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Can we find a vector if its dot product and cross product with another vector is given? If I have two vectors $\vec{b}$ and $\vec{v}$, and I know that
$$
\vec{b} \times \vec{v} = \vec{c}
$$
and
$$
\vec{b}\cdot\vec{v} = \lambda
$$
can I find the $\vec{v}$ vector in terms of the $\vec{c}$ vector, $\vec{b}$ vector, and $\lambda$? I have been struggling with for quite sometime. And I hope it's time I asked for help.
| Notice, we have $\vec b\times \vec v=\vec c$ or $ \vec v\times \vec b=-\vec c$ & $\vec b\cdot \vec v= \lambda$
now,
$$\vec b\times \vec c=-\vec b\times(\vec v\times \vec b )$$$$\vec b\times \vec c=-\vec v(\vec b\cdot \vec b)+\vec b(\vec b\cdot \vec v)$$
$$|\vec b|^2\vec v=\lambda\vec b-\vec b\times \vec c$$
$$\vec v=\frac{\lambda\vec b-\vec b\times \vec c}{|\vec b|^2}$$ $\ \ \ \forall\ \ \ \ |\vec b|\ne 0$ & $\lambda\vec b\ne \vec b\times \vec c$
| {
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Does voltage always cause current? I'm wondering if the following statement is fully correct:
Voltage causes current through a closed circuit, but through an
inductor it is the change in current that causes a voltage.
Obviously there is no current without voltage. In a simple DC circuit there's no doubt that voltage causes the current to flow.
However in AC circuit with an inductor, the voltage drop across the inductor is proportional to the rate of change of current. So we can have 0 voltage + peak current and vice versa.
But I wouldn't say that current causes voltage in this case, because there would be no current change without a voltage source connected to the circuit!
| The magic here is Jefimenko's equation (of causity). It is charge, and moving charges that produce and respond to a field.
Let's suppose you have a voltage field. These can exist, but if the charge is fixed, then little current flows. This is what a resistor does. On the other hand, if the field can cause a displacement of charge, then a current will flow until a counter-field exists to stop it.
Note for example, no current flows across an inductor (transformer) or capacitor. The energy is transferred, and creates a new current on the other side of these devices.
But there are situations where voltage does not cause current, sometimes the charge and field must build high enough for the charge to discharge through a spark, or bolt-of-lightning.
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What are good references for learning about Biophysics at graduate level? I am graduating this year on both Physics and Mathematics and I want to pursue a career in research, more concretely I want to study Biophysics.
I've been recommended Boal's Mechanics of the Cell and I think that it is a good book after giving it a look but I would like to know if there is any sort of canonical reference on the field.
I would like a book as mathematical as possible but at the same time my background is very limited when it comes to biology so I would like it to explain the basic and fundamental biological concepts behind the physics.
| Quantitative Human Physiology: An Introduction By Joseph Feher (link to google books)
Although the content is (obviously) designed towards physiology rather then biophysics, his mathematical treatment of the some of the main concepts in biophysics (e.g. the Nernst equation) is the best that I have come across. The downside is that it does not cover a wide range of topics in biophysics.
Biophysics Problems: A Textbook with Answers P.Maroti, L.Berkes, F.Tolgyesi
This is simply a collection of problems with solutions (thats it), which depending on your learning style may be helpful. It covers a wide topics from membranes to medical imaging, but the number of questions on each topic is limited.
Biochemistry By D.Voet, J.Voet
Don't be put of by the title. This book covers a wide range of topics in biophysics, and was the set book for my biophysics course. In my opinion this is not as readable* as Feher, but does cover a wider range of topics.
* That said I am dyslexic, so I wouldn't really take my word on if something is 'readable'.
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How should I interpret a Chi-Squared Result? I've got a Model A with a reduced chi-square of 1.28. I've got a Model B with a reduced chi-square of 0.70. Which is a better model? The model closest to 1 or the model closest to zero?
(Yes, I know this is probably better on the math site, but I got no answer there. Besides, the chi-square is ubiquitous in astronomy and it would be useful to have a definitive answer so we can read the papers and understand the results)
| The one closest to zero is the best fit, but depending on the conditions you can't rule out the model with 1.28. Most often you cannot rule out anything where Chi-squared is closer than 1 to the value of your best fit - but it does depend in reality on a bunch of things including the number of variables you used for your model fit, for example.
Numerical Recipes has a good description of Chi-squared fitting.
in response in comments....
0.7 and 1.28 are both reasonnably close to zero (and to 1) and the Chi-squared test indicates that both are reasonnable fits. --- if one fit gave 0.7 and another gave 325.6 then you could rule out the second model. Normally for 68% confidence we allow chi-squared to increase by 1, but here with several paramters we may need to increase by more than 1 - so here the Chi-squared test says that both are plausible fits to the data
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Induced EMF dependent on terminal wire connection? Figure(a):
When a conductor moving inside magnetic field, of a given length at a certain velocity the induced EMF is:
$$\epsilon = vBL$$
However, what if we changed the position where the bottom wire is connected to the wire like so:
Is the induced EMF now:
$$\epsilon = vBL_2$$
I'm not sure how that can be true, when the conductor's length has not changed just the position of where the circuit wire is connected "shortining" the current medium(or path) I agree, however, how would it change the induced EMF? The charges are still at the top & bottom of the conductor, would the terminal wire's connection reduce the induced EMF?
| You need to look at the rate of change of flux through the circuit, including the flux being added by the slanted section of wire (stretchable?). A more realistic (and simpler) arrangement would be to have a rod sliding on parallel rails which can be moved closer together.
| {
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How do we find the number of bounded states in this potential? for the potential
$$V(x)=-\frac{1}{1+\frac{x^2}{m^2}}$$
we can approximate the wave function and bounded state accurately for $x << m$ as simple harmonic oscillator, so what are we gonna do if $x$ is large compared to $m$? Is it the number of bounded state in this exact potential is no more than the bound state energy that is great than 0? How do we find the exact number?
| You can get an estimate that is correct (up to order one) from WKB. Given the Hamiltonian $$H = \frac{p^2}{2m} - \frac{1}{1+x^2/x_0^2}$$
the $n$-th level is characterized by
$$\frac{1}h \oint_{H= E_n} p\,dx = (n+\tfrac12). $$
The last bound state is at $E=0$, so the number of levels $N$ is given by
$$N = \frac{1}h \oint_{H= 0} p\,dx - \tfrac12.$$
For $H=0$, we have $$p = \frac{\sqrt{2 m}}{ \sqrt{x^2/x_0^2 +1}}$$ and the integral can be evaluated as
$$ \oint_{H=0} pdx = 4 \int_{0}^\infty \frac{\sqrt{2 m} dx}{ \sqrt{x^2/x_0^2 +1}}.$$
It diverges at the upper limit of integration (as the integrand behaves as $1/x$). Thus the number of bound states is infinite.
| {
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Why should gluons move at a speed determined by $\mu_0$ and $\varepsilon_0$? I understand that the speed of light can be derived from Maxwell's equations, giving $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$
I furthermore understand how the principle of invariance of laws w.r.t. inertial reference frames gives rise to special relativity in order to preserve the above equation.
I am aware that gluons are theoretically massless and also travel at $c$.
I am also aware that the speed of light is considered to be the "speed of massless particles" or "the speed of information", but I'll get to that in a moment.
My question is: why should the speed of gluons by given by the electric and magnetic constants $\mu_0$ and $\varepsilon_0$? This connection seems sensible in the case of the photon, an electromagnetic particle, but why should this apply to the gluon as well?
I reject the "all massless particles" and "speed of information" answers as an explanation because they don't actually explain anything -- the situation is just as mysterious after these "answers" are given as before.
*
*If "all massless particles" is really the answer, then $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ needs to explain how $\mu_0$ and $\varepsilon_0$ are derived from $c$, not the other way round. This is two new mysteries: firstly, how do we obtain $\mu_0$ and $\varepsilon_0$ from $c$ in a philosophically sound manner, and secondly, why should the classical derivation coincidentally obtain the same answer?
*If "speed of information" is the answer, we need both to supply a sensible fundamental definition of "information" and furthermore show that photons and gluons actually satisfy that definition. Then we still have the problem in the bullet point above.
Can anyone shed some light (ha ha) on this? How can we present these results such that the speed of the gluon is naturally given by electronic and magnetic constants, or how do we derive $\mu_0$ and $\varepsilon_0$ from some common concept?
| I suggest that you write Maxwell's equations in (the physicists') Gaussian units, rather than (engineers') SI:
$$\nabla \cdot \mathbf{E} = 4\pi\rho$$
$$\nabla \cdot \mathbf{B} = 0$$
$$\nabla \times \mathbf{E} = -\frac{1}{c}\frac{\partial \mathbf{B}} {\partial t}$$
$$\nabla \times \mathbf{H} = \frac{4\pi}{c}\mathbf{J}_\text{f} + \frac{1}{c}\frac{\partial \mathbf{D}} {\partial t}$$
I think this should shed some light on your question.
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For the transition metals, how does counting the number of up-spins and down-spins still give you a non-integer magnetic moment? The transition metals like Fe, Co and Ni have magnetic moments of 2.2, 1.7 and 0.6 Bohr magnetons, respectively.
The band theory says that you get this when you calculate the density-of-states of the 3d band and you subtract the number of spin-up electrons from spin-down electrons.
How does this subtraction nonetheless net you a non-integer value for the magnetic moments? Shouldn't there be nothing but integer values of available states?
The references where I found that the magnetic moment can be calculated by the difference of spin-up and spin-down electrons in the 3d band of the density of states, are:
*
*Stöhr, J. and H. C. Siegmann. “Magnetism: From Fundamentals to Nanoscale Dynamics” (2006) <WCat>.
*Chikazumi, S. “Physics of Magnetism” (1964) <WCat>.
| When the Fermi level cuts through some bands, only the fraction of the Brillouin zone for which the band is below the Fermi level counts toward occupancy of that spin.
Those fractional occupancy values are averages. It's not that an individual electron will flip spins instantly and change bands, but out of many unit cells, that fraction will find that band occupied.
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Operators and addition of angular momenta Consider a two particle system with one particle having spin 1/2 and the other spin 1.
One state of the system is $||\frac{3}{2},\frac{3}{2}\rangle\rangle$ where a double ket means this is in the coupled basis. So $S=\frac{3}{2}\text{ and } m_s=\frac{3}{2}$. This state can also be written in the uncoupled basis as $|\frac{1}{2},1\rangle (\equiv~|\frac{1}{2},\frac{1}{2};1,1\rangle )$.
If I want to act on this state with the $\hat{S^2}$ operator do I operate on the state represented in the coupled or uncoupled basis? Why and why not?
| It doesn't matter on which side $S^2$ acts. It should be equivalent shouldn't it?
here is the math.
$$\hat{S}=\hat{S_1}+\hat{S_2}$$
$$\implies \hat{S^2}=\hat{S_1^2}+\hat{S_2^2}+2\hat{S_1}.\hat{S_2}$$
$$S_1=S_x^1 \hat{i}+S_y^1 \hat{j}+S_z^1 \hat{k}$$
$$S_2=S_x^2 \hat{i}+S_y^2 \hat{j}+S_z^2 \hat{k}$$
$$\hat{S_1}.\hat{S_2}=S_x^1.S_x^2+S_y^1.S_y^2+S_z^1.S_z^2$$
Applying this on either side,
$$S^2||\frac{3}{2},\frac{3}{2}\rangle\rangle$$ $$=\hbar^2 (\frac{3}{2}. \frac{5}{2}) ||\frac{3}{2},\frac{3}{2}\rangle\rangle$$ $$=\hbar^2 \frac{15}{4} ||\frac{3}{2},\frac{3}{2}\rangle\rangle$$
Using the results above.
$$(\hat{S_1^2}+\hat{S_2^2}+2\hat{S_1}.\hat{S_2})|\frac{1}{2},\frac{1}{2};1,1\rangle$$ $$=\frac{15}{4}\hbar^2|\frac{1}{2},\frac{1}{2};1,1\rangle$$
Exactly the same!
Hope this helps!
| {
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Velocity of a leak in a closed water tank Bernoulli's equation states
$P_1+{1\over2}\rho v_1^2+\rho g h_1 = P_2+{1\over2}\rho v_2^2+\rho g h_2$
In a classic "water tank with an open top and a leak" scenario, "point 1" is the surface water in the tank, and "point 2" is the leak. The equation could be rewritten for $v_2$ as
$v_2= \sqrt{2g\Delta h}$
This is simple, but suppose it involves a tank where the top is closed off. The above simplification will no longer yield the correct velocity.
How do I apply Bernoulli's equation for "water tank" scenarios in which the water tank is closed?
| Depends on your assumptions. In the extreme case, with vaccum on the top of the fluid you can set $P_1=0$. This yields $$v_2=\sqrt{2g\Delta h-\frac{p_{\infty}}{\rho}}$$ with $p_{\infty}$ as ambient pressure. Be aware that $v_2$ is only an approximation since in reality $\Delta h$ changes as fluid leaves the tank.
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What is the purpose of taking coefficients as 1 in numerical solutions? How can we recover the real solution after getting a solution by solving with parameters set to 1?
For example, on my case to solve the Shrödinger equation via finite difference method, the author took the coefficients like $h$ and $2m$ as 1, and gets result with this form.
How do the result of this new equation relate to the solution for the original one?
| Another way to think of it is that you're dividing the equation by $\frac{\hbar}{2m}$. Any solution to
$$ \hat{H} \psi = 0 $$
is a solution to
$$ \frac{2 m \hat{H}}{\hbar} \psi = 0 .$$
Since the second equation is just the first multiplied by something on both sides. You are interested in finding the space of solutions of the differential equation, so either one is okay. In a general differential equation, you are right, you can't necessarily just set things to one and get the same set of solutions.
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Supersymmetry as a solution of hierarchy problem Hierarchy problem is the statement of why the weak force is much stronger than gravity.
In terms of coupling constant, weak force (Fermi coupling) is much larger than gravity (Newton's constant).
I want to know how supersymmetry can be a solution to Hierarchy problem.
| It is all about appropriate cancellations in the expansions when calculating the Feynman diagrams:
The hierarchy problem
Supersymmetry close to the electroweak scale ameliorates the hierarchy problem that afflicts the Standard Model. In the Standard Model, the electroweak scale receives enormous Planck-scale quantum corrections. The observed hierarchy between the electroweak scale and the Planck scale must be achieved with extraordinary fine tuning. In a supersymmetric theory, on the other hand, Planck-scale quantum corrections cancel between partners and superpartners (owing to a minus sign associated with fermionic loops). The hierarchy between the electroweak scale and the Planck scale is achieved in a natural manner, without miraculous fine-tuning.
| {
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What's the difference between charge density wave and charge ordering for superconductors So far, my understanding is that they are the same. Charge ordering is a phase transition and the material will have charge density waves once it's in a charge ordered state...? This sounds too simple though... And a similar question would be the difference between SDW and spin ordering? Thank you!
| I agree with Meng Cheng, charge ordering is a broader notion than CDW. My understanding is that CDW means charge ordering with a non-zero value for the wavevector $q$ : if your order parameter is some $\Phi(q)$, you develop a nonzero expectation value for $<\Phi(q)>$ with $q \neq 0$.
On the other hand, you could also get charge ordering for $q=0$ : it will be a uniform order, so I guess you can't call it a wave in this case.
| {
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Is it feasible to lift a human off the ground with a suit of fans?
So, we live in a world that is technologically developed enough where it is reasonably within reach to attach small rockets to someone/something to lift it off the ground.
One serious problem we encounter with this is damage to the surrounding environment. Even if you manage to protect yourself, your "rockets" could cause damage to the surrounding objects and terrain (since it is shooting fire after all.)
My question is this: is it possible to use enough small sources of lift, say, a suit of fans, spread out over a human body, to lift the human without causing damage to the environment or the human. Assume for the purposes of this question that these fans can be pivoted and controlled to point in any direction and remain there using a perfect control system, we're just worried about the physics of lift here.
I am also open to other sources of lift, but the objective would be to have something that will not hurt anyone or anything (too much. )
| lift is achieved when you push down on the air with a force equal to(hovering) or greater than your weight this means that you can In theory create the system of fans that you want in fact these machines have already been created tested and they do work but the main problem is flight time and a system of fans addresses that problem more efficiently than a rocket propulsion system since by the time you have enough rocket fuel to carry you and enough fuel to carry that fuel ...you'll end up with a huge rocket and only minutes of flight time
| {
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Laser-induced electrical discharge Attracting lightning with an ion beam has been done in a lab but how strong of a laser is needed to reach the clouds to redirect the lightning bolt?
| I can talk about the discharge experiment since I am not aware of any experiment to trigger real lightning with lasers. Note Discharge in lab is different than the lightning however the effort of all experiments is towards taming the lightning.
Laser filaments are low density plasma channels (order is $10^{16}cm^{-3}$). This channel could be achieved by focusing a several mili-Joules pulses of 800nm and hundreds of femtosecond duration into air. A variety of energies and pulse durations work as well, however the plasma density remains the same. In laser physics this orders of energies are a lot and you can do many cool stuff with them.
Some technical stuff for future lightning experiments are:
Since the ionization draws energy from the pulse, so, couple of meters after the focusing lens, the pulse no longer has enough energy to ionize the medium. For such short and strong pulses even without a lens the beam starts to collapse on its axis of propagation as if there is a lens. This effect is called Kerr Lensing which depends on the intensity of the beam and causes your beam to focus before the clouds and hence the ionization starts too early before reaching the desired altitude and as a result the ion channel is established prematurely. In order to avoid this unwanted lensing effect you can make your beam bigger and focus it with a very long focal length lens.
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What is a state in physics? What is a state in physics? While reading physics, I have heard many a times a "___" system is in "____" state but the definition of a state was never provided (and googling brings me totally unrelated topic of solid state physics), but was loosely told that it has every information of the system you desire to know. On reading further, I have found people talking of Thermodynamic state, Lagrangian, Hamiltonian, wave-function etc etc which I think are different from one another. So in general I want to know what do we mean by state in physics and is there a unique way to describe it?
| Our physics prof once put it informally that way:
A state is a set of variables describing a system which does not include anything about its history.
The set of variables (position, velocity vector) describes the state of a point mass in classical mechanics, while the path how the point mass got from point $A$ to point $B$ is not a state.
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What is a quasi-probability distribution? What is quasi-probability distribution? Why is it important in quantum mechanics? What does "quasi" mean?
| A quasi probability distribution relaxes an axiom of probabilty. In the context of Quantum Mechanics,it is specificly the axiom of probability that requires $p_{i} \geq 0$. So the sum of the distribution can include negative terms!
Quantum mechanics allows for events with a negative expectation values, to acount for phenomena like destructive interference. Intuitively The negative expectation values make it possible for events to "cancel out" another event with different sign. This would not be possible with non negative numbers. A distribution of these expectation values, that is normalized to one can be seen as a quasi probability.
Im sorry if this answer is a bit untechnical, but its all i could make up quickly and i hope for a quick read its enough
.
I think Scott Aaronson has an excellent and pedagogical in depth post, exactly telling you why quasi probabilites come up in QM, better than i could ever do. Refer to: http://www.scottaaronson.com/democritus/lec9.html
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Polarized light in single mode fiber In single mode fiber the light propagates in two orthogonal planes. Input will be linearly polarized light, which state of polarization will be on output and why? And if there will be some different state of polarizatin on output what will happen?
| In standard single-mode fiber, the polarization will tend to drift as the signal propagates (due to slight and varying birefringence of the glass, possibly stress-induced, coupling one polarization to the other).
For short lenths (1 m or so) polarization is typically maintained fairly well. For long distances (1 km or more, I'd guess, but I don't have a lot of experience with this issue) the output polarization is typically pretty well randomized.
There is also readily available "polarization maintaining" (PM) single mode fiber, that is designed to allow a signal to propagate while maintaining its polarization. It does this by having a deliberately introduced birefringence in a well-defined direction.
This does mean that if you launch a signal into PM fiber but not aligned with the preferred axis, it will have strong dispersion between the two polarizations, which could even lead to a single input pulse generating two output pulses.
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What parts of a string stretch most when a wave passes through it? So there's a (transverse) traveling wave on an (ideal) string under tension. Why does the stretching occur around zero displacement? Why not at the crests?
| When a string is pulled tight to allow a transverse wave to pass, there will be some tension everywhere, not just in the bits near the crests or the zeros. However, if you consider a sine wave as a distorted straight line, then you can see that the amount of distortion (stretching) will depend on the gradient of the line. Consider a line element of initial length $dx$ that is horizontal, and add a small vertical displacement $dy$ at one end. Keep the ends apart by the same horizontal distance $dx$; then the total length of the element is now $\sqrt{dx^2 + dy^2} = dx\sqrt{1+\left(\frac{dy}{dx}\right)^2}$
From this it follows that the stretching of the string is a function of the slope - and that the string will be more stretched where the slope is greatest. Which is, of course, at the zero crossings.
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Is it always possible to determine whether or not one is accelerating? Consider the following two situations: | You wake up in an elevator that is in free fall in a gravitational field.
B: You wake up in an elevator that is floating in a vacuum.
Is it possible to distinguish between these two situations?
It seems to me that Newton's second law formulated in a local coordinate system would look the same whether or not situation A or B is the truth: If P is an object in the elevator (which we denote by E), Newton's second law would be:
$$F_P = m_P\ddot{x}_P \Rightarrow$$
$$(-m_Pg + F'_P) = m_P(\ddot{x}_E+\ddot{x}_{P\backslash E})$$
where $F'_P$ is the force applied to the object in addition to whatever gravitational force is present, $x$ denotes position relative to an inertial frame, and $x_{P\backslash E}$ is the position of the object relative to the elevator.
Since the elevator is in free fall,
$$F_E = m_E\ddot{x}_E \Rightarrow -m_Eg = m_E\ddot{x}_E \Rightarrow \ddot{x}_E = -g$$
Thus, the equation of motion for the object P becomes
$$F'_P = m_P\ddot{x}_{P\backslash E}$$
But this would be true regardless of the value of $g$! Or am I missing something? If not, is there any other way of knowing whether or not situation A or B is the truth?
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Globally defined solutions in bc CFT system Consider $bc$-system which is 2-dimensional CFT of fermions:
$S = \int_\Sigma d^2 z \ b \bar{\partial} c + h.c. $
where $\Sigma$ - 2-dimensional manifold of genus $p$, fields $b, c$ have dimensions $(\lambda, 0)$ and $(1-\lambda, 0)$ respectively.
The question is to find number of globally defined solutions of equations of motion
$\bar{\partial} b = 0, \bar{\partial} c = 0$ and for Sphere and Torus find these solutions explicitly.
The problem is that I don't even understand, what does "globally defined" mean, and why solutions may be not globally defined, what is the problem here.
I've been told that it's somehow connected to Riemann-Roch theorem, but I don't understand how.
| Comments to the question (v2):
*
*Given a manifold $M$, the word locally is associated with an open neighborhood $U\subseteq M$, while the word globally refers to the whole manifold $M$.
*If $E\to M$ is a fiber bundle, let $E_{|U}\to U$ denote the restriction of the bundle $E\to M$ to the neighborhood $U\subseteq M$. For instance, if $s\in\Gamma(E_{|U}\to U)$ is a locally-defined section, it may or may not be possible to extend $s$ to a globally defined section $\tilde{s}\in\Gamma(E\to U)$ such that the restriction $\tilde{s}_{|U}=s$.
*Now the ghost fields $b$ and $c$ are sections in appropriate vector bundles over a Riemann surface $M=\Sigma$.
*The Riemann-Roch theorem (and more generally, the Atiyah–Singer index theorem) yields information about the dimension of the space of globally defined sections over certain vector bundles.
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In most physical cases, the elements of a group can be represented by unitary matrices. Why no time-reversal? In Dresselhaus's group theory page 19, a theorem writes:
Every representation (of a Hamitonian's group) with matrices having non-vanishing
determinants can be brought into unitary form by an similarity
transformation.
On page 21 writes:
On the other hand, not all symmetry operations can be represented by
a unitary matrix; Time reversal operator is represented by an
anti-unitary matrix.
My question is, since time reversal operator represented by a matrix, which has the determinant of a non-zero value(?). Is this a contradiction?
| I think this passage is not very good (By the way, here is an online version with slightly different paging: http://web.mit.edu/course/6/6.734j/www/group-full02.pdf).
If you have a look at the proof, what they actually does is the following:
Theorem: Every representation of a finite group with matrices having non-vanishing determinants can be brought into unitary form.
I don't see how they can get rid of the finiteness assumption (unless he uses some sort of compactness/boundedness), so they should state this - it's at best sloppy. In particular, the counterexample of katz is then invalid, since the group of integers is clearly infinite.
Anyway, there are well-known theorems along these lines such as:
Theorem: Every representation of a finite group is unitarizable. (see e.g. here)
Theorem: Every finite dimensional representation of a compact group is unitarizable. (follows from e.g. here)
However, when they speak about representations, they always mean linear representations. Which brings me to the time-reversal symmetry:
This is antiunitary and therefore in particular antilinear. It can therefore not be represented by a matrix - since a matrix represents by definition a linear operation. Therefore, the theorems don't apply.
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Why quantum fluctuation existed before big bang? I read somewhere that quantum fluctuation can give rise to a big bang and thus the creation of a universe which we know today, where do these quantum fluctuations come from if space is only created after big bang?
| The Big Bang is a purely classical concept. If we make a couple of assumptions about the distribution of matter in the universe then Einstein's equation tells us how the scale factor of the universe evolves with time. We can take our current universe and evolve it backwards in time, and if we do so we find the geometry becomes singular about 13.8 billion years ago. This is what we refer to as the Big Bang.
You'll hear statements like space was created at the Big Bang or time started at the Big Bang but these are at best imprecise and at worst meaningless. What the equations actually tell us is that we can't tell what happened at the Big Bang because the geometry was singular there. Therefore any statements about what happened at the Big Bang are not founded on known physics.
When we introduce quantum mechanics we get effects that mess up the nice simple classical picture. For example it's now widely believed that some quantum effect drastically changed the energy density around $10^{-33}$ seconds after the Big Bang and that caused cosmological inflation.
At earlier times it becomes unclear that we can usefully talk about a scale factor because spacetime may no longer be adequately described as a smooth manifold. This is typically the time at which people have suggested quantum fluctuations become important, and the idea that a quantum fluctuation may have been responsible for the entire visible universe has been around for a long time.
The point of all this is, well, there are two points:
*
*This is all based on speculation with no observational and precious little theoretical evidence that there is anything to the idea
*Assuming there is anything to the idea, spacetime still existed at this period but it was radically different to the smooth classical manifold we see today.
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Why are the jets of the "light saber" star slightly curved? Why don't the jets of the HH-24 object follow a straight line?
In the image below, notice how they bend towards left from the expected straight line.
Is it an optical distortion, or some nearby massive object exerting a gravitational field?
| Pretty picture, but your reference is full of misnomers and errors with a reference to "Star wars" to capture attention.
"Beams of light at supersonic speeds"???, light speed is 'pretty' constant.
Young stars do not emit beams of light, they emit light in all directions. Some or most of the light, (or any other EM energy, or matter ["material falling onto a newborn star"]) may be blocked by debris, in a vortex, creating a beam effect. Although the geometric idea of a vortex looks perfect, in nature (think of a tornado) it is not perfect, so this vortex does not necessarily eject energy or matter ( yes, supersonic speeds are possible here ) in the exactly same direction all the time. so the matter can be scattered or curved as it appears in this picture. If it is energy, even light, it is not pointed at us, so it has to be reflected ( i use the term loosely ) towards us off matter that is not necessarily in the straight line determined by the beam, which has spread out some over the 1/4ly it has traveled.
| {
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Elliptical orbit changing as a star's mass increases I'm studying Kepler's Laws, specifically the orbit of the Earth around the Sun. I know that if the Earth was more massive, the orbit would not be significantly affected. If the Sun was more massive, I know the velocity of Earth's orbit around the Sun would increase, but how would the shape of the orbit change? This is a theoretical case, neglecting conservation of mass and assuming the Sun's radius and volume don't change.
| Kepler's laws are not correct:
No area law, no elliptical orbits, nor period. Starting with Newton's mechanical laws, Kepler's area law does not exist. Newton's universal attraction force is radial. $F=F_r$. A side force component, a perpendicular force does not exist $F_p=\frac {m dV_p} {dt}=0$. Then when integrated , we get $V_p=Ct$ . Kepler says $r V_p=area=Ct$. Newtons equation is a prove. Kepler's area law is an estimation.
When using $V_p=Ct$ in the differential form of the energy conservation equation,we get
$r=-4*{t^2}+4 t T-4 \frac {T^2} 6 $ as the equation of celestial motion. This is a spiraled orbit and not an ellipse.
When to the validity of the period law: Newton says period law is valid for circular motion with non accelerated velocity. Kepler says period law is valid even for elliptical orbits with accelerated velocity. High velocity at perihelion low , velocity at aphelion. And no body feel this acceleration?
| {
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Are orbitals observable physical quantities in a many-electron setting? Orbitals, both in their atomic and molecular incarnations, are immensely useful tools for analysing and understanding the electronic structure of atoms and molecules, and they provide the basis for a large part of chemistry and in particular of chemical bonds.
Every so often, however, one hears about a controversy here or there about whether they are actually physical or not, or about which type of orbital should be used, or about whether claimed measurements of orbitals are true or not. For some examples, see this, this or this page. In particular, there are technical arguments that in a many-body setting the individual orbitals become inaccessible to experiment, but these arguments are not always specified in full, and many atomic physics and quantum chemistry textbooks make only casual mention of that fact.
Is there some specific reason to distrust orbitals as 'real' physical quantities in a many-electron setting? If so, what specific arguments apply, and what do and don't they say about the observability of orbitals?
| The general answer is that when there is electron correlation the picture of each electron occupying an orbital is no longer adequate. In this case a single Slater determinant is no longer sufficient.
The Hartree-Fock or selfconsistent field approach to atomic and molecular problems approximates the many electron wave function by a single Slater determinant. Although a Slater determinant is invariant under an orthonormal transformation of its orbitals, as remarked above, the eigenfunctions and eigenvalues of the Hartree-Fock operator have special meaning. These can be used to estimate energies and other properties of excited states through the Hellmann-Feynman theorem.
https://en.wikipedia.org/wiki/Hartree–Fock_method
https://en.wikipedia.org/wiki/Hellmann–Feynman_theorem
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How many fixed points does a Kelvin scale have? I have a book that says:
In the absolute Kelvin scale, the triple point of water is assigned the value of 273.16 K. The absolute zero is taken as the other fixed point.
But, then another section in the same book says:
On the Kelvin scale, the lower fixed point is taken as 273.15 K and the upper fixed point as 373.15 K.
What does this all mean? Can anyone clarify?
All I want to know is: How many fixed points does a Kelvin scale have?
| The Kelvin scale is based on the Celsius scale in which 0 °C and 100 °C were defined to be the freezing and boiling points of water.
When the absolute minimum temperature was discovered to be -273 °C, scientists began simplifying matters by simply adding 273 and using the Kelvin scale in which 0 °C becomes 273 K. By defining the Kelvin scale in terms of absolute zero and a 273 K melting point, the 373 K boiling point is fixed incidentally.
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Magnetic field due to stationary electric dipole As we know from Maxwell's 3rd equation the magnetic field is given as
$$\nabla \times \mathbf{E} = - \frac {\partial \mathbf{B}}{\partial t}$$
Now, if we consider an electric dipole which is stationary, there will be electric field lines
like this:
I want to know what will be the curl of this electric field, if it exists. And will there be a magnetic field associated with this electric field too?
| It would be interesting if you wrote our each vector component of the curl-E equation, and equate each of them to zero. Try to intuit what the derivative of each E-field component means with respect to your drawing.
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Change in acceleration due to gravity because of rotation of earth
The formula above is the equation for acceleration due to gravity when earth rotates. G is the original acceleration.
Can someone explain how this formula came?
| In the above formula $g'$ is the component (directed towards centre of earth) of the net acceleration due to centrifugal force which is due to rotation of earth and acceleration due gravitational force of earth.
i.e. $g'=g-R\omega^2\cos^2\lambda$ where $R\omega^2\cos^2\lambda$ is the component of centrifugal force away from the centre of earth.
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If you lift something does the internal energy increase or remain unchanged? So change in internal energy $\Delta U=Q+W$; so in lifting an object you do work on it, thus increasing its internal energy. Is this correct?
Sorry if this is a stupid question.
| Not a silly question! Although your formula is the way that we often memorize the First Law, the true First Law is actually
$$
\Delta (U + KE + PE) = Q + W_\text{non-grav}
$$
where $KE$ and $PE$ are, respectively, the kinetic and potential energy and $W_\text{non-grav}$ is the work done by all forces except gravity. When elevating an object, work is done on the object, but its potential energy changes by the same amount (i.e., $\Delta U$ is indeed $0$). The First Law then breaks down to
$$
\Delta PE = W_\text{non-grav}
$$
...which we know to be true from physics. The reason that the simplified form of the first law is so commonly referenced is that we often encounter systems in which $\Delta PE = \Delta KE = 0$
Edit: there is actually more than one way to look at this
*
*View Gravity as a force. In this view, the first law is still written as
$$
\Delta (U +KE) = Q + W_\text{incl. grav}
$$
but $W_\text{incl. grav}$ includes both the work done by the external force which elevates the object and by the force of gravity. These effects cancel out to zero.
*View gravity as a potential - i.e., track its effect by using potential energy rather than by considering it as a force which does work. In this view, the first law is as I wrote it above (first equation) and $W_\text{non-grav}$ the work done by all forces except gravity.
The two perspectives are equivalent because gravitational potential energy is defined by $\Delta PE_\text{grav} = - W_\text{grav}$.
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Conservation of angular momentum in a collision Suppose I have a stick hinged to a pivot and it is released from its horizontal position and just after it becomes completely vertical, it strikes a ball completely stationary as in the given figure below.
The collision is completely elastic.
QUESTION: Will angular momentum be conserved during the collision?Why?
| Angular momentum of an isolated system is always conserved. But it does require you to define what you consider part of your system - the hinge will provide a reaction force on the stick at the moment of collision, and that means there is an "external force" (that is, external to the stick and ball) to be taken into account. Your "system in which angular momentum is conserved" has to be something that has no external forces on it.
In other words - the question as posed cannot be answered unless you define your system more precisely. But if you define it as including just the ball and stick, the answer will be "no" because at the time of impact there is an external force that doesn't pass through the center of mass.
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Current from Middle Battery in a Two-looped Circuit
With this question, as with many tutorials of similar questions I’ve found online, my textbook only mentions three currents: $I_1$, which flows through the left loop from and to the 19 V battery, $I_2$, which flows through the right loop from and to the 19 V battery, and $I_3$, which flows through the middle section.
However, why can’t there be an $I_4$ which flows from and to the 12 V battery through the left loop, and an $I_5$ which flows through the right loop? In this diagram, it seems pretty clear that all the current comes from the 19 V battery. But about the current coming from the 12 V battery? Is there no current? (The subsequent analysis completely discounts the presence of any current from the 12 V battery, so I’m thoroughly confused.)
| The labelling of the voltages and currents does not matter as long as you stick to a convention.
If you had used the passive sign convention you might have labelled the voltages and currents as shown in the circuit diagram below.
In this case using Kirchhoff's current law one gets these relationships between the currents
$$I_1=I_2,\, I_3=I_4\, {\rm{and}} \,I_2+I_3-I_5=0$$
Without doing the calculations suppose that in this case the current $I_5$ was found to be $-50\,\rm mA$.
Using the labels in your circuit diagram you would have found your current $I_2$ to be $+50\,\rm mA$.
So both methods produced the same magnitude for the current $(50\,\rm mA)$ and flowing in the same direction (into the top of the $200\,\Omega$ resistor and out of the bottom of the resistor).
You could have chosen the other currents and other current directions that you have suggested in your question to label your diagram butin the end you would end up with the same magnitude currents flowing in the same direction whichever set of labels you used on your circuit diagram.
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Which elementary particles are behind magnetic field, similar as photons behind radio waves? I see, there are photons behind radio waves.
As Wave–particle duality said: the radio waves are waves and at the same time are fluxes of particles called Photons.
I'm wondering, what is behind magnetic fields?
Means, magnetic fields - are waves too. Shall this waves consists of photons too? Then how does photons moves by such trajectories??
| A particle must contain some amount of energy. Otherwise it wouldn't exist. There is no energy transfer with a steady-state magnetic (or electric) field, so there are no particles involved in transferring energy. Only when a magnetic field changes (the producer of that field moves) can there be energy transferred. In that case, the particle transferring the energy is the photon.
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Exact meaning of "pi/2 pulse" In studying Mach-Zehnder and Ramsey interferometers, I came across the expression "$\pi/2$ pulse". What does it mean exactly? I am working with a Bloch vector representation $(u,v,w)$ of a 2 state system. We have a Rabi frequency $\Omega_0$ and a detuning parameter $\delta$ to the $|1\rangle\rightarrow|2\rangle$ transition frequency. In those conditions, I think the "$\pi/2$ pulse" is a rotation around the $(-\Omega_0,0,-\delta)$ axis for a duration $\tau=\frac{\pi/2}{\sqrt{\Omega_0^2+\delta^2}}$. Is that correct?
| $\frac{π}{2}$ pulse means that all the particles in the system have gone to the higher level. $π$ pulse excites all particles in the first half time and de-exites in the second, so all particles are in lower level.
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Closing a switch in series with a capacitor Suppose we have the following circuit:
Such that for $t<0$ the switch M was open. If we close the switch at $t=0$ what will the voltage on the capacitor, $V_C$, be at $t=0^+$? What about $\dot V_C$ at $t=0^+$? Will there be a current passing through $R$ the moment the switch is closed?
I need to solve an ODE for a more complicated circuit which has this sub-circuit as a part of it, and I need initial conditions to solve for the ZIR case. I'm trying to figure out these initial conditions but I'm not sure. Here are my thoughts:
Before closing the switch, there will be a steady finite current $I$ in the circuit.The moment we close the switch, there can be no current passing through $R$, else there will be some finite voltage on the capacitor $\longrightarrow$ $\dot V_C$ will be infinite $\longrightarrow$ $I_C$=C$\cdot \dot V_C$ will be infinite which cannot be.
Therefore $V_C(t=0^+)=0$ and $\dot V_C(t=0^+)=I/C $. I'm not sure if what I said I correct, I mean we learned that if there is no impulse current (like Dirac's Delta function), the voltage on the capacitor will be continuous.Does this apply in the case too? I would really appreciate any help.
| You cannot solve the problem presented.
An ideal battery connected to a capacitor will have an initial current equal to infinity, which is obviously not possible. To make the problem solvable, you need to know the internal resistance of the battery.
Once you add the resistor in series with the battery, you will simply need to calculate the voltage across the parallel resistor (the maximum voltage you will across the capacitor).
$V_{t} = V_{max} e^{\frac{-t}{RC}}$
Where R is the resistance of the battery.
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How are the protons for collisions in the LHC made? I heard that the LHC smashes two protons together to research the universe.
But how does it create the protons for collision? If we strip off the electrons won't there be neutrons along with protons?
A "Duoplasmatron source" is used by the LHC, it ionises atoms into negative or positive ions but how is the proton separated from that?
| See this article on the LHC proton source.
Hydrogen gas is ionised using a strong electric field, and the resulting protons are accelerated and focused then injected into a linear accelerator.
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Dirac equation, $\alpha_i$, $\beta$ hermitian The argument I've seen is the one given here: http://epx.phys.tohoku.ac.jp/~yhitoshi/particleweb/ptest-3.pdf under (3.10):
$$H=\vec{\alpha}\cdot(-i\vec{\nabla})+\beta m$$
$H$ is hermitian, $-i\vec{\nabla}$ is hermitian, so $\vec{\alpha},\beta$ are hermitian.
This is not convincing, because a hermitian operator being a sum of two operators does not imply that the two are also hermitian, for example in the harmonic oscillator, the position and momentum operators can be written as a sum or difference of the annihilation/creation operators, which are non-hermitian.
So is the presented argument wrong? Why are the $\alpha_i, \beta$ really hermitian?
| The Hilbert space in which the Dirac equation acts is a product of the infinite-dimensional space corresponding to position (or momentum) and a finite-dimensional spinor space, in which the matrices $\alpha_{j}$ and $\beta$ act. We know the momentum operator is Hermitian, so we may look at a subspace for fixed $\vec{p}$; in order for $H$ to be Hermitian, the matrix $\vec{\alpha}\cdot\vec{p}+\beta m$ must be Hermitian in the finite-dimensional spinor space for every value of $\vec{p}$. This means that an arbitrary linear combination of the $\alpha_{j}$ and $\beta$ matrices is Hermitian, which is only possible if all four matrices are Hermitian themselves.
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Description of transparency of a Faraday cage to different frequencies? A Faraday cage within a static electric field leaves an internal electric field equal to $\vec{0}$ (as long as there are enough mobile electrons in the conductor to counteract the external $\vec{E}$).
It could be said as "a Faraday cage is fully opaque to static electric field".
On the other hand, it doesn't act on an external magnetic field.
It could be said as "a Faraday cage is fully transparent to static magnetic field".
Wich equations describe transparency (behaviour) of a Faraday cage submitted to different frequencies electro-magnetic fields?
Ideally I would like to find an absorption spectrum of a Faraday cage from
0 Hz to visible light (1 PHz).
| According to What is the relationship between Faraday cage mesh size and attenuation of cell phone reception signals? for a Faraday cage with a characteristic mesh size of $l$, then the cut-off frequency $f_c$ corresponds to a wavelength of $2l$, i.e. $f_c = c/2l$. In these circumstances, the wavevector
$$k = \frac{2 \pi}{c} \sqrt{f^2-{f_c}^2}$$
Thus when $f<f_c$, the electric field is exponentially attenuated as $\exp(-\alpha x)$, for a cage of thickness $x$, where (if I've done my sums right)
$$ \alpha = \frac{\pi}{c} \sqrt{2(f_c^2 - f^2)}$$
If there are no holes at all (i.e. a metal box) then the frequency dependence is explored in Faraday cage in real life
It is shown that
$$\frac{E_t}{E_i} \simeq 4 \frac{\eta_{\rm c}}{\eta_0} \exp(-x/\delta) = 0.47 \omega^{-1/2} \exp(-22 \omega^{1/2} x),$$
where $\eta_c$ is the impedance of the conductive material and $\delta$ is the frequency-dependent "skin depth". The transmitted power fraction would be the square of this. The numbers in the above formula are appropriate for aluminium, but the frequency dependence would be the same for any good conductor. Note that at high frequencies attenuation in the medium is important, but even if the box was very thin, that reflection from the surface is extremely important even at quite modest frequencies.
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What will a glass look like in 500 years? The glass is in a metastable state. It is changing constantly. So what will a piece of glass look like in 500 years in room temperature?
| It's a common misconception that glass flows appreciably over time. So the answer is that your glass will have exactly the same shape as it does now, to within under a wavelength.
You can debunk the myth of flowing glass with simple experiment / observation. There are on Earth several very large refracting telescopes over 100 years old with huge lenses ground to precise optical specifications. Astronomy, as with microscopy or any precision imaging, is an application that is extremely sensitive to optical aberrations, and even sub-wavelength deformations in the refractors of these telescopes over time would degrade the telescopes' optical performances. Yet even these sensitive devices have suffered no plastic deformation that would shift their surfaces more than a significant fraction of a wavelength, which they would do over 100 years if glass were in the state of flow that is commonly claimed.
You can make this simple observation quantitative if you can borrow of the order of a week's time on a decent interferometer. Set up a large, low cost lens (say of the order of 0.1m diameter or more) to focus into the focus of a spherical retroreflector and light the system with the probe beam of a Fizeau interferometer. Cement a heavy piece of steel or lead to the top of the lens (you'll need to do this so that you can still pass an appreciable beam from the interferometer through the unobscured part of the lens) and record the fringe pattern. You'll find that after a week, the fringe pattern has not shifted AT ALL. You can then derive an upper bound to the amount of "flow" that the lens has undergone from this observation. If the top surface hasn't moved 20 nanometers (which shift would be highly observable on the interferometer) in a week (the result I can assure you you will observe), its motion is then less one micron every year, i.e. less than than 1 millimeter deformation every thousand years!
| {
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Why is it always foggy on new years eve? I live in Hamburg in northern Germany and a few years ago I started to notice this: It is always foggy on new years eve. Yesterday there even was a sight distance of about 10 meters at some point. This doesn't happen usually in the city.
I have some basic meteorology knowledge and know about the process of how fog emerges. But the fact that it happens every new years eve with extreme intensity got me thinking. It can't be a coincidence.
My theory is that there are a lot of particles in the atmosphere from the fireworks and that water condenses on them. Am I right or is there another reason? Or am I completely wrong and this is just coincidence?
| Because of the extensive fireworks on new years eve a wide variety of particles enter the lower part of the atmosphere (boundary layer). Among those are trace gases but also black carbon [1]. Black carbon acts as a strong cloud condensation nuclei [2] which enhances condensation and therefor fogginess.
Also, "SO2 is a precursor to the formation of sulfate
aerosols that are considered important in forming and modifying
aerosols that could become CCN as well as modifying
the number concentrations and size distributions of cloud
droplets" [3]. SO2 is also a very common compound that is emitted in any form of burning.
The fact that the atmosphere is very stable during nighttime and mixing hardly occurs makes that the fog stays until the next morning/day until it is mixed over a larger volume.
| {
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Does quantum randomness measurably affect macro-sized objects? I understand that while it is believed that there is no true randomness on the macro scale, there is true randomness on the quantum scale. A previous theory that quantum processes could be determined through "hidden variables" has been disproven (through polarizing photons and radioactive particle decay), confirming that true randomness does exist.
Now for my question. Does quantum randomness measurably affect the macro scale such that true randomness actually does exist outside quantum mechanics, or will rolling a die in identical conditions always yield the same result even after factoring quantum randomness?
| There is no randomness on the quantum scale, there is only uncertainty, which describes the information that is available to a macroscopic observer. An easy way to see that quantum processes are not random is by looking at starlight. That light has been coming to us from up to billions of lightyears away, but it doesn't show any distance dependent "random" effects that you would expect from a truly random phenomenon like scattering on particles (like in fog). We have even done interference experiments on "old light" and it interferes just as well as light from local sources.
Now, as to your question... yes, uncertainty does permeate the entire macroscopic world, but probably not in the way you might expect. It is only because of uncertainty that stable matter exists. We can not model matter correctly with any other theory than quantum mechanics. Neither classical mechanics nor electrodynamics even permit stable atoms. Light, matter, magnetic fields... these are all macroscopic quantum phenomena. The force between two magnets, that's a macroscopic quantum field at work. You can hold "it" in your hand, if you like.
| {
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Sliding along a circular hoop: work done by friction
Assume a point object of mass $m$ slides along a hoop of radius $R$, starting from a position which makes 90 degrees with the line of radius connecting the center and the ground. Let the coefficient of kinetic friction between the hoop and the object be $\mu$. Assuming that the object starts at rest, what is the total work done by the friction when the object comes to the ground level?
My idea: the normal force at any instance is given by $$N=mg\sin\theta+\frac{mv^2}{R},$$ where $\theta$ is the angle between the radial line connecting the present position and the intital position of the object to the center of hoop. With this we have the frictional force as $$f_k=\mu\left(mg\sin\theta+\frac{mv^2}{R}\right),$$ so that the total work done by friction is $$W_k=\int_0^{\pi/2}\mu\left(mg\sin\theta+\frac{mv^2}{R}\right)R\mathop{\mathrm{d\theta}}.$$
The problem I am having is to figure out $v$ as a function of $\theta$, i.e $v(\theta)$. Any ideas?
| For people interested in this problem, you can also define a differential equation for the frctional work itself. Just express the differential work as a function of the angle, apply again energy conservation for the kinetic energy and then solve the one-degree differential equation that appears. Details can be found in an old publication:
Franklin, L. P., & Kimmel, P. I. (1980). Dynamics of circular motion with friction. American Journal of Physics, 48(3), 207–210. doi:10.1119/1.12306
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How is $ \left(1-\frac{p^2}{2mE}\right)^{3N/2-2} =\; \exp\left(-\frac{3N}{2}\frac{p^2}{2mE}\right)\;?$ How is
$$ \left(1-\frac{p^2}{2mE}\right)^{3N/2-2} = \exp\left(-\frac{3N}{2}\frac{p^2}{2mE}\right)$$
(Karder, Statistical Physics of Particles, Page 107)
in the large $E$ limit. Here $N$ is particle, of the order of $10^{23}$, $E$ is the total energy.
I roughly guess that it should be $\exp(-\frac{p^2}{2m})$ since both $N$ and $E$ can be treated as infinitely large.
Update: a hint to solution is provided in the comments.
| Add a comment needs 50 reputation, and I got only 46 now. So I write my opinion here.
I have read the textbook, the original formula is
$$p(\vec{p_1})=(1-{{\vec{p_1}^2}\over {2mE}})^{3N/2-2}\cdots\cdots$$
So $\vec{p_1}$ is the momentum of only one particle in the ensemble. Considering the system has very large $N$, that is only a tiny proportion of total $E$, which makes the ${{\vec{p_1}^2}\over {2mE}}$ term approaches 0.
Then with the above comments of other guys, you can get the results. Here I think $3N/2$ makes no difference with $3N/2-2$ because $N$ is large
| {
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Minimum Power Required to Maintain a Population Inversion According to many sources, the minimum power required to maintain a population N in the upper level of a two level laser system is $$P_{min}=fA_{21}Nh\nu$$ where $f$ accounts for the lack of 100% efficiency, $A_{21}$ is the Einstein A coefficient and $\nu$ is the frequency of the transition. This is just the energy required to counteract the spontaneous decay in the system.
However I don't really understand why this is a minimum energy? A possibility is that if some external radiation field is applied we should also get an imbalance in the rates of spontaneous emission and absorption and so is this the extra contribution that we are missing in the above calculation (if not, why not)? Thanks in advance!
| It is not possible to have a population inversion in a system with just two states.
The trick with any laser is that you need three possible states - let's call them 1, 2 and 3 (in increasing order of energy - see image from wikipedia):
If we can make it so that the 2->1 transition is much slower than the 3->2 transition, then you can create a population inversion, but you need quite a lot of power (since you are exciting from the ground state, you need to excite more than half of all the atoms).
For this reason, in practice, most lasers actually are four-level: this means that the two levels between which transition occurs contain a minority of the atoms, and it's possible to create a population inversion between these two levels with minimal power. At that point, "minimal" is however much power is needed to counter the spontaneous decay taking place (which is the expression given in your question) multiplied by some factor describing losses (for example, this will include the energy lost in going from state 4->3 and from 2->1)
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Do free-electron lasers actually lase? Free-electron lasers are devices which use the motion of highly energetic electron beams to produce bright, coherent radiation in the x-ray regime. More specifically, they start with a high-energy electron beam and feed it into an undulator, which is an array of alternating magnetic fields designed to make the electron beam move in a 'zigzag' path with sharp turns on either side, emitting synchrotron radiation during each turn.
The radiation thus produced is added up over successive turns, and it is produced coherently via self-amplified spontaneous emission.
One common question frequently posed of this setup is: is this actually a laser? That is, does it have population inversion, and is the radiation actually emitted via stimulated emission in some suitable understanding of it? People in the know tend to answer in the affirmative, but I've yet to see a nice explanation of why - and definitely not online. So: do FELs actually lase?
| FELs produce a coherent, monochromatic, intense light beam that can be collimated with an iris (basically a hole in a large lead block).
An optical cavity can be arranged by putting two mirrors around the undulators, spaced so each pass of the electrons constructively interferes (go here and click Watch a Movie on How HIGS Works to see a movie demonstrating the Duke FEL doing something like this, with an added Compton backscatter).
Is this stimulated emission in the atomic physics sense? No. Does it produce a beam of light so similar to a laser than it can be classified as such? Most physicists I've interacted with say yes.
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Symmetry of the Polyakov action? Let us look at the Polyakov action for a string moving in a spacetime with metric $g_{\mu \nu}(X)$:$$S_P = -{1\over{4\pi \alpha'}} \int d^2 \sigma \sqrt{-\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X^\nu g_{\mu\nu}(X) \tag{1}$$ and suppose there exists a Killing vector $k_\mu$ in spacetime satisfying Killing's equation
$$\nabla_\mu k_\nu + \nabla_\nu k_\mu = 0.\tag{2}$$
Does this lead to a symmetry of the Polyakov action?
| Killing vector fields correspond to infinitesimal isometry generators of the spacetime manifold and any physical action including the Polyakov action should be preserved under it. In fact, any physical action should be invariant under the (infinitely) larger group of diffeomorphisms of a manifold. Isomotry transformations are just a finite subset of these diffeomorphisms.
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Why is the Fourier transform more useful than the Hartley transform in physics? The Hartley transform is defined as
$$
H(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty
f(t) \, \mbox{cas}(\omega t) \mathrm{d}t,
$$
with $\mbox{cas}(\omega t) = \cos(\omega t) + \sin(\omega t)$.
The Fourier transform on the other hand is defined very similar as
$$
F(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty
f(t) \, \mbox{exp}(i \omega t) \mathrm{d}t,
$$
with $\mbox{exp}(i \omega t) = \cos(\omega t) + i \sin(\omega t)$.
But although the Fourier transform requires complex numbers it is used much more in physics than the Hartley transform. Why is that? Are their any properties that make the Fourier transformation more "physical"? Or what is the advantage of the Fourier transformation over the Hartley transformation?
| Short answer: The Hartley transform is a subset of the results given by Fourier transforms, which is only the real part (assuming your signal is real, which is almost always the case in physics).
Long answer: Practically, you need the amplitude and phase of the signal, and the Fourier transform gives you both amplitude and phase by taking the
magnitude: $A=\sqrt{x^2+y^2}$ and
the phase: $\phi=\arctan{\frac{y}{x}}$
This all is information contained in the signal. If you use the Hartley transformation, then you only have the $x$ component of your signal samples. This is only useful for very special cases, while I imagine that the computational cost is probably the same.
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Would a mass of bare iron nuclei be visible? As I understand it most of how objects look is because of how photons interact with electrons and photons emitted when excited electrons fall to lower energy levels producing photons.
So if one traps a completely ionized mass of say iron or nickel and drew off the associated electrons what would the remaining plasma look like? To my way of thinking, it would have to be colorless. There is another bit of my brain telling me it can't be so. enlighten me.
| It is extremely hard to reach any kind of density of "plasma" of just nuclei - the thought experiment of "drawing off" the electrons would be surprisingly hard in practice, as the net charge gets larger and larger so the energy required to remove one more electron gets astronomical very quickly.
Thus you would have a very low density "plasma" and if there are zero electrons, the mean free path of photons traveling through the medium would be such that there is no reflected light and it would look "black" (very dark grey) in reflected light. Most of the photons would travel right through (it would be mostly transparent).
So yes - most likely no "color" can be observed.
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Why is chemical potential, μ=0 when calculating critical temperature of BECs? How do we justify taking the chemical potential, $\mu$ as $0$ when calculating the critical temperature of Bose-Einstein Condensates (BECs)?
I apologise as I do not how to use LaTeX, for if I did the elegance of mathematics would’ve allowed me to construct my question with ease...
I understand to calculate the total number of particles in a system comprised of non-relativistic bosons of mass m at thermal equilibrium at temperature $T$. One must simply some over the occupancies for each energy state, the occupancies is given by the bose-einstein distribution...
For some reason during the derivation setting chemical potential to zero within the bose-einstein distribution gives us the largest possible number of particles for a given temperature, can someone explain why this is true?
Edit: Also I know the within the bose-einstein distribution, the energy of the states must always be greater than the chemical potential, this confines the distribution to a range of $$ 0<\text{bose-einstein distribution}<+\infty$$ I can say that the lowest energy state (ground state) has an energy of 0 and thus chemical potential < 0, but if my ground state has an arbitrary non zero energy would the chemical potential = 0?
| You can think of the chemical potential as the amount of free energy needed to add one additional particle to the system. Because the ground state of a BEC is degenerate and can hold an infinite number of particles, there's no energy cost to add another particle to that state. So, $\mu = 0$.
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Why is the speed of sound lower at higher altitudes? At sea level the speed of sound is 760mph, but at altitudes like the Concorde would fly at (55,000ft) the sound barrier is at 660mph, so 1000th slower. Does it have to do with lower pressure?
| A sound wave moving through a gas requires a small scale bulk movement of gas molecules back and forth as pressure at any locations builds or falls. Therefore, the sound wave can not possibly move through the gas at a speed greater than that of the individual molecules themselves, and in fact must move at a lower speed than that due to the random nature of molecular movement. Since molecular speed in a gas is a direct function of average molecular kinetic energy and that is a direct function of temperature, the speed of sound in a gas will also be a function of temperature.
At the cruise altitude of the Concorde the atmospheric temperature is below that of the standard atmospheric temperature at sea level, so the speed of sound is also lower. That is the simple answer.
One might also note that the temperature is lower as one ascends primarily because the pressure is lower as one ascends and then conclude that the speed difference is equally attributable to lower pressure (or lower density for that matter). However, it is primarily and more directly related to molecular speed and thus temperature than it is to anything else.
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Hamiltonian of a quantum harmonic oscillator On page 286-287 of Nielsen Chuang's Quantum Information and Quantum Computation (10th edition) book, the Hamiltonian for a quantum harmonic oscillator is approximated as $H=a^\dagger a.$ What are the assumptions involved in such an approximation and why is this approximation needed?
| I have a lot of experience with this particular book, but unfortunately it is in my office right now so I can't reference the exact page you're on.
In QIQC, and this book in particular, you're going to be doing a lot of manipulating the Hamiltonian of the QSHO using commutators. Since constants always commute, the constant term of the QSHO falls out of every one of these operations and would be extremely burdensome to include. You'll see what I mean when you start doing problems with:
$$[H,a]$$
$$[H,a^2]$$
$$[H,\left(a^\dagger\right)^2\left(a\right)^2]$$
$$...$$
To clarify one thing: this is not an approximation but more of a shorthand for these types of algebraic problems. It is very important to keep the $1/2\hbar\omega$ for some problems in the book, like those involving expectation values!
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visible light spectrum Why do we see black objects? Colors of objects are formed when the spectrum of that color is reflected. Example Green objects are green because they reflect the green spectrum of light, red objects are red because they reflect red spectrum of the visible light and white objects because they reflect all the visible spectrum of light. When an object absorbs all the visible spectrum of light, it becomes black. But if nothing is reflected from the object, shouldn't it be invisible instead of black?
| We don't actually see light itself. We see objects which emit, reflect, absorb or refract light. Where the light strikes our retina and which cone is triggered allows the brain to create images because of contrast. This visual representation of reality is what we see. Because we have binocular vision this image appears 3 dimensional. Of all the species on the planet, ours is the closest to objective reality but we only actually see about 1% of the universe.
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Acceleration of log moving without slipping with/under a plank So I have this image:
I know how to solve the accompanying physics problem once I have determined the relative accelerations of the logs and the plank. According to my physics textbook, the cm of each log will move half the distance that the log will and thus the $a_{plank} = 2 a_{log}$. While this does seem to make sense and I can see it when I test it out with an eraser, I still can't seem to model it in my head. How does the cm of each log only move half the distance? If the log moves $2\pi R$ won't the log cm move $2R$? That clearly isn't half the distance. (There is friction between the logs and the plank and the logs and the ground).
| The point where the log touches the ground is the point where the log has zero velocity. If you now draw a small displacement, you see (from equal triangles) that if the center of the log moves $x$ the top moves $2x$ .
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Electron-Positron Annihilation in a Gravitational Field When an electron and positron annihilate what happens to any gravitational energy they have?
In a way the energy is 'shared' between these two particles and all the matter in space. It must take a finite time for it to vanish. I realise of course that the gravitational energy of the two particles is very small.
| We all learned in school that gravity is associated with mass. For example if we have two bodies with masses $m$ and $M$ then Newton's law tells us that the gravitational force between them is:
$$ F = \frac{GmM}{d^2} $$
So if either object has zero mass the force goes to zero. And since photons are massless it's entirely reasonable to ask if the gravity just disappears when an electron and positron annihilate into two photons.
However Newton's law is only an approximation and to understand gravity better we need to move to general relativity. In general relativity Einstein's equation tells us:
$$ G_{\alpha\beta} = 8\pi G T_{\alpha\beta} $$
The $G_{\alpha\beta}$ on the left hand side is related to the curvature of spacetime and from it we can work out how freely falling particles will move. So the $G_{\alpha\beta}$ is sort of related to the gravitational force. On the right hand side the $T_{\alpha\beta}$ is called the stress-energy tensor and it takes the place of mass.
In the stress-energy tensor we don't distinguish between mass and energy - we consider them equally and interconvert between them using the well known equation $E = mc^2$. So before the annihilation the masses of the electron and positron would go into the stress-energy tensor and after the annihilation the energies of the two photons would go into the stress-energy tensor. The result is that the gravitational field doesn't just vanish the moment the electron and positron annihilate.
However the stress-energy tensor for two photons moving in opposite directions is very different to the stress energy tensor for two (effectively stationary) 511keV mass particles. This means that as soon as the particles annihilate the spacetime curvature will start to change and that change will propagate outwards at the speed of light.
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Heisenberg's uncertainity principle In the Heisenberg uncertainty principle,
$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$
The values of $\Delta x$ and $\Delta p$ are the standard deviations which we get from the probability distribution function of the particle and I heard that it has nothing to do with the measuring instrument.
Actually while measuring, the probability distribution function of a particle also changes, Does this means that the measuring instrument has some effect?
| These are standard deviations of a probability distribution indeed. The probability distribution is that of getting a particular value while the system is in the state prior to measurement. So we imagine measuring a state - it collapses to some value, and then somehow resetting time back before measurement and measuring again. If we repeat this process, rewinding the clock and measuring, this is what is meant by the probability. Equivalently, and perhaps more accurately you could think of preparing a bunch of identical quantum systems, and the distribution would be of the values you'd get by measuring all of them. So THIS is what the standard deviations apply to - these probability distributions. So if you measure x with complete certainty, then the standard deviation for the momentum distribution becomes infinite. It has nothing to do with the accuracy of measurement or something.
Furthermore measuring does have an effect - causing the system to randomly select a value from the previous probability distribution and making it choose that exact value to be.
Hope this helped!
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What's the reason double-slit experiment can't be explained by edge effects rather than quantum interference? Say we had exactly this...
But instead, it was a PING PONG GUN (imagine as table tennis players use to train),
throwing out PING PONG BALLS. The two slits are say 20 cm wide, and the observing screen is say 5m distant.
If the ball goes through the EXACT MIDDLE of a 20cm slit, it will travel in a perfectly straight line and make a "dot" on the observing screen.
If the ball travels nearer and nearer to the left or right edge of a slit, the flight path will bend slightly towards that side. For example, due to electrostatic force (rather like how a vertical pour of water from a faucet will bend slightly as your hand approaches).
Note that this is not some sort of fantasy; you could very easily organise for the ball path to bend slightly when near an edge, using either electrostatic force, magnetic force, aerodynamic factors or other forces, with the correct material of balls and slits (substitute small metal balls and slits of magnetic material .. whatever).
Indeed, you could trivial arrange so that precisely this famous image
is the outcome.
This is the "trivial mechanical bending" explanation of "all this interference pattern stuff".
Can you help me understand in a clear way, What is the explanation of why this is not at all the explanation?
| Your explanation has a tennis ball that is a body and you are influencing it with forces , but in reality particles like electrons , photons even complete atoms show this behavior because everything behaves like a wave of some wavelength , it is just that as the particles become macroscopic , the effect is minimized .
The "famous photo" when you notice is just like a wave where the white regions being crest and the dark being the trough
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What would be the view like from inside a black hole looking towards the event horizon? Ignoring the fact that we would be torn apart by gravitational gradient and assuming we get some time to make some observations before hitting singularity, what would we see looking towards the event horizon or in any other direction away from the singularity?
| To answer this question we have to raytrace the lightrays hitting the observer, for rotating black holes the coordinates of choice are Doran raindrop coordinates which make it possible to calculate the view from both outside as well as inside. The first image shows the outside perspective of an observer at a latitude of θ=45° free falling with the negative escape velocity (relative to a local ZAMO) v=-c√(1-1/gtt) into a rotating Kerr Newman black hole with an equatorial accretion disk whose inner radius is at the ISCO at r=1.6367 and the outer radius at r=7 (full panorama 360°×180° in equirectangular projection):
The second image shows the view when the observer is crossing the outer horizon:
In the third image we see the perspective from inside the black hole:
For different angles see here. Color code for the frequency shift:
When you fall into a nonrotating black hole, the shadow of course stays spherical, for a raytraced simulation of a free fall into a Schwarzschild black hole see the first animation in this link; when you are falling with the negative escape velocity, right before you hit the singularity, exactly half of your viewing field (180°×180°) is black.
For a stationary observer this already happens when he is hovering at the photon sphere, with the difference that the light he receives from behind would be blueshifted, while for a free falling one it would be redshifted.
When you cross the horizon with the negative escape velocity, the light that hits you from behind is shifted to half its original frequency which means the visible light would shift to infrared while ultraviolet would shift to the visible spectrum, while the light hitting you perpendicular to the direction of motion would not be shifted because the gravitational and kinematic time dilation would cancel out.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is the sound channel in the ocean especially good for low frequency sound? Why does not the high frequency sound propagate as far?
The dispersion curve $\omega(k)$ is almost linear, right?
| The basic phenomenon is that high frequency sound is more strongly attenuated than low frequency sound. The mechanism for sound attenuation is viscous damping. The absorption coefficient is
$$
\gamma= \frac{\omega^2}{2\rho c^3}\left[ \frac{4}{3}\eta + \zeta
+ \kappa\left(\frac{1}{c_v}-\frac{1}{c_p}\right) \right],
$$
where $\omega$ is the frequency, $\rho$ is the density, $c$ is the speed of sound, $\eta$ is shear viscosity, $\zeta$ is bulk viscosity, and $\kappa$ is thermal conductivity.
Physically, what is happening is that sound is a longitudinal oscillation of the fluid velocity. Viscosity tries to equalize velocity, and damps any oscillation. The damping force is proportional to gradients of velocity $\eta\nabla u$, and the energy dissipated scales as $\eta(\nabla u)^2$. In Fourier space this is $\dot E \sim \eta k^2 u^2\sim \eta\omega^2u^2/c^2$. Since $E\sim \rho u^2$ the damping $\dot{E}/E\sim \eta\omega^2/{\rho c^2}$. If we write this in terms of the absorption length there is one extra power of $1/c$, which gives the formula above.
Bulk viscosity and thermal conductivity work analogously. Again, the damping is proportional to $\omega^2$.
| {
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Given a current velocity and a fixed input energy, how much faster will a relativistic particle be? The relativistic kinetic energy of a particle with mass $m$ and velocity $v_0$ is $$m c^2 (\gamma_0 - 1) \textrm{ where } \gamma_0 = \frac{1}{\sqrt{1 - \frac{v_0^2}{c^2}}}$$
I would like to know how quickly the particle will be moving ($v_1$) after energy $E_i$ is added to the system in the direction of positive acceleration.
I presume that I can use the following equivalence:
$$E_i = E_1 - E_0 = m c^2 (\gamma_0 - 1) - m c^2 (\gamma_1 - 1)$$
I believe my goal is to solve for $v_1$ (from $\gamma_1$ by analogy) given $E_i$ and $v_0$.
I have two questions:
First, is this approach correct, or am I misunderstanding how kinematics works?
Second, how can I solve the equation for $v_1$? I find that the algebra is beyond me and I haven't been able to leverage online solvers to improve my situation. Any suggestions as to how to approach this would be much appreciated.
| You do not need the kinetic energy. Working with the total energy $\gamma m c^2$ produces the same result.
Assuming both the total initial energy $\bar E_0 = \gamma_0 m c^2$ and the additional energy $E_i$ are known, write $\gamma_1 mc^2 = \frac{mc^2}{\sqrt{1-\beta_1^2}} = \bar E_0 +E_i$ for $\beta_1 = \frac{v_1}{c}$, then
$$
\sqrt{1-\beta_1^2} = \frac{mc^2}{\bar E_0+E_i} \;\;\Rightarrow \;\; \beta_1 = \sqrt{1-\left(\frac{mc^2}{\bar E_0+E_i}\right)^2}
$$
| {
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Does momentum space have a speed limit? In ordinary $xyz$ space, the maximum velocity of propagation for mass-energy and/or information is $c$.
So, my question: Is there also a maximum velocity of propagation in momentum ${p_x}{p_y}{p_z}$ space, one that would for example place a maximum limit (not necessarily approached) on how quickly an electron can fall from the top of a conduction band into a newly opened hole at the bottom of the band?
If not... well, why not? It's easy to say "no," but I'm not aware of any actual study of such a question. Given some of the remarkable symmetries and mathematical links between the $xyz$ and ${p_x}{p_y}{p_z}$ spaces, I suspect that some sort of specific theoretical answer should be possible.
| For a transition like electron-hole recombination, its probability is linear with time with some characteristic time scale that depends on the system: the probability of having a transition between $0$ and $dt$ is $\frac{dt}{\tau}$. So there is a non-vanishing transition probability at arbitrarily small times.
In momentum space, the evolution of a wavepacket looks something like this: it is nonzero near a point $\vec p$ for a while, until it scatters and jumps to some other $\vec p'$. I don't see how $c$ comes into this.
In any case, if you want to look into those sorts of limits you're better off studying relativistic quantum mechanics / field theory. Otherwise you are going to find holes.
| {
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Newtons third law with two charged particles Imagine we have two charged particles, $q$ and $Q$.
$q$ is at rest at a point and $Q$ is moving with a velocity. Now $q$ is exerting an electrostatic force on $Q$ and $Q$ makes a magnetic field but because $q$ is at rest there is no force from $Q$ to $q$.
So, is the third law of Newton violated in this case?
| $Q $ doesn't generate just a magnetic Field, but also an electric field: Just because $Q $ moves, that doesn't mean that it doesn't generate a field, this field just has to satisfy the Maxwell equations. $\mathbf{\nabla\cdot E} = \rho$ still has to hold, and also the magnetic field that is created is time dependent, so $\mathbf {\nabla\times E} = \dot{\bf B}\;.$ This field creates a force on the resting charge $ q.$
| {
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Understanding of dipole moment and its vector property I have a trouble understanding the electric dipole moment.
The electric dipole moment formula is
$${\bf p}= \int {\bf r}' \rho({\bf r}')d\tau '$$
I'm interested in the coordinate, the origin of which is changed into $\bf a$.
$${\bf r}' = {\bar {\bf r}}' + {\bf a}$$
Now calculate dipole moment in the new coordinate
\begin{align}
{\bar {\bf p}} &= \int {\bar {\bf r}}' \rho ({\bar {\bf r}}') d\bar\tau' \\
&= \int ({\bf r}'-{\bf a}) \rho ({\bar {\bf r}'}) d\bar\tau'
\end{align}
In Griffiths, it says
$$\rho ({\bar {\bf r}}') = \rho ({{\bf r}}') $$
so that yields ${\bar {\bf p}}= {\bf p} - Q {\bf a}$.
I don't understand how to verify $\rho ({\bar {\bf r}}') = \rho ({{\bf r}}') $.
| I think you are confused with the new coordinates system.
You sould correct the above equations like this:
\begin{align}
\bar{\mathbf p} &= \int \bar{\mathbf r} \bar\rho(\bar{\mathbf r})d\bar\tau \\
&=\int ({\mathbf r}-\mathbf a) \rho(\mathbf r)d\tau\\
&=\mathbf p - \mathbf a\int\rho(\mathbf r)d\tau = \mathbf p - Q\mathbf a
\end{align}
In other words, you should consider a new density function $\bar\rho$.
| {
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Identifying an ideal gas I'm pretty confused. So I've been told that the equation of state for $n$ moles of some type of gas is $P(V-b) = nRT$. That's not quite like an ideal gas. But then the relations $C_p - C_v = R$, and $\gamma = \frac{C_p}{C_v} = 5/3$ hold for this gas, and at the very least the last equation is definitely for ideal gases.
So if a gas has these properties, and only one mol is considered, is it fair to assume it is an ideal gas? Using the above relations gave me $C_v = \frac{3}{2} R$, which again is for ideal gases.
| The gas you are describing is not precisely an ideal gas, but is pretty close. In an ideal gas, the molecules are dots, they don't have volume ; moreover, threr are no interactions except for the elastic collisions that allows the gas to thermalize.
The gas you describe is is a gas with no interactions, but with molecules of finite volume. Having molecules of finite volume reduces the space avaliable for molecules to move around, hence the $V-b$ factor instead of just $V$ for an ideal gas.
| {
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Do we know why there is a speed limit in our universe? This question is about why we have a universal speed limit (the speed of light in vacuum). Is there a more fundamental law that tells us why this is?
I'm not asking why the speed limit is equal to $c$ and not something else, but why there is a limit at all.
EDIT: Answers like "if it was not.." and answers explaining the consequences of having or not having a speed limit are not -in my opinion- giving an answer specifically to whether there is a more fundamental way to derive and explain the existence of the limit.
| Why is there a speed limit in our universe? This might have something to do with the principle of locality in physics. Note that in Fredkin 's universe as a cellular automaton , there always is a speed limit, for any emerging pattern (just an example). So the existence of a speed limit in our universe is an endorsement (consequence) of the principle of locality in physics. As a side note , quantum non-locality arguments (based to quantum entanglement experiments) must be explained in terms of synchronized chaotic systems, but not rejecting the principle of locality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does a trumpet play at a tritone lower without lip vibration? My trumpet teacher noticed that if you blow into a trumpet for warm-up, without any lip vibration, there is still a slightly audible pitch which is a tritone lower than "expected" in the following sense:
For example, if none of the vents are pressed on a Bb trumpet (e.g. G with lip vibration), and air is blown through it, there is a slightly audible C# pitch. If you depress the middle vent, instead of Gb, a C pitch is audible, and so on.
The effect is consistent on several Bb trumpets he tested. Is there a mechanical explanation?
| There are three points to be noticed:
*
*If you just blow without closing the lips, you would change the boundary condition.
*The trumpet waveguide is not "nicely predictible", the approximation of an open tube does not work cause the bore variations $S(x)$. You need to solve this kind of beasts for reasonable 1D propagating pressure approximation:
$$
\frac{\partial^2 p}{\partial x^2} + \frac{\partial (\ln S(x))}{\partial x}\frac{\partial p}{\partial x} = \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2}
$$
and the change in boundary conditions can move the mode peaks in various manner.
*
*Most importantly: The driving mechanism is in this case the vortex sound. Therefore the instrument resonates not on its impedance maxima but minima (generall difference between flute-like and trumpet-like instruments). That might do the impression of sounding in between the usual trumpet natural tones.
| {
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Can one write down a Hamiltonian in the absence of a Lagrangian? How can I define the Hamiltonian independent of the Lagrangian? For instance, let's assume that i have a set of field equations that cannot be integrated to an action. Is there any prescription to construct the Hamiltonian of a such system starting from the field equations?
| Comments to the question (v2):
*
*First of all, let us stress that OP is correct, that a given set of equations of motion does not necessarily have a variational/action principle, cf. e.g. this Phys.SE post and links therein.
*On one hand, if there exists a Lagrangian formulation, then one may in principle obtain a Hamiltonian formulation via a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2.
*On the other hand, if we have a (possible constrained) Hamiltonian formulation, of the type discussed in Refs. 1 and 2, then it is possible to give a Hamiltonian action formulation, which in itself can be interpreted as a Lagrangian formulation, e.g. after integration out momentum variables along the lines indicated in my Phys.SE answer here.
*In other words, Lagrangian and Hamiltonian formulations traditionally go hand in hands. Thus it is unclear what precisely OP is looking for.
References:
*
*P.A.M. Dirac, Lectures on QM, (1964).
*M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.
| {
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Perturbative QCD application in FAIR In FAIR CBM scenario i.e at high density will be perturbative QCD applicable?
| PQCD is only applicable at really high energy and FAIR CBM experiment will have quite low energy and finite baryonic density.
As it is said on the official doc :
http://www.fair-center.eu/fileadmin/fair/publications_FAIR/FAIR_BTR_3a.pdf
Low energy model will be available to compare data, but in my opinion, lattice QCD is the best tool available to discuss results at FAIR and to probe the phase transition between hadronic matter and the QGP.
However, it might still be possible to use pQCD for some initial collision and to try to understand energy loss mechanisms at finite density.
| {
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When sunlight bounces off the Earth, why isn't the entire spectrum reflected rather than just the infrared portion? I've read that greenhouse gases absorb and reemit sunlight, and that the infrared portion is what bounces off Earth back to space. When sunlight bounces off the Earth, why isn't the entire spectrum reflected rather than just the infrared portion?
| It does. Consider this: you can see the Earth from space. Therefore, not just infrared light gets reflected but also light on the visible spectrum.
Here's a graph (by NASA) of various planet's radio emissions. The ways that Earth can release radio waves is a bit limited. Because of that, it is safe to assume that at least some come from the Sun.
| {
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What does a zero topological S matrix element mean? I realize that for nonabelian anyons, their S matrix elements could be zero (eg. the Ising anyons). I'm confused by the meaning of a zero S matrix element. Does it mean that the corresponding braiding process results in a zero amplitude?
| Let us consider the S matrix of Ising anyons as an example: $S_{\sigma\sigma}=0$. To be more precise, suppose we have four $\sigma$'s, labeled as $1,2,3,4$. We assume $1$ and $2$ are in a definite fusion channel. Then we braid $3$ around $1$ (or $2$, no difference). Then the state of $1$ and $2$ must be orthogonal to the initial one: they must be in the opposite fusion channel! This is the meaning of $S_{\sigma\sigma}=0$. One can do a simple calculation using Majorana modes as a "model" of Ising anyons: say we have $\gamma_1,\gamma_2,\gamma_3,\gamma_4$. Braiding $\gamma_3$ around $\gamma_1$ results in the following transformation: $\gamma_{1,3}\rightarrow -\gamma_{1,3}$. Therefore, the fermion parity $i\gamma_1\gamma_2$ which represents the fusion channel of $1$ and $2$, is flipped.
This observation was first made by Bonderson et al in http://arxiv.org/abs/cond-mat/0508616. They devised an interference experiment to see this zero (vanishing of interference pattern). So the physical meaning of this zero is worth a PRL!
| {
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What is the minimum force required to move this block Please don't report. It's not a homework question. Yesterday on my physics test there was this question. there is a block of mass $m$ connected to a spring as shown in the figure. the spring constant is $k$ and the friction coefficient between the block and the floor is $\mu$. they have asked what is the minimum horizontal force $F$ applied as shown in the figure so that the block starts to move. I answered $\mu mg$ considering the whole spring mass system as a single system of mass $m$. but a friend said afterwards that it would be $\frac12 \mu mg$ because the force is being applied on the spring and if the force elongates it by $x$ length, $$kx=\mu mg$$ and, $$Fx=\frac12kx^2$$ so solving, $$F=\frac12µmg$$ . Is he right ? please explain in detail why he is right or wrong. And please point out the problem in my thinking if I am wrong.
what he is saying is that he is equating the increase in potential energy to the work done by the force on the spring. I don't understand his point.
| F(x) = kx, according to Hooke's Law. This means that your friend is incorrect, and you got the question correct.
Note: the potential energy of the spring is (kx^2)/2, so your friend confused potential energy with force. A bit of dimensional analysis would greatly decrease the chance of mixing the units in this way.
| {
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Quantum computing, NP complexity Hi I have a very limited knowledge of quantum physics and its bothering me trying to understand quantum computing.
I want understand how a qubit can return usable data faster then a regular bit, how is the NP complexity not an obstacle?
| The best way to understand this is to work through an example. Here is how you factor the number 15 using Shor's algorithm. (If you prefer, view the problem not as factoring 15 but as solving $2^r=1 \hbox(mod 15)$, and skip Step Seven). As you work through this, imagine replacing 15 with a much larger composite number, and you'll see the advantage of having the quantum algorithm.
Step One. Prepare a register in
the state
$$|1>+|2>+|3>+|4>+|5>+|6>+|7>+|8>
+|9>+|10>+|11>+|12>+|13>+|14>+|15>$$
Step Two. Apply the function
$x\mapsto 2^x$ (mod 15) to the above
register and put the result in a
second register. The second register
now contains:
$$|2>+|4>+|8>+|1>+
|2>+|4>+|8>+|1>+
|2>+|4>+|8>+|1>+
|2>+|4>+|8>+|1>$$
which is the same as
$$|2>+|4>+|8>+|1>$$
Step Three. Observe the second
register. With equal probabilities,
we might observe $2$, $4$, $8$, or
$1$. For illustration, let's suppose
we observe $2$. (The other cases are
similar.)
Step Four. Now the first register
contains
$$1>+|5>+|9>+|13>$$
These differ by 4, but we don't know
this because we haven't done the
computations necessary to carry this
out (actually we {\it have/} done
them in this example with $N=15$, but
they'd be prohibitively costly if $N$
were huge.) If we knew that the
difference was 4, we could skip to
Step 7. As it is, let's call the
difference $r$ because it's still
unknown to us.
Step Five. Apply the function
$$x\mapsto \sum_ye^{2\pi i x
y/15}|y>$$
to the first register and put the
result in the second register.
The second register now contains
$$\sum_y \Big(e^{2\pi i y/15}
+e^{2\pi i 5 y/15}
+e^{2\pi i 9 y/15}
+e^{2\pi i 15 y /15}\Big)|y>$$
Numerically, the norms of the
coefficients are given approximately
by the following table:
$$\matrix{
|y>&\hbox{ Norm of Coefficient}\cr
1&.9\cr
2&1.7\cr
3&2.6\cr
4&2.5\cr
5&1\cr
6&.4\cr
7&2\cr
8&2\cr
9&.4\cr
10&1\cr
11&2.5\cr
12&2.6\cr
13&1.7\cr
14&.9\cr
15&4\cr}$$
which peaks (this is no accident) near
$y=3,4,7,8,11,12,15$---the values
where $4y/15$ is close to an integer.
Step Six. Observe the second
register. If you get $y=15$, you
learn
nothing. Otherwise, you probably get
$y=3,4,7,8,11$ or $12$---which allows
you to
guess---after a couple of trials---
that $r=4$, based on the knowledge
that $yr/15$ is probably close to an
integer.
Step Seven. Now we know that $r=4$,
which tells us that $2^4=1$ (mod
$15$). So 15 divides $2^4-1=(2^2-
1)(2^2+1)=3\times 5$. Choose either
factor (say 3) and compute its
greatest common divisor with 15 using
euclid's algorithm. This gives 3.
Therefore 3 is a factor of 15.
| {
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How do two metal balls of radius $r$ kept at a large distance form a capacitor? So this is something really new to me. I have learnt of plate capacitors, cylindrical capacitors, single spherical capacitors, etc. but that to, of finite distance between them. Also please explain if the distance is not very large and comparable to their radius how would the capacitance of system differ.
| Here is how:
And the expression is (from wikipedia):
$$2\pi \varepsilon a\sum_{n=1}^{\infty }\frac{\sinh \left( \ln \left( D+\sqrt{D^2-1}\right) \right) }{\sinh \left( n\ln \left( D+\sqrt{ D^2-1}\right) \right) } $$
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Charge inside a charged spherical shell
*
*If I were to put a negative charge inside a negatively charged spherical shell, will it move to the center?
*Electric field inside the shell due to the shell is zero (Gauss's Law), would that mean the charge inside the sphere faces no force?
But, that doesn't make intuitive sense to me. If the negative charge was near the walls of the sphere, wouldn't the charges on the near wall push the negative charge to the centre as the force due to the charges on the wall closest to it is higher than that form the walls further away from it.
*What about in the case of a ring? Will the charge move towards the center?
|
If I were to put a negative charge inside a negatively charged spherical shell, will it move to the center?
Electric field inside the shell due to the shell is zero (Gauss's Law), would that mean the charge inside the sphere faces no force?
The answer depends upon whether the spherical shell is conductive, and if not, whether or not the charges on the spherical shell are uniformly distributed.
If the charges on the spherical shell are uniformly distributed, then there will be no net electric field due to the shell anywhere within the shell.
However, if the spherical shell is conductive, and the charge within the cavity is not perfectly centered, the charge within the cavity of the shell will cause the charge on the shell to re-distribute itself. The resulting charge distribution on the shell will not be uniform. Consequently, the shell will exert a net force upon the charge in the cavity. The direction of the force will be away from the center, not toward it. That is, an uncentered charge in the cavity will want to move to the surface.
| {
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Does matter stacks up as it approaches Black hole? When something approaches Black hole it'll experience time dilation with respect to a frame away from the black hole.
So to an observer away from the hole the object would seem to slow down until it finally appears to have stopped .
Would that mean all the matter that is falling into Black hole since its birth would still be observable stacked on its surface to an observer from earth if he visits it?
| Yes and no. Remember in special relativity whenever someone asks a question, they always are told to draw a spacetime diagram. The same thing happens in general relativity. If you want to see what is possible, consider drawing a Carter-Penrose diagram.
For a black hole you can draw the event of a test particle crossing the event horizon. The past light cone of that event includes parts of the spacetime and not others. So anything outside that past light cone can't see that event without themselves crossing the horizon. And that event only sees the thing in the past light cone before it crosses.
When you stay outside, you only see the event on the outside. So if you have a black hole that formed by collapse you still see the center of the star, just from way back before the event horizon formed.
It will be slow, and it will be red. And if it is the center the outside of the star might block your view (just like for any star).
So, it is not stacked up on the outside. If you dug a mineshaft on your star and supported the sides then you'd still be able to look inside. The outside is only more visible since there isn't anything farther out to stand in your way.
The whole stat, center, medium layers, and outer layers. Is all red shifted, and you see the events from before the horizon formed. And you can see it except for how slow and red it is.
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Symmetry and group theory book I would like to start learning about symmetries in physics and how they affect physical quantities. As far as I know, the mathematical language that describes symmetries is the Group Theory. So, I think the best start would be getting deep in this theory. Although there are many books on that topic, I would like to find one that would be written in a very simple way, for beginners, with many examples, preferably in the physical context. Could anyone suggest such a book?
Note: I would like to study graphene and other 2D materials such as transition metal dichalcogenides.
| opinion based question, so it may be closed.
The author Vincent (family name) has a very good introduction to group theory for molecules.
I like this book as it has questions for you to answer as you go along so you really learn it as you read.
If you are interested in solid state then you will have to go further to space groups with another text - this text deals with point groups.
| {
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Does light interact with electric fields? We know that light is an electromagnetic wave and it does interact with charges.
It contains magnetic field and electric field oscillating perpendicularly but when we apply an electric or magnetic field in any direction to the wave the applied electric field or magnetic field vector doesn't alter the magnetic or electric field in the electro magnetic wave (according to vector addition rule)....why?
| An applied electric or magnetic field doesn't alter the field of an electromagnetic field because, as you said, the superposition principle holds. This principle is a principle of linearity, and comes from the linearity of electromagnetic equations : there is no interaction between photons at low energies.
You can see it from a field theory point of view, as there is no bare interaction vertex between photons in QED.
On the other hand, in other theories such as QCD, gauge bosons (the gluons) carry a colour charge and can interact.
| {
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Does potential difference or electric field change with distance between parallel plates? Say you have a set of parallel plates, one is positive and one is negative, if you change the distance between them would electric field strength change or potential difference, given the equation
$E=dV/dx$
From pure intuition, I think electric field strength will change as the field line bulges out from the sides, and as the plates are far enough from each other, each plate can be treated as a point charge and the system becomes a dipole, and E decreases at 1/r^3:
picture: (sorry I didn't upload the picture here in case I violate copyright)
http://www.electrobasic.com/uploads/3/2/3/4/32342637/9893979_orig.jpg
But I read from this post:Why does the potential difference between two charged plates increase as they move further apart?
and it says E must remain constant, and I am not sure why that is, and I do not see how simply moving the plates apart will increase the potential difference, as
$V=kQ/r$
and as r increases, v should decrease, and I don't understand how simply moving them apart will store energy in the system.
So why must E remain constant and V increase as the plates are separated further?
Thanks.
| $C={kA\over d}$. When $d$ increases, $C$ decreases.
$Q=CV$. $C$ stands for capacitance, and $V$ stands for potential difference. $C$ decreases, $V$ increases. That's why the potential difference increases.
| {
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Why is meteor speed what it appears to be? Is the speed of a meteor through our sky because of the speed of the earth's axis rotation, or because the meteor is speeding towards us at that speed?
| All speed is relative. But an object that starts from rest at infinity will reach a velocity of about 11 km/s when it hits Earth, if Earth is the only thing pulling on it. At the same time, Earth is moving with an orbital speed of about 30 km/s. Their relative importance will depend on the direction from which the meteor is approaching - but on the whole Earth's orbital speed is pretty significant. The rotation of the earth's surface (and atmosphere) matters much less - at the equator, it's 465 m/s, orders of magnitude smaller than the other two.
If you take into account that the meteorite is moving in an orbit around the sun, then it can have a velocity up to $\sqrt{2}v_{earth}$ when it crosses Earth's orbit (if it's falling straight towards the Sun) - about 42 km/s. At the kinetic energy associated with that velocity, the additional energy due to the Earth is not so significant.
Depending on their relative directions, these two velocities (30 km/s and 42 km/s) can either add for a blistering 72 km/s head-on collision, or be reduced to a meagre 12 km/s if the meteor is catching up on the Earth from behind. However, in that case the additional attraction by Earth would not be negligible and the final approach velocity would be affected by Earth's gravity - so you'd reach about $\sqrt{11^2 + 12^2}\approx 16~\rm{ km/s}$. In fact, if the meteorite starts out in an orbit close to earth's, there is no telling how slowly it will approach... but I suppose that it could be as low as 11 km/s.
These are just ballpark numbers - it says that the velocity of Earth and the meteorite both play a significant role in determining the velocity of impact, and that the rotation of the Earth does not.
| {
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Why do materials show plastic behaviour for large stress? As the stress is increased, the strain increases proportionally up to elastic limit and the material regains its original dimension within elastic limit. When the stress is increased further the material shows a plastic behaviour. What change in the internal structure causes the transformation from elastic to plastic behavior?
|
What change in the internal structure causes the transformation from elastic to plastic behavior?
It depends on the material. For metals, small elastic strains are just the result of very small changes in the interatomic spacings. When more stress is applied, pre-existing dislocations in the metal start to move, causing re-arrangements in the atomic positions which result in plastic flow behavior. Dislocations are also generated with increasing strain by mechanisms such as Frank-Read sources. As the amount of plastic strain increases, the dislocation density tends to increase.
| {
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Balancing Utensils: Center of Mass If you have a cork piece on top of a nail, it is extremely hard to keep it stable, and the slightest action will make the cork fall off. However, when you balance it on top of a nail but put forks into it, apparently it becomes extremely easy to keep stable. I looked in my textbook and it said that this is because the center of mass is lowered from its original position (and the center of mass was below the point of contact in the first place), which is confusing because if you are adding more mass towards the top, shouldn't the center of mass move upward and not down? It also says that when you try to topple the cork with the forks in it, the center of mass moves upward, increasing its potential energy.
Can someone really good at this stuff explain the entire process to me? I am truly confused about this.
| The experiment involves sticking the tines of the forks into the cork so that the long heavy handles of the forks extend downward. Take a look at the photo in this link: https://www.kecksci.claremont.edu/physics/demo/corkfork.htm.
Now, the cork and the forks are bound together as one object, and the center of mass of that object is down toward the middle of the forks, well below the top of the cork. You've effectively lowered the center of mass of the cork by making it part of a cork/forks object. This link explains the process: http://scienceblogs.com/principles/2007/12/17/the-twofork-toothpick-trick-ex/, though using a toothpick rather than a cork.
With the center of mass lower than the point of contact, gravity works to stabilize the object, because if the object tilts (raising the center of mass), the center of mass wants to drop down again and the object will automatically balance.
But if the center of mass is higher than the point of contact, gravity tends to de-stabilize the object because any slight movement lowers the center of mass, beginning the process of falling.
| {
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Why is thermal energy not a state variable? It is written on Wikipedia: (https://en.wikipedia.org/wiki/Thermal_energy)
The thermal energy of a system scales with its size and is therefore an extensive property. It is not a state function of the system unless the system has been constructed so that all changes in internal energy are due to changes in thermal energy, as a result of heat transfer (not work). Otherwise thermal energy is dependent on the way or method by which the system attained its temperature.
My question is: Why thermal energy is dependent on the way or method by which the system attained its temperature? You only need to give me an example proving that. Thanks in advance.
| An example you asked for :
Consider a system of an ideal gas where volume of the gas is constant (call it system A) and one where the pressure is constant (System B). Because the internal energy of the system is a state variable, same change in temperature would cause same change in internal energy. Now, $$ \Delta U = Q + W$$
In A volume is constant so W=0 and hence Thermal energy $Q$ is equal to $\Delta U$. But in B the gas will expand, doing some work (say $-W$) and hence for the same change in $U$ we need to supply more thermal energy, equal to $\Delta U +W$.
| {
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Coefficient of friction and practical experience of sliding The classical model of friction has a coefficient of friction depend only on the materials, but not area, and the force proportional to the normal force and coefficient of friction. So a given object on the same surface has the same friction whether it is supported by full bottom area or small legs as long as the materials are the same.
However every child knows that on a slide one goes faster if one lays down on their back compared to sitting on their butt. The slide is obviously still the same and since jackets usually extend below butt, the other material is also the same. So the friction should be the same as well, but it clearly isn't. So what is going on here?
Note: I mean typical stainless steel or fibreglass laminate slide, not ice, which is soft enough to complicate the matter further.
| Going faster on a dry playground slide if you lie down seems not to be well-documented, so I shall discount the effect of posture. The amount of friction depends on your weight, which is the same lying down as it is sitting up.
If there is any difference on a dry slide it is probably due to the reduced coefficient of friction provided by clothing, especially wool, nylon, polyester and polythene. Skin and rubber have relatively high coefficients, so lifting shoes and bare legs and arms off the slide makes a big difference. It makes no difference whether you lie down, sit up or stand (surf- or skate-board style) if the material beneath you is the same.
(I once almost slid off a steep rock face while mountain climbing by lying down on nylon clothing to avoid losing my balance, instead of standing upright on rubber-soled boots as my experienced companions did.)
On a water slide the key difference is the lubrication provided by the stream of water which flows down with you. Lying down spreads your weight over the greatest area, reducing the pressure beneath you, which allows a relatively thick layer of water to lubricate your motion over the plastic slide. Sitting up squeezes the water beneath you into a much thinner layer which offers less lubrication.
| {
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Variant of the Sokhotski–Plemelj theorem I am aware of the Sokhotski–Plemelj theorem (I have also heard people referring to it as the "Dirac identity") which states that in the limit $\eta\rightarrow 0^+$
$$\frac{1}{x\pm i\eta}=\mathcal P\frac{1}{x}\mp i\pi\delta(x) \, .$$
Now, I am reading the book "Solid State Physics" by G. Grosso and G. Pastori Parravicini which states on page 430 that using the above formula it can "easily be proved" that
$$\frac{\hbar\omega}{E_j-E_i-\hbar\omega-i\eta}=
\frac{E_j-E_i}{E_j-E_i-\hbar\omega-i\eta}-1\, .$$
However, I fail to see how the latter formula follows from the former. Is there a trick that I am missing here?
| I don't think we need Sokhotski-Plemelj for this.
Think of $E_j - E_i$ as a fixed value $E$.
Then the formula is re-written as
$$\frac{\hbar \omega}{E - \hbar \omega - i \eta}\, .$$
Now let $x \equiv \hbar \omega$ and you get
$$\frac{x}{E -x - i \eta} \, .$$
This integral is dominated by the part where $x \approx E$ so let's try shifting the variables $y \equiv E - x$,
$$\frac{E - y}{y - i \eta}$$
and then expand the numerator and put the original variables back in:
$$\frac{E_j - E_i}{E_j - E_i - \hbar \omega - i \eta} - \frac{E_j - E_i - \hbar \omega}{E_j - E_i - \hbar \omega - i \eta} \, .$$
The first term already matches the first term in the target expression, so we only need to worry about the second term.
The second term is 1 because, well, in the limit $\eta \rightarrow 0$ it's identically 1.
| {
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Why are the closed and open ends of an organ pipe nodes and anti-nodes? Here is a diagram of a wave in an organ pipe you'll find in most physics books
*Waves in air are longitudinal (not traversal), so what do the curves represent?
*Why are the open ends always anti nodes and the closed ends always nodes?
| Sound waves are made of alternation of compression (higher density) and rarefaction (lower density) regions in the air. However, this can be somewhat difficult to visualize. Because of this, textbooks often show the wave like it's a string in the organ pipe. Really what the curves are showing you in the amplitude of this compression wave. It's also drawn this way to help make the connection to waves on a string which is likely easier to grasp.
The reason the open ends are always antinodes instead of nodes is because a node is where you can't have any movement. This corresponds to the closed end of the pipe. The air at the very end of the pipe can't go any further.
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Why does a laser beam stay collimated? I am looking for a simple way of explaining the collimation of a laser beam. The typical discussion of the two slit experiment of quantum theory relies heavily on the Huygens principle. Its application to a laser beam would tend to predict spreading. From a purely electromagnetic field point of view, how can one visualize what happens at the edge of the beam?
| Just as Jon Custer wrote in his comment, even a perfectly collimated laser beam with a planar wavefront will diverge. The way it happens is determined by the Huygens principle, and depends on the beam profile:
When the light intensity is abruptly cut by a sharp flat obstacle, the light will indeed diffract in almost all angles. A razor blade cutting a laser beam, even though viewed from the geometric shadow, will still have a bright rim. (In wave optics, no absolute shadows exist!)
When the intensity of the beam is modulated smoothly, it will diffract in small angles only. Intuitively it can be imagined that at the edge of a beam passing through a "soft aperture", the Huygens elementary source with slightly higher amplitude than its neighbour forces the resulting wavefront to be bent away by a minute angle, outwards from the beam axis. It is still all about the superposition of spherical waves.
Edit: To illustrate this, I employed my open-source simulation scripts, using the excellent MEEP Maxwell-equations solver, to a classical edge-diffraction experiment. In the following three animations, I computed for you a wave diffraction on a sharp edge (left), and on "soft apertures" with characteristic transition width of 50 % and 100 % of the full image width.
| {
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Why is amorphous classified solid? Because it does not have a crystal structure, it is hard to find physical similarities with a solid. Why isn't it then another state other than solid?
The physics of amorphous is also quite different from crystal solid.
|
"Because it does not have a crystal structure, it is hard to find physical similarities with a solid"
simply proves that you're using the wrong definition of "solid". Relatively few solids have crystalline structure, so a good definition of solid simply can't depend on that.
Let me see if I can offer a better definition.
Rigid body (Idealization) The parts of a rigid body maintain exactly the same distances between one another over time.
Solid The parts of a solid obey an approximate version of that rule for rigid bodies (subject to deformation under stress). The atoms and molecules are not free to move past one another but remain near a set of fixed equilibrium positions in some body-fixed coordinate system, thereby maintaining approximately the same distance between one-another over time.
Under that rule glass-like materials are obviously solids and obviously fluids at higher temperatures. Unlike normal solids, however the transition between states happens gradually aver a range of temperatures.
| {
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Non-Euclidean mechanics; is it useful? Special relativity has the following single-particle Lagrangian:
$$S = \int_{t_0}^{t_f}\sqrt {\langle \mathrm d\vec{s},\mathrm d\vec{s}\rangle}.$$
Clearly it is based on Euclidean norms; it is in Minkowski or Riemannian-geometry norm, but both norms are only a generalization of the Euclidean norm.
Now I can formulate another Lagrangian that Looks like this:
$$S = \int_{t_0}^{t_f} ({\langle \mathrm d\vec{s},\mathrm d\vec{s},\mathrm d\vec{s}\rangle })^{\frac{1}{3}}\;.$$
I have generalized the Standard Lagrangian of a relativistic particle to the 3-norm and tried to concept a generalized scalar products for 3-norms.
Are such field theories developed now and can such field theories be constructed? Is there any evidence to construct a physical theory based on 3-norms?
| OP's proposal (v2) is a special case of Finsler geometry with $n=3$. The main idea is to replace the quadratic metric tensor $g^{(2)}_{\mu_1\mu_2}$ for pseudo-Riemannian manifolds, which defines (infinitesimal, possibly imaginary) distance on the manifold via
$$ds ~=~ \sqrt[2]{g^{(2)}_{\mu_1\mu_2}dx^{\mu_1}dx^{\mu_2}},$$
with (possibly a sequence of) higher metric tensors $g^{(n)}_{\mu_1\ldots\mu_n}$ with a Finsler distance formula
$$ds ~=~ \sum_{n\in\mathbb{N}} \sqrt[n]{ g^{(n)}_{\mu_1\ldots\mu_n} dx^{\mu_1}\ldots dx^{\mu_n}}.$$
There exists already a huge literature on Finsler geometry and its applications to physics.
| {
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Would the Moon be able to take water from Earth? I know that if you add mass to the moon, it would get closer to the Earth. We all know that the moon causes the tides because it's gravity pulls the water. So, my question is: If the moon gained more mass and got closer to the Earth, could it have enough pull on the water that it actually pulls it into space?
| Gravity acts on all matter, not just water (it just so happens that water flows with less resistance than rock) which is why we get noticeable water tides but not very noticeable earth tides. However, if you were to bring a very large gravitating body too close to earth, you would find that the earth isn't quite as solid as it feels.
The answer to your question is yes, but along with the water that would go into 'space', the earth itself would get ripped to shreds by tidal forces bringing large chunks of earth up into 'space' as well, though the name space doesn't really apply here since it's not more of a very messy debris field or accretion disk.
| {
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Why is the Pythagorean Theorem used for error calculation? They say that if $A = X \times Y$, with $X$ statistically independent of $Y$, then
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
I can't understand why that is so geometrically.
If $X$ and $Y$ are interpreted as lengths and $A$ as area, it is pretty easy to understand, geometrically, that
$$\Delta{A} = X\times\Delta{Y} + Y\times\Delta{X} + \Delta{X}\times\Delta{Y}$$
Ignoring the term $\Delta{X}\times\Delta{Y}$ and dividing the both sides by $A$ ($= X \times Y$), that expression becomes
$$\frac{\Delta{A}}{A} = \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y}$$
which is different from
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
which looks like a distance calculation. I just can't see how a distance is related to $\Delta{A}$.
Interpreting $A$ as the area of a rectangle in a $XY$ plane, I do see that $\Delta{X}^2+\Delta{Y}^2$ is the how much the distance between two opposite corners of that rectangle varies with changes $\Delta{X}$ in $X$ and $\Delta{Y}$ in $Y$. But $\Delta{A}$ is how much the area, not that distance, would vary.
| The square root is there as a better estimator of the error than just adding the errors together. If you add the errors together you are finding the maximum possible error which will happen when both quantities are a maximum(or minimum) together. This is an unlikely event compared with all the other domination of errors. The square root formula you you quote has been deduced by statisticians assume, I think, a normal distribution of random errors. You might look up "theory of errors" to get a more detail answer to your question.
| {
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Deriving Wave Function for Scattering States with Delta-Function Potential I am following the Griffiths Book on Quantum Mechanics, and am following the derivation for the wave function for Delta-Function Potentials.
$$V(x) = -\alpha \delta(x)$$
In the scattering states, where $E > 0$, when solving the Schrodinger Equation in the range $x < 0$, we are left with
$$ \psi(x) = Ae^{i\kappa x} + Be^{-i\kappa x}, \text{ where } \kappa \equiv \frac{\sqrt{2mE}}{\hbar} $$
Whereas in the boundstate, we had the $Ae^{-\kappa x}$ term having to be zero due to the blow-up at -infinity, why do we not have the same problem here? The book states: "this time we cannot rule out either term, since neither of them blows up." Isn't the -infinity still blowing up for the $B$ term?
| Neither one of the terms blow up because of the complex exponential. This exponential is real for the bound state case.
| {
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Can the linear momentum of a system be constant, even though its centre accelerates? My instructor says if the velocity of center or mass is constant, it means that the linear momentum of a body is conserved. So if no external force acts on a body, there is no change in the linear momentum of the system.
*
*Now, if I know the linear momentum of a system to be a constant, does it not imply that the center of mass of the system doesn't accelerate?
*My instructor thinks otherwise. Where am I going wrong?
| It's not difficult to show that the total linear momentum of a system can be found by multiplying the velocity of the center of mass by the total mass. If the momentum is constant (in the absence of a net external force), then the velocity of the center of mass is constant. But be careful with your statements. The center of mass of an object (or a system) may not be at its “center”. If you throw a ball which has a non-uniform mass distribution, the center of mass will follow a smooth curve, but the center of the ball may revolve around that.
| {
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How far do we need to be removed from the earth to show the curvature with a viewing angle between 42 and 48 degrees? I have seen already a couple of answers but none of them give an exact number of what should be the minimum height where we would be able to record the curvature of the earth
All I could find is minimum of 10km but you need a 60 degree viewing angle to see it... if that is true there is some amateur rocket footage out there with non fish eye lenses that show no curvature at 32km height
do we need to go higher? if so how high exactly?
Here are the video's i'm talking about both go around 120,000 feet high.
A high altitude balloon which uses some lens (most probably fish eye) and gives distort so not conclusive if what we see as a straight horizon is real http://youtube.com/watch?v=tvhFbvY_99o
the other is an amateur rocket launch at 2:13
http://youtu.be/qY7W3EMfrgc?t=133
The lens from a FlipHD camera as far as I can find out is between 42 and 48 degrees so much smaller then the 60 mentioned before
the horizon is straight from the beginning of the launch but also when it reaches the top the horizon appears straight.
| In order to see the curvature, you need a 60 degree field of view and a cloud free day. From what I've read, you need to be about 35,000 feet above the surface. Find more information in this article here: http://www.howitworksdaily.com/how-high-do-you-have-to-go-to-see-the-curvature-of-the-earth/
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Scattering amplitude Green's function integral On page 208 of Weinberg's QM book, he calculates the following integral
\begin{align}
G_k (\vec{x}-\vec{y})
=& \int \frac{d^3 q}{(2\pi \hbar)^3} \frac{e^{i\vec{q} \cdot (\vec{x}-\vec{y})}} {E(k)-E(q)+i\epsilon} \\
=& \frac{4\pi}{(2\pi)^3}\int_0^{\infty} q^2 dq \frac{\sin(q|\vec{x}-\vec{y}|)}{q|\vec{x}-\vec{y}|}\frac{2m/\hbar^2}{k^2-q^2+i\epsilon} \, .
\end{align}
It is clear that he is integrating in spherical coordinates. However, I don't see how where the
$$\frac{\sin(q|\vec{x}-\vec{y}|)}{q|\vec{x}-\vec{y}|}$$
comes from. Can someone explain?
| Write $d^3 q = dq q^2 d\theta d\phi \sin\theta$ and integrate over the angular variables. The only angular dependence in the integrand is in $e^{i \vec{q} \cdot ( \vec{x}-\vec{y}) } = e^{i q r \cos\theta}$ where I've defined $r = | \vec{x} - \vec{y} |$. Then, we have
$$
\int_0^{2\pi} d\phi \int_0^\pi d\theta \sin\theta e^{i q r \cos\theta}
$$
There is no $\phi$ dependence so that just gives $2\pi$. For the $\theta$ integral defined new integration variable $t = \cos\theta$. Then, the above becomes
$$
2\pi \int_{-1}^{1} dt e^{i q r t} = \frac{2\pi}{i q r } e^{i q r t} \bigg|_{-1}^1 = \frac{2\pi}{ i q r } \left[ e^{i q r } - e^{- i q r } \right] = \frac{4\pi }{ q r } \sin(qr) = \frac{4 \pi \sin \left( q | \vec{x} - \vec{y} | \right) }{ q | \vec{x} - \vec{y} | }
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the electric field exactly on the surface of a conducting sphere? Within a conducting sphere, the electric field is 0, but is the electric field still 0 exactly on the surface?
| The point about the E-field being zero inside is a conductor is that you must be writing about electrostatics, the study of charges when they are not moving.
In the ideal world every bit of metal inside your surface (an infinitely thin shell) has no E-field within it.
In the real world when the surface is particular in nature it must be very difficult to decide exactly what is going on.
I do not think that discussion of an infinitely thin sheet is useful in that it is a totally abstract idea whereas discussion of what happens when you have a one or two or three atom layer of metal probably is because such a sheet can be made. It is then highly likely that then E-field is not discontinuous as defining the surface is not possible.
| {
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Determine resultant couple moment by summing moments The question states:
Determine the resultant couple moment by (a) summing moments about point $O$ and (b) summing the moments about point $A$.
I used scalar analysis for solving this question, wherein
$$M=\sum{F_xd_y}+\sum{F_yd_x}$$
When completing part (a), I found the following answer:
$$M_O=\sum{F_xd_y}+\sum{F_yd_x}=9.686\mathrm{ kNM}$$
for $$\sum{F_xd_y}=(-8\sin(45)+2\sin(30)) \mathrm{kN}(0 \mathrm{m})+(-2\sin(30)+8\sin(45)) \mathrm{kN}(-0.3 \mathrm{m})=-1.3971 \mathrm{kNm}$$
and $$\sum{F_yd_x}=(-8\sin(45)-2\sin(30)) \mathrm{kN}(-3.3 \mathrm{m})+(2\sin(30)+8\sin(45)) \mathrm{kN}(-1.8 \mathrm{m})=11.083 \mathrm{kNm}$$
I solved part (b) similarly, and got the same result for $M_A$, with the caveat that both $\sum{F_xd_y}$ and $\sum{F_yd_x}$ have opposite polarity (negatives, i.e. $\sum{F_xd_y}=1.3971 \mathrm{kNm}$).
Where I'm having trouble is two-fold: I'm uncertain that I'm solving for the moments correctly, and I'm not sure what how to calculate the final resultant moment.
If I did solve for the moments correctly, would the the resultant couple moment be $\sum{M}$, which would result in a resultant moment of zero?
Thank you in advance for any clarifications or explanations you can provide.
| The net force acting on the beam is zero but there is a couple acting on the system. A couple has the nice property that the moment about any point is the same.
So you should have found the same answer for both parts.
So find the vertical component of the two forces on the left which will equal the magnitude of the vertical component of the two forces on the left.
The moment of a couple is force $\times$ perpendicular distance between the two forces.
Later
Moment about $A$ is $Fa$ clockwise
Mpment about $O$ is $F(a+b)$ clockwise $+ Fb$ anticlockwise
$\Rightarrow F(a+b)$ clockwise $- Fb$ clockwise = $Fa$ clockwise the same as about $A$
| {
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How can a product of Bra and Ket be a scalar if they are matrices? I am trying to teach myself Quantum Mechanics and am currently on Complex Vector Space Arithmetic. According to Wikipedia, product of a Bra and Ket is a scalar (which, I think, means a complex number). But then, on the same page, it also says that both Bras and Kets can be represented by 1xN and Nx1 matrices respectively.
The product of to such matrices should be a 1x1 matrix, not a scalar as answered here. My questions are:
*
*Can we really represent Bras and Kets by matrices in multiplication, or is this just a analogy taken too far? If yes,
*can we reliably replace a 1x1 matrix by a scalar is every situation? Or is it a shortcut applicable only in some contexts?
*If it is not universally applicable, where can we do this replacement and why?
| This is just the dot product of two vectors. If you want to get more into dot products I would suggest migrating to the math site.
| {
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Are isentropic closed rigid thermodynamic systems also isolated? $\newcommand{\d}{\, \mathrm d}$Currently it seems to me that isentropic closed rigid thermodynamic systems have to be also isolated. Is this correct?
The reason why it seems to me like this is the following.
The energy-based fundamental equation is: $$\d U = T\d S - P\d V + \sum \mu_{i} \d n_{i}$$.
Because the system is closed, $\d n=0$.
Because the system is rigig, $\d V=0$.
And because the system is isentropic, $\d S=0$.
Thus $\d U=0$, which means there is no energy change possible. And if there is no exchange of energy possible between the system and the surrounding, then the system is essentially isolated. And therefore also thermally isolated.
Is this argument correct, or am I getting something wrong here?
| No it doesnt necessarily mean that. In general, the internal energy is of the form,$$d U = Td S - Pd V + \sum \mu_{i} d n_{i} + J.dx$$
where $J$ refers to a generalised force and $dx$ to a generalised displacement, and even $-P.dV$ can be written as product of a generalised force and displacement with pressure serving as the former and volume serving as the latter. Suppose the system is immersed in a magnetizing intensity $H$ and the system develops a magnetization $M$. Then the net magnetic field is given by
$$ B = \mu_0(H + M)$$
and the work done excluding the mutual field energy is given by
$$ \delta W = - MdB$$
which can now be included in the expression for internal energy as
$$d U = Td S - Pd V + \sum \mu_{i} d n_{i} - M.dB$$
In general, there are many other possible scenarios too where you can prove that isentropic closed rigid thermodynamic systems are not totally isolated.
| {
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"timestamp": "2023-03-29T00:00:00",
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Navier Stokes: what about angular momentum? I play with CFD for a while, and suddenly, a transcendantal question raises: :-)
Navier Stokes is basically Newton applied on a continuum in Eulerian.
For solids, we would consider linear, but also angular momentum.
Why don't we have to do that for fluids ?
Conversely, you can take the curl of Navier Stokes and have an equation expressed in vorticity, which looks like our angular momentum world.
Does it mean that the equation with velocity somehow embed the both kind of momentum, and they are totally correlated for fluids ?
But how it's not the same for solids ? i.e., where is the intrinsic difference that makes it different degrees of freedom in one case and equivalent in the second ?
| For the exact equation, that's almost true (one should look at a derivation of the NS equation from the more fundamental Boltzmann equation, where the conservation laws are build into the particle-particle interactions -- but the closure in the approximation of the continuity and NS equations might spoil this if not done correctly).
There is but also a practical question behind here for CFD, imho, where the NS equations are solved only approximately on discretized regions/equations - e.g. using Finite Differences, Finite Volume, or Finite Elements (or, more generally, some other sort of ansatz function Galerkin-like discretization or so). With none of those, you really can conserve of angular momentum (currently). This usually does not matter on large scare vortices. But for turbulence e.g. in DNS simulations, it matters, where a lot of small vertices are generated and transported.
So, even if the NS equations are made angular momentum conserving, the practical (discretized) numerical solution is most probably not exactly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Relation of conformal symmetry and traceless energy momentum tensor In usual string theory, or conformal field theory textbook, they states
traceless energy momentum tensor $T_{a}^{\phantom{a}a}=0$ implies (Here energy momentum tensor is usual one which is symmetric and follows conservation law) conformal theory. (i.e, see page 3 )
I wonder how they are related to each other.
I found similar question Why does Weyl invariance imply a traceless energy-momentum tensor?
and get some idea about weyl invariance.
and get some another useful information from Conformal transformation/ Weyl scaling are they two different things? Confused! which dictates that conformal transformation and weyl transformation is totally different things
.
| Note that under an infinitesimal change in the metric of the form $g \to g + \delta g$ the action changes to
$$
\delta S = \int T^{ab} \delta g_{ab}
$$
Now, under Weyl transformations we have
$$
g_{ab} \to e^{2\omega} g_{ab} \qquad \implies \qquad \delta g_{ab} = 2 \omega g_{ab}
$$
For Weyl transformations $\omega$ is completely arbitrary. If we consider a conformal transformation then, the metric also transforms as above except that $\omega = \frac{1}{d} \nabla_a \xi^a$ where $\xi^a$ is a conformal killing vector, i.e. $\omega$ takes a specific functional form.
Either way, for both conformal or Weyl transformations $\delta g_{ab} =2\omega g_{ab}$. Thus, for either of these transformations, the variation in the metric is
$$
\delta S = 2 \int \omega T
$$
Thus, if the trace of energy momentum tensor vanishes, $T = 0$, then
$$
\delta S = 0
$$
and we have a symmetry of our theory!
OK. So we have shown that if $T = 0$, then the theory is invariant under Weyl and conformal transformations. What about the inverse statement? Can we infer from Weyl and conformal invariance that $T = 0$? The latter is a more subtle question.
Weyl or conformal invariance implies
$$
\int \omega T = 0
$$
Now, when talking about Weyl invariance, the above is true for arbitrary $\omega$. In this case, we can most certainly conclude that $T = 0$ (for instance take $\omega \propto \delta^4(x)$ or some smoothed out version thereof and we immediately reach this conclusion.
When talking about conformal invariance, $\omega$ is not arbitrary and we cannot conclude that $T$ must vanish. For instance, in a flat background, $\omega$ takes the form $\lambda + a_\mu x^\mu$ where $\lambda$ and $a_\mu$ are arbitrary constants. Thus, all we can conclude is that we must have
$$
\int T = 0 ~, \qquad \int x^\mu T = 0
$$
These two conditions no longer imply that $T = 0$. Thus, as per this argument the inverse statement is not necessarily true in conformal field theories. I'm not sure if there is any other argument that can be used to justify that $T$ must vanish in CFTs, but so far, all the CFTs we study have $T = 0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Effect of paint on drag force Aerospace engineering as well as automobile engineering gives a particular significance to the shape of a vehicle to enable proper and more effective transportation.What I want to know, is, about the impact that the protective covering aka the paint on an automobile etc. has on the drag force.While I am assuming it is not much, but, can our using a more slick coating, say Teflon paint,have an impact on the motion, thereby increasing fuel economy. Do let me know the quantitative figures involved.Thanks.
| Teflon has a low surface tension. What counts in viscous drag is the mechanical roughness of the surface. If the "peaks" of this roughness stick out into the boundary layer, they decelerate the flow more than a smooth surface would.
Normally, the drag coefficient drops with increasing Reynolds number. The Reynolds number is the ratio of inertial and viscous forces and can be calculated by dividing the product of flow speed and length by the kinematic viscosity of the fluid. Depending on the relative roughness (roughness relative to length), at some point the drag coefficient stays constant with a further increase in the Reynolds number. Only then will roughness on the scale of different paint surfaces become a factor in drag.
Friction drag of tube flow for different values of roughness (picture source)
German air force units in WW II would routinely test-fly new attack airplanes and select the ones with the highest top speed and then polish those. Normally, the camouflage paint on the aircraft was matte, but when polished, they would fly a little faster. Those airplanes were selected for missions where top speed would be important.
Roughness on the scale of rivet heads or larger is another matter. Here, the drag effect can be felt at all speeds. Therefore, most rivets are flush on aircraft.
A teflon coating would only be helpful if the structure is soiled by insects. Gliders have a serious problem on summer days with bugs which trip the boundary layer early and increase drag.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/235314",
"timestamp": "2023-03-29T00:00:00",
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