of space dimensions to two), it is said that for every model (closed, flat , or open) the developing balloons are identical in the early universe (around $T=1$). I'm not so sure what is meant by essentially flat means that the balloons locally don't differ from a flat spacetime (though their overall form is that of a balloon). This is also the case for the three types of universe in the distant future.
So while the balloons don't differ much from each other now, they will differ from each other in the future.
If you take accelerated expansion seriously you can nothing but conclude that we are living in an open universe (the balloon will not only expand forever, but it will do so in an accelerated way (though it remains to be seen if the accelerated expansion will continue forever). Before the accelerated expansion started the universe looked locally flat.
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Why is there no kinetic term in the Hamiltonian of the Ising model? I am used to the Hamiltonian formalism in the context of (quantum) field theory, where as far as I can remember it always has the form of a kinetic term + a potential term. For me the absence of kinetic terms means a theory without dynamics. In Wikipedia the Hamiltonian of the Ising Model reads:
$$H(\sigma) = \sum_{i,j} J_{ij} \sigma_i \sigma_j + \sum_j h_j \sigma_j\,, \tag{1}$$
where the first term corresponds to interactions and where the sum runs over nearest neighbors, so I suppose $i \neq j$. The second term corresponds to an external potential.
Why is there no kinetic term? Can the system evolve in time, e.g. by seeing the 2d Ising Model as a (Euclidean) (1+1)d model? How should I picture this Hamiltonian as a system in my head?
| For what its worth, recall that a kinetic term in the Hamiltonian for a rigid body in classical mechanics is $H=\sum_{i=1}^3\frac{L_i^2}{2I_i}$.
Or more abstractly: A kinetic term of the form $H=\sum_{i=1}^n\frac{p_i^2}{2m_i}$.
In that sense a kinetic term for the Ising model would be $\propto \sum_i\sigma^2_i$, which is just a constant that can be dropped, since the spin is $\sigma_i\in\{\pm 1\}$.
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What is the entropy change of the universe for a rock if it falls from a height into a lake? The rock and the lake are at the same temperature According to my textbook, the entropy change of the universe is $+mgh/T$. I'm confused about why this happens.
after falling (without air resistance), wouldn't the rock possess $K_E = mgh$, which would then be transferred to the lake in the form of heat. Wouldn't this mean that the lake absorbs the same heat energy ($mgh$) from the rock to bring it to a standstill.
Would this not result in change in entropy of universe being $= 0$?
| The rock dropping all by itself (without friction) is a completely classical (non-thermodynamic) process, so the stone alone dropping causes no entropy change. The change in entropy comes from excatly the heat absorption process you describe.
At first, the stone has some potential energy. During falling, this is converted into kinetic energy. By hitting the water, the kinetic energy of the rock is then converted to heat and absorbed by the whole system of lake + stone. This heat absorption is what causes the entropy change. For constant $T$, we have
$$\Delta S = \frac{\Delta Q_\text{rev}}{T}$$
and in our case $\Delta Q_\text{rev} = mgh$.
| {
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Does this research paper prove that warp drives are impossible? Does this preprint prove that warp drives are impossible?
J. Santiago, S. Schuster and M. Visser, "Generic warp drives violate the null energy condition"
It states that the NEC (Null Energy Condition) is violated in this paper and many others: E. W. Lentz, "Breaking the Warp Barrier: Hyper-Fast Solitons in Einstein-Maxwell-Plasma Theory"
| As Santiago et al say, there have been no-go theorems ruling out warp drives for decades. In fact, as far as I can tell, those earlier results already disprove the claims in Lentz and the other recent articles that Santiago et al are primarily responding to.
This doesn't mean that warp drives are impossible, as you can evade no-go theorems by violating their assumptions, but I think that no one, including Lentz, has managed to do that.
Earlier no-go results usually show violation of the weak (or sometimes strong) energy condition. Santiago et al show violation of the null energy condition, which is a stronger result. On the other hand they consider only warp drive geometries that can be written in a certain form. It is (they claim) general enough to include all or almost all of the warp drives proposed in the literature, but it seems less general that the definition of warp drive used in some previous no-go results.
| {
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Einstein Equivalence principle & Locally inertial frame Weinberg's cosmology book stated that Einstein Equivalence principle implies that at any spacetime point in an arbitrary gravitational field there is a “locally inertial” coordinate system in which the effects of gravitation are absent in a sufficiently small spacetime neighborhood of that point.
So if I underatand this correctly, irrespective of being free-fall, one can always find a locally inertial frame. I have two questions:
*
*How is EEP related to Weak Equivalence principle relating inertial & gravitational mass? It seems to me that the most intuitive way to think of EEP is to imagine a tangent surface at an arbitrary point on curved manifold. However it is purely geometrical and has nothing to do with mass.
*Based on EEP, human sitting on Earth can also come up with a locally inertial frame around him/her (assuming the size of human can be considered as local). I forgot its name, but one experiment proved that a rest frame on Earth is not inertial since we can observe redshift (which makes sense because at rest in the presence of gravity is not inertial in GR). How can we reconcile these two? Is it because the experiment setup was not small enough to be considered local or something more profound?
Thanks
| There have been some misunderstandings. The 'local inertial frame' is the freely falling frame, as in GR,only freely falling frame can be considered as inertial frame,albeit it has to be confined in a small region. Besides,it is the implications that WEP leads to that is interesting: universality of free fall. Then gravity can be considered as inertial force and cancelled by it as well. This lead to EEP naturally.
| {
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What can be derived from the metric tensor? I am working on a computational project about General Relativity. In this process, I want to code 'the stuff' that can be derivable from the metric tensor. So far, I have coded Riemann Tensor, Weyl Tensor, Einstein Tensors, Ricci Tensor, Ricci scalar. What are the other essential/needed quantities in the GR calculations that can be coded?
Some notes to answer the comments:
*
*It's not precisely numerical. I will not solve the Einstein field equations for a given energy stress-energy tensor etc.
*The program's purpose is to obtain possible mathematical objects that belong to GR (tensors, etc.) If the user only knows metric tensor and nothing else.
*I am using / will use python to calculate these things. It's also kind of a relativistic tool, yes.
| *
*Metric $ds^2$ in Cartesian/Spherical/... coordinates
*Inverse of the metric
*Angle between $d^{(1)}x^{\alpha}$ and $d^{(2)}x^{\alpha}$
*Christoffel symbols
*Geodesic equations
*Geodesic equations in Newtonian limit
*Components of generalized momentum
*Riemann tensor
*Ricci tensor
*Traceless Ricci tensor
*Ricci scalar
*Einstein tensor
*Weyl tensor
*Some of the identities (e.g Bianchi) and properties
| {
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Elastic collision with one moving object hitting a stationary object In an elastic collision, I understand that momentum is conserved and kinetic energy is conserved. If billiard ball of silver (with velocity $v_{(Ag)}$ impacts a stationary billiard ball of aluminum, I am trying to calculate the velocity of the aluminum ball after the collision, $v_{(Al)}$. After an elastic collision, the impactor is at rest and the impactee has the motion.
Using momentum, $= m \cdot v$
$$m_{(Ag)} \cdot v_{(Ag)} = m_{(Al)} \cdot v_{(Al)}$$
Assuming silver is 4x denser than aluminium, then using momentum, the aluminium ball should have velocity
$$v_{(Al)} = 4\cdot v_{(Ag)}$$
But if we use kinetic energy, $1/2 m \cdot v^2$
$$\frac12m_{(Ag)}\cdot v_{(ag)}^2=\frac12m_{(Al)}\cdot v_{(Al)}^2$$
$$v_{(Al)}^2=\frac{m_{(Ag)}}{m_{(Al)}}\cdot v_{(Ag)}^2$$
$$v_{(Al)}=\left(\frac{m_{(Ag)}}{m_{(Al)}}\right)^{\frac12}\cdot v_{(Ag)}$$
$$v_{(Al)}=2\cdot v_{(Ag)}$$
Somewhere I have lost some neuron connections in my brain because I cannot resolve this conflict. This is a perfectly elastic collision so both momentum and kinetic energy should be conserved.
I have read multiple threads including:
When is energy conserved in a collision and not momentum?
How to calculate velocities after collision?
How can I calculate the final velocities of two spheres after an elastic collision?
Calculating new velocities of $n$-dimensional particles after collision
Velocities in an elastic collision
Summation of the velocities before and after an elastic collision
| If the objects have different masses, then there isn't a way to start the collision with object 1 moving and object 2 at rest and then end the collision with object 1 at rest and object 2 moving while also having the collision be elastic. You have over-constrained your system, and so you will find contradictions like the one you found here.
Using the equations from this answer in one of your linked questions, if we are setting $v_{A,f}=v_{B,i}=0$, then we end up with the system of equations
$$0 = \dfrac{m_A - m_B}{m_A+m_B} v_{A,i}$$
$$v_{B,f} = \dfrac{2m_A}{m_A+m_B} v_{A,i}$$
Which you can see is only consistent if $v_{A,i}=v_{B,f}=0$ for $m_A\neq m_B$ (which is the case of no collision), or if $m_A=m_B$.
| {
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How do you differentiate this differential equation? I have to differentiate this equation (Gravitational force between N-Bodies)
$\begin{align}
\frac{d^2}{dt^2}\vec{r_i}(t)=G
\sum_{k=1}^{n}
\frac
{m_k(\vec{r}_k(t)-\vec{r}_i(t))}
{\lvert\vec{r}_k(t)-\vec{r}_k(t)\rvert^3}
\end{align}$
where $\vec{r_{i/k}}(t)$ is the position of a body in 3D space and $m_{i/k}$ is its mass
How would you calculate $\frac{d^3}{dt^3}\vec{r_i}(t), \frac{d^4}{dt^4}\vec{r_i}(t) ...$?
Edit: I know that solving for $\vec{r_i}(t)$ is very complicated but is that also the case for the third derivative of $\vec{r_i}(t)$? I'm asking because since $\vec{r_i}(0)$ and $\frac{d}{dt}\vec{r_i}(0)$ are given (and therefore $\frac{d^2}{dt^2}\vec{r_i}(0)$ is also known) one could make a Taylor series with $\frac{d^3}{dt^3}\vec{r_i}(t), \frac{d^4}{dt^4}\vec{r_i}(t)$ and so on
| Even for three bodies it's really complicated
https://en.wikipedia.org/wiki/Three-body_problem
Unless someone else knows different, it would seem like a job for computer simulation.
| {
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Why does the intensity of the bright fringes decrease as we move away from the central maxima in Young's Double Slit Experiment? I studied that in Young's Double Slit Experiment the variation of intensity ($I$) of the fringes on the screen with respect to the phase difference ($Φ$) is given by :
$I = 4I_{0} \cos^{2}\frac{Φ}{2}$
$I_{0}$ is the intensity of light coming from each slit.
At maximas or constructive interference, $Φ = nλ$, where $n$ is any whole number and hence we get $I = 4I_{0}$ Below I have given the image of an interference pattern from a laser beam passing through double slit. As you can see as we move away from the central maxima, the intensity decreases and eventually it becomes zero. But how is this possible? According to our equation, the intensity of the centre all the bright fringes should be $4I_{0}$ and hence we should get equal brightness in all the maximas. But why does the intensity decrease and become zero at some point? Shouldn't the interference pattern extend upto infinity and there should be equal brightness at all the maximas? Please explain. I am so confused.
| Why do you imagine that the intensity would be the same all the way to infinity? That would require an infinite amount of energy to illuminate an infinitely wide screen.
The best way to think about the observed phenomenon is to imagine what you would see if you had only one slit. You would find that the incident light from a single slit would not be constant everywhere on the screen, stretching in infinity in either direction, but would vary, having a maximum directly opposite the single slit and quickly fading away either side of it.
The mistake you have made in your thinking is to assume that the intensity of the incident light is constant everywhere after it passes through the slit.
| {
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Why a rotating ball (at the end of a rope) does not fall down? Is the explanation shown in the diagram right?
This is: the net force F1 = tangential + tension is way much bigger than the weight of the ball and, therefore, the resulting force F1 + weight is F1 so that the ball does not fall down.
UPDATE
Thank you very much for your answers.
Actual situation:
| If you were to spin a ball about a vertical axis using a string, it would never rise up to a level above the horizontal level. In fact the angle upto which it will rise could also be calculated. Suppose the length of the string is $l$ and at equilibrium the angle it makes with the vertical axis is $\theta$, $\omega$ be the angular velocity of the axis, and the tension in the string is $T$. Then, we have:
$$T\cos \theta =mg$$
$$T\sin \theta =m{\omega}^2 l\sin\theta $$
From these two equations, we have:
$$\cos\theta=\frac {g}{{\omega}^2 l}$$
Clearly, if $\omega\to \infty$, $\theta\to 90°$, which means $\theta$ can never exceed $90°$. Also note that $w>\sqrt {\frac gl}$ is required if ball is to rise up even a little.
| {
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Book recommendation: Does anybody know a book adopting a more intuitive approach to the topic of Crystal Vibrations (phonons) than the Book by Kittel? I have tried Simon's 'Oxford Solid State basics' and Kittel 8th edition but I am not impressed by both (I mean the content covered through Chapters 4 and 5 in Kittel)
| I cannot give an explanation but can point to some other sources. Hook and Hall have a decent book on Solid State Physics, here is the link
https://www.wiley.com/en-gb/Solid+State+Physics%2C+2nd+Edition-p-9780471928058 .
There are also numerous sets of lecture notes that come up when you google 'Solid state physics pdf'. Hope this helps a bit!
| {
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Since the speed of light is constant and also the speed limit; would you, in your reference frame, have no upper bound on your speed? Let us imagine you are in a vacuum and after having maintained a speed of 0 km/s (standing still) you accelerate to 297,000 km/s (99%). You know this is now your speed because you have a speedometer telling you so. You then decide to maintain that speed for a while.
With the speed of light is always ~300,000 km/s faster than you, what is preventing you from (again in your reference frame) increasing your speed, as shown by a speedometer, an arbitrary amount faster than ~300,000 km/s? After all, the speed of light will always be always faster.
I feel like length contraction even backs this since it will make your space wheels tinier. You're essentially scaled down and your tiny wheels would have to rotate many more times to go the distance just 1 rotation would have taken you with your non-contracted length. This then would cause the speedometer to relay speeds faster than the speed of light.
| In your reference frame you are always at rest, so you can always accelerate to a new reference frame.
Suppose you were at rest in some reference frame. Then suppose you were accelerated to 0.9999999999999999999999999c in that initial reference frame.
In your new reference frame you are stationary. You can now accelerate yourself to 0.9999999999999999999999999c in your new frame.
You are now stationary in a third frame, in which you can also accelerate yourself to 0.99999999999999999999999999c.
And so on endlessly.
However, when you add your series of velocity increases relative to the original frame, you must use the relativistic rules for adding velocities. No mater how many velocities of 0.999999999999999999999999c you add, you will never exceed c.
| {
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An algebra step in the Quantum Partition Function for the Harmonic Oscillator On page 183 of Altland Simons, we are told:
$$ \prod_{n = 1}^{\infty} \Big[ \Big( \frac{2\pi n}{\beta} \Big)^2 + \omega^2 \Big]^{-1} \sim \prod_{n = 1}^{\infty} \Big[ 1 + \Big( \frac{\beta \omega}{2\pi n} \Big)^2 \Big]^{-1} \sim \frac{1}{\sinh(\beta \omega / 2)}. $$
How do we get the first relation?
This physics SE question suggests that it is a multiplication and division by $$\prod_{n = 1}^{\infty} (\beta / 2\pi n)^2 ,$$ but I'm confused how multiplication and division by the same factor switches the $\beta, n$ from denominator/numerator to numerator/denominator:
$$\Big[ \Big( \frac{2\pi n}{\beta} \Big)^2 + \omega^2 \Big]^{-1} \times \frac{(\beta / 2\pi n)^2}{(\beta / 2\pi n)^2} \sim \Big[ 1 + \Big( \frac{\beta \omega}{2\pi n} \Big)^2 \Big]^{-1}.$$
I understand that one answer in the linked post addresses this in a different way (zeta function regularization) and why we can multiply by $\prod_{n = 1}^{\infty} (\beta / 2\pi n)^2 $. I'm stuck on the step before that -- why would multiplication and division of that product help in the first place?
| $$
\newcommand{\qcl}{q_{\rm cl}}
\newcommand{\ket}[1]{| #1 \rangle}
\newcommand{\bra}[1]{\langle #1 |}
$$
This is a source of very sloppy work which appears in many textbooks. You are completely correct that it makes no sense to divide by this diverge factor ad hoc. The reason they are doing this is because they weren't careful enough with the measure of the path integral which essentially provides the perfect divergent factor you need.
When calculating this quantity, you decomposed variations around the classical path into Fourier modes. This change of variables in the path integral comes with an associated (divergent) Jacobian factor $J_N$. Before getting into the Harmonic oscillator, let's start with the free Hamiltonian when $\omega = 0$. Because this Jacobian factor doesn't depend on the Hamiltonian, we can use the well known expression for the heat kernel of the free Hamiltonian to solve for it. After we extract this factor we will then move to the Harmonic oscillator and use it to compute the transition amplitude exactly.
(While I will present this work in terms of real time and a transition amplitude instead of Euclidean time and the traced partition function, it is a simple matter to go between the two.)
We will follow Nakahara's "Geometery, Topology, and Physics second edition" section 1.4.
Let us begin with the well known expression for the Heat kernel of the free particle Hamitlonian $H_{\rm free}$.
\begin{align}\tag{1}\label{eq1}
\bra{q_f}e^{- i H_{\rm free} T / \hbar} \ket{q_i} &= \left( \frac{m/\hbar}{2 \pi i T} \right)^{1/2} \exp( \frac{i}{\hbar} \frac{m}{2 T} (q_f - q_i)^2 ) \\
&= \left( \frac{m/\hbar}{2 \pi i T} \right)^{1/2} e^{i S_{\rm free}[q_{\rm cl} ]/\hbar}.
\end{align}
Recall also that the measure of our path integral is defined as
\begin{equation}
\int \mathcal{D} q(t) = \lim_{N \to \infty} \left( \frac{m/ \hbar }{2 \pi i \epsilon}\right)^{(N+1)/2} \int dq_1 \ldots dq_N
\end{equation}
where
\begin{equation}
\epsilon = \frac{T}{N+1}.
\end{equation}
For the free particle, the action is
\begin{equation}
S_{\rm free}[q] = \frac{m}{2} \int_0^T dt \; \dot q^2.
\end{equation}
If we break up our paths as
\begin{equation}
q = \qcl + y
\end{equation}
then, for this simple quadratic action,
\begin{align}
S_{\rm free}[q] = S_{\rm free}[\qcl] + S_{\rm free}[y]
\end{align}
and
\begin{align}
\bra{q_f}e^{- i H_{\rm free} T / \hbar} \ket{q_i} &= \int_{q(0) = q_i}^{q(T) = q_f} \mathcal{D} q(t) e^{i S_{\rm free}[q] / \hbar} \\
&= e^{i S_{\rm free}[\qcl] / \hbar} \int_{y(0) = 0}^{y(T) = 0} \mathcal{D}y(t) e^{i S_{\rm free}[y] / \hbar}.
\end{align}
Combining this with our expression \eqref{eq1}, we get
\begin{equation}\label{heatkernel2}\tag{2}
\left( \frac{m/\hbar}{2 \pi i T} \right)^{1/2} = \int_{y(0) = 0}^{y(T) = 0} \mathcal{D} y(t) e^{i S_{\rm free}[y]/\hbar}
\end{equation}
which is our starting point.
We now consider $y(t)$ in terms of its Fourier coefficients.
\begin{align}
y(t) = \sum_{n = 1}^N a_n \sin( \frac{n \pi t}{T}).
\end{align}
We have only allowed $N$ coefficients for the following reason: our integral is an integral over the $N$ variables $y_k$ for $k = 1 \ldots N$, and When we change the variables of integration to $a_n$, we must preserve the number of variables we are integrating over. The Jacobian of this transformation, which our main object of interest, is
\begin{equation}
J_N \equiv \det( \frac{\partial y_k}{\partial q_n} ) = \det( \sin(\frac{n \pi t_k}{T} ) ).
\end{equation}
The action is now
\begin{align}
S_{\rm free}[y] = \frac{m}{2}\sum_{n=1}^N\frac{T}{2} \left( \frac{a_n n \pi}{T} \right)^2
\end{align}
and Eq. \ref{heatkernel2} becomes
\begin{align}
\left( \frac{m/\hbar}{2 \pi i T} \right)^{1/2} &= J_N \left( \frac{m /\hbar}{2 \pi i \epsilon} \right)^{(N+1)/2} \int d a_1 \ldots d a_N \exp( \frac{i m}{\hbar} \sum_{n=1}^N \frac{a_n^2 \pi^2 n^2}{4 T} ) \\
&= J_N \left( \frac{m /\hbar}{2 \pi i \epsilon} \right)^{(N+1)/2} \prod_{n=1}^N \left( - \pi \frac{ 4 T \hbar}{i m \pi^2 n^2} \right)^{1/2} \\
&= J_N \left( \frac{m /\hbar}{2 \pi i \epsilon} \right)^{(N+1)/2} \left( \frac{i 4 \pi T \hbar}{m \pi^2} \right)^{N/2} \frac{1}{N!}.
\end{align}
After some clean up, this becomes
\begin{equation}
J_N =2^{-N/2} \pi^N (N+1)^{-(N+1)/2} N!
\end{equation}
which happily is dimensionless.
Now let's see how we can use $J_N$ to get a finite result for the transition amplitude of the Harmonic oscillator, which has the action of
\begin{equation}
S[q] = \frac{m}{2} \int_0^T \left( \dot q^2 - \omega^2 q^2 \right) dt.
\end{equation}
The simple and quadratic nature of this action still means that
\begin{equation}
S[\qcl + y] = S[\qcl] + S[y].
\end{equation}
The action of our Fourier expanded paths is
\begin{align}
S[y] &= \frac{m}{2} \sum_{n = 1}^N \frac{T}{2} a_n^2 \left( \left(\frac{n \pi}{T}\right)^2 - \omega^2 \right) \\
&= \frac{T}{4} \sum_{n = 1}^N a_n^2 \lambda_n
\end{align}
where we have defined
\begin{equation}\label{eigenvaluefirst}
\lambda_n = m \left( \left(\frac{n \pi}{T}\right)^2 - \omega^2 \right).
\end{equation}
Therefore,
\begin{align}
\bra{q_f} e^{- i H T/\hbar} \ket{q_i} =& e^{\tfrac{i}{\hbar} S[\qcl] } \int \mathcal{D} y(t) e^{\tfrac{i}{\hbar} S[y] } \\
=& e^{\tfrac{i}{\hbar} S[\qcl] } J_N \left( \frac{m/ \hbar }{2 \pi i \epsilon}\right)^{(N+1)/2} \int d a_1 \ldots d a_N \exp( \frac{i T}{4 \hbar} \sum_{n = 1}^N a_n^2 \lambda_n ) \\
=& e^{\tfrac{i}{\hbar} S[\qcl] } J_N \left( \frac{m /\hbar}{2 \pi i \epsilon} \right)^{(N+1)/2} \left(\frac{i 4 \pi \hbar}{T} \right)^{N/2} \left( \prod_{n = 1}^N \lambda_n \right)^{-1/2} \\
=& e^{\tfrac{i}{\hbar} S[\qcl] } \left( \frac{m/\hbar}{2 \pi i T} \right)^{1/2} \left( \frac{m \pi^2 }{T^2} \right)^{N/2} N! \left( \prod_{n = 1}^N \lambda_n \right)^{-1/2}.
\end{align}
Note the terms divergent in $N$. If we were using the free action, with $\omega = 0$, then
\begin{equation}
\lambda^{\omega = 0}_n = m \left( \frac{n \pi}{T} \right)^2, \hspace{1 cm} \left(\prod_{n=1}^N \lambda^{\omega =0}_n \right)^{1/2} = \frac{m^{N/2} \pi^N}{T^N} N!
\end{equation}
and thus
\begin{equation}\label{ratioeigenvalues}
\bra{q_f} e^{- i H T/\hbar} \ket{q_i} = e^{\tfrac{i}{\hbar} S[\qcl] } \left( \frac{m/\hbar}{2 \pi i T} \right)^{1/2} \frac{\left( \prod_{n = 1}^N \lambda_n \right)^{-1/2}}{\left( \prod_{n = 1}^N \lambda_n^{\omega = 0} \right)^{-1/2}}.
\end{equation}
We now compute
\begin{equation}
\prod_{n=1}^N\frac{\lambda_n}{\lambda_n^{\omega=0}} = \prod_{n=1}^N \left( 1 - \left( \frac{\omega T}{n \pi} \right)^2 \right) = \frac{\sin(\omega T) }{\omega T}
\end{equation}
making
\begin{equation}
\bra{q_f} e^{- i H T/\hbar} \ket{q_i} = \left( \frac{m \omega}{\hbar \pi} \frac{1}{2 i \sin ( \omega T) }\right)^{1/2} e^{\tfrac{i}{\hbar} S[\qcl] }.
\end{equation}
| {
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What happens to entropy when half the particles are removed? Curiosity question. What happens to entropy in the following situation?
A gas fills an entire container and is in equilibrium. Suddenly all particles are removed from half the container. As such, there are now 1/2 the original number of particles, but all occupy one side of the container.
What happens to entropy? Does it rise or fall?
| With the thermodynamic definition of entropy:
$$dS = \frac{dQ}{T},$$
where $Q$ is heat and $T$ the temperature, the answer of what is the entropy of half the volume of your container will be , half of the entropy of the full one.
Now saying
Suddenly all particles are removed from half the container.
the state stops being "in equilibrium" and has to be studied dynamically, as it will be not an isothermal process to treat entropy as $Q/T$.
Depending on the mechanism of "removed" the entropy can increase, over the nominal half, given its definition in terms of "Entropy as a Measure of the Multiplicity of a System" as shown in the link. counting the microsystems of the changing gas conditions.
| {
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How to calculate the centroid of a simple shape but rotated? Question:
Locate the centroid $y$ for the cross-sectional area of
the angle.
I tried to use four triangles, two of them are the bigger ones $(a+t)$ and two of them are the smaller ones $(a-t)$. I then used the idea of proportionality and Pythagoras, but I am now lost on the algebra. Is there a nicer way of doing this like using rotation? I am thinking finding the centroid normally and apply some rotational matrix to the values. Is this even feasible or am I just complicating things? I did it in an arduous way and still didn't get the result. Any guidance would be appreciated. Thanks -Sarah
PS:
Answer
| The rotation matrix is overkill here. Simple geometric reasoning helps.
If you add a dotted square to L-shape, you will get a bigger square. Thus, if you put a pivot at $O_2$, then the small square (with CoM at distance $O_1O_2$) would be balanced by L-shape (with CoM at distance $O_2C$). Can you know find the distance $\bar y$ from the top right corner to $C$?
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What is the force pair for the normal force? Clarification on Newton's 3rd Law In the process of trying to wrap my head around Newton's 3rd law I've come across 2 definitive statements.
*
*Forces must occur in pairs
*Forces must act on different bodies
This is confusing to me when applied to the classic box on a flat plane scenario (assuming the flat plane is Earth). I've been taught in school that the present forces are like this:
So accordingly I always assumed that the gravitational force of the box and the normal force were pairs. However, after watching a few videos explaining the concept, namely this one, the impression I was left with is that the paired forces are FBE and FEB.
However, I'm aware that the normal force still exists, but if it can't exist without a paired force, what would be its paired force?
Additionally, can a force be paired with multiple forces?
| The reaction to the normal force is the normal force generated by the box pushing into the table. The table experiences a downward normal force from the box, just as the box experiences an upward normal force from the table.
| {
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How to measure the speed of an electric current? We all know that the definition of a current is the amount of charge flowing per second, that is often expressed by the equation $i=dq/dt$. But is it possible to measure the speed of an electric current in m/s? And also how can we measure such speed?
| From two other answers you know how the speed, $v$, of the charge carriers is related to the current, provided that you know $n$, the number of charge carriers per unit volume.
The Hall effect gives you an independent way to measure $v$. Send a current in the $z$ direction through a cuboid of the conducting material of dimensions $\Delta x \times \Delta y \times \Delta z$. Apply a known magnetic field $B$ to the cuboid in the $x$ direction. A 'transverse' pd $\Delta V_y$ will appear across the faces separated by $\Delta y$, given by
$$\Delta V_y = Bv \Delta y.$$
Hence $v$, though there are various practical difficulties to be dealt with.
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Does a object gain heat faster the colder it is? Does an object at $-273°{\rm C}$ gain heat faster than an object at $-1°{\rm C}$
| Heat transfer by conduction, convection and radiation is a function of the difference in temperature between two objects, or an object and it's surroundings (for radiation it's the difference between the fourth powers of the temperatures, but the effect is the same). So if you have two objects, one at $0^o$C and one at $-100^o$C, sitting in the same room at $20^o$C, the heat transfer to the object at $-100^o$C will be much greater than to the object at $0^o$C. This means that the $-100^o$C object will warm up at a faster rate, but won't ever become warmer than the $0^0$C object.
| {
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How to translate from a state/density matrix formalism to matrix product state representation? From what I understand, MPS is just a simpler way to write out a state, compared to the density matrix. But how do I get those $A_i$ matrices? From all the examples I read, people just somehow "have" the matrices in their pocket already. Is there a way to generate them from some random state, say $|\Psi\rangle = \alpha|\uparrow\rangle^{\otimes N}+\beta|\downarrow\rangle^{\otimes N}$, or some random density matrix? Feel free to set $N$ to small numbers if it becomes too complicated, I just really want to see an example of how this works in general, not just a GHZ state! Thanks a lot!
Crosspost from: https://quantumcomputing.stackexchange.com/questions/17590/how-to-find-the-a-i-in-the-matrix-product-state-representation?noredirect=1#comment25914_17590
| Any state can be written as a matrix product state. There are systematic procedures to construct such a description, based on sequential SVDs, see e.g. Section 4.1.3 of this review.
On the other hand, this description is usually of interest if the resulting MPS description has much less parameters than the $2^N$ parameters needed to describe a general quantum state. This is, e.g., the case for the example you give.
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Regarding the charge configuration on a spherical shell based on the atomic level behavior With regards to the fundamental theories in Electrostatics, when a spherical shell of inner radius $a$ and outer radius $b$ with a $+q$ charge at the center is considered, there will be no electric field lines within the region $a<r<b$. Theoretically, this is agreeable as $Q_{enclosed} = 0$ within this region.
However, is it possible to understand this phenomenon in a physical aspect? What physical cause (atomic-level behavior) blocks these electric field lines entering this region? As I understood, the charge separation at the inner and outer radii of the shell is due to the movement of electrons caused as a result of the attractive and repulsive forces caused by the $+q$ charge at the center, resulting the net enclosed charge to be zero. Yet, I am unable to verify the reason for why these electric field lines cannot enter this region. What atomic level constraints are acting on this system.
| You don't even need to consider a specific charge in the center. (Reasonably low) static electric fields cannot penetrate any material containing free electrons (i.e. metals). Free electrons, by definition, are free to move if exposed to a force. If you try to put an electric field through such a material, the electric field exerts a force on the electrons, which they follow. This results in the charges collecting at the surface (because usually they don't have enough energy to leave the metal). This process continues until the surface charges completely compensate for the electric field trying to penetrate, resulting in zero field inside the metal.
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Wick's theorem, contracting field operators at the same point I want to calculate the amplitude for nucleon meson scattering $\psi \varphi \to \psi \phi$ in scalar Yukawa theory, with interaction term:
$$H_{I} = g \int d^{3}x \psi^{\dagger} \psi \varphi.\tag{3.25}$$
This involves dealing with a time ordered string of operators
$$T\{\psi^{\dagger}(x) \psi(x) \varphi(x)\psi^{\dagger}(y) \psi(y) \varphi(y)\}.\tag{3.46}$$
We can apply Wick's theorem to the above expression. Tong's lecture notes identify two relevant terms:
$$:\psi^{\dagger}(x) \varphi(x)\psi(y) \varphi(y): \overbrace{\psi(x) \psi^{\dagger}(y)}\tag{3.53}$$
and
$$:\psi(x) \varphi(x)\psi^{\dagger}(y) \varphi(y): \overbrace{\psi^{\dagger}(x) \psi(y)}.$$
My question is, what about the terms:
$$:\psi^{\dagger}(x) \psi(x) \varphi(x) \varphi(y): \overbrace{\psi^{\dagger}(y) \psi(y)}$$
and
$$:\varphi(x) \psi^{\dagger}(y) \psi(y) \varphi(y): \overbrace{\psi^{\dagger}(x) \psi(x)}.$$
They are not mentioned in the notes and I cannot come up with a Feynman diagram that would correspond to those. Is it even correct to contract two field operators defined at the same point?
| There is an implicitly written normal ordering symbol $:~:$ in the Hamiltonian (3.25), and therefore 2 implicitly written normal ordering symbols in the expression (3.46). The answer to OP's title question is that there are no contractions among operators within the same normal ordering symbol, i.e. belonging to the same spacetime point. This is explained in e.g. this related Phys.SE post.
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Where is energy in energy density? I was learning about energy density and it seemed to be defined as the potential energy per unit volume in an electric field
$\frac{dU}{dV} = \frac{1}{2}\epsilon E^2$
But how can just the electric field have a potential energy on its own without presence of any charge? What is causing this energy to be present in an electric field?
| There are some reasons for believing that fields are real as well as being a mathematical convenience; and that they do contain energy. A prime example would be the electromagnetic wave, which consists of fields that can carry energy from one point to another. Another example would be a magnet, which can push or pull another magnet. Where does the energy come from to do that work? The only thing that changes is the configuration of the fields.
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Why intensity of light reaching the sensor or film with a particular lens directly proportional to $\frac{D^2}{f^2}$? The following is quoted from my book:
"The intensity of light reaching the sensor or film is proportional to the area viewed by the camera lens and to the effective area of the lens. The size of the area that the lens "sees" is proportional to the square of the angle of view of the lens, and so is roughly proportional to $\frac{1}{f^2}$. The effective area of the lens is controlled by means of an adjustable lens aperture, or diaphragm, a nearly circular hole with variable diameter $D$; hence the effective area is propor tional to $D^2$. Putting these factors together, we see that the intensity of light reaching the sensor or film with a particular lens is proportional to $\frac{D^2}{f^2}$."
My question is how did they conclude that the area that the lens "sees" is roughly proportional to the square of angle of view of the lens and $\frac{1}{f^2}$ and how is the effective area proportional to $D^2$? Ultimately, my question is how is the intensity of light proportional to $\frac{D^2}{f^2}$? Can someone please explain? I did not understand what the paragraph explained. Please help.
| Since we are talking about a camera, it's more appropriate to talk about the illumination of an image which translates to the amount of light per unit area of the image. The dependence on D$^2$ is straightforward as that is proportional to the area of the lens aperture.
The magnification due to the camera lens is given approximately by the image distance divided by the object distance. For a far away object, the image distance is approximately equal to f. So the linear extent of the image depends on f and the area of the image depends on f$^2$. The smaller the area, the more concentrated is the light on the image.
So the image brightness depends, for a distant object, on D$^2$/f$^2$. This is just the inverse of the f/number and as we know the smaller the f/number (hence the larger D$^2$/f$^2$), the brighter the image.
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Why is the number of isotopes of an element bounded? Is there a known reason why any given element has finitely many isotopes? Here I mean both stable and unstable isotopes.
If we know this, do we have a reason why, for a given element, are the isotopes limited to that particular number?
| I think this is a good question -- after all, if there's no extra Coulomb repulsion penalty for adding more neutrons, unlike for protons, why can't nuclei have lots of neutrons?
One model for the nucleus we use is called the Semi-Empirical Mass Formula (SEMF), which has a bunch of terms describing the energy contributions to the nucleus. See wikipedia for the full formula. The main term that answers your question is the "Asymmetry Term", given by
$$a_\text{A}\frac{(N-Z)^2}{A}$$
where $a_\text{A}$ is some constant we can find empirically, $N$ is the number of neutrons, $Z$ is the number of protons and $A=N+Z$ is the nucleon number.
This is a penalty term in the energy of a nucleus. If there is a large difference in $N$ and $Z$, this term is large. If $N$ is similar to $Z$, the term is not as large. The rational for this is the Pauli Exclusion Principle, which tells us identical particles cannot occupy the same energy state. If we're adding lots of identical neutrons, we must put them in different energy states. We can get a cheaper energy cost by filling in some protons instead for a given nucleon number $A$.
To answer your question in the comments: why do isotopes often have more neutrons than protons, I think the answer there is it is somewhat favourable to add nucleons to the nucleus, because that increases the strong force present, but its cheaper to use neutrons than protons, at least for cases where the ratio $N/Z$ does not deviate too far from 1.
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Why is power transmission carried out at low current high voltage? My textbook states that power is transmitted at high voltage and low current since $P=I^2R$ and as the current has a small magnitude, the heat dissipated across the transmission lines is less than when we carry it out at high current and low voltage. But $P=I^2R$ can also be written as $P=V^2/R$ and hence a discrepancy would occur. Where am I going wrong?
| Here are the possible ways of writing power
Look at how resistance exists in wires. The power is inversely proportional to resistance if one uses the voltage as a criterion, the more resistance in the way the smaller the power transferred for the same voltage. The smaller the current the less power is dissipated by a given resistance on the way and can be transferred where it is needed.
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The nice Veritasium video about a wind-powered vehicle that can go downwind faster than the wind itself Today Derek Muller posted a video about a wind-powered vehicle that can go downwind faster than that downwind itself.
The vehicle is custom made for that case only: the case of going straight downwind.
The vehicle has three wheels, I estimate about 5 meters from front wheel to back wheels, and and a two-bladed propellor is mounted about 5 meters up.
The mechanical connection between the propellor and the wheels is such that when the vehicle is rolling forward the propellor is moving air from the front of the vehicle to the rear of the vehicle.
The vehicle is as lightweight as possible for its size, so it's quite flimsy. This is definitely not a practical device, it is a proof-of-concept device.
In the video Derek himself indicated that he was not quite confident that he understood the physics of it. I anticipate that questions will start coming in on physics SE, so I present this case as a self-answer.
How can this vehicle, when going straight downwind, go faster than the wind itself?
Previous question about that vehicle:
Details about mechanics of directly-downwind-faster-than-wind vehicle
| One should note the difference between reaching faster-than-the-wind travel and being able to maintain faster-than-the-wind travel (i.e. is the vehicle in steady-state when going faster than the wind).
The difference is in whether energy can be continuously harvested from the air (i.e. the wind) once the vehicle has reached the downwind speed, and passed it, or whether the vehicle depends at the point on energy that it had accumulated when it was traveling slower.
IMHO the energy at that point is no longer coming from the air stream, it is coming from the kinetic energy stored in the propeller. As long as it keeps rotating it is able to accelerate the vehicle, much like a propeller accelerates an aircraft. However, since this propeller is not powered, it is losing speed, due to mechanical friction and air resistance. Additionally, the propeller may transfer energy directly to the wheels through a proper gear ratio to allow the wheels to "extract" the kinetic energy of the rotating propeller and accelerate the vehicle through traction with the ground.
TL;DR: It probably goes faster than the wind due to the propeller acting as a flywheel once the vehicle reaches wind speed.
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Is it possible to replace the welding machine with a light beam? Using mirrors, lenses, and a battery of any power?
Yes, it can be expensive, yes, it can be inefficient, but it is absolutely not important.
Not about laser
| That was the basis of a rather known Russian sci-fy novel from 20s: The Garin death ray (at least known in Russia). Notably, the novel discusses the docusing system based in a parabolic mirror, which is mistakenly called hyperbolic (hence the Russian name of the novel is literally The Garin's hyperboloid). Russians physicists inevitably discuss it when making a point about what can be done with a laser but not with mirrors.
Of course, if we could increase the power infinitely, we could always end up melting something. The meaningful question is really about the efficiency of the power use, but oen has to be clear about what the efficiency means in this context: it is the efficiency of the power transmission from its source to the point that we heat, using a ligh beam. The two obvious problems with using non-coherent light source are:
*
*The light ray diverges, i.e., it is in practice impossible to focusit to a tiny spot at significant distance.
*The destructive interference between the incoherent light waves eats most of the power.
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Do materials sort in order of density in space the way they do near Earth? Obviously, if you drop materials into water near the surface of the Earth, they sort according to densities, with the least dense materials near the surface, and the most dense materials near the bottom. Common sense suggests this is due to gravity, because the more dense a material is, the more mass it has per unit of volume, and therefore, the more it interacts with any gravitational field. This should cause the most dense materials to experience a greater force due to gravity, causing them to sink more than others.
So my question then has two parts, which is, (1) is this the correct explanation for the apparent sorting of densities, and (2) if so, then what happens in space when you don't have any appreciable gravity?
| Yes that is more or less the correct explanation. The configuration with highest density materials at the bottom and lower density materials at the top is the one with the lowest potential energy, and thus the one favoured at equilibrium.
In space, far away from massive objects, there is no direction called “up” or “down”. Materials don’t sort themselves in order of decreasing density. To get a feel as to how they behaves, you can imagine what would happen if you filled a shallow plate with water, oil, rocks and so on, and look at their 2D movement from above. No direction in the plane of the fish is favoured and you get interesting shapes depending on happenstance and the dynamical properties of the materials.
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Why doesn't current flow through an open branch? I know that current doesn't flow through open branch because current can't flow through air due to its high resistance .But i was thinking , what's the problem if current flows through an open wire (assumed 0 resistance for matter of circuit solving). I mean, isn't there a possibility that the charge flowing through the wire keeps accumulating at its end as it can't flow through air as battery is incapable of flowing it through the air. Why do we have to entirely abandon that wire?
| Current (or more properly, conduction current) can flow into an open wire and charge accumulate at the end IF there is displacement current to continue the circuit. Displacement current is not the flow of charge, but a change in the electric field. Two situations where this might happen is in an antenna, where the displacement current is provided by an electromagnetic wave,. The other situation is where the end of the wire is (in effect) one half of a capacitor, and the voltage across the capacitor is changing.
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In LED's do the number of charge carriers (electrons and holes) decrease with time? According to page-1268-69 of Halliday, Walker & Resnick's Fundamentals of Physics (10th edition),
To emit enough light to be useful as an LED, the material must have a
suitably large number of electron-hole transitions....What we need is
a semiconductor material with a very large number of electrons in the
conduction band and a correspondingly large number of holes in the
valence band. A device with this property can be fabricated by placing
a strong forward bias on a heavily doped p-n junction, as in Fig.
41-16. In such an arrangement the current I through the device serves
to inject electrons into the n-type material and to inject holes into
the p-type material. If the doping is heavy enough and the current is
great enough, the depletion zone can become very narrow, perhaps only
a few micrometers wide. The result is a great number density of
electrons in the n-type material facing a correspondingly great number
density of holes in the p-type material, across the narrow depletion
zone. With such great number of densities so near each other, many
electron-hole combinations occur, causing light to be emitted from
that zone.
Now, if electron-hole pairs are ceasing to exist due to recombination and resulting in a greater number of "gridlocked" electrons and light, who will continue to conduct electricity? Won't current flow stop after a while due to the absence of electron-hole pairs?
| You won't run out of electrons and holes. There are two main processes through which you will get more.
Thermal energy will naturally generate electron/hole pairs in your semiconductor. That's where they come from in a semiconductor at thermal equilibrium. However, this process is slower than we would like for an LED, or any diode. Fortunately, we can help it along by injecting carriers from outside the p-n junction. We can imagine a simple p-n junction with metal contacts set back a bit from the p-n junction. You can directly inject electrons into the n side, replenishing the lost electrons. At the p side you will have a greatly increased electron/hole generation rate allowing you to effectively inject holes into the p-side (by generating an electron/hole pair and removing the electron into the metal).
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Can information travel faster than speed of light in this situation? I know the answer is no but I have a thought experiment that seem to be violating that. Imagine two persons living on two different planets namely A and C which are 10 light years apart. There is a planet in between, B, which is located exactly at the same distance from A and C. If two persons from A and C get on their hypothetical spaceships at the same time, that can travel around .99C, and move towards B, they can talk to eachother in around 5 years. Didn't information just convey between two persons at the speed of ~2C in this case?
| The principle is that information cannot move from Point 1 to Point 2 faster than the speed of light. In the example you give, information moves from the initial positions of the two participants to Planet B at less than the speed of light, so the principle is not violated.
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Trying to show new matrix elements of momentum operator after unitary transformation of $|x \rangle$
The problem is this. Given that I have a new x basis
$$|\tilde{x} \rangle = e^{i g(X)/ \hbar} |x\rangle$$
I have to show that
$$\langle \tilde{x} | P | \tilde{x}' \rangle = \left(-i \hbar \frac{d}{dx} + \frac{dg}{dx}\right) \delta (x-x').$$
Here is my attempt at a solution
$$\langle \tilde{x} | P | \tilde{x}' \rangle= \int \int \langle \tilde{x}| x \rangle \langle x | P | x' \rangle \langle x' | \tilde{x}' \rangle dx' dx$$
$$= \int e^{-ig(x)/ \hbar} \int -i \hbar \delta'(x-x') e^{i g(x')/ \hbar} dx' dx $$
But this doesn’t give the correct result. I suspect my mistake is thinking that the projection of $\tilde{x}$ in the old x basis is $e^{ig(x)/\hbar}$. If that is really my mistake, then what is the projection? If that isn't, well then what am I doing wrong and how do I do this correctly?
| We know that for any function $f$, we have :
$$[P,f(X)]= -i\hbar f'(X)$$
In particular :
$$\left[P,e^{ig(X)/\hbar}\right] = g'(X)e^{ig(X)/\hbar}$$
Now, we can compute :
\begin{align}
\langle \tilde x |P |\tilde x'\rangle &= \langle x|e^{-ig(X)/\hbar}Pe^{ig(X)/\hbar}|x'\rangle \\
&= \langle x |g'(X) + P|x'\rangle \\
&= \left(g'(x) - i\hbar \frac{\text d }{\text dx}\right)\delta(x-x')
\end{align}
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A cylinder doing pure rolling in the front collides Consider a cylinder rolling in the forward direction with velocity of $v$ and angular velocity of $v/r$. There is a wall in front and the cylinder collides with the wall. The collision is perfectly elastic.
Here I am unable to judge what will happen next.
Will it stop pure rolling and moves in translational for a while, or will it continue to pure roll in opposite direction after the collision?
| It should stop rolling instantaneously and then start rolling in the opposite direction.
When the cylinder collides with the wall, the point on the cylinder's surface in contact with the wall is moving instantaneously in the downward direction. Elastic collisions reverse velocities, so the instant after the collision, the velocity of that point becomes diametrically upward. Same thing happens with the translational velocity. Before the collision, the velocity of the center-of-mass of the cylinder is left-to-right, whereas after, it is right-to-left. The overall motion is such that the rolling reverses, with the cylinder at rest very briefly before the motion flips.
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How to generate electric current without a permanent magnet? The question is pretty simple:
Can we build a device that coverts mechanical work in electric current1 without employing a permanent magnet and without access to any external source of current?
The restrictions in place seem to rule out the possibility of current generation via induction; and I cannot think of another practical method. I have heard that industrial alternators sometimes work with electromagnets, but we don't have access to any external source of current, so this path doesn't seem viable.
Do we really need stupid magnetic rocks to produce current? Unacceptable.
To be more specific and minimize to risk of misunderstandings: my question is more or less equivalent to the following one
Can we build a device, powered by hand via some sort of rotating lever, that produces electric current, crucially without employing any external current and without any permanent magnet?
[1]: Usable electric current, let's say sufficient to properly power up a lamp; doesn't matter if AC or DC.
| Piezo electric cells convert mechanical energy to electric energy
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Can it be shown in experiment that the momentum (or position) states of the electron and proton in hydrogen are entangled? The states of the electron and proton in hydrogen are entangled. Which means that the momentum and position of both are entangled. Can this be shown in an experiment, so if you measure the momentum (or a small range of them) of the proton you will know in advance what the momentum of the electron will be?
If you first have to separate them, to do a measurement on them, will the information of the entanglement show itself in the measurement? As the information of entanglement of two spin states can show itself in measurement?
| In a two body bound state, (quantum mechanical or not) from conservation of energy and momentum, once the masses are known, measuring the four vector of one particle , the four vector of the other is known.
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Does energy of a capacitor means energy stored in both plates? I've a doubt in this,
Does the term potential energy of a parallel plate capacitor means the energy stored in both the plates or a single plate, since the formula $E=Q^2/2C$ , $Q$ is the charge of only one plate?
Please help me in this.
| It is the total energy stored in the entire system.
Consider that you have an uncharged parallel plate capacitor and you decide to charge it up such that the charge on one plate is $+Q$ and the charge on the other plate is $-Q$ where $Q>0$. At any point in time, the charge on the positive plate is $q$, where $0\leq q\leq Q$, which gives a potential difference between the plates of $\frac{q}{C}$.
We know that $dW=Vdq$ and so $$W=\frac{1}{C}\int_0^Qqdq=\frac{Q^2}{2C}$$
As you can see, the derivation of the formula above utilized the potential difference between both plates and so the state of the second plate was also taken into account. Therefore, $\frac{Q^2}{2C}$ is the total energy of the entire system.
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Are the accelerations the same at either end of a moveable pully? Given a moveable pulley with a fixed pulley on either side,
Is the acceleration of the left weight (m1) the same as that of the right weight (m2)?
Intuitively, I would imagine it to be, since if m1 drops by 10 metres, then m3 would rise by 5 metres, and thus m2 would drop by 10 metres. But mathematically the accelerations don't appear to be the same... Perhaps there is a mistake in my reasoning.
Assume the system is released from rest, an inextensible and massless string is used for the pulleys, and there is no friction forces acting on the system.
| Hint: The tension in the string is the same throughout.
Hope this helps.
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A K-theory isomorphism I found this identities in a paper on Floquet topological classification which the author mentioned as a "well-known K-theory isomorphism"
$$K_{R}^{0,n}(S^1\times X, \{0\}\times X) = K_R^{0,n+1}(X).$$
Here $S^1$ is a circle (corresponding to time) and $\{0\}$ is a point in the circle (the initial time), $X$ is the Brillouin zone. The relative K-group implies that at $t=0$, the unitary must be identity for all $k\in X$. I initially thought this was just suspension but it didn't add up. Does anyone have any ideas?
| I found an identity in Karoubi (4.11 page 87)
$K^{-n}(X,Y):= K((X-Y)\times \mathbb{R}^n)$
Applying this,
$K^{-n}(S^1\times X,\{0\}\times X) = K((S^1-\{0\})\times X\times \mathbb{R}^n)=K(X\times \mathbb{R}^{n+1})=K^{-(n+1)}(X)$
I just blindly applied the identity so I'm not completely sure.
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Are charged particles cold? Are charged particles colder than neutral ones?
If a charged particle is vibrating due to temperature, it will release some of its energy as electromagnetic waves. So that means it's losing energy, cooling itself off. Is there an error in my logic?
| Temperature is defined for a system in (at least local) thermal equilibrium. The electromagnetic field is present everywhere and, when in thermal equilibrium, has a blackbody spectrum. Thus, a body made of charged particles "vibrating" at temperature $T$, in the presence of an electromagnetic field at the same temperature $T$, will on average gain as much energy from the field as it loses to the field.
A body "will release some of its energy as electromagnetic waves" at a net energy loss only when the electromagnetic field is initially colder than the body. So yes, a body can cool (here by radiation), but only because it is not yet in equilibrium with its environment.
Meanwhile, uncharged particles simply lack a coupling to the electromagnetic field, so they can maintain a different temperature from that of the field. They do not necessarily stay warmer than charged particles. For example, if placed in a hot oven, uncharged particles would stay cooler because they are not heated by radiation.
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If photons have also particle properties why should they not collide with each other? If photons have also particle properties why should they not collide with each other? Collisions between fermions are possible as collisions between fermions and photons(bosons) except collisions between photons (that are described by Albert Einstein as particles in the photoelectric effect experiment). Why?
| If photons are neutral particles with a very small size, the interaction may be very improbable (small cross section). But there is a more long-distance interaction of photons - they are bosons and thus somewhat "attract" each other, roughly speaking. Lazers are typical devices using this effect.
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Is energy really quantised? I'm currently doing an introductory course to quantum mechanics, and came across an assumption that Planck used in solving the UV catastrophe. From what I understand, he essentially stated that that the change in energy cannot be smaller than $hf$. So generally $\Delta{E}$ = $n*hf$ where $n$ is a real number. This makes sense, but I never understood how energy is truly then quantised, as can't light take an infinite number of possible frequencies? (not at the same time but just generally). Maybe I'm just misunderstanding the statement or making it overcomplicated - in any case please do shed some light.
| The crucial insight made by quantum mechanics is that electromagnetic waves in the cavity can be described by simple harmonic oscillators. If the angular frequency of the mode of oscillation is $\omega $ (FIXED) , then the energy associated with this mode is given by
$$E_n=\hbar \omega \left(n+\frac{1}{2}\right)\ \ \ \ \ n=0,1,2,\cdots $$
which are quantized.
NOTE that we are looking at a particular mode of frequency, not the whole spectrum.
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Mechanical energy in a body moving upwards Why is it that mechanical energy is always conserved, I mean when an object is thrown in air, why does the kinetic energy convert to potential energy and not any other form of energy?
| Energy forms are associated with forces. Gravitational forces cause the presence of gravitational potential energy. Therefor we might expect this energy form to be involved.
Had other types of forces, such as electrical forces, elastic forces etc., been involved then we would have expected electric potential energy, elastic potential energy etc. to be involved.
In fact in your scenario there might be other forces involved. Such as air drag. Then energy would have been lost to air resistance (heat lost to the air). This is typically a tiny amount so we often ignore it for low speeds.
Finally, note that the statement "mechanical energy is always conserved" is incorrect in general. Mechanical energy is only conserved when only mechanical forces are involved.
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Graph Interpretation of Gravitational Waves In the image is the data recorded by the LIGO's 2 observatories in USA. What is its interpretation? I mean what does the zig zag lines represent? Similarly, what does the blond red and blue lines (that seem like exponentially increasing up) represent? Could you please clarify? (Note: I am a highschool student so please make sure a high schooler like me too understands it.)
| LIGO works essentially by monitoring the separation of two large mirrors.
The zig-zag lines could be thought of as by how much the separation of the mirrors changes over the course of the 0.2 seconds represented along the x-axis. The separation oscillates in response to the passage of the gravitational wave. Confusingly, there are no indications of the size of this signal on the y-axis (the frequency numbers refer to the other plot, see below).
If you look carefully at these zig-zag lines you will see that the oscillation appears to grow in strength (the amount by which the signal oscillates) and in frequency (the peaks are getting closer together in time) before dying away at the end of the data sequence. This is the classic "chirp" signature of the merger of a binary black hole system.
The coloured plots show how the frequency of the oscillation (the numbers on the y-axis) increases with time, whilst the intensity of the colour of the signal represents the growing strength of the signal.
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Why do I feel centrifugal force If I move with constant speed on a turn? Title sums it up pretty much, I'm studying for my physics exam right now and I just can't wrap my head around this
| Acceleration is the change in velocity $$\frac{d\vec v}{dt}$$. Importantly $\vec v$ is a vector, meaning that it has both magnitude and direction. So changing the direction of your velocity vector is indeed an acceleration. In uniform circular motion your velocity vector is not constant because its direction is changing, even though the magnitude stays the same.
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Scalar field displacement from the minimum of the potential gives rise to particles/dark matter, why? In This paper (Kobayashi et al -- Lyman-alpha Constraints on Ultralight Scalar Dark Matter: Implications for the Early and Late Universe) it says, at the beginning of Section 3.1:
A light scalar field stays frozen at its initial field value in the early Universe. Hence, any initial displacement from the potential minimum gives rise to a scalar dark matter density in the later universe.
I don't understand this statement. Can someone explain its meaning? Why would such a configuration give rise to matter later in the universe? Is it due to the fact that later in the universe the scalar field would oscillate and oscillations can be seen as particles?
Sorry if the question is not clear, I studied physics quite a long time ago and study these things in my free time so there are many gaps in my understanding of fundamental physics and Cosmology. Feel free to be as technical as you wish but please remember I'm not expert or anything
| If there is a non-zero particle field in the early universe, this means there are particles present. These particles are driven away so fast from each other (by inflation) that they can't return to the zero-field configuration anymore. So they are "frozen".
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Why don't the Earth's oceans generate a magnetic field? Many questions have been asked here about why the Earth has a magnetic field, e.g.,
*
*What is the source of Earth's magnetic field?
*How does Earth's interior dynamo work?
*How can an electrically neutral planetary core be geodynamo?
*Why does the Earth even have a magnetic field?
At the risk of oversimplifying a bit, the answer is the dynamo theory. Convection in an electrically conductive, rotating fluid – in this case, the molten metal in the planet's core – creates electric currents that, in turn, generate a magnetic field.
Why doesn't the same thing happen in the oceans? A large ocean like the Pacific would appear to have all of the general properties required for a dynamo. It is made of conductive saltwater; it has significant bulk flows (indeed, ocean currents are much faster than convective currents in the core); and it rotates with the planet. Is the higher resistivity the key difference? If so, would a saltier ocean be able to generate a magnetic field?
| Maybe it's good to point out how this field comes about. You would expect no field as the currents are both equally positively and negatively charged. On their own they don't produce a magnetic field. This only comes about if tboth currents interact with magnetic field of the Earth. These currents are separated because they move through the magnetic field. When they are separated they can produce their own local magnetic fields not canceled by the opposite current (which would be the case if they were not separated),
So if the Earth itself didn't produce a magnetic field the oceons would be neutral.
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Explain the theory behind this problem(if possible using number line or graph or any other method of pictorial representation) In a situation in which data are known to three significant digits, we write 6.379 m = 6.38 m and 6.374 = 6.37 m. When a number ends in 5, we arbitrarily choose to write 6.375 m = 6.38 m. We could equally well write 6.375 = 6.37 m, “rounding down” instead of “rounding up,” because we would change the number 6.375 by equal increments in both cases. Now consider an order of-magnitude estimate, in which factors of change rather than increments are important. We write $500 m$ ~ $10^{3}$ m because 500 differs from 100 by a factor of 5 while it differs from 1 000 by only a factor of 2. We write 437 m ~ $10^{3}$ m and 305 m ~ $10^{2}$ m. What distance differs from 100 m and from 1000 m by equal factors so that we could equally well choose to represent its order of magnitude as , $10^{2}$ m or as , $10^{3}$ m?
| When we are using order of magnitude estimates we are typically using logarithmic scales, and when we are dealing with powers of ten we specifically use base ten logarithms. So $\log_{10}(10) = 1$, $\log_{10}(100) = 2$, $\log_{10}(1000) = 3$ and so on. Then the "mid point" of $100$ and $1000$ would be the number given by:
$$ \log_{10}(x) = 2.5 \tag{1} $$
That is any logarithm less than $2.5$ is rounded down to $2$ and any logarithm greater than $2.5$ is rounded up to $3$. So the number you are looking for is the $x$ in equation (1), and rearranging this gives:
$$ x = 10^{2.5} = 100\sqrt{10} \approx 316.23 $$
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A space ship has maximum proper acceleration of $a_0$. How close can it fall freely towards a black hole before it can no longer accelerate away? Edit: To clarify, all motion is radial only.
Classically, the answer to this is pretty obvious. You just find the distance from the black hole at which the gravitational acceleration matches $a_0$ and then determine that once the spaceship gets closer to the black hole than this radius, it cannot escape the gravitational pull. Unfortunately though the Universe isn't classical.
But that doesn't change the fact that we feel gravitational acceleration in our non-inertial frames of reference. Now, this question has made me realise that most textbooks and courses on general relativity tend to deal with things following geodesics, so we never actually worry about problems like this.
In this case, we have a spaceship along a non-geodesic world line that is in a curved spacetime. Is there any limit (aside from the Schwarzschild radius) on how close it can get to the black hole before it can safely accelerate away? How do we account for the gravitational force on the spaceship in GR?
| Assuming you are talking about radial motion the limit is when the proper acceleration for a stationary observer is equal to $a_0$. This is calculated in twistor59's answer to What is the weight equation through general relativity? The proper acceleration at a distance $r$ is given by:
$$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$
Note that this is similar to the Newtonian result but modified by the extra factor of $1/\sqrt{1-\frac{2GM}{c^2r}}$. Equate this to the maximum acceleration of your spaceship, $a_0$, and solve for $r$ to get the smallest distance from which the spaceship can escape.
Note that the proper acceleration goes to infinity at the event horizon i.e. when $r = 2GM/c^2$. At the event horizon it is impossible to escape no matter how powerful an engine you have.
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How do sound waves cause air molecules to oscillate? I would like to clear up some confusion about the mechanics of air particles that are propagating a sound wave.
I understand that there is no net movement of air molecules when a sound wave passes through air. Instead, the particles oscillate and the wave is propagated through various elastic collisions between air molecules which cause the compression to keep moving forward.
What I don’t understand is how the air molecules move back to approximately their original position after colliding with the other particles to keep the wave moving further. Doesn’t this seemingly violate the laws of conservation of momentum? (This can’t be the case since sound exists) If a particle hits its neighbor, and that neighbor molecule now has the momentum from the wave to keep moving forward, how can the original particle have the momentum to move backward and to its original place.
I would also like to clarify that I understand gas molecules have their own random motion in addition to the wave motion and was wondering if this had something to do with the aforementioned phenomena.
| Without the random movement of molecules, it is true that a vibrating membrane for example would create a depleted region around. It is like a fight in the middle of a crowd. Suddenly there is a clear region around the event and a wave of pressure that extends for some distance.
But because of that random movements, any depleted area is filled almost immediately. The same reasoning is valid along the propagation. The net displacement of molecules in the wave front is like a vibrating membrane pushing the next layer, and the random movements fill the gaps behind.
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Why is difference of points not a valid definition for a vector in curved space? In page-49 of MTW (1973 edtn), the following picture is shown:
After seeing this picture, the question which arose in my head is why exactly can we not define a vector as difference of points in curved space?
| For one thing, vectors obey linear structure, i.e. we can write $\vec{v}$ as sum of two vectors $\vec{v_1}$ and $\vec{v_2}.$
So if we want to have operation $A=B+\vec{v},$ we should also demand that operations $$A=B+\vec{v}_1+\vec{v}_2=(B+\vec{v}_1)+\vec{v}_2=C+\vec{v}_2$$ and
$$A=B+\vec{v}_2+\vec{v}_1=(B+\vec{v}_2)+\vec{v}_1=D+\vec{v}_1$$
produce the same point.
In curved space however, going first in the direction of $\vec{v}_1$ and then $\vec{v}_2$ does not produce the same point as going first in the direction of $\vec{v}_2$ and then $\vec{v}_1$. The difference between these two paths actually defines Riemann tensor, which characterizes curvature of space and is zero only if space is flat.
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What is the drag coefficient of a semi-cylinder travelling at laminar region but with high Reynolds number?
For Laminar flow along a cylinder, the drag coefficient is about 1.
Question:
Can we assume that for Laminar flow along a semi-cylinder, the drag coefficient is about 0.5 because the pressure drag dominates greater than frictional drag at laminar region but with high Reynolds number that is just before $10^5$?
| The variation of drag coefficient of a semi-circular cylinder with orientation angle is shown below, taken from Yamagata et al. 2016. The case you are interested in is $\theta=0$. The measurements were taken at $Re=6.4\times10^4$. Taking the drag coefficient to be $0.5$ is a reasonable approximation.
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How to reconcile time dilation in relativity with what you actually see?
If a traveler is moving at $0.5c$ towards a clock which is located one light year away, his relativistic time dilation is $1.15$. But in the time he takes to arrive at the clock, he must catch up the 1-year lag that he saw when the clock was 1 light year away, in 2 years travel time. This means he must see the clock moving at twice its normal speed, not $1.15$.
What am I doing wrong?
Why was this question marked as "unclear"? I don't know how else to ask it.
If this question is so unclear, then how is it that I got three perfectly clear answers?
Anyway, thanks to those who took the time to answer.
Apparently, what I was doing wrong was to ask what I was doing wrong? Go figure.
No, this isn't my homework, as I have not attended any university since 1999.
| You are being downvoted unfairly. Your answer is logically OK (in the sense that there is more to it than just the Lorentz Transform), but the calculation is wrong. Use the doppler effect, and you will see the clock appearing to move $\sqrt {1.5 / 0.5}$ times (1.73) normal speed as you approach (it will also appear to run slower - by a factor of $\sqrt {0.5 / 1.5}$ - if you recede at 0.5$c$).
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What is a bulk state and bulk bands? I am a bachelor student and I started studying topology and I came across two terms I have never seen before: Bulk band structure and bulk states.
Can someone explain these two terms or provide me a reference where I can check? Everything I found on the internet is for people who already know what are they.
thanks
| The bulk states are the quantum states of electron inside the volume (bulk) of a crystal, obtained from Schrödinger equation. These states are usually labeled by the two indices, n and k representing respectively the band index and wave number. The plot of the energy of these states versus k for various values of n is called bulk band structure. In contrast to bulk states are edge states, which are localized on the surface (boundaries) of crystal and not in the bulk.
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If $E=mc^2$, then why do different substances have different calorific values? Today during a classroom discussion, I realised that if we consider the equation $E=mc^2$, then we are establishing a relation between energy and mass but we often observe that different substances produce different amount of energy when they are burned. For example: Burning a kilogram of wood will not produce same amount of energy as 1 kilogram of petrol.
| Although often used, I would not use the term "conversion to energy". $E=mc^2$ says that energy and mass are the same things (in the center of mass frame).
If a system at rest has some mass $m$, it has, at the same time, a corresponding total energy $mc^2$. No conversion is going on. I would rather say that in any reaction (chemical or nuclear) a certain amount of energy is released as radiation and then it is considered lost by the system. As a consequence of such a change of energy of the system, the energy of the reaction products, and therefore their total mass is decreased.
The calorific value of a material corresponds to the difference of energy between products and reactants after the energy corresponding to some interaction has been released in the form of radiation.
From a pound of oil one can get $2.4~10^7~$J. From a pound of $~^{235}$U, $3.7~10^{13}$ J. The corresponding mass difference between the original system and the final atomic/molecular products is $2.7~10^{-10}$ kg in the case of oil and $4~10^{-4}$ kg in the case of Uranium. Below our ability of measuring in the first case, small, but measurable, in the second one. But in both cases the variation of mass is a small fraction of the total mass of the system.
In order to transform the whole system into radiation, obtaining the same calorific value per unit of mass, one would need to have half of the system made by matter and the other half by anti-matter. Something which is presently possible only in the high energy physics labs, by using beams of particles/antiparticles.
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How is Noether's theorem applied in fluid dynamics? As the title says, I'm trying to understand the applications of Noether's theorem in fluid dynamics. I was looking for references in this context, but I only find very old papers, where explanations are very limited. Are there any recent studies for the application of symmetries and conservation laws (momentum, energy conservation ..), for example, in fluid simulators?
| You can find a nice discussion about the virtual (added) mass force that arises when a submerged body accelerates relative to the its surrounding fluid in Gregory Falkovich’s Fluid Mechanics book. Apparently, this force is a consequence of the conservation of quasimomentum, which is the conserved quantity associated to the symmetry of the fluid equations with respect to linear translations of the particle in an infinite fluid, while keeping the fluid (bulk) still. This is not a systematic study of the application of Noether’s theorem in the context of fluid mechanics, but I think you will still find this application very relevant.
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How does stroking metal with a magnet magnetize the metal? How does stroking metal with a magnet magnetize the metal? I have thought about it for a while and got absolutely no clue as to how the magnetic domains in the steel rod got shifted to the same direction (regions of metallic ions pointing the same direction, in an unmagnetised metal these domains cancel each other).
related link showing the stroking of metal with a magnet. Its particularly the manner in which we are stroking that could be causing the magnetisation. The video shows it being repeatedly stroked the same direction
| If a ferromagnetic material is placed in, and then removed from, a magnetic field, the “remnant” magnetization within the material will depend on the strength of the field. Stroking a bar will bring the strong field from the pole of the magnet into close proximity with each part of the bar. Then the magnetic “domains” within the bar tend to drop back into alignment with nearest crystal axis.
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Can you exit the event horizon with a rocket? The reason given in most places about why one cannot escape out from an event horizon is the fact that the escape velocity at the event horizon is equal to the speed of light, and no one can go faster than speed of light.
But, you don't really need to reach the escape velocity to get away from a massive object like a planet. For example, a rocket leaving earth doesn't have escape velocity at launch, but it still can get away from earth since it has propulsion.
So, if a rocket is just inside the event horizon of a black hole, it doesn't need to have the escape velocity to get out, and it should at least be able to come out of the event horizon through propulsion. Also, if the black hole is sufficiently large, the gravitational force near the event horizon will be weaker, so a normal rocket should be able to get out easily.
Is this really theoretically possible? If it was just the escape velocity being too high was the problem of getting out, I don't see any reason why a rocket cannot get out.
This is a similar question, but my question is not about a ship with Alcubierre drive.
| It should be nearly impossible to do. You have to be roughly the speed of light to escape. But If your position is far from the black hole, it looks like you can still escape.
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Quantum mechanics and logical statements I am a math student and currently working on my bachelor thesis with a philosophy professor. The subject is paraconsistency and thus also dialetheism which is the believe that a statement can be true and false at the same time.
I had a general/introductory course to QM and one experiment I recall is the Stern-Gerlach experiment where you measure spin with a magnetic field.
Consider the statement (I choose $\varphi$, $J$ to specify a atom and measurement, is that a good idea?):
The next specific silver atom $\varphi$ I send trough the Stern-Gerlach experiment has spin up during that specific measurement $J$.
As I understood the validity of this statement can only be determined after executing the experiment. Because of our understanding of QM and the current accepted theory this is inherently probabilistic and not predictable?
What do you think about the formulation of the statement? Do you think I can say that the above statement is neither true or false (or arguably both) before actually sending the atom through the experiment? Or do i have some misunderstanding about QM and what statements you can make about it?
| Let us say that the setup is such that you can get up or down with some probability. Then the truth of your statement depends on who do you ask, that is, on the interpretation of QM that that person has. If he is a local non-realist, the statement, before the measurement, is false because the particle has no determined value of J, the value does not exist until you measure. For someone with a non-local hidden variables interpretation, the value of S does exist before the measurement, and the statement would be true. I dont know if there are other interpretations of QM in which the statement could be both false and true, or any other combination you like.
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Will there be potential difference across a resistor if the circuit is open?
In such an open circuit, will there be potential difference across the resistor? More specifically, will charges accumulate on the left side of the resistor? or will it be able to flow through the resistor and stop at the open end, resulting in 0 potential difference. Because if charges were to accumulate on the left of the resistor, that would mean there can be a potential difference across the resistor even though there is no current?
| Both sides of the resistor will be at the same potential, so there will be no potential difference. We use the term potential difference because potentials are only defined up to the addition of an arbitrary constant. Thus, if we say the potential on each side of the resistor was zero before the cell was connected then the potential on both sides of the resistor will still be zero after the cell is connected and if the positive side of the cell was connected as in your diagram then the potential of the negative side of the cell would be -5V (assuming the 5V cell). Alternatively, if only the negative side of the cell was connected then the resistor would still have zero potential on either side after the cell is connected and the positive side of the cell would be at 5V.
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Why is everything not invisible if 99% space is empty? If every object is $99$% empty space, how is reflection possible? Why doesn't light just pass through?
Also light passes as a straight line, doesn't it? The wave nature doesn't say anything about its motion. Also, does light reflect after striking an electron or atom or what?
| Yes, atoms are mostly empty, comprising a nucleus with electrons surrounding the nucleus$^1$.
But electrons are not to be considered point particles that follow fixed orbits. In fact electrons are probability clouds that are smeared around the nucleus, and fill the volume of the atom.
It is similar for bonds between atoms and in molecules, and the electron clouds fill the space in between the atoms and around them.
Also, because of the Pauli exclusion principle, no two electrons in an atom can simultaneously be in the same state, so that atoms with increasing numbers of electrons, will have these electrons with average distances further and further away from the nuclei, meaning these probability clouds must sweep an ever increasing physical volume.
Light does indeed interact with these clouds, and because of all these reasons, "seeing through" matter is not possible.
$^1$ A great majority of the matter in an atom is concentrated in the nucleus, which is very small compared to the region where the electron clouds reside. And the volume of an atom is on average about $15$ orders of magnitude larger than the volume of a nucleus. So the electrons do indeed occupy a huge portion of the volume of an atom.
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Why is acoustic intensity inversely proportional to density of the medium? The definition of sound (acoustic) intensity is given by
$$ I = {p^2 \over {\rho c}} \;\;\;\; \text{or} \;\;\;\; I = {p^2 \over {2\rho c}}$$
I've seen both definitions in different textbooks and am not sure which equation is more accurate. But in either case, the relation of intensity and density of air is:
$$ I \propto {1 \over \rho} $$
This contradicts my intuition.
I would hear no sound in a vacuum because there is no molecules in the medium to scatter the vibration from the source. If I start adding some molecules in the medium (low density), the source will vibrate these molecules in the medium, acoustic energy can be transferred and sound may be heard if there are just enough molecules hitting by eardrum (and within the audible frequency range). If I add a whole lot more molecules in the medium (high density), many more molecules will vibrate and hit my eardrum, and I would therefore hear a louder sound (or detect a higher acoustic intensity).
So why the equation suggests the opposite relation? Could it be that the instantaneous pressure is also a function of density, or $ p(t, \rho) $, in a way that pressure is proportional to density of higher order to offset the drop of $ 1 \over \rho $ have on intensity?
| Your intuition is in the right direction. Nevertheless, you are only considering the amplitude and the intensity is the power per unit area. So, you need to consider that when a material is more dense you need a higher increment in pressure to get the same increment in particle speed.
If we write it mathematically, we get the following relationship between the pressure and particle velocity
$$v = \frac{p}{\rho c}\, $$
where $\rho c$ is the characteristic acoustic impedance of the medium. Thus,
$$I = \frac{1}{2}\operatorname{Re}\{p v^*\}\, ,$$
or
$$I = \frac{1}{2}\frac{|p|^2}{\rho c}\, .$$
| {
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Do Newton's laws of motion hold true in non-inertial frames of reference? My book derived the formula for the acceleration of a rocket at any instant in the following way:
$v_r=$ velocity of gas released from the nozzle relative to the rocket
$dt=$ infintesimal time interval
$dm=$ mass of the gas released from the nozzle in time interval $dt$
$dP=$ momentum of the gas released in time interval $dt$
$F=$ thrust force acting directly opposite to the direction of the release of gas
$M=$ mass of the rocket after time interval $dt$
We know from Newton's 2nd law,
$$F=\frac{dP}{dt}$$
$$\implies F=\frac{dm}{dt}v_r$$
$$\implies a=\frac{1}{M}\frac{dm}{dt}v_r$$
In this derivation, the velocity of the released gas, $v_r$, has been calculated from the perspective of the rocket, which is constantly accelerating, making it a non-inertial frame of reference. Newton's laws don't hold true in non-inertial frames of reference, but we used newton's 2nd law in this derivation. So, how is this derivation correct?
PS: A similar derivation can be found in Fundamentals of Physics by Halliday, Walker & Resnick
| This derivation makes the most sense in momentarily comoving reference
frame.
Say you want to derive the equation at some arbitrary time $t$.
At this particular time, there exists an inertial reference frame where
the velocity of the rocket is exactly zero. In this reference frame, the
rocket was moving backwards and decelerating at times $t'<t$, and will
be moving forward at increasing speed at times $t'>t$.
Since this frame is inertial, Newton's second law
$$F = \frac{dP}{dt}$$
does hold. Here, the change in momentum is
$$dP = v_r dm$$
where $v_r$ the velocity of the released gas. But since the velocity of
the rocket is zero, $v_r$ can also be interpreted as the velocity of gas
relative to the rocket.
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How can I find the angle after the last lens from in the last lens?
I want to find the angle after the last lens when the beam diffracts. How can I do that? Is it the field of view? But for the field of view formula it takes account the size of the sensor. I want to find the relation of this angle and the origin of the point source. So when I launch the beam from a different distance the angle will change.
| If this is a camera with a lens, the angle $\alpha$ you are interested in is really the angle of view, which can be calculated as
$$\alpha = arctan \left( \frac{d}{2f} \right) $$
You may have a look here for details about this equation https://en.m.wikipedia.org/wiki/Angle_of_view (for angle of view in photography you have to scroll down a bit). In this equation, $f$ is the focal length of your system and $d$ the physical dimension of the camera chip (width, height or diagonal, depending on what is of interest - your camera has different angles of view in the horizontal and vertical as long as the chip has different width and height). You could even calculate the angle in which a specific pixel is pointing by replacing $\frac{d}{2}$ by the distance of the repective pixel from the center of the chip.
The above equation assumes that the camera chip is in distance $f$ behind the principal plane of your lens, meaning you focus to infinity. If you focus to closer objects it might be necessary to increase the distance between camera and lens in order to get a sharp image. You need to use the real distance between lens and camera in the above equation then.
Actually you could test the correctness of the formula by taking an image of some object of known length (or you measure the object's length).
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Why is the absolute magnitude of a source negatively correlated to its luminosity? The absolute magnitude $M$ of a source of luminosity $L$ is given by $M = -\frac{5}{2}log(L) + constant$. Why is $M$ defined with a negative correlation to the luminosity? Wouldn't it be more intuitive to have a larger absolute magnitude for a larger luminosity of the source?
| The definition of the magnitude $M$ is such that smaller values correspond to brighter objects. Hence, the larger the luminosity $L$ the smaller the absolute magnitude $M$.
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Does real life have "update lag" for mirrors? This may sound like a ridiculous question, but it struck me as something that might be the case.
Suppose that you have a gigantic mirror mounted at a huge stadium. In front, there's a bunch of people facing the mirror, with a long distance between them and the mirror.
Behind them, there is a man making moves for them to follow by looking at him through the mirror.
Will they see his movements exactly when he makes them, just as if they had been simply facing him, or will there be some amount of "optical lag"?
| As the speed of light is finite, sure enough there is some lag, but let's evaluate how big that lag is. Considering that the mirror is 100 meters away, than the lag will be $$2\times 100\: \mathrm{m}/(3\times 10^8\:\mathrm{m/s}) = 667\:\mathrm{ns}.$$ Comparing it to average human reaction time of about $0.1\:\mathrm{s}$, one can conclude that it is impossible for a naked eye to notice any lag at all.
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What does it mean for a nucleus to be more stable? When we say that a nucleus is more stable than another nucleus, are we saying that the nucleus is at a lower energy level than the other nucleus? In chemistry, a more stable compound is one that is at a lower energy level, is this the same when we talk about a nucleus?
If yes, wouldn't this make all nuclei with a larger mass more unstable since they will be at a higher rest mass energy.
But clearly, this is not the case as the binding energy per nucleon graph increases initially before decreasing, suggesting that a heavier nuclei can be more stable than a lighter one.
| Wnen a nucleus (or a particle) is called stable, it usually means that it does not decay, i.e., it is not radioactive. In practice one would describe as stable the nuclei and particles that decay over very long times, so long that their decay rate can be considered negligeable for practical purposes. The stabilities of two nuclei are compared in the same sense: the more stable nucleaus is the one having longer lifetime (smaller decay rate).
This is not quite identical to the stability of chemical compounds. The nuclei can be often treated as isolated, so that the value of their decay rate is dominated by the strength of their coupling to the environment. Temperature and hence the excitation energy play minor role in this respect (although they certainly matter, e.g., when considering many nuclei in a nuclear reactor). On the other hand, chemical compounds are nearly always a part of a solution or a gaseous mixture, and their decay rate varies significantly with temperature according to the activation law:
$$
\Gamma\propto e^{-\frac{E_b}{k_BT}},
$$
where $E_b$ is the binding energy.
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Can we cool Earth by shooting powerful lasers into space? In a sense, the climate change discussion revolves around the unwanted warming of the earth's atmosphere as a whole.
It seems a bit too obvious to be true, but could we cool the atmosphere by simply shooting that unwanted energy somewhere else?
Energy might be collected from remote expanses where it would otherwise be somewhat pointless to harvest it due to lack of habitability and resulting anticipated losses due to transmission (ocean surface, ???)
If so, what would be a good place to shoot it?
| Like other answers say... Your problem is that the act of forcibly pumping energy one place to another, beyond what nature itself would do, takes additional energy.
So for example, moving the warmth from inside your fridge to outside it, costs extra energy. Informally (not literally to scale), you can think that the fridge interior loses 100 units of heat energy, so it does cool down. But the outside - the room and ultimately the planet as a whole - gains 110 units of heat energy: 100 units moved from inside the fridge, but also an extra 10 units from the work the pump must do, to move that 100 units from inside to outside. Overall the planet heats up.
Your lasers will do the same. They will send 100 units of heat energy into space, but to do so will require an extra 10,000 units of heat into the room, or the planet generally. (Not literally to scale). Big lasers take a lot of energy to fire up.
Similarly with anything collecting low grade heat and concentrating it as high density heat, or collecting solar energy to pump heat around. These all pretty much will add more heat to the planet than they remove, as they move heat round.
If you want to remove heat from earth, the best ways are 1) stop adding heat, or be more energy efficient and reduce the energy you use on earth, 2) make it easier for heat to escape (reduce heat-retaining mechanisms: CO2, methane, etc), 3) move energy consumption into space.
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Gravitational wave and 1st law of thermodynamics Introduction:
A prediction of the general relativity is that any moving mass produces fluctuation in the space-time fabric, commonly referred as Gravitational-Wave.
This prediction was recently confirmed by the LIGO experiment.
The generation of such gravitational waves requires energy, as stated on the wiki article linked above:
Water waves, sound waves, and electromagnetic waves are able to carry
energy, momentum, and angular momentum and by doing so they carry
those away from the source. Gravitational waves perform the same
function. Thus, for example, a binary system loses angular momentum as
the two orbiting objects spiral towards each other—the angular
momentum is radiated away by gravitational waves.
The first law of thermodynamics states that:
The first law of thermodynamics, also known as Law of Conservation of
Energy, states that energy can neither be created nor destroyed
Given that, one can imply that any moving object having a mass would create gravitational waves - even ever so tiny -, thus having a drag.
Question:
How does a system, for instance earth-moon orbit, can be stable and not decaying over time over the model of the general relativity? (Where does the energy comes from?)
Is this question solved?
| The energy comes from the binary system itself and ultimately from the mass-energy of the binary components.
The binding energy of a binary system (the sum of its kinetic and potential energies) is negative. The acceleration of these masses, or more precisely, the accelerating mass quadrupole moment, produces gravitational waves that carry away energy and angular momentum. The binary components move closer together and the system binding energy becomes more negative and thus total energy is conserved.
This is happening in the Earth-Moon system but the flux of gravitational waves, which depends roughly on mass to the power of 5 and inversely on the component separation to the power of 5, is pitifully weak (about $10^{-5}$W) compared to a binary black hole system. The evolution of the Earth-Moon system is instead governed by the interplay of orbital and rotational energies caused by tidal forces.
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Does the physical singularity of the Reissner-Nordstrom metric have a ring structure? The physical singularity of the Kerr metric has a ring structure due to the axi-symmetric nature of the metric.
The Reissner-Nordstrom metric is the solution for a non-spinning, electrically charged black hole, and has two horizons: an event horizon and a Cauchy surface, the locations of both depend on the black hole's mass and charge.
Question: mathematically what is the structure of the physical singularity in the Reissner-Nordstrom metric? Is it a point like the Schwarzschild case? Is it a ring like the Kerr case? Or is it different from both? And why?
I've done some digging but cannot find a concrete explanation. The answer at this SE question is helpful, since it shows that even theoretically, the existence of the Kerr singularity could be a mathematical artifact. But I'm curious about if such a mathematical feature exists for the Reissner-Nordstrom metric, and why?
| It's worth noting that the Reissner-Nordstrom singularity is timelike, like how the Kerr singularity is. (and unlike the Schwarzschild singularity, which is spacelike)
but the metric is spherically symmetric, so the singularity cannot be a ring, because that would choose a special plane in the spacetime.
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Can a material have high permittivity and low permeability or the other way around? Since the electric and the magnetic fields are interrelated, can a material have high permittivity and low permeability or the other way around? Would it be possible to have a material that would be highly magnetized upon an applied magnetic field but have low levels of electric polarizability, or the inverse?
| It all depends from the temperature of the material and of course from the existence of unpaired electrons in the outer shell of the atoms.
For permanent magnets below the Curie temperature a stable magnetic field do to the alignment of the magnetic dipoles of the involved subatomic particles exist and on higher temperatures this alignment gets destroyed by the kinetic energy (vibrations) of the subatomic particles. Such magnets could be an isolator or a good electric conductor.
For other materials immobile electrons in the outer shell are in interrelation with the subatomic particles of the atom or molecule and this determines the alignment of their magnetic dipoles. For some materials the interrelation can be changed by an external magnetic field easily and for others not. Independent from this the mobility of outer electrons can be good or bed or be zero.
In short, search for insulators with electron configurations which magnetic dipoles are easy to influence and for conductors with high interacting dipole moments which are hard to change.
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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?
| They would form a hydrogen atom.
They would not merge, because merging would result in a neutron, which is heavier than a proton and an electron combined. However, if they are sufficiently forcibly "slammed" into each other, they might form a neutron (and an electron neutrino, which has an negligible mass). The neutron would later decay into proton, electron and electron anti-neutrino (it has a half-life of about 10 minutes).
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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?
| The correct answer is that energy is conserved. It is not pedantic, but simply correct, to insist that if the questioner meant kinetic energy, or mechanical energy, which would be conserved only in an elastic collision, then they should have said so. The conservation of energy is such a fundamental property of nature that any wording which risks confusing a student's understanding of it should be strictly avoided.
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What forces act on a droplets hanging from surfaces? What are the different forces acting on a hanging water droplet (tension of the droplet itself, surface adhesive forces, meniscus formation, etc)? What is the direction of each force, and how does each affect droplet stability?
Will these forces be affected if the surface is flat vs patterned with some tiny structures? (assuming the surface material is the same)
| *
*Adhesion, normal to the solid-liquid interface, found only very near
the solid-liquid interface, increasing with proximity to the
solid-liquid interface. Makes the droplet stick to the solid surface.
*Cohesion, radially inward and uniform
throughout - each little bit of water is pulling on all the adjacent little bits. Tends the droplet towards spherical/hemispherical shape.
*Surface tension, tangent to the surface in all
directions, found only at the liquid-gas
interface. Not really its own force, this is just a consequence of the bulk force imbalance between inside the droplet where the cohesion force is present and outside the droplet where there is no such force. Tends the surface towards spherical / hemispherical shape and opposes penetration of the surface.
*Gravity, downward, uniform throughout. Tends the droplet to an elongated shape for the configurations depicted.
If we slowly dialed up gravity, the one on the left would drip first, since the total adhesive force is smaller.
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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.
| A point cannot produce any magnetic field, it should have some length for it to produce any field.
Now coming to the second question. As pointed out by @Angry Refrigerator in their answer, dlXr shows the direction of the magnetic field, while I and distance from wire show its magnitude. It is not correct to say dB is proportional to dL. As you rightly thought, adding something to an infinitesimal number isnt correct and it is better for u to not think of it in that way. Maybe you can think of it qualitatively where length of wire doubled then mag. field double keeping everything same. But @Angry Refrigerator's interpretation is correct and think of it in that way.
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Negative energy solutions in Klein Gordon and Dirac equations It is often read that Dirac equation poses the problem of negative energy solutions (later solved by antiparticles and second quantization/QFT schema). My question is: negative energy solutions were obviously present also in the previous KG equation. Why they were not considered a problem? I read somewhere that this was related to the second order derivative in KG but it is not entirely clear to me.
| In QFT, when making a field operator, negative frequency solutions correspond to annihilation operators while positive frequency solutions correspond to creation operators. Not worrying about normalizations and things like that,
$$
\hat \phi(x) \propto \int d^3 k \left( \hat a(k) e^{- i k \cdot x} + \hat a^\dagger(k) e^{ i k \cdot x} \right).
$$
Furthermore, Dirac's initial worry, that the Klein Gordon current isn't positive definite, was really a misconception stemming from the fact that people didn't understand how to quantize spin $0$ particles at the time. It is only the negative frequency solutions which have a negative probability current, but those don't correspond to the probability current of actual particle states. Only positive frequency solutions, i.e. particles created on the vacuum, have probability currents which are indeed positive.
Dirac's initial interpretation was that special relativity required particles to have spin, but really this was just because there was a lot of confusion over how to interpret the different mathematical objects in QFT. There are certainly particles without spin, like the Higgs boson for instance.
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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 calculating the EPE as if there is no tension in the string before it is stretched from extension $\frac l {10}$ to extension $\frac l {5}$. But we know there is already tension $Mg$ In the string, and so the EPE initially stored in the string is $\frac {Mgl}{20}$.
The additional energy stored in the string by stretching it a further distance $\frac l{10}$ is
$\displaystyle \text {EPE}\left(\frac l 5\right) - \text {EPE}\left(\frac l {10}\right)
\\ = 4 \times \text {EPE}\left(\frac l {10}\right) - \text {EPE}\left(\frac l {10}\right) = 3 \times \text {EPE}\left(\frac l {10}\right) = \frac {3Mgl}{20}$
Another way of looking at this is to see that the mass has a lower gravitational PE at the end of the stretching, but this difference in gravitational PE is not lost - it is stored as an additional EPE of $\frac {Mgl}{10}$ on top of the work $\frac {Mgl}{20}$ done to further stretch the string. So two thirds of the extra EPE in the string comes from the work done by gravity and one third comes from the extra force required to further stretch the string.
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Why do physicists like to put the imaginary unit $\:i=\sqrt{-1}\:$ everywhere? There are many disagreements of convention between mathematicians and physicists, but a recurring theme seems to be that physicists tend to insert unnecessary factors of $i = \sqrt{-1}$ into definitions.
I understand this is only convention, but I’m curious about why this seems so widespread. Does anybody know about the “etymological” reason(s) for physicists’ $i$-heavy conventions?
$\newcommand{\dd}{\mathrm{d}}$
Thing
Mathematics convention
Physics convention
Lie algebra structure constants (Ref.)
$[L_a, L_b] = f_{ab}{}^cL_c$
$[L_a, L_b] = if_{ab}{}^cL_c$
Lie group transformations in terms of generators (Ref.)
$R_z(θ) = \exp(θJ_z)$
$R_z(θ) = \exp(-iθJ_z)$
Covariant derivative with $\mathbb{C}$-valued connection 1-form
$\nabla V = \dd V + A V$
$\nabla_μ V^a = ∂_μ V^a -iqA^a{}_{bμ} V^b$
Curvature of connection or gauge field strength (Ref., §7.4)
$F = \dd A + A ∧ A$
$F_{μν} = ∂_μA_ν - ∂_νA_μ \pm iq[A_μ, A_ν]$
I have a vague guess: physicists read and write $e^{iωt}$ a lot, and an exponential with an $i$ in it screams “rotation”. Fast forward to describing $\mathrm{SO}(n)$ rotations in terms of matrix generators, and an expression like $e^{iθJ_z}$ just “feels more familiar” so much so that an extra $i$ is pulled out of the definition of $J_z$.
Can that guess be supported?
Not sure about the third and fourth rows, though.
| Real classical observables are quantized as Hermitian operators, and $\partial_\mu$ is anti-Hermitian. So:
*
*Structure constants We seek Hermitian generators.
*Transformations We seek real $\theta$, unitary $R_z$ and Hermitian $J_z$.
*Covariant derivative We seek Hermitian $A,\,V$.
*Curvature/field strength We seek Hermitian $F$.
| {
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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.
| A charged sphere stays charged in vacuum because it takes a minimum energy for an electron to exit the physical material. This is called the work function.
Indeed, if you would just have a spherical shell distribution of truly free charges this would fly apart.
| {
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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
| The strength of gravity does vary from point to point on the Earth's surface, because the earth is not of even mass distribution. The images in the link you have provided are evidence of this.
The strength of the earth’s gravitational field is given by $$a=\frac{GM}{r^2}$$ where $M$ is the mass of the earth, $G$ is the gravitational constant, and $r$ is the distance from the center of the earth.
Since it is inversely proportional to the distance from center squared, the further you move away from earth, the weaker this field strength is.
| {
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Energy stored in the electric field of an electron Energy stored in the electric field or self energy of a solid sphere of radius $r$ and uniform charge distribution $q$ would be
$$\frac{3}{5}\frac{1}{4\pi\epsilon_0}\frac{q^2}{r}$$
This result could be derived if we take electrostatic energy density $u = \frac{1}{2}\epsilon_0{E^2}$ where $E$ is the electric field intensity. From work-energy theorem, this energy would be the total work done to configure this uniformly charged solid sphere from infinity. Since the charge is quantised, $q=ne$ where $e$ is charge of an electron. If $dq=e$ quantised charged units are assembled from infinity, work done on the system gets stored in the electric field.
But when $n=1$ and if we consider it to be an electron with its classical radius, we get the value of self energy as
$$\frac{3}{5}\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$$
Or at least we get something very finite.
I don't understand if this is the energy stored in the electric field and if so, is it intrinsic because with one $e$ charge, work-energy theorem doesn't make sense here. Or what about the error(s) in the assumptions leading to it?
Edit: I have come across this Wikipedia article (that I should have stumbled upon before asking here) and it happens to address most of my doubts (and inconsistencies).
| It's simply wrong!
An elementary particle or fundamental particle is a subatomic particle with no (currently known) substructure, i.e. it is not composed of other particles.
That means you can't consider an electron as a sphere of the finite radius with charge $e$. This is question is beyond the range of classical electrodynamics.
| {
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Why does $dG < 0$ imply that processes involving chemical reactions are spontaneous? Here is a short proof/derivation of why $dG < 0$ implies that a process is spontaneous (for constant temperature and pressure):
But this derivation assumes that only mechanical work is done on the system. If the process involves chemical changes the expression for internal energy becomes:
With this expression for $dU$ the last step in the derivation no longer holds: since $dG = dQ - TdS + \sum_i \mu_iN_i \neq dQ - TdS$, $dQ < TdS$ is not equivalent to $dG < 0$. It seems to me like there could be a hypothetical process for which $dG < 0$ but where $dQ > TdS$. With a process like that, the total entropy change (the environment plus the system) is negative (i.e. it is impossible). Yet, $dG < 0$ is commonly used by chemists as a criterion for the spontaneity of chemical reactions. What am I missing?
| We have, for the Gibbs free energy $G=\sum_i \mu_i N_i$ for species $i$ with numbers $N_i$ and chemical potentials. For a set of reactions denoted in stoichiometric notation
$$
\sum_i \nu_i X_i -\sum_f \nu_f X_f = 0
$$
and imposing conservation of species: $\frac{dN_i}{\nu_i} = dN = -\frac{dN_f}{\nu_f} = $ constant for all $i$ and $f$, yields
$$
dG = dN \left(\sum_i \mu_i \nu_i - \sum_f \mu_f \nu_f \right).
$$
For reactions that proceed "from left to right"
$$
\sum_i \mu_i \nu_i > \sum_f \mu_f \nu_f.
$$
where $i(f)$ denotes initial (final) reactants. For these reactions $dN<0$, implying $dG<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 structure of ice comes from hydrogen bonds. These occur because the electrons are more strongly attracted to the oxygen atom, so this atom is slightly negative, whereas the hydrogen atoms become slightly positive. You can think of the electrons as spending more time near the oxygen atom. When you put this together with the shallow V shape of a water molecule, the result is that the molecules come together with the hydrogen atoms on one molecule near the oxygen atoms on other molecules.
The resulting structure is not the smallest possible packing for the molecules. Water is denser than ice because in water the molecules can move into positions where there is closer packing as they move around. As the temperature of the water rises to approx 4°C the effectiveness of the hydrogen bonding decreases as the molecules move faster and so the water continues to contract. Above that temperature the speed of the molecules and collisions between them cause expansion, in the same way as for most substances.
Under very high pressures the hydrogen bond forces in ice can be overcome, and other, denser, forms of ice can exist.
| {
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Law of conservation of energy using work done vs maxwell theory of electromagnetic radiation Consider the motion of a charged particle in a uniform magnetic field. $\vec{B} = B_0(-\hat{k})$. Let the initial velocity with which it enters the field be $\vec{v_i} = v_0(-\hat{i})$. It is well known that it follows a circular path of radius $R = \frac{m v_0}{qB_0}$.
*
*Using Work Energy Theorem
$$∆K = W_B = \int \vec{F_B}.\vec{v}dt = \int q(\vec{v}×\vec{B}).\vec{v}dt = 0$$
$$ K_f = K_i $$ Therefore speed of the charged particle and radius of the path remains constant.
*Using Theory of Electromagnetic Radiation
Direction of velocity changes continuously. Therefore, the charged particle is in accelerated motion. Therefore, it continuously loses energy in the form of electromagnetic radiation. Therefore it must follow a spiral path.
Which of the following is correct?
N.B. I am high school student. So, please limit the answer within high school mathematics.
| There is a formula for the Power of the emitted radiation due to an accelerating charge, the Larmor formula
$$P=\frac{q^2a^2}{6\pi\epsilon_0c^3}$$
where $a$ is the acceleration of the charge also $\frac{Bqv}{m}$.
In calculations the energy lost by radiation can often be ignored, you can put in various numbers to see the size of the effect, but technically you are right, the particle would follow a spiral path.
| {
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"timestamp": "2023-03-29T00:00:00",
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