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Why does the kinetic energy of a photo-electron vary? Provided it is above the threshold frequency of the metal, when electromagnetic radiation is shone onto a metals surface photo-electrons are emitted. This occurs because 1 photon is absorbed by 1 electron giving it enough energy to be ejected.
We know that the energy of the incident photons are all equal from the equation E = hf. If this is so why does the kinetic energy of the emitted photons vary? Why is there a maximum kinetic energy, is it not the same amount every time?
| Look my friend you can consider incident photon as a bigger ball having more energy(in motion) and the electron of the metal surface be a smaller ball kept at rest in a muddy area.
when the both balls collide whole energy from bigger ball is tranferred to the smaller one.
The energy of the bigger ball gets divided into two parts;
one part of energy will help the smaller ball to overcome resistance of muddy area and the other part of energy will provide him kinetic energy.
The energy required to overcome resistance is always constant no matter how much the energy the big ball is providing to it, while the rest of energy will help the smaller ball to attain varying speed.
The smaller ball will absorb all the energy of the bigger ball and ultimately its kinetic energy will vary.
Similarly the muddy area is nothing but work function of the metal which is holding the electron at its place.
When the bigger ball i.e. photon strike electron, it tranfer whole energy into it, so the energy required to beat work function will remain constant and the rest of energy will be utilised in increasing kinetic energy of the photoelectron.
ENERGY OF PHOTON VARIES WITH VARYING FREQUENCY OF LIGHT ...
Hence kinetic energy varies..
Hope you got it!!!
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Why don't the molecules in the air "observe" the electron in the double slit experiment? Question says it all. The air should also interact with the electrons which should result in wave function collapse.
| The double-slit experiment with electrons can only be done in vacuum, unless the dimensions of your double-slit experimental set-up is much smaller than the free mean path of the electrons in air, which is below $0.5\mu m$ in air under normal conditions. Otherwise only very few electrons will reach the screen. Inelastic scattering destroys the coherence of the electron waves and thus also the interference pattern on the screen.
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Why is Helium 4 so stable? I've been looking at stuff to do with binding energies and was wondering why Helium 4 is so stable. The fact everything up to carbon is less stable seems a bit odd. Is there a reason for this or another, that's how the universe works?
| Yes as Pieter mentions being fermions with Pauli exclusion you can get a spin up and a a spin down in the same ground state for both n and p. Like in atomic physics you can put 2 electrons in the s orbital.
In addition a geometric effect that 4 particles can be arranged in a tetrahedron shape with each ball touching the other 3 giving a very high binding energy per nucleon compared to say 3 nucleons. So that may explain why He4 is ejected in preference to He3 or He2 or H3 or dineutron or often even in preference to a single proton or neutron.
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Why should the vacuum be unique? Among the Wightman axioms is the requirement that there is a unique Poincare-invariant state called the vacuum.
We know that QFT vacuua are not necessarily unique, for example in situations where spontaneous breaking of symmetry is relevant.
So, why is that? Why do we demand uniqueness of the vacuum when it is clearly not true?
| Let $G$ be a group, represented on the C*-algebra of quantum observables. By Haag's theorem, if in the C*-algebra of quantum observables there are two $G$-invariant states that are also $G$-abelian (an asymptotic condition that it would be too long to explain here), then they are either equal or disjoint (the latter meaning that one cannot be written as a trace class operator in the GNS-representation of the other).
Therefore, since the Wightman axioms are set in a given representation of the algebra of observables, only one ground state (i.e. Poincaré- invariant and abelian state) can exist as a trace class operator. There may be however more than one $G$-invariant state, provided only one is $G$-abelian. I think that an example would be given by the free vacuum and the free Gibbs state of a scalar theory: they are both trace class in the Fock representation, and invariant (at least w.r.t. time translations).
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Separation distance of Michelson Interferometer Suppose I would like to use Michelson Interferometer to observe fringes of equal thickness by creating an angle between the mirrors. Why is it vital for the path difference between the mirrors to be small in order to observe the fringes?
| Imagine the laser beam used for the interferometer. Picture it shining onto a distance wall. The beam expands with distance, like a cone. You can then see how the center of the beam has the shortest distance to the target. Now picture them in the interferometer. When the beams are at the exact same length you will get a perfect match - complete constructive and destructive interference, that is one bright spot or one black. As the mirror differences become less and less equal more rings will appear, but they will be smaller. There will also be less contrast between them. Eventually, if the difference between beam lengths become too great and the classic pattern is lost.
Note: The image that appears on the screen always has an opposite image that is projected back onto the laser. So the “light” never really goes away.
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Why can't the Lagrangian for a free point particle depend on distance? I have looked thorough the derivation of Lagrange equations in Landau and Lifshitz, Vol 1, $\S 3$, p.5. They argues that the lagrangian of a free particle cannot explicitly depend on position vector $\vec{r}$ because of the isotropy and homogeneity of space. But why can't it depend on $r^2$? (They do not explain why this cannot be possible.) Or say a power of $r^2$?
| Lagrangian of a free point particle may depend on $\vec{r}$, as a matter of fact, but this dependence is of no meaning and normally is reduces to a full time derivative of something that contains this dependence.
Lagrangian seves to obtain differential equations supplied with the initial conditions. Through the initial conditions you obtain solutions depending on the initial distance, initial velocity and time.
Free motion is an approximation of a motion in a relatively weak external force, which is just neglected for simplicity within a given period of motion in question.
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Why we can assert, in general, that physical processes have the behaviour of low-pass filter? Consequently, why is it not allowed to produce physically some controllers for processes that are described by a transfer function that is an improper function?
A simple example is the driven harmonic oscillator. So the equation in this case is $mx'' +Ax'+k=f(t)$, where $x$ is the variable that stands for the position of the mass, $m$ is the mass, $A$ is a certain constant for the friction and $f$ is a force that pull the mass.
Omitting some passages we get a transfer function (in the frequency domain) that is:
$$W(s)=\frac {1/m}{s^2+2\alpha s+w_0^2}$$
where $\alpha = \frac Am$ and $w_0 = \sqrt\frac km$.
So the harmonic function is:
$$W(jw) = \frac {1/m}{(w_0^2-w^2)+2\alpha jw}$$
Now the module of the function is $M=\frac {1/m}{\sqrt{(w_0^2-w^2)^2+4\alpha^2 w^2}}$ and $\lim_{x\to+\infty} M = 0$. So the process has a visible response just in "low" values of the frequency of the force ($w$). The professor added that in some controlling problems of some process we could obtain a transfer function of the controller that is an improper transfer function, and that a controller of this type is impossible to produce physically. I'd like just some kind of intuitive explaination for this.
| Exponential fall-off in frequency domain is equivalent to smoothness in time domain. Since physical processes usually happen smoothly, they act as a low-pass filter.
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If gravity isn't a force, then how are forces balanced in the real world? Consider a simple situation like this- an object is sitting on a table. In classical mechanics, we say that the net force on the object is zero because gravity (treated as a force) and normal reaction force are equal and opposite to each other, and hence, it's acceleration is zero. But according to Einstein's General Theory of Relativity, gravity isn't a force at all, but instead curvature created in spacetime by a massive object, and objects near it tend to move towards it because they are just moving along the geodesic paths in that curved spacetime. So if an object kept on a table gets acted only by the normal reaction force (as gravity ain't a force), how is the net force on it zero?
| You can describe the physical situation either in a Newtonian way or according to GR (general relativity), but you have to be consistent in each description.
In a curved spacetime a free-falling body follows a geodesic, that is a path with nil acceleration. In the example the object on the table does not follow a geodesic, in fact its four-acceleration is not zero. So the body is acted upon.
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Magnetic field due to a current carrying wire at the wire Why do we take zero magnetic field at a point on the axis of a current carrying wire.
When i use $B =\frac{\mu_0 I}{2\pi r}$ formula for magnetic field calculation at a distance $r$ from the wire then for $r\to 0$, $B$ becomes infinity. What I am missing here?
I am assuming that wire has zero thickness which we normally use for problem solving.
| For an infinitely long wire of infinitesimal thickness carrying a steady current $I$ the magnetic field at a distance $r$ from the wire is $$B =\frac{\mu_0 I}{2\pi r}.$$ This result can be derived from Ampere's Law $$\int {\bf B . d s} = \mu_0 \times \text{current enclosed by path},$$ where the current enclosed by the path (for infinitesimal thickness the enclosed current is always the total current $I$ flowing through the wire. For a wire of finite thickness you can still make use of Ampere's Law though for a distance $r \lt s$ the enclosed current is now only a fraction of the total current flowing through the wire. Typically this is taken to be $$I \frac{r^2}{s^2}.$$ You can then determine the magnetic field for the two cases (i) $r \le s$ (current = $I \frac{r^2}{s^2}$) and (ii) $r \ge s$ (current = $I$).
For distances inside the wire you only have a fraction of the current that contributes to the magnetic field and the magnetic field has a finite value at a point on the wire itself.
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Would a magnet be able to attract an object through a sheet of stainless steel? For example, if there were a neodymium magnet at position x=0, a sheet of stainless steel at position x=1, and a magnetic object at x=5, would the magnet still attract the object?
Is the attraction force less than if the stainless steel sheet were absent?
Would a thicker sheet of stainless steel dampen the attraction force?
| Most stainless steel is austenitic alloys of iron, which is not ferromagnetic. In practice, there are always some other phases, and that will results in a relative magnetic permeability that is a bit larger than unity. This will result in a force that is slightly less than if one had wood instead of stainless steel. But the difference is small and may be difficult to notice.
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Electricity generator We know in an electricity generator , electrons move from negative voltage to positive voltage of the stator winding and we can use that electric power on its way , so is it possible for the stator to loss all of its electrons((because we consumed it )) so it can not produce more voltage and electric current in the generator?i mean does it finally turn to positive ions ?
| the condition you describe is analogous to the state in which the generator is spinning but the load resistance to which it is connected is extremely large, so the current being pushed by the generator through the load goes to zero. In this case, the generator still applies its full rated voltage to the load, but this is not going to be sufficient to ionize the metal wire in its windings.
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Why external chemical potential? My textbook Thermal Physics by Kittel and Kroemer [1, p. 124] says:
When external potential steps are present, we can express the total chemical potential of a system as the sum of two parts:
$$\mu = \mu_\mathrm{tot} = \mu_\mathrm{ext} + \mu_\mathrm{int}\tag{16}$$
Here $\mu_\mathrm{ext}$ is the potential energy per particle in the external potential, and $\mu_\mathrm{int}$ is the internal chemical potential* defined as the chemical potential that would be present if the external potential were zero.
I am very confused since
$$\mu = \frac{\partial F}{\partial N},$$
where $F$ is Helmholtz free energy.
As there is only internal energy in Helmholtz free energy, why can we define something related to the external potential energy (such as gravitational potential energy) from the internal energy? Would anyone explain this to me?
Reference
*
*Kittel, C.; Kroemer, H. Thermal Physics, 2nd ed.; W. H. Freeman: San Francisco, 1980. ISBN 978-0-7167-1088-2.
| The energy (per particle) is defined up to a global constant/reference. If we scale the energy per particle (of all particles!) by a constant $\mu_{\text{ext}}$, then if we add another particle to our system while holding temperature and volume constant, the free energy raises by
$$\begin{align}\Delta F&=\mu_{\text{int}}(1)+\mu_{\text{ext}}(1)\\
&=\mu_{\text{int}}+\mu_{\text{ext}}\\
&\equiv \mu_{\text{tot}}
\end{align}$$
where the $(1)$'s are meant to be $\Delta N$, i.e. the addition of one additional particle. With this, we can see that the chemical potential is by definition $\mu_{\text{tot}}$.
If the external potential $\mu_{\text{ext}}$ depends on position (like placing the system in a force-field, possibly with an appreciable gradient), then we would break up the system into a bunch of smaller systems (grand canonical ensembles) each in thermodynamic equilibrium with their neighbor, and apply the same formalism.
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The path difference when a block covers one slit in Young's double slit experiment A modification of the simplest case of Young’s double slit experiment is when the path length for one of the slits is changed.
I've been told that if a strip of material of thickness $ t $ and refractive index $ n $ is placed over one slit then it adds a path difference of $(n − 1)t$, which results in the fringes being shifted. However, I am not sure how $(n − 1)t$ is derived and why this gives the path difference?
| Earlier that light had to travel the distance $t$ in vacuum with refractive index 1. Now the refractive index is $n$ so the path will be $n*t$ The additional path is $n*t - t$ which is $(n-1)*t$. So if the path difference between the two rays was some $x$, $(n-1)*t$ would be added to it.
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Velocity of light in Galilean transformation What is the velocity of light in Galilean transformation? Is it infinity?
| Maybe this reading of your question gets to what you are really asking.
It's not really about the "speed of light" [electromagnetic radiation], but about the maximum signal speed.
The eigenvectors of the Lorentz boost transformation are light-like [null] vectors in spacetime. In special relativity, they represent a finite invariant speed that is unattainable by particles with timelike worldlines.
The eigenvectors of the Galilean boost transformations are space-like [and null] vectors in Galilean spacetime. In Galilean relativity, they could represent an infinite invariant speed that is unattainable by particles with timelike worldlines. [My definition of spacelike is "orthogonal to timelike" (i.e., tangent to the "unit circle", which is orthogonal to the radius).] This seems consistent with the idea of the light-cone opening up in the Galilean limit to be a spacelike plane of simultaneity in Galilean relativity.
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Does the mass of a bicycle directly affect stopping distance? In this answer on the cycling SE, the claim is made that adding more mass to a bicycle increases the stopping distance. I was under the impression that mass should not affect the stopping distance so long as all the other factors remain the same (balance, coefficient of friction, etc.).
What factors in this scenario contribute to increasing stopping distance on a bicycle? If the bicycle is balanced the same but weighs more, will the stopping distance be equal?
| when we stop a bike,we convert its kinetic energy(energy due to motion) into heat due to friction of brakes,now kinetic energy is dependant on mass
(1/2mv^2) so increasing the mass increases the kinetic energy and thus the stopping distance should INCREASE,assuming a constant braking force is applied,because the amount of energy it needs to disipate is more,consider a more intuitive explanation:imagine that the same force of the brakes is now applied on the pedals of a bike with more mass,it would take longer to reach the same speed as that of a bike with lesser mass.similarly you could conclude with respect to slowing down,cheers.
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How strong would a magnetic field need to be to go nuclear A sufficiently strong magnetic field would rip electrons out of their orbits triggering a chain reaction causing a nuclear reaction, with that said, how strong would it need to be before I would have to start worrying about this?
| There has been modelling and simulation work on atoms in magnetic fields up to $10^{13}$ Gauss, the kind of environment you get around neutron star surfaces. As was first noted by Ruderman, superstrong magnetic fields make electron orbitals near-cylindrical and enable superstrong atom chains - quite the opposite of a nuclear chain reaction. Later research has questioned whether this works for all atoms, but it appear that infinite chains are possible for light atoms.
The reason this does not rip apart atoms is that electrons are not accelerated further away, but into looping orbits (Landau levels). In an electric field a wayward electron will gain energy the further it moves towards the positive pole. That might lead to it hitting other atoms, ionizing or heating them up leading to further breakdown - but this does not happen for magnetic fields even at very high densities (it might cause other weird instabilities, but not anything explosive).
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Concept regarding Venturi Tube-Bernoulli application I was recently studying applications of Bernoulli Equation and came across the Venturi tube. This is diagram I have used to analyse the venturimeter. I understand how we obtain the first equation using bernoulli theorem which is
$$P_1 - P_2 =(1/2)ρ(v_2^2 - v_1^2) \tag{1}$$
and also the continuity equation
$$A_1 v_1 = A_2 v_2. \tag{2}$$
However, I am unable to process how to obtain the following third equation
$$ P_1 - P_2 = ρgh $$
where $h$ is difference in the heights of the liquid level in the two tubes and $ρ$ is density of fluid.
Lets say the atmospheric pressure at the top of each tube is $P$. Now since the fluid in the two VERTICAL tubes are at rest and not moving, their velocities are 0. Hence, if I proceeded by applying Bernoulli equation.
$$P_1 +ρ(v_1^2)/2 = ρg(h_1) + P \tag{3}$$ and
$$P_2 + ρ(v_2^2)/2 = ρg(h_2) + P \tag{4}$$
Here $P_1$ and $P_2$ are the pressures at the points in the tube and constriction respectively and the points are at the SAME HORIZONTAL LEVEL.
Subtracting (3) and (4) and even using (1) does not yield me
$$P_1 - P_2 = ρgh $$
rather gives me $0 = ρg(h)$ which makes no sense whatsoever.
What am I missing here and how do I obtain the right result ?
| Let the atmospheric pressure be $P_0$
Therefore
$$P_1=P_0+\rho gh_1 \quad \text{and}\quad P_2=P_0+\rho gh_2$$
Subtracting we get the required equation
$$P_1-P_2=\rho gh$$
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Understanding the statcoulomb conceptually I've just learned about the statcoulomb, which is basically a way to express charge when we didn't have units of charge (I think), with the definition $$1 \mathrm{statC} = 1 \mathrm{dyne}^{1/2} \mathrm{cm} = \mathrm{cm}^{3/2} \mathrm{g}^{1/2} \mathrm{s}^{-1},$$ but I don't understand where this comes from. I've read in various texts, and in all of them they explain the correspondencies and why this definition works, and I understand it, but I don't understand it conceptually, I mean, how can something like a Coulomb, which for me it couldn't be more far from units of length, mass and time, be explained in terms of these (not entirely, I know, since a Coulomb is different from a statcoulomb, but with a very close relation).
| As you probably know, it comes from Coulomb's Law written without any constants: $F=q_1q_2/r^2$. Since we've already defined force and distance, the units of charge are fixed. There's nothing more to it.
Well, almost. There are at least four "cgs" systems of units: those based on magnetic force, those based on electric force, with and without a factor of $4\pi$. While these systems are good for theoretical developments, they are very confusing for experimental work, because devices generally display SI units. And confusion exists when comparing equations in the various cgs systems.
And then there is the conceptual problems, such as you are having. There's nothing to conceptualize. It's just a unit of charge.
It's little wonder that cgs systems are relegated to specialty fields.
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Why is it said that density of nucleons in a nucleus is constant? Question:Why is it said that density of nucleons in a nucleus is constant?
I am studying an introductory course in nuclear and subnuclear physics. Based on the context in which it is cited (I cannot cite the notes because they are private notes) I do not understand if it is a theoretical assumption or hypothesis derived from an experiment.
I searched in literature,
especially on the book Krane-Introuctory Nuclear Physics but my doubt has not been clarified.
| Robert Hofstadter received a Nobel Prize for his research regarding The electron-scattering method and its application to the structure of nuclei and nucleons
Hofstadter fired high energy electrons at the nuclei of atoms and from the resulting scattering and diffraction effects he was able to map out the charge density within a nucleus and hence show that the charge density was approximately constant within a nucleus and he also was able to estimate the radii of nuclei.
What he found was that the radius of a nucleus $R$ is related to the mass number of the nucleus $A$ as follows $R = R_0 A^{\frac 13}$ where $R_0$ was a constant approximately equal to $1.2 \,\rm fm$ which is good evidence for nucleii being of the same constant density.
$R_0$ is not actually constant but and varies a little by about $0.2 \,\rm fm$ depending on the nucleus being considered.
The relationship $R = R_0 A^{\frac 13}$ can be thought of as coming from the idea that a nucleon has a fixed volume $V$ and there are $A$ nucleons in a nucleus.
So the volume of a nucleus is related to $VA$ which means that a linear dimension (the radius) of a nucleus is related to $(VA)^{\frac 13} \propto A^{\frac 13}$ if $V$ is constant.
Since Hofstadter's paper was published in 1957 much more research has been done on the structure of the nucleus and of the constituent particles.
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Goldstone modes $\hat a^\dagger(k) \left|0\right>$ for small $k$ or $\hat a^\dagger(0) \left|0\right>$ Consider the Hamiltonian:
$$H=\sum_{\vec k} \varepsilon (\vec k)a_{\vec k}^\dagger a_{\vec k}$$
with $\varepsilon(\vec k) \rightarrow 0$ as $|\vec k|\rightarrow 0$. I know that this has gapless excitations and therefore Goldstone modes but I am confused about the actual definition of what counts as a Goldstone mode. Do the refer to the states $\hat a^\dagger(k) \left|0\right>$ for small/infinitesimal $k$ which do still have some energy or to the ground states $\hat a^\dagger(0) \left|0\right>$ which have no energy.
| The massless field is called the Goldstone 'mode'. The term 'mode' does indeed make one think of a particular momentum mode and in that it is a misnomer. For simplicity lets take the $U(1)$ case with $\phi=|\phi| e^{i\theta}$. After symmetry breaking the Goldstone part of the Lagrangian is
$$
\mathcal L = \frac{1}{2} \phi_0^2 (\partial \theta)^2
$$
with $\theta \in [0, 2 \pi)$ and for some non-zero value of $\phi_0$ corresponding to the minimum of the potential. Symmetry breaking chooses one value of $\theta=\theta_0$ arbitrarily. Linearized excitations can be quantized and they change the value of $\theta$. However, strictly speaking $a^\dagger_0 |\theta_0 \rangle$ is not defined as it is not normalizable by itself and you cannot put it in a wavepacket while keeping the nerdy zero. This is related to the fact that one cannot change the vacuum of this system if the volume is infinite and it is really only in that limit that the symmetry is spontaneously broken as for finite volume one can take linear superpositions of different $\theta$ much like one can form a superposition of position for a particle living on a circle.
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For an electron, how does QED maintain a constant rest mass and remove its classical radius from CEM? From classical electromagnetism, the classical radius of the electron is calculated to be
$$r_\text{e} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{m_{\text{e}} c^2} = 2.817 940 3227(19) \times 10^{-15} \text{ m}$$
Secondly, its rest mass is predicted to change when placed in a non-constant electric field as parts are accelerated differently to one another in its rest frame. QED on the other hand treats the electron as a point particle with a constant rest mass which so far have been increasingly confirmed experimentally.
What additional physical processes, if any, come into play within QED to maintain the constant rest mass and point-like properties of the electron compared to within CEM?
| The issue is solved, in a way, by separating the "bare mass" that appears in the Lagrangian from the "physical mass" that we observe in experiments. The bare mass diverges under renormalization, irrespective of concerns about the energy content of the electric field around a point-like electron. A standard response to this is that the Lagrangian parameters aren't physically observable, anyway, they're just computational tools.
You can look at this problem from another point of view. individual electrons are excitations in a Fermionic field, usually one that obeys the Dirac equation in the appropriate limit. The electric field produced by a point-like excitation in that field has a divergent energy content in the associated electric field, but so what? That just means that point-like excitations aren't physically realizabile in that field and must, therefore, be computational tools, kind of like the virtual paths used when extremizing the action to get the equations of motion in classical physics.
| {
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What does self-closing bra-ket mean in Robetson-Schrodinger Uncertainty Relation? I was reading:
https://en.wikipedia.org/wiki/Heisenberg%27s_uncertainty_principle#Robertson–Schrödinger_uncertainty_relations
Where an inequality is presented:
$$ \sigma_A \sigma_B = | \frac{1}{2} \langle \lbrace \hat{A}, \hat{B} \rbrace \rangle - \langle{\hat{A}}\rangle \langle \hat{B} \rangle | ^2 + | \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle ^2 $$
I found the notation hard to understand:
$$ \langle \lbrace \hat{A}, \hat{B} \rbrace \rangle$$
Is bra-ket expression, but it contains a single expression inside$\lbrace \hat{A}, \hat{B} \rbrace $ (so it can't be a dot product). How do I interpret this? Similar problem arises here:
$$ \langle{\hat{A}}\rangle \langle \hat{B} \rangle$$
where the $A, B$ are single-entities contained in a closed bra-ket expression.
| So the expression has three different pieces. First, you have the operators which are represented here by hats, e.g. $\hat A$ and $\hat B$. Second, the curly brackets or braces represent the anti-commutator operator
$$\{\hat A, \hat B\}\equiv\hat A\hat B + \hat B\hat A.$$
The final piece is the "enclosed bra-ket" notation "$\langle\cdot\rangle$" you are asking about which simply represent the average or expectation value for whatever is enclosed in the braket. So $\langle \hat A\rangle$ is the expectation value for the operator $\hat A$.
Now the expectation value generally depends on what your state is, and in quantum mechanics (especially in the Schrödinger picture), this information is contained in state vectors such as $|\psi\rangle$ and it's Hermitian dual $\langle\psi|$. So the expectation value of some operator $\hat A$, given that the state of your system is $|\psi\rangle$, is given by
$$\langle \hat A\rangle = \langle\psi|\hat A|\psi\rangle.$$
Note that in terms of notation, the left hand side looks like an abbreviated from of the right side. This is no accident, and is historically why the Dirac notation was developed to look the way it does.
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Singlet gluon interaction If a ninth singlet gluon existed (U(3) instead of SU(3)) it would propagate freely interacting strongly. My question is: only with hadrons or, perhaps, with leptons (I.e, electrons) as well?
| If there were a ninth gluon ... The precise wording of your question predetermines the answer. If the supernumerary gluon were the ninth generator of U(3), then it would couple to baryon number alone; but if it were the additional diagonal generator of a grand SU(4) broken to SU(3)xU(1), then it would couple to hypercharge. There is a somewhat esoteric reason, connected with anomalies, to think that it would have to be the latter. The renormalization group also has something to say: Even if all gluons had equal couplings at the Planck scale, radiative corrections would weaken the U(1) gluon’s coupling at lower energies, while strengthening the SU(3) gluons’ coupling.
| {
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A way to increase photovoltaic cell efficiency with fiber optics? A way to increase photovoltaic efficiency using fiber optics?
One of the claimed inhibitors of efficiency was the band gap of materials which contrasts to that of a photon. My idea is a way to possibly mitigate this loss by increasing the frequency of the electricity being generated(?) and taking it directly to coil.
In a glass fiber from fiber optics, light bounces back and forth many times before finally exiting out the other end of the fiber optic thread.
What if you were to manufacture a fiber optic thread so that it has a thin copper coil embedded in it that would have a staggered patterned coating on bare copper of n type and p type silicon? Think n,p,n,p,n,p etc for the nano coating facing the light source. The copper winding would be encapsulated at this point to prevent energy leakage. The copper coil would not be tightly wound and would leave space for some of the light to reflect to the other parts of coil.
The nano silicon of p and n types would directly transfer the photovoltaic charge to the copper coil and it would remain homogenous to the coil. A secondary coil would be used (not coated with anything save for an insulator) to transfer the energy elsewhere. So basically this is induction.
Would the back and forth motion of electrical charges be sufficient for this to work?
Would the induction frequency occur near the speed of light?
| What you looking for is cooper pairing
https://en.wikipedia.org/wiki/Cooper_pair
Means that a single photon trapped can atract electrons from the walls of the tube and create enegy (better creating heat) ? Is that what you looking for?
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Why runners lean forward? Why runners tend to lean forward prior to start running? How does it help run faster? What is the physics behind his leaning?
| Leaning allows for higher acceleration. If a runner at rest tries to run standing perfectly normal to the ground, he will never be able to move without falling over. Leaning is necessary for accelerating, for a person whose centre of mass is higher up from the ground.
Walking requires applying a force parallel to the ground to oppose friction, which generates a torque that tends to tip you over. Leaning provides a counter torque.
In a running race, there is a delicate interplay between these two torques. A runner has to balance both the torques to prevent falling.
Hence there is always a stable leaning angle, above which you fall forward and below which you fall backwards.
In a race, reaching your top speed quicker would increase your chances of winning. So it is beneficial to apply a significant force on the ground. But as you go faster, it becomes more and more difficult to exert a force on the ground. Because, doing so would require your legs to strike the ground more swiftly, which is limited by your physical capabilities.
Given below is a plot of the acceleration, velocity and distance versus time graphs of Usain bolt during his 100m race in 2008(taken from this site,). Though the scales of the axes may be different, this graph is typical for any human being.
Note that the acceleration, and hence force exerted by you on the ground, is highest at the beginning of the race, and it flattens out to zero after a while.
At the beginning of the race, the torque due to friction is maximum. So runners would have to lean higher during this time, to counter it.
Also, the torque due to friction decreases as the race progresses. So it becomes necessary for you to decrease your leaning angle (and hence the counter-torque) when you near your top speed. The same principle leads to the 'straightening out' of runners in a race. (Video 1,Video 2)
In short, leaning at the beginning of the race enables you to reach your top speed faster, increasing your chances of winning.
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How to Change Coordinate Systems in General Relativity Let me preface by stating that I have no experience with General Relativity. I am working on a project for school that requires a little knowledge of it, so I am hoping to find some help. I do have experience with Special Relativity.
On to the question. I know that one can calculate the age of the Universe using the Lambda-CDM model. After making a few simplifying assumptions, one can find the relation
$$H\left ( a \right )=\frac{\dot{a}}{a}=H_{0}\sqrt{\frac{\Omega_{m}}{a^{3}}+\frac{\Omega_{rad}}{a^{4}}+\Omega_{\Lambda }}.$$
One can numerically integrate to find $t_{0}$, the age of the Universe.
$$t_{0}=\int_{0}^{1}\frac{da}{aH_{0}\sqrt{\frac{\Omega_{m}}{a^{3}}+\frac{\Omega_{rad}}{a^{4}}+\Omega_{\Lambda }}}$$
Now, if I am correct, when performing this calculation for the age, I was working in a co-variant coordinate system (the system that expands with the universe or the system of CMB). For my project, I want to calculate the age of the Universe in a different coordinate system. More specifically, I would like to calculate the age in a coordinate system that is not expanding with the Universe. I know from other articles on here that I cannot use Special Relativity, but I am unsure how to go about this. If someone could show me how to go about this, keeping in mind my knowledge on this subject is very limited, I would be appreciative.
| You can write $ds^2$ as $a^2$ times a static metric, introducing a new time coordinate viz. $d\eta=dt/a$. We say the full metric is conformal to the simpler one (this adjective refers to angle preservation upon the rescaling), so $\eta$ is called conformal time. The $\eta r\theta\phi$ coordinate system fits your criterion.
Note that $d\eta=da/(a^2 H)$; the conformal age of the universe, in the convention where $a=1$ today, is about $46.8$ Gigayears, which is why the Hubble zone is $93.8$ light Gigayears wide. Thus the conformal age is not the quantity you want! You should still integrate $da/(aH)$ instead, but write $a,\,H$ as functions of $\eta$ instead of $t$.
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Is the superfluid state a coherent state? In the normal to a superfluid phase transition, U(1) symmetry related to particle number conservation is spontaneously broken which seems to imply that the superfluid state is a state in which there is no definite number of particles? This property is shared by that of a coherent state or any arbitrary superposition of number operator eigenstates.
Is there any property of the superfluid state (the condensate wavefunction) that is not shared by a coherent state?
| The ground state of a superfluid can be indeed (very precisely) approximated by a coherent state. More accurately by a squeezed coherent state. Please see Zhang equation (72):
$$|\{z_k \beta_k\}\rangle = \prod_k \exp\{ z_k a^{\dagger}_k - \bar{z}_k a_k\} \exp\{ \beta_k a^{\dagger}_ka^{\dagger}_{-k} - \bar{\beta}_k a_ka_{-k}\} |0\rangle$$
Where, $|0\rangle$ is the unbroken vacuum and $z_k$, and $\beta_k$ are parameters dependent on the details of the underlying many-body Hamiltonian.
This type of ground states is characteristic to collective ground states of strongly interacting systems. The squeezing is obtained due to the Bogoliubov transformation required to diagonalize the Hamiltonian in the large $N$ limit. (squeezing means "flattening" the circualr uncertainty region of an oscillator into an ellipse).
A quite transparent derivation of this type of ground states is given, for example, by: Solomon, Feng and Penna.
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How does a wire bend the electric field? The electrons in the current in a wire are constantly colliding with the metal cations which means there must be a constant cause of acceleration: an electric field. How, why is it that the electric field is identical in shape to that of the wire?
| The electric field $\vec{E}$ also exists outside the wire. There is however no current there as there are no charges. Inside the wire, for simplicity assumed to be an isotropic conductor, $\vec{E}$ is related to the current density $\vec{j}$ by $\vec{j} = \sigma \vec{E}$, where $\sigma$ is the conductivity. It is not zero unless $\sigma$ is infinite as is the case for a superconductor.
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Quantum teleportation reversible? The following figure displays a circuit implementing quantum teleportation.
It uses Haramard transform (H), controlled not (+), measurements (meas), a Not gate (N) and a phase shift gate (P) to teleport quantum state q from Alice to Bob.
I have learned that all quantum computations should be reversible (see e.g. here), but this gate does not seem reversible, because it contains measurements.
Is quantum teleportation reversible? What does being reversible even mean in this context? Most quantum algorithms I know only measure in the end. I am assuming that reversibility in this case means that the gates are reversible before the measurement. But in quantum teleportation, this is not possible (because Alice must send the result of the measurement to Bob).
It is particularly confusing that at the red line below, we have a mixture of bits and quantum bits. How should I interpret this?
Possible answer
The deferred measurement principle states that I could move the measurements to the end and still get the same result. Should I interpret reversibility as "reversibility after moving the measurements to the end"?
| Quantum teleportation is not reversible since it exhausts its resource - the entangled Bell state initially shared by Alice and Bob. At the end of the teleportation Alice and Bob don't share anymore an entangled state.
Any unitary operations (here H and CNOT) are reversible. The measurements, hovewer, are not. But the measurement is crutial for the teleportation to work, specifically - from the teleportation identity it is the measurement performed by Alice that actually kicks on the teleportation. This measurement uses the entanglement resource, since afterward Alice's and Bob's states are entangled no more.
After Alice's measurement the state is teleported but encoded. It is known that classical communication should be used to transfer to Bob the proper measurement basis (otherwise superluminal communication channel could be constructed from this procedure). Bob's measurement doesn't use the entanglement resource (it's just trying to measure a single state in proper basis), so it can be deffered.
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How to calculate thermal equilibrium temperature of three different substances? I am trying to calculate the thermal equilibrium temperature of three different substances (with three difference specific heats). The substances are:
Substance 1 - mass: 0.6 kg, specific heat: 4170 J/kg°C, initial temperature: 90°C
Substance 2 - mass: 0.2 kg, specific heat: 840 J/kg°C, initial temperature: 20°C
Substance 3 - mass: 0.1 kg, specific heat: 110 J/kg°C, initial temperature: 10°C
I have already calculated the equilibrium temperature of Substances 1 and 2, which is approximately 85.61°C. What I am trying to figure out is, when Substance 1 and 2 are 80°C, what will be the equilibrium temperature when Substance 3 is added? I am confused about if I need to add, multiply, or find the average of specific heats for Substances 1 and 2. I would like to know what is the formula for calculating equilibrium temperature of 3 or more substances (with different specific heats).
| Think about it: the specific heat is, in principle, something characteristic of the substance. If you add substances 1 and 2, you will get a resulting mixture, let's call it 1/2. Do you think that it's specific heat will be more like 1's or 2's?
I think it is intuitive to say that it will be closer to the specific heat of substance 1, since it is predominant in your mixture (you added a larger mass of substance 1 to the mixture, compared with substance 2).
You can work this out from here. Hope it helped.
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Are materials which are bad at conducting heat always bad at conducting electricity also? When defining a material's conductivity, we usually consider its conductivity of heat and conductivity of electricity separately. However, I realize that materials like metal conduct both heat and electricity well. In contrast, materials like wood and glass conduct both heat and electricity poorly. Therefore can we conclude that if a material is bad at conducting one kind of "flow of energy", then it will also be bad at conducting another kind of "flow of energy"? Thanks a lot.
| As an Engineer I can assure you that many electronic devices work just because there are materials that are good thermal conductors but excellent insulators as well.
Electronic device and equipment, especially those designed for high power, must get rid of the excess heat produced by electrical power dissipation. Most of the time because of joule effect, but there are other dissipative phenomena like the reorientation of magnetic domains in a ferro-/ferri-magnetic materials (e.g., transformers core).
Not being able to get rid of that heat will cause the device to fail because of its temperature rising above its safe maximum.
The use of electrically-insulating thermal conductors allow to improve thermal conduction inside a device/piece of equipment without causing short circuits between parts placed close together.
A classic example, used even at the dawn of electronics, is mica (Wikipedia article). Excerpts (emphasis mine):
Sheet mica is used principally in the electronic and electrical industries. Its usefulness in these applications is derived from its unique electrical and thermal properties and its mechanical properties, which allow it to be cut, punched, stamped, and machined to close tolerances. Specifically, mica is unusual in that it is a good electrical insulator at the same time as being a good thermal conductor.
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Is it possible to harvest the energy from the movements of a satellite in orbit? I was thinking about how energy is harvested on Earth from movements of certain forces like wind and ocean currents. Could similar principles be applied in space?
Satellites are virtually in perpetual motion when orbiting the Earth. Is there kinetic energy that can be extracted from this orbital motion and harvested for use on Earth?
| I do absolutely agree with @knzhou, but going to provide a couple of points here, more intuitive than scientific.
A satellite can spin for quite a long time since the kinetic energy of it doesn't decrease significantly (it doesn't mean that the energy doesn't decrease at all - it does, and for that reason a satellite will fall down eventually).
Now if we're going to harvest any energy with the aim of having a "profit", we would just "drain" it's kinetic energy, its speed will decrease at a greater rate, hence the satellite will fall noticeably sooner which would make the benefit of launching the satellite way less than the proposed benefit.
In other words, that's like trying to harvest an energy from a freely moving object. Harvesting would be pointless since the object would stop because of harvesting and you won't gain nothing.
I.e., you thought you have harvested the energy, but in fact your object has just passed all its energy to you and left with nothing, which is not what harvesting energy is supposed to be.
A side note regarding "perpetual motion" - it could be perpetual only in that ideal case when the energy is not lost. If you try to harvest energy from an object in "perpetual motion", the motion will no longer be "perpetual".
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Is Yellow a monochromatic light? I have got a serious doubt.
I have read,
"yellow light from a low pressure sodium vapour is monochromatic"
How can it be monochromatic when yellow light is a combination of red and green primary colours?
| Color is not something that happens in the physical world. Color happens in your brain. In the physical world, every light source has a spectrum. The topic of how your brain and eyes reduce the spectrum of some light source to what we call "color" is quite deep.
https://en.wikipedia.org/wiki/Color_vision
The short answer is, there are many different spectra that your brain and eyes reduce to the color that we call "yellow." The light emitted by a low-pressure sodium lamp is one such spectrum that contains all of its energy in two very narrow, and very closely spaced emission lines. But, you can experience the same "yellow" color by looking at a light source (e.g., a color computer screen) that emits a totally different spectrum containing lines that you would call "red" and "green" if you saw them individually.
you can use a simple spectroscope to see the spectra of different light sources.
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Symmetry of Spin Function I have a question concerning the symmetry of the spin function in multiple identical particle systems. In the solutions of one of the quizzes, my professor said that the $s=3/2$ spin function is completely symmetric, which is why we need an antisymmetric spatial component (by a Slater matrix) to describe three fermions.
I understand why we would need the spatial component to be antisymmetric if the spin part is symmetric. But how do we know, from the value of s, if the spin part is symmetric or not?
Thank you.
| I'll assume your system contains three spin-1/2 particles. With the notation $\vert s,m\rangle$, it is clear that
$$
\vert 3/2,3/2\rangle = \vert \textstyle\frac{1}{2},\frac{1}{2}\rangle_1
\vert \textstyle\frac{1}{2},\frac{1}{2}\rangle_2 \vert \textstyle\frac{1}{2},\frac{1}{2}\rangle_3
$$
is symmetric under permutation of particle indices. The other $\vert 3/2,m\rangle$ will also be symmetric since they can be reached from $\vert 3/2,3/2\rangle$ by applying $L_-$, where
$$
L_- = L^{(1)}_-+L^{(2)}_-+L^{(3)}_-
$$
is also symmetric under permutation of particle labels and so cannot change the permutation symmetry of the states it acts on. (Here, $L^{(k)}_-$ is the lowering operator acting on particle $k$ alone.)
Note that if you have $5$ particles, then the $\vert 3/2,3/2\rangle$ state would NOT be symmetric, and thus the entire $J=3/2$ multiplet would not be symmetric either.
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Does an event horizon happen abruptly, and if so, what would happen if I was floating in front of it and shined a flashlight in its general direction? Does an event horizon happen abruptly, and if so, what would happen if I was floating in front of it and shined a flashlight in its general direction? Would it appear as if I was standing in front of a giant wall? For the sake of this question, let's say I tossed up some dust first, and then shined the flashlight.
| A black hole's event horizon is a set of events which only has global significance. Locally, within a region of spacetime around some event on an event horizon, as long as the region of spacetime is small enough that tidal forces are negligible within that region, spacetime behaves like ordinary, flat spacetime. An event horizon isn't locally visible or otherwise detectable.
Suppose you're within such a sufficiently small region, and as you cross the event horizon you toss up some dust in the direction of the black hole, and shine a flashlight on it. What will happen will just be that you'll see the dust in your flashlight beam. You won't be able to tell if the light from your flashlight hit the dust while you were both outside of the event horizon, or while you were both inside of the event horizon, or whether the light hit the dust inside the event horizon while you were outside, but then the reflection hit your eyes after you crossed the event horizon.
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Source of the Chemical Potential I know that the chemical potential is the energy required to add another particle to the system but what is it's origin on the microscopic level? Take for example the case of the ideal Fermi gas, the Hamiltonian:
$$H=\sum_i \frac{p_i^2}{2m}-\mu \hat N$$
clearly the first term is the kinetic energy but what interaction is causing the energy associated with the second term?
| There is a misconception in this question about the nature of the chemical potential. Although the chemical potential is often described as 'the energy associated with adding a single particle' this is has to be interpreted with extreme care. In this saying 'energy' refers to the 'mean energy' and 'adding a single particle' refers to 'changing the mean particle number by one'.
The change in 'mean energy' is not due to any interaction but rather the statistical effect that changing the mean particle number changes which microscopic states are more likely then others.
That said the presence of the chemical potential in the Grand partition function is equivalent to that of an interaction which acts on all particles equally. But this interaction does not have any physical meaning.
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A doubt related to Significant Digits Could someone please explain this statement to me
"Reporting the result of measurement that includes more digits than significant digits is superfluous and also misleading since it would give a wrong idea about the precision of measurement."
Also, shouldn't the word accuracy be used instead of precision because isn't precision the closeness of various measurements for the same quantity and only one measurement is being talked of here.
| In general, it's best to give all potentially relevant information. So, in general, that quote provides poor advice.
However, it's important to write your number correctly. For example, a number understood to be approximately represented as "$1.2$" means "$1.2{\pm}0.05$" or "$\left[1.15,\,1.25\right)$".
If you want to add more information, you can't just tack the numbers on because that changes the implied precision of the value. But you can still do it. For example, Wikipedia lists a current value of the fine-structure constant, $\alpha$, as$$
{\alpha}
{\quad}={\quad}\frac{e^2}{4π{\epsilon}_{0}{\hbar}c}
{\quad}={\quad}0.007~297~352~566~4~(17),
$$in which the last "$64$" is understood to not be significant as it's modified by "$\left(17\right)$". Translated, this means:
$$
{\alpha}
{\quad}{\approx}{\quad}
\begin{array}{rl}
& 0.007~~297~~352~~566~~4 \\
{\pm} & 0.000~~000~~000~~001~~7
\end{array},
$$which if we had to write that using regular significant figures, would just be$$
{\alpha}
{\quad}{\approx}{\quad}
0.007~~297~~352~~57.
$$So, this notation can be used if you want to express additional information in a context in which the reader would assume that the number's form implies its uncertainty by codifying the difference in perceived uncertainty with an appropriate error qualification.
In general, though, the exact statistics of measurements can be pretty complicated; for example, error doesn't need to be normally distributed. So when it really matters, care must be taken to precisely specify what's meant.
In practice, most folks try to keep it simple to avoid a hassle. The basic significant figures system has been designed to be simple while working well-enough in many simple cases.
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According to Conservation of Momentum, a gun in a sealed box should not have recoil? According to the law of Conservation of Momentum, there is no way to increase the momentum of a system, except by momentum transfer from interactions with the external. If I fire a rifle while sitting on a go kart, the go kart is going to go backwards but the bullet goes forwards, conserving the momentum.
Now lets say I construct a long 1 inch thick steel box (a few meters long), and I position the gun's butt against the back of it, and fire the gun electronically. Would we not get the box flying backwards still (at least until the bullet gets lodged in the front of the box? Even if the bullet burying in the metal at the end of the box causes another force in the box at the opposite direction of the initial kick, haven't we momentarily broken the conservation of momentum?
|
haven't we momentarily broken the conservation of momentum?
No. The box/rifle is only moving backward during the period when the bullet is moving forward. If you sum their momenta, you will find it sums to zero.
Just because you cannot see the bullet from the outside does not mean that you can ignore its contribution to the momentum of the system.
From comments:
is there not a change in momentum in the opposite direction of the bullet until the moment the bullet strikes the box?
No. If during the flight of the bullet we examine it, we will find it has some momentum $p$ to the left. At the same time, the box and rifle will have a momentum $p$ to the right. In normal cases, the bullet is less massive, so it will have higher speed. But the magnitude of the momentum is equal.
If you make the bullet heavier and faster, the recoil of the gun/box will just increase to compensate.
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Why the Fermi temperature isn't zero? The fermi temperature is defined as
$$ k_BT_f = E_f$$
But the fermi energy is the energy at $T=0$, where the energy level is the highest occupied for electrons. So, why is the fermi temperature defined as $\neq 0$?, What temperature $T_f$ is measured? Over who is T measured?
| The temperature is set to zero to define Fermi energy because $T=0$ corresponds to the ground state of this electronic system. The electrons are still moving around with their zero-point motion, and the Fermi temperature corresponds to this motion. Considering that electrons don't contribute much to the heat capacity of a bulk material, the electrons can have relatively high kinetic energies without appreciably raising the temperature of the total system.
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Why polarization filter do not dim the light completely? In a circle there's infinite amount of degrees (eg. 0 deg, 0.00000000000...1 deg etc.) In a ground school we are thought that there's 360 degrees in a circle.
A landscape behind my window is incoherent light source, so it randomly emits photons with all polarization directions.
When I put a polarizer between landscape and my eye... i can still see the everything. But how is that possible if the polarizer transmits only $1/\infty$ of all photons (since there's infinite amount of directions of polarization)?
Even if we assume that there's just 360 degrees in circle... The landscape behind my window is not 360 times darker when I observe it through filter (eg. polarization glasses).
Why won't polarizer dim the light severely?
| Polarizers don’t just filter photons, they also change the polarization of the photons that make it through. If you send 1000 incoherent photons through a polarizer 500 on average will make it through and all of them will become parallel with the Polarizer. Thats why polarization filters do not dim (coherent) light completely. The closer a photon is polarized parallel with the slit the better chance it will make it through. A photon that is almost perpendicular to the slit can still make it through but has less chance. It will then become polarized as it goes through.
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How do we know yo-yo is rolling without slipping? How do we know a prior that a typical yo-yo rolls without slipping when released?
Now suppose we get a typical yoyo which is attached to the string (otherwise it would just fall and hit the ground if you released it!), then I cut the string so its not attached the to yoyo anymore. Then I dip the string in slippery oil and wrap the yoyo again with the string unattached. Could slipping occur after release as the string unravels?
How do you test experimentally whether we have rolling without slipping?
| Rolling without slipping occurs if the rotational velocity of the yo-yo matches that of its angular velocity i.e. v=wr.
This; however, is complicated by the fact that for a yoyo the radius will change as the spooled string unwinds.
In this case you want to determine a relationship between r and the length wound.
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Air driven piston creating vacuum I am looking at generating a vacuum from two interconnected pistons, a driver piston from high pressure and the drivee creating a negative pressure in a container.
I am looking at improving the amount of vacuum I can get with the use of Pascal's principle but am unsure if I am trying to get something for nothing (I think I might be)
Taking the picture below, lets say I have two pistons, with D2 being double D1. If I put high pressure air into the inlet at p1, due to the surface areas being double (and via Pascal's principle) will I create double the force at the D2 and therefore allow the piston to move the same distance, but as the force has doubled create double the amount of vacuum'd pressure at D2.
| Consider you've achieved some pressure in the left cylinder, the system has moved and reached a state of rest. Lets find the pressure in the right cylinder. I'm not sure what exactly you mean by
I have two pistons, with D2 being double D1
I assume you mean the areas of the pistons, so lets say that the area of the left piston is $A_1$ and the area of the right piston is $A_2 = 2A_1$. The pressure acting on them I will call $P_1$ and $P_2$ respectively, and the resulting forces: $F_1$ and $F_2$.
Consider the picture:
We said that the system is in rest, so all forces must cancel out, so $F_1 = -F_2$. We know that the pressure force is equal to the product of the pressure and the area it acts on, so: $F=PA$ When we substitute this in $F_1 = -F_2$ we get $$P_1A_1 = -P_2A_2$$
Then we subsitute $A_2 = 2A_1$ and we get
$$P_1A_1 = -2P_2A_1$$
With a bit of algebra
$$P_2 = -\frac{P_1}{2}$$
So we reach the conclusion that the pressure in the right tank would be negative half of the pressure in the left tank. So if you have $1kPa$ presure in the left tank, you would have $-0.5kPa$ in the right.
I think the key mistake you make is applying the Pascal's principle wrong
in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container.
For the Pascal's principle to work the medium must be a fluid. In your example the medium is a solid - the pistons and rod, connecting them.
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Can there be general relativity without special relativity? Can General Relativity be correct if Special Relativity is incorrect?
| I wouldn't have thought so.
One solution to the vacuum equations of GR without cosmological constant is Minkowski space, this is the space of Special Relativity. Thus we see that General Relativity generalises Special Relativity
And we also see why Einstein opted for the word 'General' when he named this theory - actually this is more to do with general covariance, the theory of which is category theory, at least in one of its senses, it's called functoriality there.
To have GR without SR would mean that we require some physical rule that rules out Special Relativity as a solution. We don't have such a physical rule.
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Where does the buoyant force come from? If I place a cube in water, the force at the top of the cube, $F_1$ will be $Ah\rho_wg$.
Where,
$A =$ cross-sectional area
$h =$ height at the top
$\rho_w =$ density of water
$g$ = acceleration due to gravity
All this made sense, because this downward Force ($F_1$) is actually the weight of the water above the cube.....but where did the upward force come from? is it the opposite force of the weight of the cube? or something else?
| *
*You have successfully described a downwards force on the cube, being the pressure at its top due to the weight of water above.
*Another downwards force is the cube's own weight.
*An upwards force comes from the water below the cube. Here the pressure is a bit larger than that at the top of the cube, since we are a bit deeper. This is what is called buoyancy.
If the downwards forces together equal the upwards force then the cube doesn't move. That would be the case if the cube's density equalled that of water. It would then weigh just as much as the same amount (volume) of water would.
If it's density is smaller (and thus it's weight smaller) than in this equilibrium scenario, we suddenly have less total downwards force. The upwards force due to the pressure at that depth does not change. So now the upwards force is larger, causing the cube to accelerate up.
You can think of this buoyancy effect as caused by water that really wants to move somewhere else due to the pressure. If something heavy is nearby, then that heavy object is more difficult to push away than neighboring water molecules. If something light is nearby (less dense than water itself), then it is easier for the water molecules to push this object away than to push other water molecules away. This effect is strongest where the pressure is strongest, which will be at the bottom. The net effect is thus an upwards motion.
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What is the relation between a measurement and an observable? Observables are represented by Hermitian operators. First of all, it's a little strange (to me) that some measurable physical quantity is represented by a transformation (or linear map), given that I think of a linear map as a function and I don't think of physical quantities as functions. But this is not my doubt. I try to accept this definition.
The resulting quantum state to which a system collapses after a measurement is one of the eigenvectors of this Hermitian operator. The corresponding eigenvalue is the result of the measurement.
If I understood correctly, a measurement is performed by a transformation, that is, in linear algebraic terms, you multiply by a matrix.
Anyway, I am not fully understanding the relation and connection between the measurement and an observable (or the Hermitian operator that represents it). How are measurements and observables in quantum mechanics related?
| I give here only the simplest explanation. In quantum mechanics (Schrödinger representation), the state of a physical system is completely determined by its, in general, complex (normalized) wave function $\psi$. An observable $A$ is a physical quantity that can be measured. It is represented by a corresponding Hermitian operator $\hat A$ in the sense that results of measurements of this quantity can only be the (real) eigenvalues $a_j$ of this Hermitian operator. The corresponding (normalized) eigenfunctions $\psi_j$ form a complete orthogonal set of functions so that any wave function $\psi$ representing the state of a system can be written as a (possibly infinite) linear combination of these eigenfunctions $$\psi=\sum_j c_j\psi_j \tag1$$ where $c_j$ are complex numbers. The measurement of the observable $A$ on a system in state $\psi$ yields one of the eigenvalues $a_j$ given by $$\hat A\psi_j=a_j\psi_j \tag 2$$ of the operator $\hat A$ with probability $$P=|c_j|^2 \tag 3$$ The sum of all these probabilities is 1. After the measurement with result $a_j$, the system is in the state of the corresponding eigenfunction $\psi_j$.
The measurement is not performed by a transformation, but the results are given by the eigenvalues of the corresponding Hermitian operator. More sophisticated situation like continuous or degenerated eigenvalues and systems described by density matrices are described in many quantum mechanics textbooks.
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Why do objects float in liquids denser than themselves? Why do objects float in liquids denser than themselves? I know that a balloon floats on water because it has air in it, but why?
| Objects sitting in a fluid (liquid or gas) experience a pressure on every surface, equal to the pressure of the fluid.
Horizontally these pressures cancel out - the pressure pushing on the left hand side of an object will be the same as on the right, and the object has no net sideways force.
Vertically however is a different story. The pressure of a fluid increases as you move lower (deeper) into it. Going back to our object, its bottom is deeper into the fluid than its top is, so the pressure on the bottom is larger than the pressure on the top. This difference results in a net force upwards.
We call this net upwards force buoyancy, and if you work through the maths it turns out that the upward force is equal to the weight of the amount of fluid that would fit in the same space.
If the object is less dense than the fluid, its own weight is therefore smaller than the buoyancy force upwards, and it floats.
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Red-shifting due to emitting gravitational waves Light waves exert their own gravitational pull and must be emitting gravitational waves, losing energy in the process. Does this mean that light becomes red-shifted as it travels even without the effects of universal expansion?
| First we must clarify:
*
*gravitational waves can be created by constantly changing gravitational fields, like two oscillating neutron stars.
*photons do have gravitational effects, because they have energy
*gravitational effects are because of stress-energy
*photons energy comes from their frequency, E=h*f
*a simple mass does not emit gravitational waves, like the earth, or at least not so big to have any effect.
*a traveling photon does not emit gravitational waves.
*gravitational waves are not the same as the effects of gravity (bending of spacetime), because gravitational waves are traveling gravitational fields, independently from the emitter
*a gravitational wave is a traveling change in the bending of spacetime, so like a traveling gravitational field
*a simple mass that has gravitational effects, like bending spacetime, does not emit gravitational waves (it needs to be rotating very fast around another mass, like two neutron stars rotating).
A simple mass like that will not lose energy by creating a gravitational field, and bending spacetime.
That is why a traveling photon will not lose energy by creating a gravitational field and bending spacetime locally where it is.
Photons get redshifted because the ones coming from far away galaxies that are receding, are in places where space expands, and the photons travel in expanding space that is why they get redshifted, getting bigger wavelength.
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How can a body have two axis of rotation at the same time? I m not concerned with rotation of a body with two simultaneous axis but concerned with how we choose the axis,while going through pure rolling I have observed that there are two axis of rotation one is passing through the center of mass and the other is through the point in contact with the ground,my concern is how can there be any axis of rotation through the point of contact where as m very well finding the body does not rotate in that axis of rotation that is it very well rotates only through the center of mass.
| A body will only have one instantaneous axis of rotation (Chasle's Theorem).
In your example, the center of mass translates, and thus it is not a center of rotation. The only center of rotation is the point that does not move (the contact point).
To find the location of the center of rotation relative to some point A where you know the velocity vector $\boldsymbol{v}_A$ for a body with rotational velocity vector $\boldsymbol{\omega}$, calculate the following
$$ \boldsymbol{r}_{\rm rot} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_A }{ \| \boldsymbol{\omega} \|^2} $$
Here $\times$ is the vector cross product.
So consider the motion of the center $v$ to the right, and the no-slip condition $v + \dot{\theta} R =0$. In vector form the motion of the center is
$$ \begin{aligned} \boldsymbol{v}_A & = \pmatrix{v \\ 0 \\0}& \boldsymbol{\omega} & = \pmatrix{0 \\ 0 \\ -\frac{v}{R}} \end{aligned} $$
The center of rotation is thus
$$ \boldsymbol{r}_{\rm rot} = \frac{ \pmatrix{0 \\ 0 \\ -\frac{v}{R}} \times \pmatrix{v \\ 0 \\0} }{ \| \pmatrix{0 \\ 0 \\ -\frac{v}{R}} \|^2 } = \frac{ \pmatrix{0 \\ -\frac{v^2}{R}\\0}}{ \left( \frac{v}{R} \right)^2} = \pmatrix{0 \\ -R \\ 0}$$
which is in fact the contact point.
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Doubt on an answer to a mechanics problem Question: A block of Mass $m$ is connected to another block of mass $M$ by a massless spring of spring constant $k$ . The blocks are kept on a smooth horizontal plane and are at rest. The spring is unstretched when a constant force $F$ starts acting on the block of mass $M$ (horizontally) to pull it. Find the maximum extension of the spring.
I solved it by two different methods:
Method 1:
Here I assumed that maximum extension $x$ will be produced when both the blocks would be moving with constant acceleration. Therefore constant acceleration is:
$$a=\frac{F}{m+M}\tag{1}$$
and by considering free body diagram of block with mass $m$
$$kx=ma \tag{2}$$
From 1 and 2,
$$\\x=m\frac{F}{k(m+M)}$$
Method 2:
By conservation of energy, in the reference frame of center of mass , for block $m$, we have two forces acting $m\frac{F}{m+M}$ and $kx$, in opposite directions.
For block $M$ we have three forces acting $\frac{MF}{m+M}$ and $kx$, and $F$ in the opposite direction.
Assuming $m$ moves a max distance $x_1$ from centre of mass(COM) and M moves a distance $x_2$ from COM, then work done by external force will be:
$$W=\frac{mF(x_1+x_2)}{(m+M)}$$
This will be stored as the internal energy, therefore
$$\frac{1}{2}k(x_1+x_2)^2=W$$
On solving this we get,
$$x_1+x_2=\frac{2mF}{(k(m+M))}$$
Which one is the wrong and why?
| Method 1 is correct. The error in Method 2 enters with your expression for "work done by external force". This should be a sum of the the work done on $m,$ which is $x_1mF/(m+M)$ and that done on $M,$ which is $x_2(F-MF/(m+M))$. To solve using this method, you will also need an additional relation: the ratio of $x_1$ to $x_2$, which is just $M/m.$ In your expression you are assuming that the work done on the spring is the total length by which it is stretched multiplied by the ultimate tension. But the tension in the spring only reaches that value at the end of the stretching phase, so the work is only half as much.
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Does a charged particle travelling with uniform velocity induce a magnetic field? Does a charged particle, an electron say, travelling with uniform velocity induce a magnetic field? I believe it doesn't. In primary school, we all learned how to induce a magnetic field into an iron nail by wrapping coils of wire around the nail and then hooking it up to a DC battery, but if you do not coil the wire, the magnetic nail doesn't occur. What's happening here? My only guess are the electrons are accelerating; the magnitudes of their speeds aren't changing, but rather their directions. In the coil, a force must be applying itself to the electrons in order for them to make their spiralling paths, thus, they are said to be accelerating and that is what causes the magnetic field to develop.
| Of course there is a magnetic field, otherwise the ampere would not exist. The ampere is defined by the magnetic attraction of two parallel straight wires carrying DC current.
(This answer is backwards in that it explains the existence of the field in terms of the non-silliness of the SI.)
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EM Plane wave, the changing electric field is in all directions right? I just want to confirm this, because this type of diagram seems pretty popular.
The electric field and magnetic field actually surround in all directions orthogonal to x axis, right? It is not just 2d pointing only in the y direction and z direction respectively.
| The EM wave in this diagram is just a ray, a geometrical visualization of the EM wave.
The EM wave is in reality a 4D wave, or at least you have to visualize it in 3D, where the wave rotates, that is, it goes from E wave to a M wave, because in the Maxwell equations, a moving E wave creates a moving M wave and this is how light propagates in the classical framework.
According to QM, the photon is the excitation of the EM field. So the E field creates an excitation in the M field, that creates an excitation in the E field, and that is how EM waves propagate in 4D.
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Difference between pure quantum states and coherent quantum states In the post What is coherence in quantum mechanics? and the answer by udrv in this post it seems to imply that a pure quantum state and coherent quantum state are the same thing since any pure state can be written as a projector onto the pure state when written as a density operator.
Are they equivalent? If these two concepts are not equivalent, what is a simply counterexample to illustrate the difference?
Then there is also the definition of a coherent state which defined it as a quantum state of the harmonic oscillator, quite confusing as to how these concepts are related and distinct, could someone provide some clarity on these distinctions?
| Coherent quantum states are special types of pure quantum states. The term coherent is only meaningful when there is an Fock algebra of ladder operators. Coherent states will be eigenvector of the (non-Hermitian) annihilation operator
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AP physics 1 rotation problem could someone help me with this problem?
the correct answers are a and d. one issue i have with it is that i just don't understand what the problem is asking. like what spool? what table? i tried making some sense of the question and the answers, but i can only see d moving the wheel clockwise. the answer explanation mentions,
The key is knowing where the “fulcrum,” or the pivot for rotation, is. Here, that’s the contact point between the surface and the wheel.
but why is that? isn't the pivot where the axle meets the wheel?
| Not a very clearly worded question! The spool consists of the two wheels and the axle, as shown, and these are rigidly connected. Whoever wrote this will have assumed their readers were familiar with the spool that a camera film came on - not so common today! It is assumed to be resting on a table, which isn't shown. Draw the table underneath the circle of the wheel, and think about the forces of gravity, reaction and friction. As the wheel rotates it must move across the table, without slipping.
But it's really quite pretty once it's got going. Yes, B and C obviously make the spool (axle and wheels together) roll anticlockwise and move left, and D will send it right. A is more subtle, as it looks as if it wants to roll left but the force is pulling it to the right. Which it will do, winding up the rope as you pull on it.
There's a great chance out there for somebody with a spool, a piece of string, and a video camera.
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A gravitational field is path independent. Why does a rocket not fly in serpentine lines? in theory a gravitational field is path independent, a gravitational field is a gradient field and so conservative. why doesn't a rocket fly in serpentine lines to exit the gravitational field of the moon, as said the gravitational field of the moon is path independent and the moon has no atmosphere and so there is no aerodynamic drag?
https://en.wikipedia.org/wiki/Gradient_theorem
https://en.wikipedia.org/wiki/Conservative_vector_field#Path_independence
i know that the reason is that flying in serpentine lines would require more energy ( fuel ) however where is this surplus of energy for a serpentine line path reflected in the theories?
|
In theory a gravitational field is path independent, a gravitational field is a gradient field and so conservative.
Correct. But what this means is that the total work done by gravity is constant over the path.
why doesn't a rocket fly in serpentine lines to exit the gravitational field of the moon, as said the gravitational field of the moon is path independent and the moon has no atmosphere and so there is no aerodynamic drag?
Because the rocket is using its engines. And the engines are not a conservative force on the rocket. The work done by the engines on the rocket depend on the speed and direction of the rocket.
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Does increasing tension on a string reduce or increase the harmonic wavelength for a standing wave? I had thought that increasing tension on a string increases the frequency and thus decreases the wavelength. My book says otherwise. Which is correct?
| This question is about standing waves on a string of length $L$ with fixed ends. A string like this will behave differently when it is excited with external vibrations depending on the frequency. Because frequency and wavelength are connected via $c=\lambda f$, there is a corresponding wavelength for each excitation frequency. If $\lambda = \frac{c}{f} \stackrel{!}{=} 2L$ or $L = \frac{\lambda}{2}$ a standing wave with large amplitude will occur with one antinode and two nodes at the ends. If the frequency will be increased by a small amount, the standing wave will collapse. If you keep increasing the frequency, the point where $L = \lambda$ will be reached and another standing wave with two antinodes and three nodes (ends + middle) will form. In general there will be standing waves if $L = k \cdot \frac{\lambda}{2}$ or $\lambda = \frac{2L}{k}$, $k = 1, 2, 3, \dots$.
Now you have to use $\lambda = \frac{c}{f}$ to get $$ \frac{c}{f} = \frac{2L}{k}$$
or (multiplying both sides with $f$ and $k$)
$$kc = 2fL$$
The book seems to assume that you know that $c$ will increase with increased tension. The right hand side of the equation is fixed, therefore k must decrease to keep the left hand side constant.
Note that $k$ is still limited to the natural numbers. Your initial situation has $k=3$ for the given conditions of $c$, $f$ and $L$.
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Will the weight of a man standing on a scale changes if he throws something straight up? If a man enclosed inside a chamber and the chamber is on a weighing scale and if he throws a ball straight up in air will the weight of the whole system changes or not? If yes, then what is the weight change at
1) just moments after throw;
2) before reaching the highest point of flight;
3) at highest point of flight;
4) on descending downwards; and
5) if there is any change in above observations in vacuum.
Supposing that scale is fast enough to register the changes before ball comes back to the man.
| The force on the entire system (measured by the scale), must equal the weight of the entire system, plus any required to accelerate the center of mass upward.
If the center of mass is not accelerating, the scale will measure the weight of all the contents. If the center of mass is accelerating upward (such as when the person is starting to push the ball upward), then the scale will read higher than the weight.
If the center of mass is accelerating downward (such as when the ball is in the air), then the scale will read lower than the weight.
So assuming the man has had time to stop moving, then 1, 2, 3, and 4 are all identical. The center of mass of the system is accelerating downward, so the weight indicated by the scale is less than it was when the ball was held still.
A vacuum would only change the answer to the extent that we want to model air resistance. It's a small correction in this case.
| {
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Confused with heat as a form of energy I have quite a simple question. Energy can be defined as capacity to do work.
But I have read
When energy is exchanged between thermodynamic systems by thermal interaction, the transfer of energy is called heat.
I can't understand what is the work done. For example, what is the work done in cooling water from 20 to 40 celsius (1 atm)?
So, is heat a form of energy if energy is capacity to do work?
| Heat and work can both be referred to in this context as energy in transit. They are not forms of energy themselves but rather a means to transfer energy. As you have described, heat represents the flow of thermal energy by conduction, convection and radiation. Work describes the change of the energy if the system when some external action generates thermal energy. For example the compression work to push a piston or the electrical work to increase the temperature of the element of a stove. In the case where work transfers the energy, the interaction is not strictly thernal energy transferring between objects.
| {
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What additional things should I do to understand language of science I am about to complete my High school. Typically, I study text books of science but it takes a lot of time to understand the concept resulting less time to do questions and due to which I get low marks(not having English as my native language is also a factor). How should I start, please guide me from basic to advance level.
*
*How could I study science, effectively, without wasting any time.
| Well, one of the best ways is to enroll in a full time University course but of course that is a lot of commitment. If you don't want to do that then I would suggest looking for University lectures on YouTube. Here is a full quantum mechanics module by MIT, it doesn't get much better than that! Don't start with that module though. Universities' websites will have a list of the modules that they teach and use that as a guide for what you should learn first. Try to find problem sheets online as well so that you're constantly practicing the skills you learn.
| {
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What would qualify as a deceleration rather than an acceleration if speed is unchanged? The instantaneous acceleration $\textbf{a}(t)$ of a particle is defined as the rate of change of its instantaneous velocity $\textbf{v}(t)$: $$\textbf{a}(t)=\frac{\mathrm{d}}{\mathrm{d}t}\textbf{v}(t).\tag{1}$$ If the speed is constant, then $$\textbf{a}(t)=v\frac{\mathrm{d}}{\mathrm{d}t}\hat{\textbf{n}}(t)\tag{2}$$ where $\hat{\textbf{n}}(t)$ is the instantaneous direction of velocity which changes with time.
Questions:
*
*According to the definition (1) what is a deceleration?
*In case (2), when will $\textbf{a}(t)$ represent a deceleration? For example, in uniform circular motion, why is it called the centripetal acceleration and not centripetal deceleration?
| Acceleration is the correct technical term for the physical quantity you mentioned in the equations you posted (i.e. a).
The term deceleration doesn't describe a rigorously-defined standard physical quantity, it's just a term used differently in different situations that means "handwavily" that the velocity or speed is decreasing.
Sometimes it could be clear that it refers to some precise quantity (e.g. the absolute value of a scalar acceleration along a curve, like when you are driving a car and keep an eye on the odometer), but without further context it has no rigorous meaning.
| {
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Rayleigh-Taylor instability with negative Atwood number? I was reading a paper entitled "The Rayleigh—Taylor instability in astrophysical fluids" by Allen & Hughes (1984) that indicates the instability can occur for $ \rho_{01} < \rho_{02} $ which would indicate a negative Atwood number. But how is this possible? Does not the density gradient have to be opposite the direction of the effective gravity? Must not the Atwood number be necessarily positive for a Rayleigh-Taylor instability?
| Your intuition is correct; there's no such thing as a Rayleigh-Taylor instability with a negative Atwood number. That would imply that the density of the upper fluid, $\rho_{01}$, is less than the density of the lower fluid, $\rho_{02}$, which is clearly a stable situation with respect to the R-T instability.
So how did $\rho_{01} < \rho_{02}$ appear in the Allen and Hughes paper?
I'm pretty sure it was just a typo. I read through the paper and the only place I saw anything that looked like a negative Atwood number was in section 4.2.2, where there's a sentence: In conclusion, it may be seen that the growth of R—T instabilities saturates for large
accelerations, except in the limit $\rho_{01} \ll \rho_{02}$ where the growth remains of the usual form
$$
\omega = (gk)^{1/2}
$$
But this sentence refers to an earlier paragraph in the same section that says, "Again, for $\eta$ ~ 1, compressibility has little effect and
$\omega^2 \approx gk$." Since in the authors' notation, the Atwood number $\eta$ is defined as
$$
\eta = \frac{\rho_{01} - \rho_{02}}{\rho_{01} + \rho_{02}}
$$
it is obvious that $\eta$ ~ 1 implies $\rho_{01} \gg \rho_{02}$ rather than $\rho_{01} \ll \rho_{02}$.
| {
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Law of vector addition In one (Why is heat a scalar quantity?) of the questions I answered, I came across this:-
https://physics.stackexchange.com/a/404287/181020
or
A vector quantity should obey the law of vector addition.
But I don't think it is true. Consider a y junction circuit containing 3 identical wires. If i current flows through each of the 2 arms, the current through the other wire is 2i. But the current density vector of each arm doesn't add vectorially to give the current density in the third.
I want to know whether vectors should obey the law of vector addition or not.
Thanks in advance.
| For a wire, regardless of how you orient it, there's forward and backwards only. So, for any multi wire junction, you ARE adding vectorally (just in one dimension.)
| {
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What is the difference between electrons and holes in silicon? Electrons and holes behave differently in a silicon semiconductor (e.g. mobility of holes is one order of magnitude smaller than that of electrons, the collection time of holes at the same electric field is larger than for electrons... ). I was wondering, if holes are simply "a lack of electrons", they should behave in a mirrored way as electrons (if the latter move from $V_a$ to $V_b$ in a given time, the corresponding holes created when these electrons move should move in the opposite direction at the same speed). My question is: what is the origin of a different behavior between electrons and holes?
| Check that electrons move in the conduction band, whereas holes "move" in the valence band. They have different energies, but they are also at a different border.
So the difference arises that the mass of electrons depends on the second derivative of the energy with respect to $k$. This derivative isn't equal in both bands.
Electrons are mostly near the minimum energy of the CB, called $E_C$. Their effective mass is possitive.
$$m_{e}^*|_{E_C}>0$$
Holes are mostly in the upper limit of the VB, at energies close to $E_V$. That's the maximum energy of the band, which leads to negative effective mass of electrons. If electrons have negative mass, holes have possitive mass.
$$m_{e}^*|_{E_V}<0 \Rightarrow \ \ m^*_h=-m_{e}^*|_{E_V}>0$$
So basically, holes are true "mirror" particles: opposite mass, quasimomentum, and velocities. But check that their mass is opposite only when compared at the same place, but they are at different places.
| {
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Hawking radiation calculation breaking apart with mass density $M_{planck}$? Hawking radiations are predicted from semiclassical quantum field theory, and it is sometimes said that calculations break apart at mass density $M_{planck}$.
The question is, is this true only for some Hawking radiation calculations and can be overcome by calculating differently? Or is this fundamental, showing that Hawking radiation calculations cannot be properly done with semiclassical quantum field theory?
| Sadly it cannot be overcome in general. The Planck mass marks the scale at which our current theories make sense on their own, that meaning that the techniques used in QFT in curved space-times don't hold anymore. Specially when it has to do with gravity. Non-linear effects (self-interactions) and back-reaction become relevant, so that treating the problem as having some background plus perturbations on top, is no longer useful and to my knowledge, no one knows how to implement a proper quantization of gravity in those scenario.
It might also help to know the relation between the Planck mass $M_{Pl}$ and the gravitational constant $G$.
$$M_{Pl}=\sqrt{\frac{\hslash c}{8\pi G}}$$
Since the Einstein-Hilbert action is
$$ S_{EH} = \frac{M_{Pl}^2}{2}\int d^4x \sqrt{-g} R $$
and having in mind that $M_{Pl}\approx 10^{18}$ GeV/$c^2$ you will have some trouble employing perturbation theory.
| {
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Entanglement of formation of the mixture of maximally entangled states Suppose we have two spin-$S$ systems. Let $\left| \psi_{a,b} \right\rangle = \frac{1}{\sqrt{2}}(\left| a,b \right\rangle+\left| b,a \right\rangle)$ be the maximally entangled state. ($a\neq b$ and $-S\leq a,b \leq S$.)
What is the Entanglement of formation for $\rho=\frac{1}{C} \sum_{a\neq b} \left| \psi_{a,b} \right\rangle \left\langle \psi_{a,b} \right|$? $C$ is the normalizing constant.
A trivial upper bound is 1. But can we give a nontrivial upper bound or even calculate it explicitly?
| Since you say that any other entanglement measure is fine, let's compute the negativity. Let me denote by $\rho_{ab}:=|\psi_{a,b}\rangle\langle\psi_{a,b}|$.
With $^{T_A}$ the partial transpose, we have
$$
\rho_{ab}^{T_A} = \frac12 \Big[
|a,a\rangle\langle b,b|+|b,b\rangle\langle a,a|+
|a,b\rangle\langle a,b|+|b,a\rangle\langle b,a|\Big]\ .
$$
Thus (denoting by $D:=2S+1$ the number of basis states),
$$
\rho^{T_A} = \frac{2}{D(D-1)}\sum_{a>b} \rho_{ab}^{T_A}
$$
is block-diagonal with two blocks: $\rho^{T_A}_{ab,a'b'}$ for $a\ne b$, $a'\ne b'$ is diagonal with entries $\tfrac{1}{D(D-1)}$ (i.e., $D(D-1)$ entries), and $\rho^{T_A}_{aa,a'a'}$ (a $D\times D$ matrix) equals $\tfrac{1}{D(D-1)}$ everywhere except on the diagonal (which is zero). Since the latter equals
$$
\tfrac{D}{D(D-1)}|+\rangle\langle +|-\tfrac{1}{D(D-1)}1\!\!1\ ,
$$
with $|+\rangle = (\sum |a\rangle)/\sqrt{D}$, it has eigenvalues $-\tfrac{1}{D(D-1)}$ with multiplicity $D-1$ and $\tfrac{1}{D}$ with multiplicity $1$, respectively.
The sum of the absolute value of the eigenvalues of $\rho^{T_A}$ is thus
$$
\|\rho^{T_A}\|_1={D(D-1)}\frac{1}{D(D-1)}+(D-1)\frac{1}{D(D-1)}+\frac{1}{D} =
1+\frac{2}{D}\ .
$$
The negativity is thus
$$
\mathcal N(\rho) = \frac{\|\rho^{T_A}\|_1-1}{2} = \frac{1}{D}
$$
and the log-negativity
$$
E_N(\rho) = \log(\|\rho^{T_A}\|_1) = \log(1+2/D)\ .
$$
| {
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What is the velocity of a photon through space-time? What is the 4-velocity of a photon? What is the velocity of a photon through space-time? What is the 4-velocity of a photon?
| The problem with the four velocity for anything travelling at the speed of light (i.e. any massless particle) is that we define the four velocity using the proper time $\tau$. If we choose some coordinates $(t,x,y,z)$ then the four velocity in our coordinates is:
$$ \mathbf U = \left(\frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau}\right) $$
The problem is that the proper time $\tau$ is the elapsed time in the rest frame of the moving object, and photons don't have a rest frame so the proper time is not defined. That means the four velocity is not defined either.
There are some workarounds. The four velocity is the tangent vector to a world line, and null or indeed spacelike world lines still have tangent vectors. But we'd have to write the four velocity using some affine parameter not the proper time and this isn't terribly useful.
Alternatively you could consider the behaviour as the speed tends towards $c$. You find that the norm of the four velocity stays equal to $c$ but the components of the four vector tend towards infinity.
| {
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Are graphene and coal are the same? I recently read about the atomic structure of graphene, which is carbon arrange in hexagon shape but only 1 atomic thick(2d). And then I remember that coal is also made of carbon arrange in hexagon shape but it has more than 1 atom thick(3d). So is coal is made from multiple stacks of graphene?
|
As this decomposed peat is buried deeper and deeper, the pressure and temperature to which it is exposed increases causing chemical reactions to occur which reduce the lignin (remove oxygens via expulsion of carbon dioxide and water). In this process, carbon-carbon bonds form between the aromatic rings producing the hard, black carbon-rich material we call coal. Under extreme conditions of pressure and temperature, the end product or most reduced form is graphite. Graphite is a form of carbon in which the carbon layers are arranged into layers of fused aromatic rings. The process of converting organic plant material into coal is called coalification.
See the images in the section "Coal Reserves" here.
| {
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Kirchoff's radiation law So, I have some problems understanding Kirchoff's Radiation law.
My textbook, Transport Processes and Separation Process Principles, by Geankoplis, states that at the same temperature T1 the emissivity and absorptivity of a surface is equal, which holds for any black or non black solid surface.
In a problem from my professor it is given that :
The sun radiate a flat surface with 1000 W/m2. The absorptivity of the plate is 0.9 and the emissivity is 0.1. The air temperature is 20 C and the heat transfer coefficient is 15 W/Km2. Calculate the surface temperature at equilibrium if the bottom is isolated.
My question is: how is it possible that the emissivity and absorptivity in this case is not equal, which contradicts Kirchoff's law?
| To add some detail to Pieter's answer:
The absorptivity quoted is an average over all the wavelengths of light incident on the body from the Sun. This light may have an average wavelength of $\lambda \approx 0.5 \mu \text{m}$ or so which presumably has an absorbtivity of 0.9 in your case.
The emissivity is also averaged, but is given as an average relative to the blackbody radiation of the plate itself. Since its not glowing hot we conclude this average wavelength is in the infrared and according to your data the emissivity for that case must be 0.1.
Kirchoff's law only states that the two equal each other at a given wavelength. The data we are comparing are at two different wavelengths (visible and IR).
| {
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Change of wavefunction due to relativistic speed Imagine a spacecraft which is moving at a speed comparable to the speed of light relative to a reference frame with a hydrogen atom at it's origin. How would the probability distribution function of an electron in 1s orbit look relative to an observer inside the spacecraft?
| Copy/pasted from https://physics.stackexchange.com/questions/591368/galilean-invariance-of-schrödinger-equation as a reaction on @MikeStone's answer. It is too long for a comment.
The Schrödinger equation is not Galilei covariant, even in the absence of electromagnetic fields. There is a symmetry group, the Schrödinger group, that describes the symmetry of the Schrödinger equation. One reason for the Galilei non covariance is that the rest energy of matter is not included. For example, the Schrödinger equation for the hydrogen atom does not include the proton and electron rest energy. These can be included but in a manuscript I argue that this is not sufficient. In the same manuscript I argue that its elements are not coordinate transformations as they depend on mass. Indeed the factor $$e^{i(m\mathbf{v\cdot r}-m\mathbf v^2t/2)/\hbar}$$ depends on mass and it does not follow from application of a Galilei transformation to the Schrödinger equation. It has to be postulated as part of an element of the Schrödinger group. For a review of the Schrödinger group see ref 2 of my manuscript: H. Brown and P. R. Holland, Am. J. Phys. 67 (1999) 204 (behind a paywall).
| {
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Why is $ \frac{\vec{r}}{r^3} = \frac{1}{r^2} $? I know it's surely a beginner's question but I don't see why you can write
\begin{align}
\frac{\vec{r}}{r^3} = \frac{1}{r^2}\cdot \frac{|\vec{r}|}{r}
\end{align}
Could someone explain it please? It would help understand quite a few things ...
| Answer: Equality: $\frac{\vec r}{r^3}= \frac{1}{r^2}$ is not true
Explanation: A vector quantity (having both magnitude & direction) & a scalar quantity (having only magnitude) can never be equated.
Now, $\frac{\vec r}{r^3}$ is a vector quantity while $\frac{1}{r^2}$ is a scalar quantity hence
$$\therefore \frac{\vec r}{r^3}\ne \frac{1}{r^2}$$
Similarly $|\vec r|=r$ is the magnitude of vector quantity $\vec r$ so $\frac{1}{r^2}\frac{|\vec r|}{r}$ is a scalar quantity
$$\therefore \frac{\vec r}{r^3}\ne \frac{1}{r^2}\frac{|\vec r|}{r}$$
However your equality seems to probably be similar to
$$\frac{|\vec r|}{r^3}= \frac{1}{r^2}$$
| {
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Analysing decoupling channels in Hubbard-Stratonovich transformation I have an action defined in terms of fermionic fields $c$ and $d$ that looks like
$$S = - \bar{d}(t)\bar{c}(t') V(t,t') d(t')c(t)$$
where $V$ is an interaction matrix.
Then performing Hubbard-Stratonovich transformation via the exchange channel, say, (see Altland and Simons' Condensed Matter Field Theory (2nd ed.), sec. 6.2.) would introduce an auxiliary (exchange) field defined as
$$\phi_1=\langle \bar{d}(t) d(t') \rangle.$$
Alternately we may choose to let the auxiliary field be
$$\phi_2=\langle \bar{c}(t') c(t) \rangle.$$
My question is, how do we choose which $\phi$ to use? Are there any existing physical systems that would choose one over the other?
If we choose to use both $\phi_1$ and $\phi_2$, this would be like decoupling via the same channel twice. Hence would we need a multiplicative factor of 1/2 somewhere? (Again, are there any physical systems that require this procedure?)
(This is a follow-up question on my other post: Hubbard-Stratonovich transformation and decoupling channels)
| It seems that there is no reason to discreminate the $c$ and $d$ fields when just the interaction term is given.
I think a more good notation would be helpful to you.
Instead of treating $c$ and $d$ fields separately, let us introduce a two-component spinor $\psi=(c,d)^T$. Then, the interaction you give as an example is written as
$$
S=-V \psi_1^\dagger \psi_2^\dagger \psi_2 \psi_1 = -\frac{V}{2}\sum_{i,j} \psi_i^\dagger \psi_j^\dagger \psi_j \psi_i \approx -\frac{V}{2}(\psi^\dagger\psi)(\psi^\dagger\psi)
$$
where in the last eqaulity, I ignored some unimportant terms involving just two fermionic operators.
Introducing the Hubbard-Stratonovich bosonic fields $\phi$ which will be identified with $\langle\psi^\dagger\psi\rangle $ at the level of the saddle point approximation, we obtain
$$
S=\frac{1}{2V}\phi^2+\phi\psi^\dagger\psi
$$
Note that no distinction has been made between the $c$ and $d$ field up to this point.
Now, if we add the single-particle Hamiltonian of $c$ and $d$, for example, $$H_0=
\begin{pmatrix}
\varepsilon_c & 0 \\
0 & \varepsilon_d
\end{pmatrix},
$$
the distinction between $c$ and $d$ fields appear:
$$
\langle c^\dagger c \rangle = \frac{1}{\varepsilon_c+\phi} \\
\langle d^\dagger d \rangle = \frac{1}{\varepsilon_d+\phi}
$$
with $\phi=-V(\langle c^\dagger c \rangle+\langle d^\dagger d \rangle)$.
When $\varepsilon_c \neq \varepsilon_d$, $\langle c^\dagger c \rangle \neq \langle d^\dagger d \rangle$ is expected.
| {
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Method of Images and Green functions in Quantum Field Theory Traditional Laplace equation solutions techniques for a system of conductors rely on the fact that the Green function for solving Laplace's equation represents the potential due to an image charge distribution ([1] Sections 1.10 and 2.1-2.6, for example). I would like to know whether the connection between the method of images and Green functions holds up similarly in the context of QFT.
For example, for the Schrödinger equation,
$$\left[\hat{H} - i\frac{d}{dt}\right]\phi(x,t)=0$$
we suppose there is a Green function $G(x,t,x',t')$ that satisfies
$$\left[\hat{H} - i\frac{d}{dt}\right]G(x,t,x',t')=-i\delta(x-x')\delta(t-t')$$
In [2] (and many more), this Green function is identified as the propagator that evolves the quantum state in spacetime from $\phi(x,t)\rightarrow \phi(x', t')$. But this diverges from the E&M technique since $G$ does not directly recover the solution for Schrödinger's equation (but is instead used to propagate an initial state).
My question is, is there an interpretation that can explain Green functions and propagators in QFT in terms of method of images?
[1] Jackson, J.D. Classical Electrodynamics
[2] Lancaster and Blundel, Quantum field theory for the gifted amateur
| Method of images is based on the uniqueness theorem: the solution to a differential equation with specified boundary conditions is unique (up to the integration constants, which for PDEs may be functions). That is, if you guess a solution that satisfies both the equation and the boundary conditions, then this is the correct solution.
Method of images is a method for guessing solutions in certain classes of problems, where (e.g., due to the symmetry of the problem) guessing turns out to be easier than solving the differential equation with straightforward mathematical techniques.
Thus, the method of images applies equally well to quantum mechanical problem, but the class of problems is very limited.
Note that the method of images is also used for the diffusion equation with either reflecting or absorbing boundary conditions.
| {
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Inertia on a rotating disc? If I toss a ball upwards in a train moving with uniform velocity, the ball will land right back in my hand. This is because the ball has inertia and it continues to move forward at the speed of train even after leaving my hand.
Now consider I'm standing on the outer edge of a rotating disc (merry-go-round). If I toss a ball upwards, it doesn't fall back in my hand. Why? Doesn't it have a rotational inertia (is that even a term?) to continue rotating even after I let go of it?
Is the ball going to land on a new location on the disc? Or is it going to fall away from the disc? At least the ball should have inertia of tangential velocity at which I tossed the ball upwards, right? So the ball should fall away from the disc? Can someone describe what happens in this situation?
| The first thing to note is that on the train you and the ball are moving with constant velocity whereas on the disc you and the ball are undergoing a (centripetal) acceleration.
The vertical motion of the ball is the same in both cases as the only vertical force acting is the same in both cases - the gravitational attraction between the ball and the Earth.
When the ball is released if no other force acts on it in the horizontal plane then due to the ball’s inertia it will continue on with the horizontal velocity it had at the instant it was released by your hand.
The direction of motion of the ball will be at a tangent to the circle along which ball was travelling before it was released so the ball will move away from the disc.
In the case of the train that constant horizontal velocity is the same as your velocity and so the ball will return back to you.
On the disc you will still be accelerating in the horizontal plane whereas the ball will move with constant horizontal velocity, so in the horizontal plane the motion of the ball and you will differ.
The idea of a rotational inertia is inappropriate for the ball when you are on the disc as there is nothing trying to make the ball rotate after the ball has been released.
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Why can a wave be expressed with a sine function? I see many expressions which express waves with the sine function like $y=\sin(kx-\omega t)$.
Waves really look similar to the shapes of a sine or cosine function, but does this guarantee that expressions that show wave-like movement are sine or cosine functions or is this just an approximation?
| This is a slightly different interpretation of the question...
Waves really look similar to the shapes of a sine or cosine function, but does this guarantee that expressions that show wave-like movement are sine or cosine functions or is this just an approximation?
These aren't always approximations. A lot of waves actually follow a sine graph, so we can prove that sinusoidal motion is a real thing in nature.
The easiest example to prove this is a pendulum in simple harmonic motion:$\hspace{175px}$.
You can easily find a derivation of the corresponding equation $x=A\sin(\omega t)$ online. But the important thing is that we resolve the gravitational force into the components, using trigonometry, to get $F=mg\times \sin(\omega t)$. This is directly proportional to negative displacement, hence displacement in SHM truly shows sinusoidal motion, no approximation needed.
Several waveforms, including those for standing waves, are developed through the manipulation of this formula, but for some things, we need to approximate several sine waves together using Fourier transforms; other answers explain that idea.
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Can air make shadows? I have read about schlieren photography, which uses the ability of non-uniform air to create shadows. Is it really possible that air makes shadows?
| In addition to V.Joe's excellent answer about refraction, I want to add that air does directly block (absorb, not refract) some infrared and ultra-violet light.
If your eyes could see light in the infrared and UV spectrums, air would appear partially opaque in those wavelengths. The surface of the earth is in the infrared and UV 'shadow' of the atmosphere.
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Why is Ampère's law violated if there are no fringe fields? What's wrong with the following diagram?
Image source: Page 183, NCERT Physics Textbook for Class XII Part I
The reason stated in my textbook is as follows:
Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric field lines.
I don't understand how is Ampere's law violated when fringe fields are absent.
Can anyone please explain how Ampere's law is violated?
| Consider the two paths $ABCDA$ and $EFGHE$.
Path $AB$ contributes a positive value to the $\vec B\cdot d\vec l$ integral but the other parts of the loop contribute nothing, so overall there is a finite value for the $\vec B\cdot d\vec l$ integral but no enclosed current which violates Ampere's law.
Again path $EF$ contributes a positive value to the $\vec B\cdot d\vec l$ integral but now all the other parts of the loop contribute a negative value to the $\vec B\cdot d\vec l$ integral with the result that the total integral is zero as is the enclosed current.
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Do quantum fluctuations in the inflaton field lead to fluctuations in the potential energy density? During inflation there are quantum fluctuations in the inflaton field. Do quantum fluctuations in the inflaton field lead to fluctuations in the energy density?
|
Quantum fluctuations in the microscopic inflationary region, magnified to cosmic size, become the seeds for the growth of structure in the Universe (see galaxy formation and evolution and structure formation).
As the inflaton fluctuations are modeled to be the seeds of the cosmic microwave background radiation and the locations of the distribution of matter in the present universe, the end result is fluctuations in energy density. Thus I would answer with a yes for the energy density. Now potential in this scenario is not well defined.
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Help with D. Tong example on Noether in QFT In this lectures, example 1.3.2 on page 14 concludes that the Noether current is
But how can the current be a two index object when it is defined in eq. (1.38), which is
as a one index object? If I apply the formula I obtain something of the form $j^\mu$. Can someone make the calculations explicitly?
| $T^{\mu \nu}$ is conserved because the Lagrangian is invariant under space-time translations, meaning that it is invariant under space translations and time translations. Naturally, all the quantities conserved should be inside $T^{\mu \nu}$.
For example, $Q^0 = \int d^3x T^{00}$ is the first one, and $Q^0$ is the Energy of the system (conserved because of time translation symmetry).
Another quantity is $P^i = \int d^3x T^{i0}$ which is the conserved momentum due to space translation symmetry.
The current is expressed in the form of a tensor to put the conservation laws (plural) in the covariant way $\partial_{\mu}T^{\mu \nu} = 0$.
Writing it explicitly, we have $\partial_{0}T^{0 \nu} + \partial_{i}T^{i \nu} = 0$ which is a vector equation. The 0th component would be the conservation of energy, and the jth component, the conservation of each component of the momentum.
Note that if the conserved current had only one index, you could only express one equation (which would mean one conserved quantity). The conserved current is a tensor because it describes multiple conserved quantities (or conservation equations) in a single object.
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Does density of light power (w/m²) increase after passing a convex lens ? Does density of light power (w/m²) increase after passing a convex lens ? and if so i need the equation that represent that increase
| In general, power density of a light beam will be increasing or decreasing depending on whether the beam converges or diverges.
A convex lens is referred to as a converging lens because it converges parallel rays. But, if the rays entering a convex lens are not parallel, the light exiting the lens could either converge or diverge, depending on the position of the source relative to the focal point as well as the height of the source and the lens.
So, you can determine how much power density increases or decreases after the lens by figuring the angle of the exiting beam using ray optics and a little geometry.
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Aerodynamics equation I still need to find an equation for the following restated problem: Suppose a full-size glider passes 10 feet over my head at high speed. No doubt I will feel the downward air pressure caused by the angle of attack of the undersides of the wings. But if the same glider flies over me at a distance of one mile, I will not feel any effect at all. So obviously there is a diminishing strength of the downward air pressure that eventually lessens to a point where I cannot physically detect it. Therefore, is there an equation that, with all relevant criteria being inputted, such as air temp, air density, wing speed, wing characteristics, wing angle of attack, etc., can allow me to at least approximate how far I must be below a passing wing to be unable to detect its downward pressure wave?
| I know of no explicit equation of the sort you seek, but here is a useful estimate which may be of help.
A pilot who is flying her airplane in for a landing on a runway will notice that the proximity of the ground beneath the plane begins to affect the flow of air over the wings on her plane when the plane is within one or two wingspans above the runway. This is called "ground effect" (see wikipedia) and it corresponds to the height at which a "bubble" of air pressure gets entrained between the runway and the underside of the wing, which means that if you were standing under the plane as it flew over you at that height, you would definitely feel a sudden whoosh of air in your face as it passed over you.
There are a number of skilled aerodynamicists on the aviation stack exchange who may be able to furnish equations for ground effect, so you might want to pull this post and move it over there.
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Connecting a charged capacitor to an uncharged capacitor
I was attending a lecture about capacitors and something confused me.
If I charge a capacitor using a DC supply, the capacitor will gain charge $Q_0$.
Now, if I discharged it along an uncharged capacitor in this arrangement, according to the lecture notes, the capacitors share the total charge $Q_0$.
Now, I had a question.
Aren't there electrons on the uncharged capacitor, such that they flow between the two capacitors to cause equal p.d. on both capacitors hence the total charge in this circuit greater than $Q_0$?
| When we charge a capacitor, it gains charge q on one of the plates and loses charge q from the other plate, i.e., its total charge remains zero.
Capacitors differ, in that sense, from other objects, like our bodies or spheres and rods used in various electrostatic devices and experiments, which actually gain a net charge, when they are charged.
So the charged capacitor on the top of your diagram, initially has charge $+Q_0$ on the left plate and charge $-Q_0$ on the right plate and its total charge is zero.
When the switch closes, $Q_0/2$ worth of electrons are flowing from the right plate of the top capacitor to the right plate of the bottom capacitor and $Q_0/2$ worth of electrons are flowing from the left plate of the bottom capacitor to the left plate of the top capacitor. The resulting total charge remains zero.
| {
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Why are there only four fundamental interactions of nature? Is there an answer to the question why there are only four fundamental interactions of nature?
| The answer "because we do not need more" by @rubenvb is fine.
Studying physics, you must realize that physics is not answering fundamental "why" questions. Physics uses mathematical tools to model measurements and these models have to fit new data, i.e. be predictive. As long as the models are not falsified, they are considered valid and useful. Once falsified, modifications or even drastic new models are sought. A prime example, quantum mechanics, when classical mechanics was invalidated: black body radiation, photoelectric effect and atomic spectra falsified efforts of classical modelling.
Physics using the appropriate models show "how" one goes from data to predictions for new experimental data. Looking for "why" in the models, one goes up or down the mathematics and arrives at the answer "because that is what has been measured"
| {
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How to think of the components of 4-momentum? Suppose I have some photon with a 4-momentum $p_\mu$, with $\mu = t,r,\theta, \phi$.
There is also a point with coordinates $(x,y,z)$. The 3-vector $x^i$ describes the vector between the origin of the coordinate system and the point.
The photon is at the origin of this coordinate system, I want to find the angle that the photon makes with the vector $x^i$.
Now, if we were just dealing with normal vectors in normal 3-space, I could just take a dot product to find the angle.
How would I do it in this case? Specifically, I am having trouble understanding the components of $p_{\mu}$; e.g. can I think of $p_{\phi}$ as a $\phi$ angle?
| The angle between photon momentum and a vector is only defined in three dimensions. It is given by $\theta = \arccos (\vec p \cdot \vec x)/ (|\vec p||\vec x|))$. It has different values when observed in different reference frames. In other words, it is not a relativistic invariant.
The inproduct of four momentum with some coordinate four-vector , $\varphi = \omega t - \vec p \cdot \vec x$, however, is invariant. It is the phase of the plane wave describing the photon.
| {
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Free surface in fluid dynamics In Computational Fluid Dynamics, if the fluid does not cross the free surface, the relation between fluid velocity at free surface and free surface velocity is given by$$
{\vec{V}}_{\text{surface}} {\cdot} \vec{n} = \vec{v}_{\text{fluid}} {\cdot} \vec{n}
\,,$$where:
*
*$\vec{V}_{\text{surface}}$ is free surface velocity;
*$\vec{v}_{\text{fluid}}$ is the fluid velocity; and
*$\vec{n}$ is the normal to the surface.
How does such a relation occur?
| I think it should be $$\mathbf{V}_{surface}=(\mathbf{v}_{fluid}\cdot\mathbf{n})\mathbf{n}$$ Above equation says that the shape of the free surface of a fluid can change only due to fluid motion normal to the free surface. Fluid motion "in the surface", i.e. tangential to the surface, does not cause a change in the shape of the free surface.
This is so because a free surface is composed of fluid particles, and does not have an independent existence. Think of a flat free surface of water at rest; if the fluid particles lying in the surface move only in the plane of the free surface, then the shape of the surface does not change and consequently the free surface cannot be said to have moved.
Forming the dot-product of the previous expression with $\mathbf{n}$ gives the particular formula in your question. However the formula in your question gives only partial information about $\mathbf{V}_{surface}$, and must be supplemented by the second equation $\mathbf{V}_{surface}\cdot\mathbf{t}=0$, in which $\mathbf{t}$ is the tangent vector to the surface.
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How to visually understand that dislocation moves on a slip plane? When we look at images of edge or screw dislocation, it seems as if the direction in which that dislocation will move is already fixed by the 'way the dislocation is present'. For example, take any image of edge dislocation, and say the top half is moving with respect to the bottom.
However:
Dislocations will move in the direction of slip direction and on the slip plane. That is the dislocation motion direction is dictated by the slip system of that crystal (FCC, BCC or HCP)
Hence, only in the case when the plane between the top and bottom is also a slip plane, things work out. But what happens when the plane between the top and the bottom set of atoms is not a slip plane for that particular crystal?
| There are cases, when two mobile dislocations (dislocations with Burgers vector pointing into a slip plane, therefore, can move with gliding) contact-interact and form a third dislocation. If this two dislocations split up into two-two Shockley partial dislocations, and one Shockley partial dislocation contact-interact with the other dislocation's Shockley partial dislocation, they can form a dislocation with Burger's vector not matching any slip plane. This dislocation cannot glide, therefore, became immobile, i.e. fixed. This is called a Lomer–Cottrell junction.
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What happens while breaking a Bar magnet? If we have a bar magnet and we break it as shown below
Then as per the image we get two bar magnets But does this happen instantaneously or there is some time lag between this process.
Or simply, Can magnetic monopoles exist even for a short time period?
| As long as the two parts are together, and the boundary is not a physical separation, you may consider them as one magnet or as two. There is no difference as the two poles you inserted cancel out and nothing is created. Of course when you pull the two parts things happen but the poles already existed in this sense.
Magnetic monopoles are incompatible with electromagnetic field theory in its present form.
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Which program to analyse and graph laboratory measurements? I'm interested in which program would you recommend for drawing graphs with x and y errors. Also to be able to analyse data and then graph it. My ideal graph would look like this:
Until now I was drawing my graph reports in Excel, but I feel like Excel is made more for an accountants than scientists.
I would really like to work more in scientific based programs that could come in handy in the future (graduate and postgraduate work, research).
Any recommendation?
| Python is certainly a major language for scientific data analysis today. The three key words are: Numpy, Scipy and Matplotlib.
*
*NumPy: it's the basic numerical analysis package in Python. It allows for array manipulation, matrix calculus, Fourier transforms, statistics, ...
*SciPy: it's the extended version of numpy with more advanced built-in procedures: image and signal processing, optimization, graphs routines, more statsitic and so on. Importing SciPy makes essentially redundant importing NumPy.
*Matplotlib: here we came to the standard for visualizing data in Python: you can do almost everything in 2D or 3D with this library. Take a look at this sample plots!
There is also a useful IDE for Python called PyZo, in case you want to have the shell and the editor in a single window. It makes the user experience and the debug a little bit easier.
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Effect of greater coils on induced current in a solenoid Simple case:-
A bar magnet is inserted into a solenoid with 'n' turns at velocity 'v'. The current is measured using an ammeter that is part of the complete circuit of the solenoid.
My question is that if I increase the number of turns 'n' then will this lead to a greater induced pd and in turn, induced current detected by the ammeter.
Please could you provide a brief explanation as well because its not intuitive whether it will increase or decrease
| According to Faraday's law of electromagnetism, emf induced is given by
$\epsilon=\frac{d\phi}{dt}$
But $\phi=B.A$
Here B is the electric field and A is the effective area though which the electric flux changes.
$\therefore$effective area =$n.a$
Here a is the area of each turn and n is the number of turns of the coil.
$\therefore \epsilon = \frac{d(B.n.a)}{dt}$
$\implies \epsilon =n.a.\frac{dB}{dt}$
Therefore emf induced is directly proportional to the number of turns in the coil.
By ohm's law, the current in the circuit is proportional to emf across the conductor. Therefore the current, emf both increases with increase in number of turns.
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Why do punctured balloons fly around chaotically? If an inflated balloon is punctured, it can fly around wildly like in this cartoon @18:07.
Why is this motion so chaotic as opposed to being like a straight line or parabola as with rockets? Is there a mathematical framework for understanding why the balloon performs so much twisting and turning?
| The balloon moves because of conservation of momentum.
When the gas molecules diffuse out to conserve momentum the balloon moves. But inside the balloon the gas molecules possess random motion called Brownian Motion . Suppose at time t those molecules escape in some direction. Thus conserving momentum in that direction balloon moves in opposite direction. But after very small interval dt the direction of molecules escaping out changes due to Brownian motion. Thus there is no fixed direction in which balloon would move. Hence its path is chaotic as you say.
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What is the cause of wave impedance? As in electrical impedance,
Causes:
Resistance - collision of electrons with atoms and other electrons,
Reactance - Capacitive and inductive effects.
Likewise, what offers opposition to a wave traveling in a medium?
| The "opposition" to the wave or wave impedance or impedance of a medium to a wave is caused by characteristics of the medium analogous to the resistance, capacitance and inductance. While resistance is a pretty generic term, applicable to different types of waves, the energy storing characteristics, capacitance and inductance, could be generalized as compliance or stress and inertia or motion.
When a wave is propagated, it energizes the medium and the speed of the propagation is reduced or opposed by the resistance of the medium and by its ability to store energy. So both high capacitance or compliance and high inductance or inertia of the medium act to slow down the wave or, we can say, it takes more time and energy to energize a medium with high capacitance and inductance. This for instance, is reflected in a formula for the wave propagation speed in an ideal transmission line, $V_p=\frac 1 {\sqrt {LC}}$, where both capacitance and inductance contribute symmetrically to oppose or slow down the wave.
The impedance, on the other hand, characterizes the tendency of a medium to oppose the motion component of the wave at a given stress level or, in electrical domain, the tendency to oppose the current or the magnetic field at a given level of voltage or electric field. This is reflected in a formula for the characteristic impedance of an ideal transmission line, $Z_0=\sqrt \frac L C$. Here, capacitance and inductance are not contributing symmetrically: high capacitance encourages the current flow, while high inductance impedes it.
For EM wave in space, the formulas for the propagation speed and impedance, $V=\frac 1 {\sqrt {\mu \epsilon}}$ and $Z=\sqrt \frac {\mu} {\epsilon}$, have similar meaning and underling mechanisms. The propagation is opposed or slowed down by both greater magnetic permeability (inductance) and electric permittivity (capacitance) of the medium. On the other hand, the impedance (to the motion component of the wave) is increased with the magnetic permeability and decreased with its electrical permittivity.
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Non-zero electric field inside a conductor, when applying an large external field I'm probably missing something, or does not understand conductors well enough. But I have a question related to the title of this message.
In many places you read that there can be no electric field inside a conductor. The arguments typically go something in line with, since there is an electric field, charges inside the conductor will rearrange themselves so cancel the field. Very simply stated.
That I don't understand, is that this seams to assume that there always is "enough" charge to redistribute. To clarify my confusion, let's say we have a conducting solid sphere with some charge. If we apply an "large" external static field to this sphere, charges inside it will tend to cancel it out. But, what if the total charge inside it is not enough? The total charge in the sphere can only generate a limited field, but the external one can be arbitrary large. What if the field outside is so large that the potential it generate, from one side of the sphere to the other, is larger than what the internal charge can generate?
As I said, I'm probably missing something essential, but can someone please point out the misstake in the above argument?
| If the field gets stronger electrons near the surface of the conductor will feel a stronger residual field, until some fly off in a spark that will hit you voltage source and reduce the voltage it produces.
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Is it possible for a lightning strike to hit the ground if there are high rise buildings nearby? Say we have pointed conductors connected to the top of the high rise buildings. Will the strikes hit the nearby ground in such a case?
| Yes.
A lightning conductor on top of a high-rise structure is designed to minimize the possibility of lightning hitting the structure by the virtue of its pointed tip giving rise to very high electric fields (the reason). However this doesn't always ensure that lightning will definitely strike the conductor. This link has quite a few examples of the same.
This is because when cloud-to-ground lightning happens, the lightning goes the way of strong electric fields. Even though the pointed tips of the lightning conductors set up huge electric fields, the path in which the lightning conductor is in the way isn't necessarily the most preferred path for the strike to happen. So the nearby places can get hit too, as demonstrated in these images.
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Does photon absorption annihilate the associated EM wave instantly? My Understanding
A single photon has an associated electromagnetic wave. The wave is spread out in space, but the photon is considered a point particle. If the photon is absorbed, the entire wave disappears. Photon absorption is instantaneous, so the wave disappears instantly. In other words, the wave can no longer be detected anywhere in the universe; despite that the interaction happened at a single point.
My Question
Is my understanding correct, and if not, what am I missing?
| Light, a classical concept, emerges from zillions of photons in a continuous mathematically demonstrable way, but photons, a quantum mechanical entity, are not light.
At the quantum level photons are mathematically connected to the frequency of light with a complex wavefunction, a solution of a quantized form of Maxwell's equation, that is why it can be expressed with the average E and B fields it may build up when assembled in bulk.
and its $Ψ^*Ψ$ , as all quantum mechanical solutions, represent the probability of finding the photon at an (x,y,z,t). So the individual photon is not spread out all over space time, it has a probability of existing in space time points given by the solution.
You are missing that the wave nature of a single photon is a probability amplitude, not an energy amplitude. Once and if you learn quantum electrodynamics here is a link on how the classical emerges from the quantum probabilistic level.
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Density function in phase space What does density function in phase space physically mean? How does it indicate, the more familiar density that we are accustomed to ( an analogy may be), in phase space?
| It is a probability density, not a density of something like matter or energy.
The probability density $f$ answers the question "how likely is it (what is the probability that) the microstate $\omega$ in one of the points in some set $A$":
$$P(\omega\in A) = \int_A f(p,q)dpdq$$
The most other famous example of probability density in physics is the squared modulus of the wave function $\psi$, $|\psi(x)|^2$ being the probability density of finding the particle at point $x$. Another example is Brownian motion of a single particle.
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Quantum tunneling versus over-the-barrier ionization I was doing a quantum physics past paper and there was this question about ionizing atoms with laser. My answers made reference to the photons within the laser knocking off or transferring their energy to the electrons and allowing them to escape if this energy exceeds their biding energy. I then checked on the markscheme and all it mentioned was the laser modifying the nuclear potential to allow the electron to quantum-mechanically escape. I am confused.
I know quantum tunneling happens when a wave function meets a potential barrier, but I don't understand how the laser provides this barrier. Isn't a laser just a stream of photons basically? Also, the markscheme made reference to OTBI ionization for hydrogen (the second part of the question was specifically about hydrogen).
I did some reading and I get the idea it is the same thing as quantum tunneling, so I am not sure why in the markscheme the two seem like different ideas altogether. Unfortunately, the markscheme is not a full answer but just a string of key words to award marks so that doesn't help.
| The laser doesn’t provide the barrier, on the contrary, the laser helps to overcome the barrier. That means that when you give visible light, you achieve just an excited state of the atoms, and it’s impossible to get them ionized. But these excited electrons can easily escape through quantum tunneling, something that is not possible in classic physics. Using a laser you can give a pure wavelenght of visible light, exactly towards a target.
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"url": "https://physics.stackexchange.com/questions/409915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does plasmon have higher erngy than phonon? In my mind plasmon is movement of electrons and phonon is movement of atoms in an lattice. movement of atoms should have a large energy because atom is larger.
| Like it's said Plasmons describe collective oscillations of charge carriers. One can derive that the energy is given by (with $n$ the charge density, $\varepsilon$ vacuum dielectric constant):
\begin{equation}
E_{p}=\hbar \cdot \omega_{p}=\hbar \sqrt{\frac{ne^{2}}{m\varepsilon} }
\end{equation}
Now you can compare it to the energy of a phonon (with $n_B$ being the Bose statistic):
$$\varepsilon _{n}({\mathbf {k}})=\hbar \cdot \omega ({\mathbf k})\cdot \left(n_B+{\frac {1}{2}}\right),$$
but here you have to distinguish between optical and acoustical modes (see e.g. Kittel "Solid State Physics"). You can basically excite optical phonons (which have an higher energy than the acoustical ones) with inelastic neutron scattering.
To make a long comment short, I would suggest you can put in some numbers, but you can also argue I want to excite with same frequency (same $\omega$). Then still the Phonons have higher energy, because the Bose statistic gives you very large values for small energies. All in all the Phonons have more energy then the Plasmons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410066",
"timestamp": "2023-03-29T00:00:00",
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Which pole will electrons flow towards in a changing magnetic field (generator) In a generator a magnet spins in the middle of a coil of wire and the changing magnetic field causes current to flow, but the current changes direction whenever fields from a new pole of the magnet cross a point on the wire, so will the current move towards the north pole or towards the south pole (rather the part of the wire being touched by the pole's magnetic field).
I know that when you use current to magnetize something the north end of the magnet is the end that was negative, so I would assume that current would move away from the north. Is that correct?
To make things more clear I have added this diagram. Assume the generator is running. Would electrons be taking the blue or the red path?
| The image you presented looks like a DC motor. If thats the case, there would be a DC source connected at the terminals. When current flows through the wire, a temporary magnetic field is formed due to the current flow through the wire. This temporary field repels the magnetic field felt from the magnets and causes the wire to flip over.
What causes the wire to keep flipping is called a commutator (https://en.wikipedia.org/wiki/Commutator_(electric)). This reverses the current flow after the wire has flipped. The reversing of the current at timed intervals causes the wire to keep flipping. If there was no commutator to reverse the current the wire would only flip once where it would now be in the most stable state of the magnetic field. It's the reversing of the current that keeps the motor moving.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Energy conservation on expanding universe Due to the expansion of the universe, the photons emitted by the stars suffer redshift, Its mean that the energy is lowered a little bit. Does this mean that the energy is lost? Does the expansion of the universe violate some conservation principles according to Noether's theorem?
| Let us take the simple case, we see the iron spectrum of a star shifted, and we can use energy conservation to assign a velocity to the star that would assure the Doppler shift of the spectrum.
The expansion of space was deduced because a redshift was measured that can only be interpreted as "every cluster of galaxies is moving away from every other cluster of galaxies ". This led to an image of an "explosion", i.e. the original Big Bang model. In an explosion energy conservation comes considering the whole system, original energy transferred to parts. The complication comes from General Relativity which does not have energy conservation as part of its structure. It is only in flat spaces where Lorenz transformations can be effectively applied , as in the case of a receding star and the spectrum of iron, that one can talk of conservation of energy.
So,imo, qualitatively energy in all its forms comes from the original Big Bang, but one cannot write down general conservation of energy equations, they have to be compatible with general relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why does a mirror reflect visible light but not gamma rays? Visible light (~500 THz) as well as gamma rays (~100 EHz) are electromagnetic radiation but we can reflect visible light using a glass mirror but not gamma rays. Why is that?
| Look at the electromagnetic spectrum:
Visible frequencies have wavelengths of microns, $10^{-6}$ meters.
Gamma rays have a wavelength of $10^{-12}$ meters, picometers.
In physics, there are two mainframes, the classical frame, which includes Maxwell's electrodynamics, Newton's mechanics, and derivative theories, and the quantum mechanical frame which becomes necessary for small distances and high energies, where gammas (photons), electrons, atoms, nucleons, lattices belong.
The classical electromagnetic wave emerges from zillions of superposed photons. Maxwell's equations describe very well the behavior of light beams when scattering or reflecting or generally interacting for macroscopic distances and small energies. Reflection, classically, needs a very flat surface so that the phases of the reflected waves are retained. Depending on the material the classical beams may be absorbed, decohered in reflecting from many point sources, or reflected coherently if the scattering is elastic (mirrors elastically and coherently scatter incoming light).
Gamma rays though force us to go to the micro level, because of the very small wavelength that describes them as a light beam.
One has to look at the details of the surface, and whether a classical smooth surface for classical reflections can be modelled for gammas, and the answer is, no it cannot.
The spacing between atoms in most ordered solids is on the order of a few ångströms (a few tenths of a nanometer).
For micron wavelengths (optical light) the fields built up by atoms with angstrom distances in the lattice appear smooth and can be classically modelled.
Gamma rays considered as a classical light beam, with their picometer wavelengths see mostly empty space between the atoms of the solid.
An alternative analysis, still within the quantum frame, would be considering the photons which make up light, and the Heisenberg uncertainty $ΔpΔx$ in the location of the photon. For the small wavelengths of gamma rays, the photons see mostly empty space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 3,
"answer_id": 0
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What would be the charge distribution of a conducting sphere in front of a positive point charge? What would be the charge distribution of a conducting sphere in front of a positive point charge? I mean if it's a positive charge then it should induce negative charge in the near side and positive on the other side. But as it's conducting then it should distribute the charge all over the sphere. So it should make the sphere nutral. Or something extra-ordinary might happen. Assume the sphere is isolated.
| This electrostatic problem of a point charge $q$ in vacuum at a distance $L$ from the center of an isolated conducting sphere with radius $R$ can be easily solved by the method of images, which introduces a virtual image charge $$q'= -q\frac {R}{L}$$ at a distance $$l_1=\frac {R^2}{L}$$ from the center in the sphere's interior on the connecting line between the center and the outer charge. Using the superposition of the Coulomb potentials (or electric fields) of the charges $q$ and $q'$ one obtains the total potential and electric field outside the sphere and thus also the normal electric field $E_n$ on the surface of the sphere.
From this follows the surface charge distribution on the sphere $$\sigma=\epsilon_0 E_n$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Why acoustic glitches in stars translate into extra oscillatory components in the normal frequencies? Acoustic glitches are locations inside the star where the sound speed changes abruptly compared to the wavelength of the acoustic waves that propagate through. Examples are the ionization zones and also transition between the convective zone and the radiative zone.
Those regions of rapid change in sound speed will leave a signature on the normal frequencies that can be measured on the surface of the star. The signature is such that will add an oscillatory component to the frequencies that would be expected in the case of no glitch.
I have always thought of the glitches in stars like adding a high density dot on an oscillating rope through which sinusoidal waves propagate. However, what results from that analogy is simply an attenuation of the original wave amplitude, but no changes on its frequency.
Could someone point out the mistake on the analogy or explain straight why the glitches result in an extra oscillatory component in the normal frequencies?
Any suggestions / ideas are welcome!
| A structural glitch in the star (i.e., an abrupt or sharp variation in the stellar structure) will affect the coupling of the wave with the star at the location of the structural glitch. That impact in the coupling will translate into a change in the eigenfrequencies which are predicted in the absence of structural glitches.
How much the structural glitch impacts the coupling depends on the amplitude of the wave at the location of the structural glitch. If the wave happens to be a node at that location, there will be no impact in the coupling and the eigenfrequencies will not change. However, an eigenmode has harmonics and even if one of them has a node at the location of the structural glitch, the immediate next harmonic will be slightly shifted and will hit the glitch location with an amplitude slightly different from zero. Now the wave and the structural glitch need to couple and this will translate into the eigenfrequency of such mode changing a bit. As we move over the harmonics, an antinode will hit the glitch location and then the change in the corresponding eigenfrequency will be largest. The situation goes on for all the harmonics and that is the reason to expect an oscillatory signature when studying glitches.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/410964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
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