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Proof that a Hermitian Matrix is not defective? I am taking an introductory course into Quantum Mechanics.
To me to seems pretty simple to prove most properties of Hermitian operators. However, I am stuck at an edge case, proving that if an eigenvalue has multiplicity $n>1$, it will have $n$ linearly independent eigenvectors.
This is equivalent to proving that a Hermitian matrix cannot be defective. Can anyone give me an outline or some pointers for such a proof?
| Here is a highlight of the reasoning line:
*
*A theorem: The whole vector space is the direct sum of the generalized eigenspaces, where each generalized eigenspace is associated with an eigenvalue. The algebraic multiplicity $\mu$ is equal to the dimension of the generalized eigenspace associated. (See https://math.stackexchange.com/questions/2917617/proving-there-are-as-many-generalized-eigenvectors-as-algebraic-multiplicity-eig and http://www.math.byu.edu/~grant/courses/m634/f99/lec9.pdf)
*A generalized eigenspace associated with an eigenvalue $\lambda$ of a matrix $A$ can be defined as $G_{\lambda}\hat{=}\{v|\exists k\geq 0, \; \text{s.t.}\; (\lambda I-A)^k v =0\}$
*The eigenspace associated with $\lambda$ is a subspace of the generalized eigenspace. Thus the geometrical multiplicity $\gamma$ is not larger than the algebraic multiplicity, i.e. $\gamma\leq\mu$. The matrix $A$ being defective, i.e. $\gamma<\mu$, implies the existence of generalized eigenvectors $u\in G_{\lambda}$ such that $(\lambda I-A)u\neq 0$
*If $A=A^{\dagger}$ is hermitian, there does not exist a such $u$ stated above. Assuming there is a $u$ such that $(\lambda I-A)u\neq 0$ and $(\lambda I-A)^2u = 0$, $A$ being hermitian implies $(\lambda I-A)u = 0$, thus self-contradiction.
| {
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How does electrical energy gets converted to sound energy? I have seen that inside headphones there is a magnet with a coil of thin wire around it. There must be longitudinal waves coming out of it that is why we can listen to audio. There must be pressure waves created as sound is nothing but pressure wave. So, my question is how is the pressure actually created there in the first place and how are the variations in pressure created? There is the electric current that flows through the wires and the electric and magnetic fields due to the current and the magnet. So, how do they "exert" pressure on air?
| A permanent magnet has a fixed north/south polarity - in this example, lets say north is facing up and south is facing down. This magnet has a membrane of some kind attached to its north face.
An electromagnet beneath the permanent magnet can switch the direction of its north/south polarities by changing the direction of the electric current running through it.
When the electromagnet's south is pointing up, it pushes away the permanent magnet's south pole and therefore the membrane, creating an area of pressure in the air.
When the electromagnet's north is pointing up, it attracts the permanent magnet's south pole and therefore the membrane, allowing for the next pressure wave to be made.
| {
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Why is amorphous classified solid? Because it does not have a crystal structure, it is hard to find physical similarities with a solid. Why isn't it then another state other than solid?
The physics of amorphous is also quite different from crystal solid.
| Glass is a typical amorphous solid.
Amorphous materials typically show no melting point but do have a Glass Transition Point ($T_g$). Below it, the material behaves like a solid, with a glass-like fracture surface when fractured.
Typical amorphous materials include several types of elastomer (rubber) like natural rubber (NR), with a $T_g$ of around $-80\:\mathrm{Celsius}$. Above that temperature NR really behaves like a super viscous liquid, creating the illusion of solidity. This explains of course why NR above its $T_g$ doesn't behave mechanically like a crystalline substance.
When cooled to below their $T_g$ these materials become brittle like glass (while remaining amorphous), see the famous demonstration of the brittleness of a rubber hose when cooled in liquid nitrogen.
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Why is the Pythagorean Theorem used for error calculation? They say that if $A = X \times Y$, with $X$ statistically independent of $Y$, then
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
I can't understand why that is so geometrically.
If $X$ and $Y$ are interpreted as lengths and $A$ as area, it is pretty easy to understand, geometrically, that
$$\Delta{A} = X\times\Delta{Y} + Y\times\Delta{X} + \Delta{X}\times\Delta{Y}$$
Ignoring the term $\Delta{X}\times\Delta{Y}$ and dividing the both sides by $A$ ($= X \times Y$), that expression becomes
$$\frac{\Delta{A}}{A} = \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y}$$
which is different from
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
which looks like a distance calculation. I just can't see how a distance is related to $\Delta{A}$.
Interpreting $A$ as the area of a rectangle in a $XY$ plane, I do see that $\Delta{X}^2+\Delta{Y}^2$ is the how much the distance between two opposite corners of that rectangle varies with changes $\Delta{X}$ in $X$ and $\Delta{Y}$ in $Y$. But $\Delta{A}$ is how much the area, not that distance, would vary.
| It looks like Pythagoras, but it is only remotely related. The important concept, as presented in SteveB's answer, is that the variables are considered to be independent, i.e. one does not affect the other. In mathematics, independent parameters are said to be orthogonal , and can thus be assigned to separate axes in Cartesian N-space. It just so happens that the root-sum-square error turns out to be the diagonal of the N-cube (or rectangle in 2-D), which matches Pythagoras' trigonometry theorem.
| {
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How far do we need to be removed from the earth to show the curvature with a viewing angle between 42 and 48 degrees? I have seen already a couple of answers but none of them give an exact number of what should be the minimum height where we would be able to record the curvature of the earth
All I could find is minimum of 10km but you need a 60 degree viewing angle to see it... if that is true there is some amateur rocket footage out there with non fish eye lenses that show no curvature at 32km height
do we need to go higher? if so how high exactly?
Here are the video's i'm talking about both go around 120,000 feet high.
A high altitude balloon which uses some lens (most probably fish eye) and gives distort so not conclusive if what we see as a straight horizon is real http://youtube.com/watch?v=tvhFbvY_99o
the other is an amateur rocket launch at 2:13
http://youtu.be/qY7W3EMfrgc?t=133
The lens from a FlipHD camera as far as I can find out is between 42 and 48 degrees so much smaller then the 60 mentioned before
the horizon is straight from the beginning of the launch but also when it reaches the top the horizon appears straight.
| As a former sailor I can assure you that you can see the curvature of the earth from the crow's nest - all it takes is a calm sea.
Consider a light house that stands 50 meters above sea level. If you are in the crow's nest, say 25 meters above sea level), at what distance could you first spot the top of the light house on a calm day?
The problem is geometric - two posts sticking out from the surface of a large ball, and you want the tangent line to the ball that passes through the tips of both posts; start by considering how the central angle most close).
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Sign of Wick rotation Suppose you have the integral
$$i \int^\infty_{-\infty} L_M(t) dt$$
and that $L_M$ contains two poles: when $t>0$ the pole lies above the t-axis and when $t<0$ the poles lies below the t-axis. Therefore you can rotate the contour from the real axis to the contour going from $i\infty$ to $-i\infty$. This path can be parameterized as $z=i\tau$ where $\tau$ is from $\infty$ to $-\infty$:
$$i \int^\infty_{-\infty} L_M(t) dt=i \int L_M(z) dz=
i\int^{-\infty}_{\infty} L_M(i\tau) id\tau=
\int^{\infty}_{-\infty} L_M(i\tau) d\tau
\\\equiv-\int^{\infty}_{-\infty} L_E(\tau) d\tau
$$
However, textbooks write instead:
$$\int^{\infty}_{-\infty} L_M(-i\tau) d\tau
\equiv-\int^{\infty}_{-\infty} L_E(\tau) d\tau $$
so they get the sign wrong (or I got the sign wrong).
| Thanks for your help. I agree this was homework-like - should've posted it somewhere else.
This is an equality
$$\int^\infty_{-\infty}f(x)dx=\int^\infty_{-\infty}f(-x)dx$$
so I guess mathematically it doesn't matter the sign of your substitution.
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What is the electric field exactly on the surface of a conducting sphere? Within a conducting sphere, the electric field is 0, but is the electric field still 0 exactly on the surface?
| The answer is "it depends what you mean by exactly on the surface".
The electric field depends on the amount of charge enclosed. From Poisson's equation:
$$\nabla\cdot E = \frac{\rho}{\epsilon_0}$$
If the charge on the surface is an infinitely thin sheet of charge, then the electric field will be zero on one side of the sheet, and a finite value on the other side of the sheet - with a discontinuity.
In a real conductor, the charge sheet necessarily has a finite thickness (if only because electrons are not infinitesimal); because of this, the electric field will increase continuously from inside to outside this "band" of charge.
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Should 4 fundamental forces really be 3 because of electroweak unification? I read @ http://www.particleadventure.org/
Physicists concluded that, in fact, the weak
and electromagnetic forces have essentially
equal strengths. This is because the strength
of the interaction depends strongly on both
the mass of the force carrier and the
distance of the interaction. The difference
between their observed strengths is due to
the huge difference in mass between the W
and Z particles, which are very massive, and
the photon, which has no mass as far as we
know.
So, should all books teaching four fundamental forces change to three with third being Electroweak?
Or still there are four fundamental forces?
| A more practical answer is that in many cases it is more useful to consider them separately. You could compare with electromagnetism. If I want to design a motor, it is much easier to work with the magnetic field generated by the coils than to invoke the whole glory of Maxwell's equations. Similarly, if I want to explain the propagation of light waves, there is no need to worry about the weak interaction. If I want to study low energy beta decay, the electromagnetic force is not important. There are substantial regions of the world where electromagnetism is isolated from the weak force. Electroweak theory is beautiful and important but for most applications they are distinct.
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what is eigenvalue of $P^{1/n}$ operator if we know eigenvalue equation of $P$ ? If $P$ is an operator and $PΨ=pΨ$ ( $p$ as the eigenvalue ) then is it true to say $P^{1/n}Ψ=p^{1/n}Ψ$ ( n is an integer and positive number )
| In general, the operator $P^{1/n}$ is relatively hard to define: we have sort of the same problems we have with complex numbers (there are in general multiple $n$th roots of any nonzero $z\in\mathbb C$) except that we have them separately on every dimension, which sort of mostly kills the whole thing.
However, if $P$ is positive semidefinite, and self-adjoint, then there are good chances that it will have a unique self-adjoint positive semidefinite $n$th root. Such an operator $Q$ is then characterized by
*
*having all-nonnegative eigenvalues, and
*$Q^n\psi=P\psi$ for all $\psi$.
If $P$ has a discrete eigenbasis $\{\psi_k\}$ with eigenvalues $\{p_k\}$, then the operator $Q$ such that
$$Q\psi_k=p_k^{1/n}\psi_k \tag{$\ast$} $$
obviously satisfies these criteria, and therefore that's the operator you want. In general, it's very hard for an operator that doesn't satisfy $(\ast)$ to make any sense as an $n$th root of $P$ (although I can't discount some pathological case outright) and generally you can always assume such a relationship.
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How do I remove the negative sign from this derivation? A homework problem required me to show that the first equation below can be written in the form of the second equation. It was all fairly simple except for the negative sign. I'm not sure how this is supposed to cancel out. Might there be some conceptual way that the negative sign is removed?
1st Equation:
$$E_n=-\frac{mk^2Z^2e^4}{2 \hbar^2 n^2}$$
2nd Equation:
$$E_1=\frac{\alpha^2 mc^2}{2}$$
All I did was to substitute this and simplify:
$$\alpha=\frac{ke^2}{\hbar c}$$
| In the first equation you can replace the minus sign ($-$) by $i^2$, then factor that to be part of the $\alpha$.
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Navier Stokes: what about angular momentum? I play with CFD for a while, and suddenly, a transcendantal question raises: :-)
Navier Stokes is basically Newton applied on a continuum in Eulerian.
For solids, we would consider linear, but also angular momentum.
Why don't we have to do that for fluids ?
Conversely, you can take the curl of Navier Stokes and have an equation expressed in vorticity, which looks like our angular momentum world.
Does it mean that the equation with velocity somehow embed the both kind of momentum, and they are totally correlated for fluids ?
But how it's not the same for solids ? i.e., where is the intrinsic difference that makes it different degrees of freedom in one case and equivalent in the second ?
| The basic answer was given here: In a fluid, why are the shear stresses $\tau_{xy}$ and $\tau_{yx}$ equal?. Angular momentum conservation follows from linear momentum conservation (expressed by the Euler/Navier-Stokes equation) combined with the symmetry of the stress tensor.
Momentum conservation is the equation
$$
\frac{\partial}{\partial t}\pi_i + \nabla_j\tau_{ij} = 0
$$
where $\pi_i=\rho v_i$ is the momentum density. Using
$$
\tau_{ij} = P\delta_{ij}+\rho v_iv_j
$$
this equation is equivalent to the Euler equation, and including dissipative stresses gives the Navier-Stokes equation.
The density of angular momentum (about the origin) is $l_i=\epsilon_{ijk}x_j\pi_k$ and $l_i$ is conserved if $\epsilon_{ijk}\tau_{jk}=0$. We get
$$
\frac{\partial}{\partial t}l_i + \nabla_j m_{ij} = 0
$$
where $m_{ij}=\epsilon_{ikl}x_k\tau_{lj}$ is the angular momentum flux.
Of course, the angular momentum of the fluid can change because of external torques, and the angular momentum of a fluid cell can change because of surface stresses. (That is, I can integrate the conservation law over a volume inside the fluid, and the angular momentum of the fluid volume changes because of surface torques. Of course, the total angular momentum of the fluid is conserved.)
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Deriving Boyle's law from only the Gay-Lussac laws My physics professor during a lecture presumably mathematically derived Boyle's law from the two Gay-Lussac laws for ideal gasses. What he said is also printed in his own textbook.
He states that, given the variables $P$, $V$, $T$,
assuming only:
*
*$T \propto V$ when $P$ is constant (1st Gay-Lussac law)
*$T \propto P$ when $V$ is constant (2nd Gay-Lussac law)
then
*
*$P \propto{1\over V}$ when $T$ is constant (Boyle's law)
*$PV \propto T$ for any $P, V, T$
If this is correct, it seems very tidy and compact so why do most of more rigorous textbooks take Boyle's law as an experimental assumption instead? This way only three experimental laws (these two along with Avogadro's) are needed to justify the importance of the ideal gas model.
| I think it can be done mathematically. Let T = T(P,V). Then, if T is proportional to P at constant V, then $$\frac{T}{P}=F(V)$$. Therefore, $T=PF(V)$. Now, if T is proportional to V at constant P, then $$\frac{T}{V}=P\frac{F(V)}{V}=kP$$In the above equation, $F(V)/V$ must be a constant in order for the right hand side to be independent of V. So T=kPV.
| {
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So Black Holes Actually Merge! In 1/5th of a Second - How? I've read a lot of conflicting answers in these forums. However, today saw the awesome announcement of gravitational waves. Two black holes merged: http://www.slate.com/blogs/bad_astronomy/2016/02/11/gravitational_waves_finally_detected_at_ligo.html
Not only that, they merged FAST. In 1/5th of a second revolving around each other 250 times a second. The entire event was quicker than a heartbeat. Moreover, we observed this happening as distant outsiders. So now we can say for sure:
*
*Objects approaching the event horizon DO NOT appear to slow down
*Black holes CAN merge in a finite (and quick) amount of time
*And all this is wrt a frame of reference far, far away
To quote the NYTimes article:
One of them was 36 times as massive as the sun, the other 29. As they approached the end, at half the speed of light, they were circling each other 250 times a second.
And then the ringing stopped as the two holes coalesced into a single black hole, a trapdoor in space with the equivalent mass of 62 suns. All in a fifth of a second, Earth time.
However, everything I've read so far has let me to believe that an outside observer should never be able to measure the collision happening in a finite time. So what exactly is happening here? I must have read at lest 5 different versions of this so far everywhere in these forums over the past several years.
| The rotation of an object around the black hole does not slow down (in terms of frequency). It is not affected by time dilation. The object gets sticked on the surface of black hole, even accelerates its rotation.
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Why are Gravitational Waves so small? I'm sure you've all seen the diagrams and/or 3D visualizations of gravity; the ball sitting on a piece of fabric which makes it sink down. They've also started using it in the videos that explain gravitational waves. Two objects will be circling each other on the fabric and will emit waves.
I know this is an oversimplification of what's going on, but it is quite misleading and has left me a little bit puzzled as to why gravitational waves are infact the size of protons, not the size of planets like the videos suggest.
So I guess my question is, why are they in fact so small, and why can't we detect them from astral bodies in our solar system that we can actually detect a physical force from?
| Actually the wavelengths often are the sizes of planets. If the period of something moving at $c = 3\times10^5\ \mathrm{km/s}$ is $1\ \mathrm{s}$ (similar to the recent LIGO discovery), its wavelength is $\lambda = 3\times10^5\ \mathrm{km}$. Other phenomena could well produce waves with wavelengths larger than the solar system.
What is small is the amplitude of the waves. The recently detected waves had amplitudes of $10^{-21}$. This means that they stretched spatial lengths by one part in a thousand billion billion. LIGO in particular has interferometer arms that are a few thousand meters in length, so these arms were stretched by a few parts in a billion billion.
Think about light. There is wavelength -- radio waves are meters long, visible waves are hundreds of nanometers long, and gamma rays are fractions of a nanometer long -- and there is intensity. Even if your eyes are optimized for detecting visible light, they can't see sources that are too faint.
The wavelengths of gravitational waves are set by the typical scales in the system generating them. For example, with inspiraling stellar mass black holes, the system is a bit smaller than Earth. The reason gravitational waves are weaker in amplitude than electromagnetic waves is usually given as gravity being an intrinsically weaker force, or equivalently as most matter being very highly charged (it's mostly protons and electrons) while not very massive (it takes a lot of matter to have a noticeable gravitational effect).
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Calculate the laser heating on a crystal Let's say I'm doing an optical experiment.
I focus a laser on a crystal with a certain amount of power.
The crystal's temperature is regulated to a certain temperature but it is localy heated by the laser.
How can I calculate the local temparature at the laser spot's location?
I know in Raman spectroscopy you can use the Anti-stokes/stokes intensity ratio but here I want to use a different method. I think I could do it knowing the thermal conductivity of the crystal but I don't know how.
| Let us assume that you have continuous laser.
You need to solve stationary heat diffusion equation given in wiki, which is basically Poisson equation. For accuracy you will need to know penetration depth of your laser. It is better if you solve equation for exponential with depth source but Gaussian shape will give you good approximation. In this case solution is given by Look here.
Overall you will need to know - penetration depth, reflection coefficient and thermal diffusivity (you typically search literature for specific heat, thermal conductivity and then calculate).
Practically you can measure for instance reflectivity as a function of temperature and then measure reflectivity as a function of power. This will give you a good estimate as well.
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Why do detectors for gravitational waves have only two perpendicular arms, not three? I wonder why detectors for gravitational waves have only two perpendicular arms, not three. Having three arms appears to allow for better detection of direction, or may even increase sensitivity (I may be wrong). So far I came up with a few guesses, but I'm by no means an expert.
*
*two arms are cheaper than three
*a 600m tower is a no-go (stability), a 600m deep hole is a challenge to build and operate
*two arms is all we need, three would not guarantee significantly better results
*photons going down and up would gain and lose energy; those in the surface arms do not or much less so. Signal evaluation would be messy.
What are the true reasons?
| LIGO's arms are 4km long, which makes the problem even worse. Besides, you can just move a quarter of the way around the planet and build another 2-arm facility there, and it'll automatically be at right angles to your first facility (and as long as you do know their relative positions and orientations precisely, you don't have to put them at exactly right angles to each other).
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Gravitational waves in other dimensions I know this question is purely speculative, as we don't know if more dimensions do exist and also we do not know if gravity is indeed stronger in other dimensions (if they were to exist). But, one of the possible explanations of why gravity is so weak compared to other forces is that it exerts its strength in other dimensions, which are too small for us to detect them. However, if that were true, wouldn't the gravitational waves on those dimensions be stronger and cause larger stretching and therefore, in some cases, allow us to detect those extra dimensions? Are there any experiment that look at this case scenario?
|
But, one of the possible explanations of why gravity is so weak compared to other forces is that it exerts its strength in other dimensions, which are too small for us to detect them.
why do you say that "gravity is so weak"? if you say that because the repulsive force due to EM between two protons greatly exceeds the attractive force due to gravity, the reason for that is because the charge on the protons is approximately a natural (Planck) unit of charge (an order of magnitude) while the mass of the protons is far, far less than the natural (Planck) unit of mass.
We see that the question [posed] is not, "Why is gravity so feeble?" but rather, "Why is the proton's mass so small?" For in natural (Planck) units, the strength of gravity simply is what it is, a primary quantity, while the proton's mass is the tiny number [1/(13 quintillion].)
Frank Wilcxek in June 2001 Physics Today
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Do gravitational waves add mass to black hole? Due to the recent discovery of gravitational waves by LIGO I was wondering suppose a black hole stood between a gravitational wave then due to the fact that black hole can attract every thing then would the gravitational wave energy (that was lost from the objects producing the gravitational wave) be deposited inside the black hole?
Or would the gravitational wave simply pass through the black hole?
Finally do gravitional waves "red shift" or "blue shift" due to the gravity of another object?
| If we accept that gravity will be successfully quantized, the question can be answered in a similar way as with photons, the gravitational wave emerging from the confluence of gravitons similar to electromagnetic waves emerging from a confluence of photons.
So, depending on the wave length ( the energy of the graviton) and the crossection of graviton-blackhole scattering, some will be trapped within the horizon, contributing to the mass of the black hole, and some will be scattered away from it, depending on angle of incidence and spin quantum numbers.
Or would the gravitational wave simply pass through the black hole?
No. See above.
Finally do gravitional waves "red shift" or "blue shift"? due to the gravity of another object
Yes, similar to photons.
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What actually is the event that we think we have detected with gravitational waves? This answer shows the "event" that is creating excitement. It looks to the untrained eye like a single "blip" on a detector. It appears to last less than a second.
It is, later in the answer, referred to as a "black hole merger".
Are we seriously saying that two black holes merged in under a second?
| Yes - the black holes actually merged in less than a second. If my memory serves me from the press announcement, at the time of final collision, the two black holes were moving relativitistically, at approximately 50% the speed of light. When you have compact massive bodies orbiting one another with very small orbits, you get very high speeds.
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Is the Landauer limit reversible As I understand it, the Landauer limit, $kTln(2)$, is the minimum amount of energy to erase a bit. Is it also the minimum amount to create a bit? I'm asking statistical, like Avogadro's number of bits, i.e., $RTln(2)$.
My thinking: $RTln(2)$ is the work required, probably both ways (create and erase), but I don't know whether that is the same as the energy. By "same" I mean classical like heat, like something that can be used in $E=mc^{2}$.
| One doesn't really create bits. One simply changes the state of a physical system to "store" a bit of information. At a microscopic level, all changes are governed by reversible laws, so that the state of the physical storage system before the storage must somehow wind up encoded in the physical state of the environment. This leads to the need to input work to the computer system to keep it at a constant macrostate, as discussed in my answer here. Otherwise put, we need to input work to "forget" the storage system's state before initialization.
So, to initialize (rather than creating) a bit does indeed require the input of a minimum quantity of work of $k\,T\,\log 2$ joules, in accordance with Landauer's principle, so that the storage system and environment's state changes comply with the second law of thermodynamics.
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When I open a window to air out the room, how does the smell disperse? Let's say I'm in a room with some kind of noxious stink, possibly of flatulent nature. The quickest way to right the world that comes to mind is to open a window. When I open a window, how do the stank particles leave the room?
| That process is called diffusion, the stinky small in the room gradually leaves the room until the stinky smell is evenly distributed. when the air is evenly distributed the room assumes a new state of smell.
| {
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Relation between intensity of light and amplitude of electric field? A question in my textbook involve finding the electric field amplitude at a point in space given the intensity of light. It uses the following equation to solve it: -
$I=\frac{1}{2}\epsilon_{0}|E_{rms}|^2c$
But where did this equation come from? I am unable to find an explanation for this anywhere.
| As light is an electromagnetic wave, it is a combination of both electric field and magnetic field. So intensity of light is basically the power transmitted through electric and magnetic field divided by the cross section area of that light beam.
The energy density of the electric field is $\frac{1}{2}\epsilon_0 E^2$, and the energy density of the magnetic field is $\frac{1}{2}\frac{B^2}{\mu_0}$. The total energy density of an EM wave is then:
$$\frac{1}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right)$$
The total energy transmitted per second per unit area is then:
$$\frac{c}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right)\tag{1}$$
As we know $|\vec{E}|=c|\vec{B}|$ and $c^2=\frac{1}{\epsilon_0 \mu_0}$, so $(1)$ turns out to be:
$$\epsilon_0 E^2_{\rm RMS}c$$
$$\frac{1}{2}\epsilon_0 E^2 c$$
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Accelerating potential Will an accelerating potential accelerate a neutral atom? For example, consider an atom of hydrogen subjected to an accelerating potential of $V$. As the kinetic energy of a particle accelerated by a voltage is given by $qV$, where $q$ is the charge on the particle, won't the kinetic energy of the hydrogen atom be 0 as it is neutral? Does this mean it won't have any velocity?
| A hydrogen atom consists of an electron and a proton. In a uniform field $E$ the force on the electron will be $+eE$ and the force on the proton will be $-eE$ to the two forces sum to zero. A hydrogen atom is not accelerated by a uniform field.
However if the field is non-uniform the atom can be accelerated. This is because the hydrogen atom is polarisable and when an external field is applied the average position of the electron is shifted slightly relative to the proton.
Suppose we have a non-uniform field with a field strength given by:
$$ E = A + Bx $$
with $x$ being distance. The field will polarise the hydrogen atom and cause the average position of the electron to shift by a small distance $d$. If we take our origin for $x$ to be at the proton that means the force on the proton is $-eA$ while the force on the electron is $+e(A + Bd)$ and there is a non-zero net force on the atom of $Bd$. The atom will accelerate at $a=Bd/m$.
In practice a hydrogen atom is not very polarisable, and to achieve any measurable acceleration would require an enormous field gradient. However in principle it can be done.
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If pencil tip is heated why doesn't it write? Why doesn't a pencil write if its tip is heated in a candle flame?
| The lead (I'll call it lead for brevity even though it isn't made from lead) in a pencil is a mixture of graphite and clay pressed then sintered.
A candle flame is nowhere near hot enough to chemically change the lead. The clay requires many hundreds of degrees to sinter further and the graphite doesn't burn until getting on for 2000K. So the heat from the flame is utterly inconsequantial.
However if you put the pencil into the yellow part of the flame there will be hydrocarbons present, and these will adhere and form a film over the surface of the pencil lead. This film acts as a lubricant so when you try to write the tip of the pencil just slides over the paper instead of abrading to leave a trail of graphite.
I note the comments report mixed results from the experiment. Getting the effect is very dependent on where in the flame you put the pencil. Too high in the flame and there will be no unburnt hydrocarbons left.
To fix the problem just wipe the tip of the lead with any mild abrasive to remove the hydrocarbon layer.
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Scanning the universe - edit: expanding or shrinking I know that this may sound as a very basic question, but how come that we can
detect CMB radiation, light or gravitational waves from the big-bang era?
Shouldn't this radiation has overtaken us a long time ago?
EDIT: if i got it right , the answer to my question has to do with space-expansion , but to be honest i can't understand why.
In an hypothetical situation that universe stops expanding and starts shrinking, we won't be able to detect such radiations anymore?
| The universe is expanding - and it is expanding, as far as we can tell, in all directions, at the same rate, everywhere all at once. So over a given period of time, a distance of 1000 units will become 1100 units, and in another passing of the same time, that distance will become 1210 units long, then 1321 units, and so on. Any radiation emitted beyond a certain distance will find it difficult to catch up, given that it must travel at a fixed speed.
As a result, and due to a point in time at the beginning of the Big Bang where everything was intensely small, hot and energetic, we're unable to see beyond a particular point in time. It's this hot and energetic state we see evidence of in the background radiation. It's both the edge, and the beginning of our universe.
| {
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What is the dielectric constant of a pure conductor? Dielectric constant is the ratio of permittivity of a medium to the permittivity of free space. How to find dielectric constant of a conductor?
| Inside a metal, there is no formation of dipoles, hence there is no polarization as such. We have free electrons in metals not bound like that of a dielectric. Hence we can argue its electric susceptibility $\chi$ = 0. We know $\epsilon_r = 1 + \chi$, so it can be said that its relative permittivity($\epsilon_r$) is 1, considering electrostatics problems. For time varying fields, i.e. electrodynamics, we define complex permittivity as $\hat{\epsilon}=\epsilon \times (1+\sigma/i\omega\epsilon)$, where for metals we can have the imaginary part $\sigma/\omega\epsilon >> 1$. Thus for metals $\hat{\epsilon}=i\sigma/\omega)$, which is a large imaginary value considering high conductivity of metals.
Though, not a source for this answer,the basic idea was gained by observing the value of $\epsilon = \epsilon_0$ being used in Introduction to Electrodynamics by David Griffiths, in problems (see Chapter 9, Problem 20,bit (b)).
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Betti number of Feynman graph Let $\mathcal{L}$ be a Lagrangian, which contains polynomials of bosonic fields $\phi$. After Wick's rotation we obtain a perturbation expansion od Green's function. In this expansion there are terms of the form
$$\frac{(-1)^n}{n!}\int\hat\phi(p_1)...\hat\phi(p_N)\prod\limits_{j\in A}\xi_jd\mu,$$
where $\xi_j$ are monomials of the form $\frac{-z}{d!}\phi_E(x)^d$ ($A$ is just a set of indices). These monomials come from decomposition of interaction part of $\mathcal{L}$ onto monomials.
It it well-known that we obtain a Feynman graph from pairings in the integration by parts
$$\int\hat\phi(p_1)...\hat\phi(p_N)\prod\limits_{j\in A}\prod\limits_{i=1}^{d_j}\hat\phi\left(k_i(j)\right) d\mu,$$
where $k_i(j)$ are momentum variables carried by half-line associated to $\xi_j$.
In the Green's (or rather Schwinger - because we done Wick's rotation) function there is so-called unrenormalized value of the graph $V(\Gamma)$- multiple integral obtained from the product of terms given by Feynman rules.
My question is: How to prove that for a connected Feynman graph the number of free integration variables in $V(\Gamma)$ is equal to first Betti number $b_1$ of the geometric realization of $\Gamma$ ? ($b_1=\#\Gamma^{[1]}_{\mathrm{int}}-\#\Gamma^{[0]}+1$, where $\Gamma^{[0]}$ is a set of vertices, $\Gamma^{[1]}_{\mathrm{int}}$ is a set of internal edges. )
| *
*In the momentum representation, we should integrate over all the internal momentum variables. The number of internal momentum variables is the number $E$ of internal edges in the Feynman diagram.
*Each vertex obeys momentum conservation. (Note that the external momentum variables in a connected component are assumed to already satisfy momentum conservation.) Hence, the number of constraints is the number $V$ of vertices minus the number $C$ of connected components.
*Therefore the number of independent momentum integrations is $b_1=V-E+C$. This number is the first Betti number in graph theory, a.k.a. as the cyclomatic number or circuit rank.
References:
*
*C. Itzykson & J.-B. Zuber, QFT, 1985; p.287.
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Measuring Static Electricity Is it possible to measure the static electricity in a room using a VOM or some other digital meter? I have a lot of static electricity building up in a carpeted room, and would rather not spend the money buying a meter to measure this.
| In general, no. The input impedance of most cheap meters is too low. So, you either need a specialist meter for numbers, or a cheap and easy way to detect the static field and give you an approximate idea. This circuit, for example
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Force on current carrying square loop I'm asked to find force on square loop (side a) carrying current $I$, flowing counter clockwise, when we look down x-axis, lying in yz plane. the loop is centered at the origin. The magnetic field is given as:
$\vec{B} = kz\hat{x}$
Its solution states that force on left an right cancel each other .The force on top is $IaB=iak(a/2)$ pointing upward and the force on bottom is$IaB=-iak(a/2)$ also pointing upward .How the force on bottom is upward? (From where minus sign came?). By R.H.R it should be downward.
| first you take the direction of current vector and start turning /rotating screw from I towards the B vector ;if the rotation of the screw is clockwise screw movement will be perpendicular to the plane containing Current and Field vector and it will give you direction of the force .
if the screw rotates anticlockwise it is coming out backwards this direction will be direction of the force- naturally on your rectangular frame the two arms carrying currents which are perpendicular/or making an angle with the B field will experience a couple providing a torque let us see if it works!
for those currents which are in the direction or opposed to magnetic field direction will not experience any forces as the vector product will vanish.
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Is the local Lorentz transformation a general coordinate transformation? There is a saying in Nakahara's Geometry, Topology and Physics P371 about principal bundles and associated vector bundles:
In general relativity, the right action corresponds to the local Lorentz transformation while the left action corresponds to the general coordinate transformation.
Because the structure group right acts on Principal bundles and left acts on associated vector bundles.
But I don't think that the local Lorentz transformation is general coordinate transformation. Since for local Lorentz transformation, the structure group is $O^{+}_{\uparrow}(1,3)$ while for general coordinate transformation, the structure group is $GL(4,\mathbb{R})$.
So is the book wrong? Or I didn't understand correctly.
| By a "frame" in GR we tend to mean a tetrad or vielbein $e^\mu_i$, $\mu,i\in\{0,1,2,3\}$. See here for the precise definition. Now, we can think of this object $e$ as a matrix. It is common to refer to the index $\mu$ as a "coordinate" index and $i$ as a "Lorentz" index. Now, as Nakahara talks about in Chap. 7, a tensor transforms with a term like $\partial x^\mu/\partial\tilde x^\nu$, which is a $\mathrm{GL}(4,\mathbb{R})$ matrix. The $i$ index transforms via Lorentz transformations, as explained in the linked post. So if we imagine the matrix $e$ being written as $(e)_{\mu i}$, it is clear that we must multiply by the appropriate $\mathrm{GL}(4,\mathbb{R})$ matrix on the left to get coordinate transformations. We multiply on the right by an $\mathrm{O}^+_\uparrow (1,3)$ matrix to get Lorentz transformations.
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How precise must the energies match for absorption of photons? According to Quantum Mechanics, in order for an atom to absorb a photon the energy of the photon must be precisely that of a "jump" between energy states of the atom.
How precise must it be?
If I create a photon with an energy within an error of 0.0001% of that of an energy state, will it be absorbed by my atom?
| The linewidth broadening everyone is talking about is actually a very classical effect that comes straight from antenna theory and depends only on the size of the antenna as compared to the wavelength of light. It is well known in classical antenna theory that the bandwidth of a lossless short antenna goes as the cube of the electrical length (the physical length divided by the wavelength). For the s-p transition of the hydrogen atom, this parameter is close to the fine structure constant, 1/137. The cube of this number gives the (dimensionless) bandwidth of around 10^7.
Since the frequency of the transition is around 10^16, this gives a transition time of around 10^-9 seconds. I think this is about right for the hydrogen atom. You just treat the atom as a classical antenna and everything comes out.
| {
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How is the universe flat? I have real trouble visualising what is meant by the descriptor 'flat' when referred to the shape of the observable universe. Which one of the below is more accurate?
a) It is flat in a 2D way, like a big disk mostly spread out on one plane, similar to a solar system or disk-shaped galaxy
b) It is flat in a 3D way, in the sense that lines in space travel straight and even in all directions, e.g. like the lines of a gridded cube
c) It is not like either of the above, but something else all together
BONUS QUESTION: if the closest and most simple explanation is b), then what would be a better term than 'flat', which suggests 2 dimensionality to general audiences?
| You probably learned Pythagoras' theorem at school, and this states that if you move a distance $x$ along the $x$ axis then $y$ along the $y$ axis the distance between your starting and ending points is given by:
$$ s^2 = x^2 + y^2 \tag{1} $$
If we extend this to three directions by including motion along the $z$ axis then we get:
$$ s^2 = x^2 + y^2 + z^2 \tag{2} $$
What your teacher didn't tell you is that this is only true when the space is flat. For example if you try (the 2D version of) this experiment moving on the surface of a sphere then you'll find that:
$$ s^2 \lt x^2 + y^2 $$
When we say the universe is flat we mean that for any three displacements $x$, $y$ and $z$ equation (2) correctly gives the distance between the starting and ending points.
Incidentally, spacetime is not flat. When we say the universe is flat we mean that if we take a moment in time, i.e. a spatial hypersurface, then equation (2) applies.
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How does the drift velocity of electrons in a conductor depend on the temperature? How does the drift velocity of electrons in a conductor depend on the temperature?
I have two contradicting views for this.
*
*First, we can say that increasing the temperature of the conductor will increase the kinetic energy of the electrons. Hence, their drift velocity should increase with increase in temperature.
*Or, from the relation $v_d = \frac{eE}{m}T$ ($T$ is the relaxation time) we can say that the drift velocity is directly proportional to the relaxation time. Increasing the temperature will obviously decrease the relaxation time - as collisions will become more frequent - and thus decrease the drift velocity. Hence, an increase in the temperature will cause a decrease in the drift velocity.
So which view is correct?
| Both the cases lead to the same result: decrease in drift velocity!
In the first case the drift velocity is also going to decrease. Increase in KE will increase the speed of the electrons and not their drift velocity. An increase in speed will cause the electron to cover the same distance in a smaller interval of time, hence decreasing the mean relaxation time. This, in turn, decreases the drift velocity in accordance with the equation:
$v_d=\frac{eE}{m}T$.
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Ammeter range and shunt resistance Its said that for an ammeter to give good reading, the full current in the circuit must pass through it. But if I am right, the ammeter is basically a galvanometer connected parallel to a very low resistance called a shunt. I am aware that connecting a low resistance in parallel will reduce effective resistance to a value lesser than the least resistance.
But in an ammeter, if the shunt is a low resistance (lesser than galvanometer's resistance), then most of the current would pass through the shunt than the galvanometer. Thus, the reading given by galvanometer would decrease (as its the component which gives deflection in an ammeter), which means that the reading of ammeter would decrease.
Is my interpretation correct? If its wrong please explain me where I have gone wrong.
Also, how will range and sensitivity of a an ammeter change if we increase or decrease shunt resistance?
| In a practical ammeter there will be a number of fixed shunt resistances, selected by a switch. The galvanometer is acting as a high-resistance voltmeter, measuring the voltage across the shunt, and has little effect on the current through the circuit was a whole.
If we know the value of the shunt resistor, then the voltage read by the galvanometer is proportional to the current passing. I = V/R.
Changing the shunt resistor affects the reading greatly. It's up to the user to select the correct shunt for the current being measured.
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What carries electric field through space? A stationary charge "creates" a constant (but not uniform) electric field around it, and a moving charge "creates" a variable electric field around it. What "carries" the information about the existence of a stationary or a moving charge through space?
In particular, are photons necessarily involved in the process?
| Macroscopically we speak of electric and magnetic fields which follow the classical theory of Maxwell's equations.
A stationary charge "creates" a constant (but not uniform) electric field around it, and a moving charge "creates" a variable electric field around it. What "carries" the information about the existence of a stationary or a moving charge through space?
In the classical theory there is no necessity for "carrying" stationary electric and magnetic fields, the equations attribute fields to charges and magnetic moments which need no "carrying". The fields describe the behavior of charges and dipoles . For variable fields, classical theory's mathematics accepts that they propagate as electromagnetic waves in vacuum with velocity c, with no need of a medium. The theory fits the data perfectly.
In particular, are photons necessarily involved in the process?
Classical electromagnetism is an emergent theory from the underlying quantum electrodynamics.. There the photon is an elementary particle of zero mass and spin 1, traveling with velocity c and is the carrier of electromagnetic interactions, either in virtual form or real. An enormous number of photons build up the classcial electromagnetic wave, as demonstrated here.
So yes, photons are necessarily involved as the carrier of the EM interactions.
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In solids, is it phonons, or is it the oscillations of electrons in bands, that emit most of the blackbody radiation? In solids (most any object we see), which tends to emit most of the blackbody radiation: phonons (atomic, or molecular dipole, lattice vibrations) or oscillating electrons in their energy bands?
| A couple snippets from the web:
I can't find the actual document name, but it's here
Phonons are quantized lattice vibrations. They possess small energies
(up to approximately 100meV) and a momentum of the order of that of an
electron in a semiconductor.
Which suggests that phonon energy is far too small to emit photons (other than longwave radio, perhaps).
The wikipedia article on direct vs indirect bandgap material says that phonons are only involved in emission or absorption when the photon's energy is slightly greater than that involved in an electron transition, so the photon "makes up the difference."
But keep in mind that there is no physical phonon particle: it's just a way of describing the quantized vibrational energy in a solid. You can't directly convert vibrational energy into a photon, although the energy can affect the final energy of a photon emitted in a quantum process (such as electron level shifts). There's a pretty good discussion at this Physics.SE page.
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How is a human voice unique? Well, I am quite new to concepts of vocal sounds. From the physics point of view I believe a sound has two basic parameters i.e, frequency and amplitude.
Considering the end sound wave produced by human voice it must have frequency and amplitude as parameters. Well, when a human can speak in multiple frequencies (multiple pitches) and amplitudes (multiple volumes) I was wondering what makes every human voice unique?
Even if two persons produce a sustained note(say a same music note) their voices can be easily distinguished.So why do the voices seem different?
Are there any others parameters that distinguishes or do I have a misconception?
| As stated by the others, a sound is made up by sinus-waves of different frequencies. The tuning you hear, is determined by the lowest frequency (fundamental). The other frequencies are multiples of that ground frequency and are called overtones.
Summarising what is shown below: the amount in which the different overtones are present, determine the colour of the sound and makes the difference between your voice and mine, between a piano and a saxophone.
As an example, I examined two a's (440 Hz). One produced by a tuning-fork, the other played on an oboe (human speech is a bit more complex, but qualitatively, it's the same).
Below, the two recorded sound are displayed simultaniously:
Performing a fourier transform (looking at which frequencies are present in the sound) on the tuning-fork sound, the result is below: one frequency is very dominant: 440 Hz, the other frequencies hardly have any influence (notice the dB-scale and thus logarithmic scale on the y-axis).
The same analysis on the oboe sound reveals much more:
Several peaks at 440 Hz, 880 Hz, 1320 Hz, ... (2x, 3x, 4x, ... 440 Hz)
As you can see, the tuning you hear (440 Hz), is not the frequency that is most present in the sound (often, the first peak is the highest, but the pattern you see below is what gives the oboe its particular sound). Your hearing is trained to perceive the series of peaks as a whole and recognize the ground frequency as the pitch.
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Lorentz Force and Apparent Conservation of Momentum Violation Useful for Unidirectional Force? My understanding is that the apparent violation of Newton's Third Law by the Lorentz Force necessitates a description of the system that describes the "missing" momentum as being absorbed/carried by the magnetic field itself. What is not clear to me, however, is why this apparent violation cannot yield an apparent unidirectional force to the system.
For example: In a system with 2 magnets aligned across from one another and a current carrying rod placed parallel to the stage between them (perpendicular to the $B$ field lines), application of current to the rod can result in the Lorentz force vector pointing upwards acting on the rod. If the rod were to be affixed to the magnets themselves by a non-conducting support, preventing the rod from moving upward out of the $B$ field between the magnets, would the upward force acting on the rod not be transmitted to the magnets themselves? This seems to imply that the entire system (rod, electrons within the rod, and the magnets supplying the $B$ field) would experience an upward force.
This can't possibly be the case because that'd basically produce an anti-gravity device, but the math seems to suggest it, meaning I'm missing something fundamental somewhere. If the back-reaction from the Lorentz Force is not being transmitted to the magnets (pushing them "down" to counter the the upward vector force applied to the rod) but instead is being absorbed by the $B$ field, what prevents the system in this setup from experiencing a net upward force?
| No, in this case Lorentz force is reciprocal: current in rod creates it's own magnetic force and acts on electron current loops in magnetic materials...
The same stands for current attracting iron rod and etc...
So no violation of Newton's Laws...
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"timestamp": "2023-03-29T00:00:00",
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Front velocity and Superluminal group velocity In some cases, according to Wikipedia, the envelope of a gaussian beam can go faster than speed of light hence leading to superluminal group velocity.
However, the signal/energy still propagates at subluminal speed which is seen from the speed of the rising front of the pulse.
Do you know a practical example for which this situation arises? Is it possible to have a interactive picture of the corresponding wave?
I presume the pulse should distort quite significantly.
| Superluminal group velocity can occur in near absorption peak, known as regions of anomalous dispersion. So-called "superluminal tunneling" experiments have bee conducted in thes regions, but when carefully analyzed there is no information transferred faster than light.
Some references are given here: https://www.rp-photonics.com/superluminal_transmission.html
I'm not familiar with any applications, but everything is good for something, certainly strong absorption lines are useful.
| {
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Adiabatic compression and expansion and isothermal expansion? Please help me in understanding the concepts
*
*Why in adiabatic compression and expansion small volumetric change occurs while in isothermic compression or expansion very small pressure is applied and why the volume changes very significantly? Here, I am referring to PV diagram of Carnot engine?
*Why in adiabatic process compression and expansion should be made very fast and vice versa in that of isothermic process?
| Let us look to your first question.
In an adiabatic process, pressure change is due to two factors: change in volume (due to work done) and internal energy change (due to the temperature change), but in an isothermal change, the pressure change is due to change of volume only (since temperature is a constant). Thus, for a fixed change in volume, pressure changes deeply in an adiabatic change because there is a change in internal energy also. But in iso thermal change, there is only a small change in pressure, because internal energy change is zero. That is why the adiabatic prossess is deeper than isothermal.
Let us go to your next question.
An isothermal change is a change occure at constant temperature. So a system to remain in same temperature till the completion of a process, the system must go through infinte slow step, with each steps have the same temperature as before. For this to happen, the steps takesplce so slowly such that there is enough time to convert the heat gained/lose to the work done, without altering the internal energy. Thus an isothermal change should takesplace so slowly.
For an adiabatic change, the process should takesplace so fast that there must no time for the system to exchange the heat with the surrounding.
| {
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Uniqueness of the magnetic vector potential? I am trying to find the magnetic vector potential a distance of $s$ (cylindrical radial variable) from an infinite wire carrying current $I$. The magnetic field at a distance $s$ from a wire is $$B=\dfrac{\mu_{\circ}I}{2\pi s}\hat{\phi}.$$
Using the fact that $\nabla\times A=B$ and $\nabla\cdot A=0$, I speculated that $A=\dfrac{\mu_\circ I z}{2\pi s}\hat{s}$ satisfies the necessary conditions:
In cylindrical coordinates the curl is just:
\begin{align}
\nabla \times A
& =\left(\dfrac{1}{s}\dfrac{\partial A_z}{\partial \phi}-\dfrac{\partial A_{\phi}}{\partial z}\right)\hat{s}+\left(\dfrac{\partial A_s}{\partial z}-\dfrac{\partial A_z}{\partial s}\right)\hat{\phi}
+\dfrac{1}{s}\left(\dfrac{\partial}{\partial s}(s\,A_{\phi})-\dfrac{\partial A_s}{\partial \phi}\right)\hat{z}
\\ & =\dfrac{\partial A_s}{\partial z}\hat{\phi}=\dfrac{\mu_{\circ}I}{2\pi s}\hat{\phi}
\\ & =B
\end{align}
and the divergence is:
$\nabla \cdot A=\dfrac{1}{s}\dfrac{\partial}{\partial s}(s\,A_s)+\dfrac{1}{s}\dfrac{\partial A_{\phi}}{\partial \phi}+\dfrac{\partial A_z}{\partial z}=\dfrac{1}{s}\dfrac{\partial}{\partial s}\left(\dfrac{\mu_{\circ}Iz}{2\pi}\right)=0$.
So this potential certainly satisfies the necessary requirements, but is different from everything I've looked up. I would have thought that these potentials were unique; I have been staring at this for too long and need another opinion. Is what I have correct, or have I made a hiccup somewhere?
| One can indeed "simply let go of the unicity of the magnetic potential", but there is an alternative: to build the theory of electromagnetism from the Fermi Lagrangian, as I have done in my paper at https://arxiv.org/abs/physics/0106078. In this theory the vector potential is uniquely determined by the requirement that it should obey causality as should any physically meaningful object. Then the proposed vector potential form is incorrect, even though its rotation gives the correct B-field, and only the form $\mathbf A=-\dfrac{\mu_0I}{2\pi}\ln(s)\hat z$ is acceptable.
| {
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When is an attractor meaningful? I’m originally a computer scientist; so I hope my question is not trivial. I’m working with time series and want to reconstruct the phase space from the time series based on time-lagged versions of time series. For this purpose I need to calculate $m$ and $\tau$ which are the embedding dimension and time-lag respectively.
But: Today I was seeing some experiments based on some random $m$ and $τ$ and I saw some topology in their attractor after which this question came to my mind:
Regardless of how one chooses $m$ and $\tau$, if the attractor has a meaningful topology (i.e. following a geometrically structured shape), does it mean that this attractor could capture some meaningful dynamics?
In other words assume that I calculated $m=M$ and $\tau=T$, but I get a meaningful attractor shape for another $m$ and $\tau$. How can I interpret it?
| I don't believe the time lag is critical. Selecting a bad one might mean that you need to analyze more data to fill in the phase space, but it should still generate it given sufficient time. As for the meaningful aspect, that does depend critically on the embedding dimension. Usually one uses the human capacity for pattern recognition to extract meaning from such data. If the imbedding dimension is poorly selected, you are only seeing a slice of the phase space and that may complicate pattern recognition.
That there is any real meaning in this endeavor depends on the degree of universality in the class of nonlinear problem you are investigating.
| {
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Why we have a leak in the pipe if the fluid pressure is lower than the atmosphere? I've read in many books that if our flow pressure is lower than the atmospheric pressure then we'll have a leak in our pipe, but I already know that the flow is always from the high pressure point to the lower one. So the flow must be from outside to the pipe not from pipe to the outside, I'll be thankful if someone explains this issue by thermodynamics or fluids mechanics phenomena.
| it depends in fact.
if your fluid is in touch with air in two places(upside of fluid and hole place)
so although pressure of fluid is lower than air we have a flow.because the equation of pressure in that place would be
P(air)+p(fluid)>p(air)
the left side is the pressure which we have in the fluid in that place outwards.
the right side is the pressure which air pushes to wards fluid from out side.
so we have flow.
but what you say is right just the time when fluid is not in touch with air by upside.
the equation would be:
p(fluid)
| {
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Simplest Live Demonstration of Adiabatic Transport I have to give a presentation on Berry phase. I would like to give the simplest live demonstration of adiabatic transport. If I move an object in a loop and return that object back into its original position then the phase changes. I was thinking about a moving a simple spinning wheel, but I have trouble matching two wheels to rotate in the same frequency and phase to compare them after one experiences adiabatic transport. Both wheels did not have the same friction coefficient.
| I think that for a reliable demonstration of the spinning wheel system, a design using bearings would be essential, in order to achieve frictionless reactions.
A demonstration can be achieved by means of the linear planimeter , which measures areas enclosed by plane curves, by measuring a rotation angle of its wheel.
I think that a planimeter model can be built by elementary means. Please see, the following elementary description by: Tanya Leise.
The planimeter returns to its original state after its tracer end completes a full turn around the closed curve, only its wheel acquires a net rotation which is a Berry phase proportional to the traced area.
In fact an elementary application of Stokes theorem shows that the area of closed planar curve $\mathcal{C}$ can be written as a Holonomy of an artificial gauge field:
$$ Area = \int_{\mathcal{C}} \frac{1}{2}(x dy - y dx) = \int_{\mathcal{C}} \mathbf{A} \cdot d \mathbf{r} = \int_{\mathcal{C}} \mathbf{\nabla}\times \mathbf{A} \cdot \mathbf{n_z} dS$$ Where, the artificial gauge field:
$$ \mathbf{A }= \frac{1}{2}\{-y , x, 0\}$$
It's corresponding "magnetic field"
$$ \mathbf{B }= \mathbf{\nabla}\times \mathbf{A} = \{0 , 0, 1\}$$
Thus
$$ Area = \int_{\mathcal{C}} B dS = \int_{\mathcal{C}} dS $$
| {
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Can two photons form a gravitational bound state? I've always wondered if it's possible to bind two photons, in particular by gravitational interaction. Photons don't have a rest mass, but do nevertheless have a gravitational mass, by which they can attract each other. However, I can't imagine a bound state. My intuition of special relativity goes against it. Is it possible at all? Can photons at least bend each other's trajectories?
Interactions of other kinds (non gravitational) are also an acceptable answer, but I'm not interested in (if they exist) effective interactions in matter.
| I agree with Lewis Miller that they cannot form a bound states if the binding is thought to be of gravity essence. By definition, a quantum state is understood only in small scales which is way too smaller than the bounding distance between two photons; it will be bigger than the size of Visible Universe. However, I would like to add to it the possibility that two photons can be entangled so that they feel each other's presence even in cosmological distances given the fact they are originated from a common source, e.g. annihilation of e+ and e- resulting into two back-to-back photons. These produced photons would then be entangled and you can prove it experimentally. But, binding is different than entanglement. Thanks,
| {
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Photon emission between an electron and positron If I placed and electron and a positron at a certain distance apart in a vaccumm, they would attract each other and annihilate producing 2 or more gamma rays. But, what I would like to know is, does the electron and positron emit photons as they are attracted towards each other before they annihilate ?
I am not refering to the virtual photons that are exchanged as they are attracted to each other. I am asking if real photons are emitted by either or both particles as they move towards each other.
An electron will emit photons as it moves towards the nucleus of an atom and drops down the energy levels, but in this case the electron is losing energy in the form of a photon emission. Does the electron and positron lose any energy during their attraction ?
| My 2 cents:
If only two gamma rays are produced with each of their energy equaling the rest energy of either e- or e+ (which is 511keV) then, it could possibly mean that there is no net linear momentum, which possibly means that this e+ and e- pair are very very close to each other and after collision, the gamma rays are emitted as per the law of conservation of energy.
If more than two gamma rays are produced then it could possibly mean that these e+ and e- particles are placed at some distance to each other and it results in net linear momentum because of acceleration generated due to their opposite charges leading to emission of a photon (with varied amplitudes).
In some cases neutrinos could also be produced because of the annhilation. What i can't comment or don't know is if the annihilation could produce one or more neutrino–antineutrino pairs.
Thanks
| {
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Normalization of a wave function in quantum mechanics A more simple question, so I am watching a quantum mechanics lecture on potentials of free particles and am doing the general solution of schrodinger's stationary equation for a free particle when I was told to normalize the solution (which I can do all well and good) but I have no idea what it actually means to "normalize"
My question being what is normalization ? What does its product describe ?
| One peculiar fact about a real life wave function $\psi$ is that it can be normalized. In order to analyze and compare the various outcomes of the solution of a Schrodinger equation, one need to assign a quality that is unique to all the wave functions, that which is to transform them such that their area is always 1.
| {
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Impulse Equations A solid sphere of mass $m$ rolls without slipping on a horizontal surface and collides with a vertical wall, elastically. The coefficient of friction between the sphere and wall is $\mu$. After the collision, the sphere follows a parabolic trajectory, with range $R$. What is the value of $\mu$ to maximize $R$?
Since the collision is elastic, we can say impulse normal impulse $J = \Delta P = 2mv$.
As frictional impulse is $\mu$ times normal impulse, $J'=2mv\mu$ (upwards).
Therefore, sphere acts like projectile with horizontal velocity $v$ and vertical velocity $2v\mu$. To maximize $R = 4\mu v/g$, $\mu$ should be maximum i.e.$1$.
However, this is not correct. What am I missing here?
| If the collision is elastic, then energy is conserved; however, this cannot mean that the horizontal velocity is the same on the way to the wall, and on the way back: the ball will also have a vertical velocity, and rotational kinetic energy.
This means that the impulse cannot be $2mv$ as you stated; you have to solve instead for the rotational velocity / energy, the horizontal rebound velocity / energy, and the vertical velocity / energy - and set these equal to the inbound kinetic / rotational energy of the sphere.
Note that since the sphere loses contact on the way back, the rotational velocity will be related to the vertical velocity only.
Finally - there is no reason why "maximum coefficient of friction $\mu$ = 1". It can be higher... although in this case it may end up being lower.
In the spirit of "homework like" questions, I will leave this for you to think about.
| {
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Propagator in Quantum Mechanics What does the propagator in Quantum Mechanics mean? I mean, except from the mathematics behind it, what does it tell us? Is it something that has to do with translations in time?
| Propagator in quantum mechanics is just a different name for Green's function for time-dependent Schroedinger equation. It is a unique function that enables us to write, for any time $t_0$,
$$
\psi(x,t) = \int G(x,t;x',t_0)\psi(x',t_0)\,dx'
$$
for all $x$ and all $t$. This means the $\psi$ function of $x$ (at any time $t$) can be written as a result of linear operator acting on the $\psi$ function at any other time $t_0$ (usually $t_0<t$ but this is not necessary).
When $\psi(x',t_0)$ is put equal to $\delta(x'-x^*)$ for some $x^*$, we obtain
$$
\psi(x,t) = G(x,t;x^*,t_0)
$$
so propagator is a result of evolution of $\delta$ distribution governed by the same equation as ordinary $\psi$ functions are - time-dependent Schroedinger equation.
However, $\delta$ distribution is not admissible as $\psi$ in the sense of the Born interpretation, so be cautious when giving $G$ physical interpretation.
| {
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Examples of Bernoulli Numbers, Euler-Mascheroni Integration, and the $\zeta(n)$ in physics In Arfken's Mathematical Methods for Physicists, there is a subsection of the "Infinite Series" chapter which covers the Bernoulli numbers, Euler-Mascheroni integration (or summation), and the connection these have with the Riemann zeta function. However, apart from a few nifty math problems these solve (explicit expressions for some sums and integrals), I can't see the use of these concepts in physics. There were a few problems at the end of the chapter that said that some practice integrals showed up in QED corrections, so that's a start. I would like to know where else these ideas show up in physics, if they do.
| In quantum field theory, especially in the treatment of divergent series and divergent integrals (like Feynman integrals which arise from calculation of self energies for example) and during the process called "regularization", lots of "Euler Mascheroni" constants arise.
For example, I remember during the calculation of the Photon self-energy (vacuum polarization) in a $2\omega$ dimensiona spacetime is provided by the parametric integral:
$$\text{reg}\Pi(k^2, M^2) = (-2)\frac{\alpha}{\pi}\int_0^1 x(1-x) \left\{\frac{1}{\epsilon} - \gamma -\ln\left[\frac{M^2 - x(1-x)k^2}{2\pi \mu^2}\right]+ \mathcal{O}(\epsilon)\right\}\ \text{d}x$$
in which you see the Euler Mascheroni $\gamma$ constant arising in the integration.
Riemann Zeta Function arises instead when you are trying to regularize divergent series. I bet at least once you saw this:
$$\zeta(0) = -\frac{1}{2}$$
A regularization of the infinite series
$$\zeta(0) = \sum_{k = 1}^{+\infty} \frac{1}{k^0} = 1 + 1 + 1 + 1 + \cdots $$
| {
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How to derive the components $\Lambda^i_j$ of a Lorentz-transformation given a boost $\vec{v}$? I am stuck in deriving a specific formula concerning Lorentz-boosts.
In my Classical Mechanics skript there is a chapter dealing with special relativity. In this chapter the Lorentz transformations are defined as being those linear transformations $\Lambda_\nu^\mu:{\mathbb R}^4 \rightarrow \mathbb R^4$, $x^\mu \rightarrow x'^\mu=\Lambda_\nu^\mu x^\nu+\rho^\mu$ that satisfy the equality $g_{\nu\mu}\Lambda_\sigma^\mu\Lambda_\rho^\nu=g_{\sigma\rho}$, where $g_{\nu\mu}$ is the metric tensor for flat space time.
Furthermore a specific transformation between two inertial systems O and O' is considered. This transformation relates the system O' and O if O' moves with velocity $\vec{v}$ relatively to O.
Then, for this specific situation, the formulas $\Lambda_0^0=\gamma$ and $\Lambda_0^i=\gamma \frac{v^i}{c}$ for $i=1,2,3$ are derived. (I appended the derivation in a picture below.)
Now to my question: My script states, that if the axis of both systems are paralell to each other it is also possible to derive formulas for the remaining components of $\Lambda_\nu^\mu$. These formulas are stated to be: $\Lambda_j^0=\gamma \frac{v^j}{c}$ and $\Lambda_j^i=\delta^i_j+\frac{v_iv_j}{\vec{v}^2}\gamma$. The proof for this equations is not carried out in the script but left as an exercis. I have tried really hard but was not able to do the derivation. I consider my problems to be rooted in my inability to figure out how to use the paralellity of the axes in the derivation. I would be really glad if someone could tell me how it is done correctly.
In the following you can find the derivation for $\Lambda_0^i$ and $\Lambda_0^0$ from my skript.
| There is an error in your expression for the $\Lambda^i_{\;j}$ components. They actually read
$$
\Lambda^i_{\;j} = \delta^i_j + (\gamma -1)\frac{v^i}{v}\frac{v_j}{v} = \delta^i_j + \frac{\gamma -1}{\beta^2}\frac{v^i}{c}\frac{v_j}{c}
$$
with $\beta = v/c$, $v^2 = v^iv_i$.
Keeping this in mind, separate once again timelike and spacelike coordinates in the original transformation and use the already known components of $ \Lambda$:
$$
dx'^0 = \Lambda^0_{\;0} \;dx^0 + \Lambda^0_{\;j} \;dx^j = dx^0 \left[ \gamma + \Lambda^0_{\;j} \frac{u^j}{c} \right]\\
dx'^i = \Lambda^i_{\;0} \;dx^0 + \Lambda^i_{\;j} \;dx^j = dx^0 \left[ \gamma\frac{v_i}{c} + \Lambda^i_{\;j} \frac{u^j}{c} \right]
$$
where $ u^j/c = dx^j/dx^0$. Then take appropriate ratios to obtain the corresponding velocity transformations:
$$
\frac{u'^i}{c} = \frac{\gamma\frac{v^i}{c} + \Lambda^i_{\;j} \frac{u^j}{c}}{\gamma + \Lambda^0_{\;j} \frac{u^j}{c}}
$$
Consider now those displacements in the unprimed system that are along the direction of the relative velocity, such that
$$
\frac{u^i}{c} = a\frac{v^i}{c},\;\;\;u^iu_i = ac^2
$$
for some $a\in \mathbb R$, and observe that the corresponding displacements in the primed system must also read
$$
\frac{u'^i}{c} = a'\frac{v^i}{c},\;\;\;u'^iu'_i = a'c^2
$$
for some other $a'\in \mathbb R$. Substituting in the velocity transformations above yields a general relationship of the form
$$
a'\frac{v^i}{c} = \frac{\gamma\frac{v^i}{c} + a \Lambda^i_{\;j} \frac{v^j}{c}}{\gamma + a \Lambda^0_{\;j}\frac{v^j}{c}}\;\;\;\;\;\;\;\;\;(1)
$$
The information necessary to derive the $\Lambda$ components comes from the following particular cases:
*
*From the isotropy of space it follows that $a=-1$ requires $a' = 0$, which can only happen provided
$$
\Lambda^i_{\;j} \frac{v^j}{c} = \gamma\frac{v^i}{c}\;\;\;\;\;\;\;\;\;(2)
$$
for any $v^i$.
*
*The invariance of the speed of light further requires that for $a = \pm c/v\Leftrightarrow u^iu_i = c^2$, we must also have $a' = \pm c/v\Leftrightarrow u'^iu'_i = c^2$, and therefore
$$
\pm \frac{c}{v}\frac{v^i}{c} = \frac{\gamma\frac{v_i}{c} \pm \frac{c}{v} \Lambda^i_{\;j} \frac{v^j}{c}}{\gamma \pm \frac{c}{v} \Lambda^0_{\;j}\frac{v^j}{c}} = \frac{\gamma\frac{v_i}{c} \pm \frac{c}{v} \gamma\frac{v^i}{c}}{\gamma \pm\frac{c}{v} \Lambda^0_{\;j}\frac{v^j}{c}} = \frac{\gamma(1 \pm \frac{c}{v} )}{\gamma \pm \frac{c}{v} \Lambda^0_{\;j}\frac{v^j}{c}}\frac{v^i}{c}
$$
Solving for $ \Lambda^0_{\;j}\frac{v^j}{c}$ gives
$$
\Lambda^0_{\;j}\frac{v^j}{c} = \gamma \frac{v_j}{c}\frac{v^j}{c}\\
\left[ \Lambda^0_{\;j} - \gamma \frac{v_j}{c}\right] \frac{v^j}{c} = 0
$$
and since the boost transformation is symmetric, we can conclude that
$$
\Lambda^0_{\;j} = \gamma \frac{v_j}{c}
$$
*The above expressions for $ \Lambda^0_{\;j}$ simplify eq.(1) to
$$
a'\frac{v^i}{c} = \frac{\gamma\frac{v^i}{c} + a \Lambda^i_{\;j} \frac{v^j}{c}}{\gamma\left(1 + a \frac{v^2}{c}\right)}\;\;\;\;\;\;\;\;\;(3)
$$
and condition (2) further implies
$$
a' = \frac{a+1}{1+a\frac{v^2}{c^2}}
$$
However, eq.(3) can be rearranged eventually as
$$
\left[ \Lambda^i_{\;j} - \left( \frac{1}{\gamma\left(1+a\frac{v^2}{c^2}\right)}\delta^i_{\;j} + \frac{\gamma(a+1)}{1+a\frac{v^2}{c^2}}\frac{v^i}{c}\frac{v_j}{c}\right)\right]\frac{v^j}{c} = 0
$$
which has to hold for any $a \in \mathbb R$. Since $\Lambda^i_{\;j}$ may depend only on the relative velocity, this means that it must be of the form
$$
\Lambda^i_{\;j} = Q \delta^i_{\;j} + P \frac{v^i}{c}\frac{v_j}{c}
$$
with $Q$ and $P$ some functions of the relative velocity such that
$$
\left[ \left[ Q - \frac{1}{\gamma\left(1+a\frac{v^2}{c^2}\right)} \right] + \left[ P - \frac{\gamma(a+1)}{1+a\frac{v^2}{c^2}} \right] \frac{v^2}{c^2} \right] \frac{v^i}{c} = 0
$$
From this we have successively
$$
Q + P \frac{v^2}{c^2} = \left[\frac{1}{\gamma\left(1+a\frac{v^2}{c^2}\right)} + \frac{\gamma(a+1)}{1+a\frac{v^2}{c^2}} \frac{v^2}{c^2}\right]
= \gamma\\
P = \frac{\gamma - Q}{\beta^2}\\
\Lambda^i_{\;j} = Q \delta^i_{\;j} + \frac{\gamma - Q}{\beta^2} \frac{v^i}{c}\frac{v_j}{c}
$$
The exact form of Q may now be determined from $g_{\mu\nu}\Lambda^\mu_\rho\Lambda^\nu_\sigma = g_{\rho\sigma}$.
| {
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Crookes' Maltese Cross Tube and the anode I understand that Sir William Crookes' Maltese cross tube shows that cathode rays travel in a straight path but I am unsure of why the cathode rays aren't affected by the anode as there is an electric field and therefore bend.
Sorry if this is a silly question but I am relatively new to physic.
| The left hand image is a schematic of the Crookes' tube arrangement.
The tube had gas at low pressure inside it and electrons which original from the cold cathode were accelerated towards the anode.
A lot of the electrons were travelling so fast that they were unable to turn the corner towards the anode and there went onward towards the Maltese cross along almost straight line trajectories.
Some hit the Maltese cross but a number reached the glass and caused fluorescence and the relatively sharp image of the Maltese cross.
What is interesting is that the electrons which hit the Maltese Cross leak away and do not affect other electrons coming towards the Maltese Cross.
The voltages used were very high but even so the flux of electrons was not.
A modern tube uses an electron gun and has a screen with a phosphor on it so much lower accelerating voltages can be used.
The Maltese cross is also at the same potential as the final anode of the electron gun to try an ensure that the trajectories of the electrons are relative straight to produce a sharp shadow.
An interesting effect can be obtained by removing the high potential lead from the Maltese Cross and this is shown on the right hand image.
Because of the high electron beam currents significant numbers of electrons are stored on the Maltese cross which the repel the electron beam.
The sharp shadow of the cross being produced by visible light emitted from the electron gun.
| {
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Derivation of Displacement current term in Ampere's Law I have a quick question:
In deriving the displacement current term for Ampere's Law, my book has the line:
$$\Phi_E= \int_S \mathbb{E} \cdot \hat{n} da= \int_S \frac{\sigma}{\epsilon_0} da = \frac{Q}{A \epsilon_0} \int_S da= \frac{Q}{\epsilon_0}$$
My question is:
Here the electric field is the electric field between two conducting plates (capacitor) neglecting edge effects. $\sigma$ is the charge density of a plate given by: $\frac{Q}{A}$ where A is the capacitor plate area. My book substitues $\sigma= \frac{Q}{A}$ and arrives at: $$\frac{Q}{A \epsilon_0} \int_S da = \frac{Q}{A \epsilon_0} A = \frac{Q}{\epsilon_0}.$$ Why do the "A"'s cancel? One $A$ is the area of the capacitor plate and the other area is the area of the gaussian surface. In general, these area's will not be equal. The image to go along with the derivation is similar to the one shown below.
| I have never understood why this sort of derivation in textbooks is not made conceptually easier by choosing an appropriate bulging amperian surface as the one shown below.
Because the surface is flat and has a plane perpendicular to the electric field in the region of the electric field the flux is easy to calculate.
$\Phi_E= \int_S \mathbf E \cdot \hat{\mathbf n}\,\mathrm da= E \; A = \dfrac{\sigma}{\epsilon_0} A = \dfrac {Q}{ \epsilon_0}$
| {
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Why fuse needs to be slightly more able to bear current than the appliance? if I have a bulb with rated current of $4 A$ then why I need to connect an $5A$ fuse?( which has slightly more ability to bear current.)
| The role of a fuse is not to protect the equipment from damage. It's to ensure that in the event of a fault, the wiring doesn't overheat and cause a fire.
The fuse needs to be rated slightly higher than the maximum operating current of the protected device, so that the fuse won't blow when the equipment is used. It shouldn't be rated too much higher than the maximum operating current, because that would require wiring the equipment with thicker, costlier wire.
In case this is a homework question, I won't talk about the specific case of the lamp directly, but give a hint in the form of a question: What is the equivalent resistance of an incandescent lamp, and how does it depend on temperature?
| {
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What is the meaning of $\mathrm{d}^4k$ in this integral? From Gerardus 't Hooft's Nobel Lecture, December 8, 1999, he states the following equation (2.1):
$$
\int \mathrm{d}^4k \frac{\operatorname{Pol}(k_{\mu})}{(k^2+m^2)\bigl((k+q)^2+m^2\bigr)} = \infty
$$
in relation to weak interactions theory, where $\operatorname{Pol}(k_{\mu})$ stands for some polynomial in the integration variables $k_{\mu}$, and then goes on to say that physically it must be a nonsense.
Why is it a nonsense?
What sort of integral is this, and how does one interpret it?
Is the $\mathrm{d}^4k$ shorthand for 4th degree integration?
At what stage and subject of a physics course does one learn about it (A pre-fresher is asking)?
| The equation is a term in the calculation of a scattering probability. Obviously a scattering probability must be between zero and one, like any other type of probability. So when the calculation of a scattering probability returns a value of $\infty$ that isn't physically possible, and it shows that the method we are using to calculate the probability is incorrect.
That is what 't Hooft means by nonsense - it means the method of doing the calculation is wrong. His Nobel prize was earned showing us the correct way to do the calculation.
The parameter $k$ is a wave vector, or more precisely the special relativistic form of a wave vector. This is a 4D vector so it has four independant components normally written as $k^0$, $k^1$, $k^2$ and $k^3$. Note that the superscript is a label and doesn't mean you're raising $k$ to a power. The integration is over all possible values of each of the four components, so it's really four integrations:
$$ \int \int \int \int\,dk^0 \,dk^1 \,dk^2 \,dk^3 $$
Writing $d^4k$ is a common shorthand for this.
You are unlikely to study quantum field theory in any depth unless you do a postgraduate course in physics, though I guess some universities may offer it as a final year option.
| {
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Why is it said that standing waves do not transfer energy? The author of my physics textbook writes that standing waves, unlike travelling waves, do not transfer energy. He says that this is because a standing wave is composed of two travelling waves carrying energy in opposite directions. Is this explanation sufficient to prove that standing waves do not transfer energy? Also if standing waves do not transfer energy, then how can instruments be heard?
| In case of electromagnetic waves the energy transfer is described, in most general terms, by the Poynting theorem:
$$
-\frac{\partial u}{\partial t} = \nabla\cdot \mathbf{S} + \mathbf{J}\cdot\mathbf{E},
$$
where the Poynting vector (i.e., the energy flux) given by
$$
\mathbf{S}=\mathbf{E}\times\mathbf{H}.$$
Note that in absence of dissipation (the last term), the Poynting theorem is simply the continuity equation for the energy.
It is now a simply excercise to calculate the Poynting vector for the standing wave of your choice to convince oneself that it does not transfer energy.
Similar equations exist for the (linear) waves of any nature: elastic, waves in miquids/gases, etc. In fact, that the standing wave does not transfer energy can be taken as a definition of the standing wave.
Remark more precisely, it is the flux averaged over the period or the wavelength if the wave that is zero. The local value of the flux however also oscillates.
| {
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Putting a capacitor into a strong magnetic field, will this change the capacity? I'm wondering, does a magnetic field change the number of electrons, placed and displaced on the two plates of a capacitor. To prove or disprove this, I think the capacitor could be connected to an other capacitor outside the magnetic field and it has to be measured the current flowing between the capacitors during the increase and decrease of the magnetic field.
Edit: Was such an experiment carried out?
| It is worth recalling that a charge that is at rest with respect to a static magnetic field incurs no force from that field. From that it follows that the steady-state capacitance should be identical to that of the same capacitor outside the field.
Or at least it would follow for a capacitor with vacuum between the plates. If there is a dielectric involved it we could at ask if the presence of the magnetic field has any effect on the dielectric constant of that material.
| {
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Killing equation manipulation Why does the killing equation $$K_{\mu;\nu}+ K_{\nu;\mu} = 0$$ equal $$K_{\mu,\nu}+ K_{\nu,\mu} -2\Gamma^{\rho}_{\mu\nu}K_{\rho} = 0 $$
when in general a covariant derivative
$V_{\beta;\alpha} = (\partial_\alpha V^\lambda + \Gamma_{\alpha \nu}^{\lambda}V^{\nu})g_{\lambda \beta}$?
Where does the opposite sign of the affine connection come from and why is there not another affine connection?
https://en.wikipedia.org/wiki/Killing_vector_field
| Your expression for the covariant derivative is wrong: it should be with a minus sign (plus sign for vectors=upper index, and minus sign for covectors=lower index):
$$
\begin{aligned}
\nabla v^\alpha\sim \partial v^\alpha\color{red}+\Gamma^\alpha v\\
\nabla v_\alpha\sim \partial v_\alpha\color{red}-\Gamma_\alpha v
\end{aligned}
$$
| {
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Nearly Free Electron Model and the Reduced Zone Scheme When for example studying the vibrational modes of a one dimensional diatomic chain we find that the dispersion relation $\omega(k)$ is periodic in the one dimensional reciprocal lattice vector $\frac{2\pi}{a}$, and so the dispersion relation can all be displayed in the first Brillouin zone in the reduced zone scheme.
When studying free electrons perturbed by a weak periodic potential we find that the dispersion relation of the free electrons $E=\frac{\hbar^2k^2}{2m}$ develops gaps at the Brillouin zone boundaries. However this (almost) parabolic dispersion does not have a functional dependence that is periodic (i.e for a single value of $k$ there is only one possible energy). Books seem to suggest we can display this in a reduced zone scheme though - why is this?
Edit: Possibly related to Bloch's theorem? I guess Bloch's theorem says that we can always reduce things into the first Brillouin zone - however if this is the case why doesn't it naturally drop out of the calculations rather than us having to force it?
| Mechanical vibrations of the periodic atomic chain and electron motion in periodic fields are pretty different problems, although they have similar features related to the periodic boundary conditions. The frequency of the diatomic chain has upper bound that depends on the interatomic coupling. The energy of electrons has no upper bound, since you can excite electron at arbitrary high energy levels whose wave function satisfies periodic boundary conditions.
In other words, the diatomic chain has only two possible modes of vibrations, while electron has an infinite number of such modes, each of which corresponds to an energy level of isolated atoms that forms a periodic structure.
Thus the dispersion law for electrons moving in periodic fields is not a periodic one and possess a form illustrated in the figure below.
| {
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Differences between eigenstates, bound states and stationary states I am not very clear about the differences between eigenstates, bound states and stationary states.
| *
*Eigen state : Particular to an operator, which when operates on it, gives a scalar number (or the eigenvalue) times itself.
*Stationary state : The state of a particle that does not vary with time.
*Bound state : The state of a particle bounded by within a potential, meaning - the energy of the particle in that state is less than that of the potential.
| {
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Are residuals supposed to have error bars? Hopefully I'm asking this in the correct section. So I've got a graph with a linear trend of data and a best fit line plotted. The data points on the main graph obviously each have their own error bars. I've also made a subplot of residuals.
My question is: do the residual subplot's data points normally have to have error bars and if so, are these error bars the same ones as those on the main graph? I ask because I thought that perhaps the error on the gradient and the error on the intercept (calculated by Excel's LINEST in my case) contribute to the error on the residual data point which would mean I can't use the same error bars as those used on the main graph? Or am I wrong here?
Note: in my case, only the dependent variable has error - the independent variable, i.e. x axis does not have error.
Many thanks for any help.
| Yes, residuals should have error bars.
If your residuals are the difference between your data and your model, and your data are well-described by your model except for independent, normally-distributed errors which you have modeled correctly with your uncertainties, then your residuals should be (a) randomly distributed, without any leftover shape, and (b) about two thirds of the error bars on the residuals should touch zero. A residual plot without error bars only lets you evaluate the first of these criteria.
Sometimes you'll see plots of "normalized residuals", where each (data-model) difference has been scaled so that the associated error bar is 1. This is closely related to the computation of the $\chi^2$ statistic used for goodness-of-fit tests.
| {
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Help! An 8 year old asked me how to build a nuclear power plant I would really like to give an explanation similar to this one.
Here's my current recipe:
(i) Mine uranium, for example take a rock from here (picture of uranium mine in Kazakhstan).
(ii) Put the rock in water. Then the water gets hot.
(iii) [Efficient way to explain that now we are done with the question]
This seems wrong, or the uranium mine would explode whenever there is a rainfall. Does one need to modify the rock first? Do I need some neutron source other than the rock itself to get the reaction started?
As soon as I have a concrete and correct description of how one actually does I think I can fill in with details about chain reactions et.c. if the child would still be interested to know more.
| Well, if you have a really adventurous kid you can follow the recipe of the Radioactive Boy Scout
...and became fascinated with the idea of creating a breeder reactor
in his home. Hahn diligently amassed this radioactive material by
collecting small amounts from household products, such as americium
from smoke detectors, thorium from camping lantern mantles, radium
from clocks and tritium (a neutron moderator) from gunsights. His
"reactor" was a bored-out block of lead, and he used lithium from
$1,000 worth of purchased batteries to purify the thorium ash using a
Bunsen burner.[2][3]
Hahn posed as an adult scientist or high school teacher to gain the
trust of many professionals in letters, despite the presence of
misspellings and obvious errors in his letters to them. Hahn
ultimately hoped to create a breeder reactor, using low-level isotopes
to transform samples of thorium and uranium into fissionable
isotopes.[4]
Although his homemade reactor never came anywhere near reaching
critical mass, it ended up emitting dangerous levels of radiation,
likely well over 1,000 times normal background radiation.
However, that bit about Tritium seems suspect and is probably wrong
| {
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Change in energy during orbital decay Suppose we have a planet orbiting a star and the planet radiates away some amount of energy $\Delta E$. I want to find by how much the orbit decays (i.e the change in orbital radius).
One way to do this is to find the total energy as a function of radius $E(r)$ (it's equal to half the gravitational potential energy $U(r)$ for circular orbits) and set $E(r_f)-E(r_0)=\Delta E$ and solve for $r_f$ in terms of $r_0$ to find $\Delta r$.
I'm not sure if this is the best way possible (but I think at least it's valid). I also want to know if I can do it without knowing the total energy $E$ (that is, just by using the potential energy).
| By talking about orbital radius you're assuming the orbit is circular, but that may not be the case. In general you need both energy and angular momentum to specify your orbit, and to solve completely you'd need to know the specific mechanism of energy loss to find out what happens to the angular momentum.
You can get around this if you're willing to use the semi-major axis $a$ instead of the radius. If you know the energy $E$ you can relate them by $$a = \frac{GMm}{2|E|}$$
Differentiating we find that
$$da = \frac{GMm}{2E^2}dE = \frac{2a^2}{GMm}dE$$
Which gives the change in semi-major axis for small changes in energy.
I don't know what you mean by doing this only knowing the potential energy. The solution to the Kepler problem is known, so you don't have to work out all the formulas. You only care about the total energy radiated away.
| {
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Fourier transform of Hamiltonian for scalar field In the Srednicki notes (http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf) page 36 he goes from
$$H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x) $$ to
$$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p) $$
Where $$\tilde{a}(p) = \int \frac{d^{3}x}{(2\pi)^{\frac{3}{2}}}e^{-ipx}a(x)$$
I tried doing this by saying
$$H = \int d^{3}x \int \frac{d^{3}p}{(2\pi)^{3}}e^{-ipx} \tilde{a}^{\dagger}(p) \left(\frac{P^{2}}{2m}\right)e^{ipx}\tilde{a}(p) $$
But then I'm unsure how to proceed with commutators. Does $P^{2}$ commute with $e^{ipx}$? What about with $\tilde{a}(p)$?
Any help would be greatly appreciated.
| Srednicki goes from
$$
H = \int d^{3}x\ a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x)
$$
to
$$
H = \int d^{3}p\ \frac{1}{2m}p^{2}\ \tilde{a}^{\dagger}(p)\tilde{a}(p)
$$
where $p^2$ is just a number (an integration variable, not an operator). Therefore, it commutes with everything.
| {
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Could an asteroid land slowly on Earth's surface? The concept in my mind is that an asteroid is on a vector similar to Earth's, but slightly slower (e.g., 50kmh slower). As Earth passes it, it enters the atmosphere at a sharp angle, and since Earth was passing it, it just barely touches down due to Earth's gravity and atmospheric drag.
Given a large asteroid (e.g., 500 meters wide), is there any reason something like this couldn't happen? And, is there any evidence that it has happened?
| Well, technically, the answer is no as the other answers and comments also say.
The approach speed can not be less than escape velocity.
But in order for such a thing to happen, nature has to be really creative and totally in our favor. For example, the asteroid can have a very very lucky combination of these:
*
*The asteroid has right kind and amount of fluid in it that starts jetting out steam at just the right times and right angles.
*The asteroid is parachute shaped with appropriate strength and falls at an appropriate angle.
Again, it would be a miracle, so, please do not hit me.
As we may be lucky due to a three body interaction with moon, this is taking the luck to kind of extreme.
| {
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Units don't match in the torsional spring energy! According to Wikipedia's description of torsion springs and according to my understanding of physics the energy of a torsional spring can be written as
$$U=\frac{1}{2}k \varphi^2$$
where $k$ is a constant with units of $\rm N\,m/rad$.
I am freaking here because if the energy of a torsional spring is really $k \varphi^2$ than the units are
$\rm (N\,m/rad) \cdot rad^2=Joule\cdot rad$. ??
What on earth am I missing here?
| In the SI system of units, the radian is a special name for 1 (see SI brochure), that is,
$$\mathrm{rad}=1.$$
Therefore,
$$[k] = \mathrm{N\,m/rad} = \mathrm{N\,m}$$
and
$$\mathrm{J\,rad} = \mathrm{J}.$$
Since the last revision of the SI, the radian is no longer a supplementary unit: an angle is now defined as the ratio of two lengths, and the unit radian is now maintained for convenience. However, it's just a synonym of 1, and can be used (but it's not necessary) to convey, or to strengthen, the information that the quantity of interest is an angle.
| {
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Calculating force with two fulcrums and many forces I may not be wording the question the right way.
I have a lever. About 2/3 the way back there is a pivot point. We'll call this B
On the very front (end of the lever) there is a block that the lever rests on. Let's just assume the block is actually a point at the very end of the lever.
I have many points (e.g. 30) of force on both sides of the lever.
I am trying to calculate the down force on a hitch for a travel trailer, and so it will depend how and what I load the trailer with.
The left triangle is a pivot point.
The right block is just for the lever to rest on.
How much weight is resting on that right block?
I am using Excel, so try to keep it simple.
The different down arrows represent different weights.
(follow-up questions have been edited)
| I assume by fulcrum you mean resting points/pivot points and that the lever is not attached to the fulcra. If it is, then the mass of the fulcrum at the back counts in this equation.
The equation for torque is Torque = Force * distance from pivot. If I apply a force of 10 N 1 m away from the pivot point, then I have 10 Nm of torque. The total torque is simply the sum of all of the torque given your frame of reference. Note that since torque is rotation, forces that go down on different sides of the fulcrum cancel oppose one another.
I don't know how to use Excel, but you would take each element in your force, multiply it by its respective distance from the fulcrum (negative distance would mean to the right or left of the fulcrum, your call) and then sum it all together.
| {
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Is acceleration relative in relativity? Suppose a box A is moving relative to a Box B, then by time dilation equation if I take 1 sec passed for an observer in A then for an observer in B will be little longer. Now if I suppose that the box B is moving while A is stationary under the same condition, then by the time dilation equation time passed in B must be shorter than A. How is this issue resolved? In the twin paradox problem my book says that it's due to acceleration, but in my opinion acceleration is relative (please correct me if I'm wrong), but here it is not the case.
| Acceleration is not "relative" even in classical mechanics, accelerating frames have fictitious forces in them (like overload, centrifugal, etc.), while inertial ones do not. It is not relative in special relativity either.
So if A is accelerating and B is inertial then A and B are not "equal", and if they are both inertial then there is no way to bring them back together as in the twin paradox. Which by the way is not a paradox, there is nothing contradictory about the stationary twin ending up older than the accelerating one. "Neither Einstein nor Langevin considered such results to be problematic: Einstein only called it "peculiar" while Langevin presented it as a consequence of absolute acceleration". See more in Resolution of the [twin] paradox in special relativity.
| {
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Why does the temperature of the gas in a container moving with constant velocity not change?
Systematic changes do not affect thermodynamic equilibrium.
What does this mean? And what kind of systematic changes are allowed?
The container with gas is stationary till some time then it's given a constant velocity and the final temperature is asked; the answer says that systematic changes don't affect thermodynamic equilibrium and temperature remains constant.
| Where are you measuring the temperature?
If you measure it using a thermometer that is inside chamber, travelling with the gas, then you will not observe any difference relative to when the chamber is at rest (obviously, we are assuming some classical, global reference frame here...).
However, if you measure from outside the chamber, in the global rest frame, then you will find that it is hotter.
To see how the gas can have a different temperature depending on where you measure it from, you have to consider that temperature is just a measure of the momentum change the gas particles undergo when they hit the thermometer. If the thermometer is travelling along with the gas, then it records only their thermal motion in the moving frame. If the thermometer is at rest and the gas whizzes past and hits it, you get the thermal motion plus the relative motion.
| {
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Explanation for concept of relative density I am in class $9^{th}$, I'm reading my physics book in which it is written that: $$\mathrm{Relative\ Density} = \frac{\mathrm{Weight\ of\ solid\ in\ air}}{\mathrm{Loss\ of\ weight\ of\ solid\ in\ water}}$$
As I mention my level that I'm in class $9^{th}$, Plaese help me to understand this in easy method.
| The method can only be used to measure the relative density of something more dense than water, otherwise the object will float.
The loss of weight of the object in water is the upthrust $U$ on the body, from Archimedes' principle (the same as the weight of water displaced).
The weight can be measured by a spring balance, equivalent to the tension in the string.
The tension in the right hand diagram is reduced by $U$, so
$$\frac{mg}{m_Wg} = \frac{mg}{U}$$
where $m_W$ is the weight of the water of the same volume. The left hand side of this equation is the first formula, the right hand side is the second formula.
In practical terms the relative density is then determined by measurements of the tension from the spring balance
$$RD = \frac{T_A}{T_A-T_B}$$
since $mg = T_A$ and $U = mg-T_B = T_A-T_B$
i.e. the weight in air, divided by the loss of weight when immersed in water.
| {
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How to write BdG Hamiltonian in graphene? In Beenakker's paper:Specular Andreev Reflection in Graphene, the BdG Hamiltonian is written as:
$$
H_{BdG}=\begin{pmatrix}H-E_F&\Delta\\ \Delta^*& E_F-H\end{pmatrix}
$$
from equation (1).
Where $H$ is the Hamiltonian of pure graphene and it is:
$$
H=\begin{pmatrix}H_+&0\\ 0& H_-\end{pmatrix}
$$
Where $\pm$ denotes different valleys and:
$$
H_{\pm}=-i\hbar v(\sigma_x\partial_x\pm\sigma_y\partial_y)
$$
Moreover, $H$ is written in the basis of four dimensional spinor $(\Psi_{A+},\Psi_{B+},\Psi_{A-},\Psi_{B-})$
Question is, what basis is $H_{BdG}$ written in? What does the $4\times4$ matrix of $\Delta$ looks like?
Finnally, why the original $8\times8$ BdG equation can be valley decoupled like this (Equ.7 in the paper):
$$
\begin{pmatrix}H_\pm-E_F&\Delta\\ \Delta^*& E_F-H_\pm\end{pmatrix}{u\choose v}=\epsilon {u\choose v}
$$
It is a bit strange because paring is bewteen two valleys, how can we decouple the equation into two seperate valleys?
| It is a very beautiful paper! But as all the old Physical Review Letters a bit cryptic, the supplementary material in the arXiv (http://arxiv.org/pdf/cond-mat/0604594v3.pdf) version is helping a bit.
In the full $8\times8$ Hamiltonian, electrons from valley K are coupled with holes from valley K' via the proximized superconducting coupling $\Delta$, the same is true for electrons from valley K' that are coupled with holes in K. However, there are no other mechanisms that are coupling the two valleys. It means that the $8\times8$ Hamiltonian is composed of two identical $4\times4$ blocks that are not coupled to each other. Here it is assumed that electrons and holes in opposite valleys have also opposite spin so to fulfill $s$-wave pairing.
Regarding your specific question, I think that the $4\times4$, coupling matrix should look like:
$$
\left(\begin{matrix}
0 & 0 & \Delta & 0 \\
0 & 0 & 0 & \Delta \\
\Delta & 0 & 0 & 0 \\
0& \Delta& 0 & 0
\end{matrix}\right).$$
In this way you are coupling an electron $\Psi_\text{A+}^\dagger$ with $\Psi_{A-}^\dagger$. This is what you would expect from a mean field $s$-wave pairing.
You can find a similar calculation done with all the details in the supplementary material of the experimental verification of specular Andreev reflection in bilayer graphene: D. K. Efetov et al., Specular interband Andreev reflections at van der Waals interfaces between graphene and NbSe2. Nat. Pays. (2015). http://doi.org/10.1038/nphys3583 or http://arxiv.org/abs/1505.04812.
| {
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Is it possible to determine the slit a photon went through in the double slit experiment by measuring its flight time? In the double-slit experiment, quantum mechanics states that if you try to determine which slit the photon goes through, you won't have a resulting wave pattern.
But, knowing the time it took for the photon to go from the source to the observing screen, can you deduct the distance of the photon path and so which slit it passes through (except if the photon impacts the exact middle of the observing screen)?
|
But knowing the time it took for the photon to go from the source to the observing screen, you can deduct the distance of the photon path and so which slit it passes through
... and indeed such information will make it impossible for fringes to appear.
Interference experiments use wavepackets that have a long duration, which makes it impossible to tell from timing information which slit the particle came through, eliminating the problem.
| {
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Synthetic Photometry - Calculating a colour index I have a theoretical black body spectrum as described by plancks law. I also have the bandpass sensitivity function for various filters. I would like to calculate a colour index from this information, so I can compare it to an experimental result.
My proposed method is to take the black body spectrum and convolve it with the passband. I would then bin the resulting spectrum and convert to photons using the bin's average wavelength. Summing up the photons should give me counts that can be used to calculate a colour index. This method is described at the bottom of the page here http://spiff.rit.edu/classes/phys440/lectures/filters/filters.html
My question is - is this the correct method? Why is the passband convolved with the blackbody spectrum, rather than multiplied together? Which is the correct method?
| It isn't a convolution, you are just integrating the product. I.e. if your (normalised) filter bandpass is $b(\lambda)$ and the spectral flux from the star is $f(\lambda)$, then the thing you are trying to calculate is a magnitude, which will be given by
$$ m_b = -2.5 \log_{10}\left[ \int b(\lambda) f(\lambda)\ d\lambda \right] + 2.5\log_{10} f_0 , $$
where the $-2.5\log_{10}$ puts the integrated flux onto the logarithmic magnitude system, and $f_0$ defines the flux zero point of that system
| {
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Translation Transformation on the Invariant Interval (Spacetime) So we know that the invariant interval in a two-dimensional spacetime in special relativity is given by
$$ s = -c^2t^2 + x^2 = -c^{'2}t^{'2} + x^{'2}$$
So this scalar should hold true in all frames. I'm trying to show that it is true under a spatial translation, but am not sure how to start.
I tried letting $x' \rightarrow x+\triangle x $. But am not sure how to continue from here. Would appreciate if someone could point me in the right direction.
| If anyone else is interested, I figured it out.
Consider a 2D spacetime
$$\triangle s^2= -c^2(t_2 -t_1)^2 + (x_2 - x_1)^2 $$
A translation would mean $ x_1 \rightarrow x_1+x_0 $.
Now we go back to the unprimed frame and consider a 2-vector $\textbf{A}_i = (t_i,x_i) $. Let me also set c = 1.
To get the difference, $$\textbf{A}_2 - \textbf{A}_1 = (t_2-t_1,x_2-x_1)$$
Going to the primed frame would give
$$\textbf{A}'_2 - \textbf{A}'_1 = (t_2-t_1,(x_2+x_0)-(x_1+x_0))$$
$$ = (t_2-t_1,x_2-x_1)$$
Therefore if we take the square of it,
$$\triangle s^2 = \triangle s'^2$$
edit: didnt notice prahar's comment but yeah! thanks for pointing out my (mis)notation.
| {
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How to interpret the units of the dot or cross product of two vectors? Suppose I have two vectors $a=\left(1,2,3\right)$ and $b=\left(4,5,6\right)$, both in meters.
If I take their dot product with the algebraic definition, I get this:
$$a \cdot b = 1\mathrm m \cdot 4\mathrm m + 2\mathrm m \cdot 5\mathrm m + 3\mathrm m \cdot 6\mathrm m = 4\mathrm m^2 + 10\mathrm m^2 + 18\mathrm m^2 = 32 \mathrm m^2$$
Dimensional analysis tells me that this is in meters squared, if I understand correctly.
Doing the cross product, however, I get this:
$$a \times b = \left[
\begin{array}{c}
2\mathrm m \cdot 6\mathrm m - 3\mathrm m \cdot 5\mathrm m\\
3\mathrm m \cdot 4\mathrm m - 1\mathrm m \cdot 6\mathrm m\\
1\mathrm m \cdot 5\mathrm m - 2\mathrm m \cdot 4\mathrm m\\
\end{array}
\right]
= \left[
\begin{array}{c}
-3 \mathrm m^2\\
6 \mathrm m^2\\
-3 \mathrm m^2\\
\end{array}
\right]
$$
This doesn't make sense to me either.
I don't know if I'm thinking about this in the right way, so my question is this: when dot or cross-multiplying two vectors, how do I interpret the units of the result? This question is not about geometric interpretations.
| On dot product you get magnitude, in units of product of operands.
On cross product you get vectors with direction , in units of product of operands.
| {
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Wave intensity and superposition Let us say we have 2 point sources of sound. My question is how do we consider the intensity to vary according to position?
Let's say both have same amplitude, frequency and speed, just different phase.
Does intensity add up individually or do we calculate the net displacement in pressure due to the superposition and then relate max intensity to regions of max pressure?
Like here, will the intensity at A and B be the same? (Take any 2 points arbitrarily such that constructive interference is happening there)
Our teacher told us this, but I'm not sure about it.
| At a point in space you have two waves arriving with amplitudes $A_1$ and $A_2$ and with wave 2 in advance of wave 1 by a phase angle of $\delta$.
I have chosen amplitudes just to be able to differentiate between the two waves.
It is not unreasonable that you add displacements if you think of one wave trying to displace a particle of the medium through which the wave is travelling by a certain amount and the other wave trying to displace the same particle by another amount.
You add those two displacements to find the resultant displacement.
Since we need to add two sinusoidal functions with the same frequency but which differ in phase, phasor addition can be used.
Using the cosine rule the resulting amplitude $B$ is given by $B^2 = A_1^2 + A_2^2 + 2 A_1 A_2 \cos \delta$.
Since the intensity $I$ is proportional to the amplitude squared
$I \propto A_1^2 + A_2^2 + 2 A_1 A_2 \cos \delta$.
If $A_1=A_2=A$ then $I \propto 4A^2 \cos^2 \left (\frac \delta 2\right)$ and the intensity graph is shown above.
A phase $\delta = 2 \pi$ corresponds to a path difference (pd) of one wavelength etc.
| {
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Physical meaning of $Tr(\rho ^2)$ If $\rho$ is the density matrix of a system then $Tr(\rho ^2) \leq 1$. If the equality holds the system is in a pure state and it is in a mixed state otherwise. But, what is the physical meaning of $Tr(\rho^2)$ ?
$Tr(\rho) = 1$ for all valid density matrices. This stems from the normalization constraint that total probability must be one. Is there any such interpretation for $Tr(\rho ^2)$ ?
| If the system is in a pure state, the density operator is just the projector onto that state, and so $\rho^2 = \rho$. Since, $\mathrm{tr}\,\rho=1$, in a pure state clearly $\mathrm{tr}\,\rho^2=1$.
Since probabilities must be non-negative, $\rho$ has only non-negative eigenvalues ($\rho$ is positive semidefinite). With this and the trace/total probability condition, each eigenvalue $\lambda_i$ for a mixed state satisfies $\lambda_i < 1$. Then $\mathrm{tr}\,\rho^2<1$ for a mixed state.
| {
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Airplane on a treadmill - Variant Thought Experiment This thought experiment is in a way related to the (in)famous airplane on a treadmill problem.
If you take a ball and place it on a treadmill, will the ball:
*
*Move backwards relative to the ground at the same speed as the treadmill (as if placing any other non-circular object on the treadmill)?
*Roll in place without moving relative to the ground (the speed of the treadmill is converted directly into rolling motion of the ball)?
*Exhibit some other behavior such as rolling while also moving backwards?
For this problem assume that there is no slippage between the treadmill and the ball (sufficient friction to make full contact at all times), and assume that the ball has mass.
I know the answer is not #1. I am not sure if the answer is #2 or #3. If the answer is #3, what factors affect the movement of the ball? Is it the mass of the ball, the speed or acceleration of the treadmill, and/or other factors?
| I've thought about the steady state scenario, for a ball rolling without slipping on a horizontal treadmill surface.
The rotational velocity of the ball will be given by $\omega = v/r$, where $v$ is the velocity of the treadmill. However, the motion of the ball's centre of mass due to its rotation will be $u = \omega r = v$, but in the opposite direction. So the ball should be rotating without moving its centre of mass from the external observer's frame of reference.
On an experimental note, there are some videos of a ball on a treadmill available (eg) that show it rolling slowly along with the motion. I think that this is due to slipping of the motion.
| {
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Does rest exist? I initially thought that the concept of rest depended on an inertial frame of reference. So for example, if the Earth and everything on it were the only things in the universe, and the Earth was floating with constant velocity, we would assume anything with the same velocity from an outsider's point of view as being at rest.
However, today, while taking a quick skim at my physics book's chapter on electromagnetic induction, I saw a diagram showing how the movement of a magnet could induce a current on a coil of wire. I quickly concluded that this was just a consequence of how the electrons are moving from the frame of reference of the magnet.
Then I thought about two electrons moving at the same nonzero velocity, assuming that these two electrons, a person "at rest" (basically some person that can measure this velocity and get a nonzero value), and a person moving at this velocity are the only objects/people in the universe. From the frame of reference of the person at rest, the electrons have a net magnetic force that they apply on each other, and they must have this force, else the definition of the ampere makes no sense. However, from the frame of reference of the person who is moving at the same velocity of the electrons, the calculations show that there is no magnetic force.
So does absolute rest actually exist? Or have I made a mistake in my reasoning.
| Whether you observer a magnetic force or an electric force depends on you frame of reference. This doesn't turn out to be a way to establish an absolute rest frame. See Michael Fowler's Modern Physics page from the University of Virginia
| {
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Doubt about Motional Emf In the derivation of a motional emf, we assume that as the conducting rod moves, the free electrons also move in the same direction as the conducting rod. However, the Tolman-Stewart Experiment showed that the motion of a conductor in one direction would lead to the movement of electrons in an opposite direction due to Inertia. Why isn't the same logic applied while analysing the motion of a rod ?
PS. If you apply this idea to analysing the motional emf you will notice that the direction of induced current will change.
| The electrons lag behind. They aren't left behind at rest nor do they move backwards. They still move in the same direction as the conductor, just more slowly.
In the reference frame of the conductor, the electrons move the opposite direction, but that's not what causes a motional EMF. What we care about is the motion of the electrons relative to the magnetic field. Relative to the field, the electrons are moving the same direction as the conductor.
The Stewart-Tolman effect is also for accelerated motion. If the conductor is briefly accelerated and then continues moving at a constant speed, the electrons will eventually reach an equillibrium where they move at the same constant speed as the conductor.
| {
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Quantum versus classical computation of the density of states If I consider for instance N non interacting particles in a box, I can compute the energy spectrum quantum mechanically, and thus the number of (quantum) microstates corresponding to a total energy between $E_0$ and $E_0 + \delta E$. In the limit of large quantum numbers, the result is well known to coincide with the available volume of the phase space of the corresponding classical system of N newtonian free particles in a box, namely
$$
\Omega(E_0,V,N; \delta E)_{\textbf{quantum}} \to \frac{1}{h^N} \int_{E_0<E<E_0 +\delta E} d^{3N}x d^{3N}p
$$
in the limit of large quantum numbers.
My question is the following. Is there any proof, besides this specific example of the quantum gas in a box, that the quantum expression is always going to approach the classical one in phase space, for any given physical system (and thus for some generalized coordinates), provided some classical limit is used?
This does not seem a trivial statement to me, and I can't find the proof in textbooks.
| Well there is a reason in this case of non-interacting particles- it is the so called "Thermodynamic limit". But I can answer this question without invoking the thermodynamic limit.
One very simple way to see this is using $h$. We know $h \ll 1$ so $P = h^{-N}$ for $N \gg 1$ will give you $P \gg 1$. And in some large limit of the number of particles, you can effectively set $ P \rightarrow \infty$ i.e. $h \rightarrow 0$, which happens to give you the classical limit (This is because setting $h = 0$ gives you a classical theory).
| {
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What Color Are Black Holes Really? (Yes, a serious question) So I got into a mini-debate in science class today because I proposed that black holes aren't really black, they only look black because light can't reflect off them. But if you were to take the material that makes up that black hole and decrease it's density so that gravity isn't so strong that light can't reflect, then what color will this new object be? Think of it like this: if you put a red apple in a room that is completely pitch black, the apple will appear black but it's actually red (we just can't observe this because there's no light).
| The material part of a black hole is (classically) compressed into a zero volume area, and almost all of the information of the matter that eventually became the black hole was dissipated away, so the original notion of your question is unanswerable.
There IS another sense in which we can think of your question, though. Black holes are known to shine light through a process known as Hawking radiation, which has a blackbody spectrum. To first approximation, stars are known to also have a blackbody spectrum.
The key point there is that the blackbody spectrum's color is determined by the temperature of the distribution. In the case of stars, this means that the hotter the star, the bluer the color of the star. For black holes, the more massive the star, the redder the blackbody distribution. For black holes that have masses anywhere near that of the sun, the "color" of the star will be very, very far into the radio edge of the light spectrum, and therefore, the black hole will not be visible to the naked eye.
| {
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Are there star systems orbited by stars? I never really heard about such occurencies and now asked my self if this could be possible.
So could there be systems with a star (or black hole) that is so heavy that other less heavy stars are orbiting it?
I could imagine 2 things that would both be a no.
First, this isn't possible for so heavy objects they would just be affect each other and not one beeing a stable center.
Or the second option is this wouldn't be possible withing a galaxy since this would just form a galaxy.
So except the 2 named scenarios, could this happen within a galaxy or would this jsut form something diferent?
| There are binary stars (orbiting around their centre of mass) and there are stars orbiting around neutron stars or black holes (or rather, again, around the centre of mass of the system).
I don't think many stars would orbit a black hole, except... There is the black hole at the centre of most galaxies, including our own. Lots of stars orbit around that - in fact the entire galaxy does. It is possible that some very small stars orbit a massive star or black hole, but I am not aware of the existence of such a system. The stars would have to be very small, possibly even brown dwarfs, as otherwise the centre of revolution of the system would be way outside the primary (as it is in the Pluto/Charon system), and your requirements would no longer be met.
SO yes, there are stars orbiting other things, be they stars, neutron stars, or black holes.
| {
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How can 2 EM waves null each other at a point but continue to propagate? how can 2 EM waves (travelling in opposite directions) null each other at a point in space but continue to propagate beyond the point in space where they interact to null each other?
| You describe an impossible scenario.
If two electromagnetic waves travel in opposite directions and their electric fields point in opposite directions, then their magnetic fields point in the same direction.
If two electromagnetic waves travel in opposite directions and their magnetic fields point in opposite directions, then their electric fields point in the same direction.
More generically, when the electric and magnetic fields both pick up a minus sign, the Poynting vector remains the same. So the momentum density can't point in opposite directions when the electric and magnetic parts of the electromagnetic field are both opposite.
If you want to get super technical: a zero field can satisfy all these things. But then there are no waves and no energy and no dynamics.
| {
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A question about different intensity of a sound source in a room A few days ago, I entered a temple room with dimensions (which are only approximate) shown below in the diagram. There was a low humming sound of the exhaust fan which was at a height of approximately 6 feet from the ground in one corner of the room. Interestingly, the intensity of the sound remained more or less the same throughout the room but increased rapidly as I moved towards the diagonally opposite corner of the room (marked as A in the diagram). Also, the sound was intense only at the height of the exhaust fan and decreased vertically downwards to the level which it was throughout the room. The ceiling of the room was not flat but dome shaped extending from a height of 6 feet to approximately 9 feet. I could not understand exactly the reason as to why only in the diagonally opposite corner of the room the intensity of the sound increased. I feel that the dome shaped ceiling has to be the reason behind it. Can anyone help understand what exactly happens to the sound waves in the room?
| This is related to so called room modes. These are caused by standing waves forming between two walls or between the floor and ceiling. With a room that you described the frequencies where these waves form should be somewhere around 50-100Hz.
What this means in practice is that when standing in a position where such a wave forms, there is a strong boost to a room mode frequency. In your case it happens to be that the noise from the exhaust contains these room frequencies and the phenomenon can be clearly heard.
| {
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What is the importance of vector potential not being unique? For a magnetic field we can have different solutions of its vector potential.
What is the physical aspect of this fact?
I mean, why the nature allows us not to have an unique vector potential of a field?
| There is no "physical aspect of this fact". The physical variables are the electric and the magnetic field, not the potentials. Introducing the potential is aesthetically and technically pleasing, but it is not necessary. A gauge symmetry is not a physical symmetry.
The reason you can have a non-unique potential is that every divergence-free field such as the magnetic field has a vector potential whose curl it is, but adding any gradient to that potential still gives the same magnetic field since the curl of a gradient is zero. The equation defining the magnetic vector potential is simply underdetermined.
Note that even the effect that is usually cited as showing the potentials being "physical", the Aharanov-Bohm effect, does not make the potential unique. The quantity that is relevant is the integral of the vector potential $A$ along a closed loop $\gamma$, and if we denote the region inside $\gamma$ as $U$, we have $\int_\gamma A = \int_U B$ by Stokes' theorem, so what really matters here is the flux through the loop, not the specific value of the potential. And one has to close the loop to observe a phase difference (or, well, maybe not always, but the phase is still only dependent on the flux, not on a gauge-variant potential value). In any case, this is a quantum effect. In the classical theory, the potential is definitely not "physical" in the sense of being measureable.
| {
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Role of Cavity Resonators in continuous wavelength-electron paramagnetic resonance(CW-EPR) Why is it necessary to place the sample in a cavity resonator for obtaining EPR spectrum in CW-EPR? What role does a cavity resonator play in a CW-EPR spectrometer?
| The use of a resonator in X-band EPR is required to overcome the intrinsically low signal over noise ratio, that is of the order of 1ppm. Using a resonator, the oscillating magnetic field is amplified and the electric component is cancelled, avoiding dielectric losses.
To take advantage of a resonator, it must be properly (critically) coupled to the microwave source so that most of the microwave energy remains trapped in the resonator.
That way, when the sample absorbs energy, the coupling is disrupted and microwave energy gets out of the resonator. This reflected component is what makes the EPR signal.
For EPR in the 100+ GHz domain, simple absorption designs can be used, so you don't need a resonator.
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Ohm's law deviation In terms of superconductivities and diodes (I do not know anything else except these), Ohm's law deviate from a linear relation. I search many titles or tags for this but I did not understand properly how it becomes. I wonder somethings related with this.
*
*What is the basic of this deviation?
*How it is deviate?
*Why there is linear relation for metal conductors, if is it true?
| Ohms law states that the voltage and current will maintain a linear relationship under the pretenses that the material is kept at the same physical condition. This often falters at higher voltages/current because of the material increases in temperature resulting in micro changes to the way electrons move.
| {
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How Are Quantum Computers Able to Store Any Data at all? So if qubits can have more than two states, and according to this video,
https://www.youtube.com/watch?v=T2DXrs0OpHU
you don't know what you get until you actually "open the box", if its all randomness and probability,then how can it store anything?
Like, if you tried opening your word document, won't it show up differently everytime you opened it?
Sorry if some of my questions seem stupid, I'm a high school student who has just recently gotten interested in quantum computing.
| When you measure a qubit, you force it to be (and find out if it's) either all-On or all-Off. If it was all-On, flip it over. Now the qubit is definitely all-Off.
Use that process to zero as many qubits as you need, then run your computation.
Note that the process I described isn't creating neg-entropy (which would violate thermodynamics). It's moving neg-entropy out of the environment and into the qubit. Which becomes very clear to me when I look at a circuit diagram of the process:
If the bottom wire didn't start out in a known state (Off in this case), the circuit wouldn't do what the state displays say it's doing. Fortunately we happen to have this giant space nuclear furnace providing an ample supply of neg-entropy.
| {
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"url": "https://physics.stackexchange.com/questions/247755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Diffraction grating (finding number of emission lines) I came across a question on isaacphysics.org where youre given information about a diffraction grating, the spacing of the gaps and the colour and location (in degrees) of 8 different maxima, of four colours.
The question asks how many emission lines are evident from the information given.
My issue is that rather than not understanding the context I just have no idea what the question is asking and what it means by emission lines. Im familiar with diffraction gratings but I've never came across a question like this...
Thanks
| If they are calling the angles listed on the chart emission lines then I see eight all together. Starting with violet at an angle of 32.7 followed by 35.5, 33.1, 35.2, 42.4, 46.6 and 35.4.
After reviewing I'm wondering now if it has more to do with the way the eight points blend together. 46.6 and 46.7 combine to make one line, 35.2, 35.4 and 35.5 combine to make one line and the other three lines are 32.7, 33.1 and 42.4.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how the form factor (stress - torque) is derived? I am working on an experiment of rheology and I need to calculate shear stress in order to calculate the viscosity. After some research I found that for the type of viscometer I will be using (cone-plate), the stress is calculated by dividing the torque given by the apparatus by a form factor equal to $\tfrac{2}{3}\pi r^3$. But I wasn't able to find the explanation of how this factor is derived. Any idea?
| In a cone-plate rheometer the plate is a disc, and the geometry means the strain rate $\dot{\gamma}$ is constant everywhere. The stress is given by:
$$ \tau = \mu\dot{\gamma} $$
Since at equilibrium the viscosity is constant, that means both variables on the right of the equation are constant so the stress is constant everywhere on the plate. Now we just need to work out how the stress relates to the torque.
Let's draw a top view of the plate:
Consider the annulus I've drawn at radius $r$ and with width $dr$. The area of this annulus is:
$$ A = 2\pi r dr $$
Stress is the force per unit area, so force is stress times area giving:
$$ F(r) = \tau 2\pi r dr $$
And torque is just force times (perpendicular) distance so the torque due to the annulus is:
$$ T(r) = \tau 2\pi r^2 dr $$
The total torque is now calculated by integrating:
$$ T = \int_0^R \tau 2\pi r^2 dr $$
where $R$ is the radius of the plate, and this gives:
$$ T = \tau \tfrac{2}{3} \pi R^3 $$
It's conventional to write:
$$ \tau = C_1\,T $$
where $C_1$ is the form factor you mention in your question. So we get the expression for the form factor that you mention:
$$ C_1 = \frac{1}{\tfrac{2}{3} \pi R^3} $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/248028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why can't I use conservation of energy to find ratio of final velocities on different planets? The question is
An object of mass $m$ is allowed to slide down a frictionless ramp of angle $\theta$, and its speed at the bottom is recorded as v. If this same process was followed on a planet with twice the gravitational acceleration as Earth, what would be its final speed?
My idea is to use conservation of energy to solve this.
So,
$K_i + U_i = K_f + U_f$
I'm going to plug in what I think I can (and this might be where I'm going wrong):
$0 + mgh = \frac{mv^2}2+0$
Solving for v:
$\sqrt{2gh}=v$
So if $g$ doubles, then v is multiplied by a factor of $\sqrt 2$. However, the answer is that v doubles as well.
Note that the question here is not what the correct solution is, because I have that. My question is where I went wrong.
EDIT: Since you all say that I'm not doing anything wrong, I guess the question becomes what is the book doing wrong?
Here is the solution that the book gives.
The normal fore will cancel out the perpendicular component of gravity, leaving $mg \sin \theta$ as the net force on the object.
$F_{net} = mg \sin \theta = ma$
$a = g \sin \theta$
This shows that $a$ is proportional to $g$. Then, using $v = v_0 + at$ we can see that the final speed is proportional to a. So if this planet has double the value of $g$, the object will experience double the acceleration, leading to double the final speed.
| The velocity does not double if the acceleration is doubled. The relevant SUVAT equation is:
$$ v^2 = u^2 + 2as $$
where in this case $u=0$ so we get:
$$ v = \sqrt{2as} $$
A doubling of acceleration means that the velocity would double if the travel time was kept constant. However in this case it's the travel distance that is held constant. The greater acceleration means the object covers the constant distance in less time, so the doubled acceleration acts for a shorter time. That's why we get the square root dependence of velocity on acceleration.
| {
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"url": "https://physics.stackexchange.com/questions/248270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Explanation of crystal formation in a Fitzroy's storm glass with the change in weather conditions I was recently reading about Admiral Robert FitzRoy's storm glass and how it was used to predict the weather conditions based on the formation of camphor crystals in a sealed glass tube containing distilled water, ethanol, potassium nitrate, ammonium chloride, and camphor.
I started searching for the possible explanations of its working. The first one I encountered stated that it only worked on the changes in atmospheric temperature and was nothing but a thermometer. But there were other articles (like How to Make Fitzroy's Storm Glass) stating that there are also some electromagnetic and quantum (quantum tunneling) explanations for the appearance of crystals and how the atmospheric pressure affects the crystal formation even though the glass tube is SEALED.
I searched for these explanations but couldn't find them. Does anyone have any idea how one can apply the concepts of electromagnetism and quantum tunneling to explain the formation of crystals with the change in weather and how atmospheric pressure might influence the contents of a sealed glass tube.
| Recently, as part of the debate on climate change, it has been proposed that cosmic rays have an influence on the nucleation of water particles that initiate the formation of clouds. It is possible that a similar effect aids initiation of camphor crystals. As clouds mean rain then perhaps this explains the clouding of the solution. Clearly temperature will also have some effect as the combination of clouding and lower temperature would result in more crystal formation indicating snow. One wonders if siting the storm glass in full sun or in shade would make a difference on its performance as one would expect no crystals if the glass is in full sun. Another issue is that the weather tomorrow is more likely to be the same as the weather today than it is to change so the predictive capability of the glass must be questionable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/248377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Methods for handling close approaches in $N$-body simulations In direct gravitational $N$-body simulations, what are the preferred methods for handling close approaches between bodies in order to preserve the accuracy of the evolution of the system?
| I assume you're talking about the numerical instabilities that arise from having an infinite potential at $r=0$. Here are three common solutions:
*
*Use a soft-core potential that behaves like $1/r$ except very close to $r=0$ where it levels off to a finite value. For example, $1/\sqrt{\epsilon+r^2}$ instead of $1/r$ is common.
*Add hard sphere collision detection (ideally this would incorporate an event-driven integration step, so it can be quite tricky to implement if you want to do it properly).
*Use a dynamic integration time-step that is a function of the distance between the nearest pair of particles. When they're far away you can use a large time-step, when they're (very) close you use a (very) small time-step.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/248522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Can lasers lift objects? I have been fascinated by a very intriguing question - Can lasers push objects up?
I have done the below math to find out
Lets say we have a $1000~\text{mW}$ laser and we would like to lift an object of weight $100~\text{g}$.
By definition: $1~\text{W} = 1 \frac{~\text{J}}{~\text{s}}$
That means the laser is emitting $1~\text{J}$ of energy per second.
On the other hand energy required to lift an object off the ground is given by $m \cdot g \cdot h$.
Putting in the number and lets say we want to solve for
$0.1~\text{kg} \cdot 9.8 \frac{~\text{m}}{~\text{s}^{2}} \cdot h = 1~\text{J}$
So, $h \approx 1~\text{m}$.
You see, if we had a $1000~\text{mW}$ laser we could lift an object of $100~\text{g}$ weight up to 1 meter in one second.
I can't see anything wrong with the above math. If this is correct, can anyone tell me then why on Earth we use heavy rockets to send objects into space?
| Laser is stimulated emission of highly energetic photons. Fundamental use of laser is heating, propulsion is very distant aim which lasers can achieve.
Few kW rating lasers can actually lift the mass (very small values though) because incident energy beam has momentum associated with it. Your assumption is not correct as you are comparing heat energy with potential energy (with 100% conversion efficiency). One simply can't achieve mechanical power of scale you specified using lasers.
The simple fact that process efficiency is very low in case of laser limits our aim to use lasers for rocket propulsion. Typical laser efficiency is nearly 5-10 % of input energy (typically electricity). Converting heat energy to mechanical energy will also have many losses, extra systems will also be needed.
Video for reference: https://www.youtube.com/watch?v=3F1FDwg4XRc
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/248649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 3
} |
What is the moment of inertia really? Is moment of inertia or second moment of inertia, simply the resistance of a body to rotate it over an axis? What is radius of gyration? What if the axis is via the center of mass or somewhere different? can you give me please an overview of these issues with SIMPLE words, and without nonsense, like maths who nobody will ever remember. I need the SENSE how the brain comprehends these stuff in simple terms.
| Basically, it is how hard it is to spin an object. If you know what regular inertia is, moment of inertia is the rotation equivalent of it.
Regular inertia is how hard it is to push an object, and only depends on its mass. For instance, picture an ice rink with a hockey puck. If it is very light, then it is very easy to push it across. However, for a very heavy puck, it will be hard to push it across.
Similarly, rotational inertia depends on its mass, but it also depends on its shape and where the it rotates. Imagine spinning a top. It should spin in nice circles. If you glue some stuff onto it unevenly, it will be harder to spin because you have changed the shape of it, increasing the moment of inertia.
Imagine a well shaped wheel. It will rotate nicely. If you put axle of the wheel somewhere other than the center, it will be harder to turn, because you increased its moment of inertia.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/248752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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How to determine the direction of arrow on Feynman diagram for $W$ boson line? I am somewhat confused. Looking through these slides (especially the 11th), which show Feynman diagrams involving $W$-bosons, I can't figure out which way to draw the arrow near the $W$ boson? How do people determine if it is to the "right" or to the "left".
| In general in Feynman diagrams an incoming particle can be read as an outgoing antiparticle and W+ is the antiparticle of W- and vice verso. Quantum number conservation holds at the vertices. (charge , lepton number..)
The reaction studied in 11 is a change of a proton to a neutron through the weak interaction. The charge of the proton has to go to the right . The diagram has a W- going to the left, i.e. a W+ which is what is necessary for charge conservation on the second vertex.
For 13, the reaction is a neutron turning into a proton by colliding with a neutrino. It has an arrow to the left and when read towards the lepton vertex it is a W- , mathematicaly, which is what is needed for charge conservation.
For 15, the reaction is antineutrino proton , turning into e+ neutron. The arrow correctly conserves charge at the vertices.
Seems to me whoever wrote the site has been playing games to make students think?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/248840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which coordinate is to be considered for the energy of simple pendulum? For an simple harmonic oscillator energy can be represented as in picture. Consider in particular picture (b) with the energy as a function of the coordinate $x$.
Consider now a simple pendulum. The coordinate $x$ in (b) is the coordinate of an horizontal axis (as in picture 1) or the coordinate or the circular trajectory, as in picture 2.
The motion of the pendulum is indeed a one dimensional simple armonic motion, but the path followed is circular, so I guess that the "$x$ coordinate" of the graph (b) is the one in picture 2. Is that correct?
| The potential energy of the pendulum is $U(θ)=mgl(1-\cos θ)$. For small angles, $U(θ)≈mglθ^2\!/2$ and you get a harmonic oscillator. So $θ$ (or equivalently $lθ$) may be taken as the oscillating variable. However, since we are considering small angles, we may as well use $x=l\sin θ≈lθ$.
Addendum: You may wonder which approximation is better.
Let's look at the next term of the Taylor series: $$1-\cos θ≈θ^2/2-θ^4/24.$$
Thus, taking $1-\cos θ≈θ^2/2$ overestimates by $θ^4/24$. Now, $$\sin θ≈θ-θ^3/6 → (\sin θ)^2/2≈θ^2/2-θ^4/6$$ so that taking $1-\cos θ≈(\sin θ)^2/2$ underestimates by $θ^4/6-θ^4/24=θ^4/8$. So, approximating by $θ^2/2$ is better than approximating by $(\sin θ)^2/2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/248957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why doesn't a swing make a full revolution on a swingset in a park? I'm familiar with the concept of swinging of a swing in a park, but I'm confused why it doesn't it swing at a complete circle around the center bar?
| why it doesn't it swing fully ?
I think the question does point to the fact that swings which are usually available in the park does not provide free swinging , low amplitude of the swing and needs constant pushing.
All the above is related to energy dissipation of the initial potential energy provided to the swing- and the dissipation is at the hinges provided at the support point from where it is hung .
If the load is heavier the frictional force operating at the hinge becomes substantial and a damping /slowly decreasing amplitude is observed.
As the park swing is made for children its in a way safe also as larger amplitude may lead to accidents by falling off from the swing as braces to hold the children is not normally provided.
the swing made out with iron chains is far from an ideal simple pendulum hung from a rigid support.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/249124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Deviation of free falling objects (Coriolis effect) using conservation of angular momentum I read this pdf on non inertial frame, in particular I have a question on the deviation of free falling object due to Coriolis effect.
Consider a ball let go from a tower at height $h$. The displacement due to Coriolis effect, calculated with formulas in Earth system, is $(4.19)$, after it there is explanation of the effect that uses the conservation of the angular momentum of the ball in a inertial frame.
$$x =\frac{2\sqrt{2}ωh^{3/2}}{3g^{1/2}} \tag{4.19}$$ Just before being
dropped, the particle is at radius $(R+h)$ and co-rotating, so it has
speed $(R+h)ω$ and angular momentum per unit mass $(R+h)^2ω$. As it
falls, its angular momentum is conserved (the only force is central),
so its final speed v in the (Eastward) direction of rotation satisfies
$Rv = (R+h)^2ω$, and $v= (R+h)^2ω/R$. Since this is larger than the
speed $Rω$ of the foot of the tower, the particle gets ahead of the
tower. The horizontal velocity relative to the tower is approximately
$2hω$ (ignoring the $h^2$ term), so the average relative speed over
the fall is about $hω$. We now see that the displacement $(4.19)$ can
be expressed in the form (time of flight) times (average relative
velocity) as might be expected.
But $$v_{average} t_{flight}=h \omega \sqrt{\frac{2h}{g}}$$
Which differs by $\frac{2}{3}$ from $(4.19)$. Is that due to the approximation made?
I also don't understand completely why the average relative velocity $v_{average}$ is taken to be half the relative velocity found. Isn't this valid only for constant accelerated linear motions?
| Considering conservation of angular momentum for the dropped ball, $\omega(z)$, the angular velocity of the ball as a function of z, is not constant for the dropped ball. $\omega(z) = {(R + h)^2 \over (R + z)^2} \omega_e$, where $\omega_e$ is the angular velocity of the earth. As the ball falls, $z$ decreases and its angular velocity increases. The answer by @Diracology assumes $\omega(z)$ is constant at $\omega_e$ for the dropped ball; this is good approximation for $h << R$ and hence $z << R$.
Eqn. 4.19 assumes the ball is dropped at a latitude of zero degrees where $\vec v = \vec \omega \times \vec R$ has magnitude $\omega R$. In general, $\vec \omega \times \vec R$ has magnitude $\omega R \enspace cos\lambda$ where $\lambda$ is the latitude. Considering the latitude, eqn. 4.19 should be multiplied by $cos\lambda$.
See the textbook, Fowles Analytical Mechanics, for the derivation of eqn. 4.19 in the non-inertial frame considering latitude and you will find the factor $cos\lambda$ in the result.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/249423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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