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Regarding the Boltzmann entropy formula, is the Boltzmann constant really arbitrary? In the top answer to this question (Is the Boltzmann constant really that important?) I read that the Boltzmann constant is just a dummy factor which converts energy to temperature.
But that allows me to put another numerical value in place of the Boltzmann constant but keep the dimension J/K.
I.e. what if, in
$$S=c\ln W,$$
I put $c=56 \, \mathrm J/\mathrm K$ in place of $c=k\approx 1.38\cdot 10^{-23} \, \mathrm J/\mathrm K$?
On the page Thermodynamic beta, the Boltzmann entropy using the Boltzmann constant implies thermodynamic beta, which implies (according to Derivation of Boltzmann Distribution Law) the Boltzmann distribution.
So the Boltzmann distribution depends on the numerical value of Boltzmann constant. Then why is the Boltzmann constant just a dummy factor?
For example, the mean speed of molecules depends on
$$S=k\ln W.$$
Changing the numerical value of $k$ would make the speed totally different.
| In thermodynamics, the temperature is actually defined in terms of the entropy:
$\frac{1}{T} = \left(\frac{\partial S}{\partial U} \right)_{V,N}$.
Therefore, if you would change the definition of the bolzmann constant to $k' = a k$, the temperature would scale accordingly with a factor $1/a$, and the combination $kT$, which appears in the Boltzmann distribution, would remain the same.
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How are atoms supported on each other in a material? Suppose we have a ball made up of iron. There are a "lot" of atoms in the ball.
My question is "how" are the atoms supported on top of each other?
And, is it due to the repulsion of electrons the atoms maintain distance between themselves?
| The atoms in any object never come into contact with one another for exactly the reason you stated - the outermost electrons repel. Furthermore, the boundaries of an atom are ill-defined in quantum theory, meaning that to discuss contact between them does not make much sense.
In the case of iron, a metal, the positive metallic ions are strongly attracted to free elctrons within the metal. In this way the arrangment is very similar to that of water, with a "surface tension" cause by the continuous electric force between the ions and the electrons. The balance between repulsion and attraction is a very stable one in metals.
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Physical intuition behind torque converter A torque converter (also here) is a device used in some cars. It uses several "fans" coupled through a liquid (transmission fluid) in order to perform the function of a clutch, but more importantly it acts as a liquid gear in the sense that it multiplies the torque going from the engine to the wheels.
Is there an intuitive way to explain what is happening in the liquid? In particular, is it possible to explain the torque multiplication effect without resorting to numerical analysis?
| Check this video below that perfectly describes the operation of torque converters.
Fluid Coupling: Principles of Operation (1953)
Torque converters (TC) are defined by a torque ration, which is in turn defined by the turbine design withing the converter. There are two turbines, as described in the video. If you make a TC out of two identical turbines you will get 1:1 ratio. The torque remains the same. This is very similar to the operation of the transformer.
However, if you change the design of the output turbine so that is increases or decreases the torque (this is done by changing the number and/or shape of turbine blades) you will get different torque ratios too.
One turbine propels the fluid and serves as a pump. The other turbine is driven by the moving fluid. In this way, two turbines in TC are coupled by the liquid. BY changing the design of secondary turbine you can make changes in the value of the output torque. Read Euler's pump and turbine equation.
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Why does the potential difference across a type of parallel circuit not act like a potential divider?
Image credit (Q3)
In this attached circuit, when $R_1=0\Omega$, I am failing to understand how the two cells affect the potential difference across the central resistor R3. I understand that potential difference is constant across different strands in parallel, and that so 12 volts should be distributed between R2 and R3 as a potential divider, and so 9V should be across R3 from V2. Likewise, from V1, 10V will cross R3 from V1, and so the total potential difference across R3 should be 19V.
However, according to the answer sheet, the potential difference across R3 is 10V. Is my misunderstanding here conceptual, or something more basic, and why does potential difference act in this way?
|
so 12 volts should be distributed between R2 and R3 as a potential divider
If you're trying to solve this with superposition, you've forgotten a step or two.
When solving for the effect of the 12-V source, you need to zero out the other source. So you set $V_1$ to 0 V. Then R1 is in parallel with R3, and you need to use this parallel combination as the lower half of the voltage divider, not just R3.
If R1 is 0, this means that the voltage across R3 due to $V_2$ is zero.
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What are “vibrations” in a field in Quantum Field Theory (QFT)? For example, in a 2013 article for NOVA Don Lincoln writes:
Everywhere in the universe there is a field called the electron field. A physical electron isn’t the field, but rather a localized vibration in the field. In fact, every electron in the universe is a similar localized vibration of that single field.
https://www.pbs.org/wgbh/nova/article/the-good-vibrations-of-quantum-field-theories/
He doesn’t really talk about where he came up with the word “vibration,” but uses it as if it’s a normal term. Is this a pop-sci metaphor, or is this a real thing? What solid intro to QFT can I read for a reliable translation of the science to a college grad who took two semesters of physics (basic Newtonian physics, and electromagnetism) ages ago?
Will such an intro likely use the word “vibration” and explain why we use that instead of “particle,” or is “vibration” a terrible approximation for the real concept, and a reliable intro would never use that term?
| The word vibration has more of a historical sense, and we use excitation more often. In QFT, real particles are excitation of the underlying field.
Now the reason why we use vibration is because historically, these fields were modeled through waves, mathematically, and if you imagine a guitar string, and create a vibration, that will create a sound, a real thing, and so historically we use the expression vibration because it is very similar to a water surface where if it gets disturbed (excitation), it will create a wave, and these waves are modeling real particles in our currently accepted theories, the SM, GR, and QM, together with QFT.
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Information content in black holes Bekenstein-Hawking formula for entropy of a black hole tells us that information content in a black hole is proportional to its area which is in fact proportional to the mass^2 of the black hole. The information content before the formation of the black hole can be different which has nothing to do with the mass. Is there any kind of information loss during the formation so that we get two black holes with an equal amount of information. Is the information due to the fields lost in the process? So we get just the bare information which comes from the mass.
|
Is there any kind of information loss during the formation so that we get two black holes with an equal amount of information.
If I'm understanding correctly, this sentence is describing two different sets of initial conditions leading to two black holes with the same area, and therefore the same entropy.
You're using "information" as if it were a synonym for "entropy," but they aren't the same thing. For example, information is always conserved in quantum mechanics, in the sense that the time evolution is unitary, but entropy increases. Entropy can be interpreted as a measure of the amount of information in a system, but only when we take into account the coarse-graining of the phase space.
Gravitational collapse to a black hole doesn't involve a loss of entropy, it involves a gain in entropy. It's not particularly mysterious per se for two different initial systems, with two different amounts of entropy, to evolve into final states with the same amount of entropy. For example, suppose we take two different samples of an ideal gas, in two different nonequilibrium states, but both with the same number of molecules and both confined to the same volume. They will both gain entropy, and will both end up in the same maximum-entropy macrostate.
We don't have a theory of quantum gravity, but the rough picture is probably that after the black hole has formed, there is a large amount of energy present in its microscopic degrees of freedom, and this is where the entropy resides.
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Does light take the path of least time because it travels in straight lines or vice versa? My question is which of these two feats is a consequence of the other?
Light travels in straight lines, mostly. Does it do that as a result of Fermat's principle of least time? and if so, is there a reason as to why it follows the path of least time? or is this another "that's the way the universe works" question?
And by reason I mean a physical explanation not mathematical deduction.
Or is it the other way around? meaning light taking the path of least time is just an obvious manifestation of the fact that it goes in straight lines?
| The fact that light travels at straight lines has the same reason as any particle moving on straight line if there is no external agent to change its path - the fact that spacetime is homogeneous and isotropic in any inertial frame of reference (actually that is the definition of inerial frame of reference).
In general relativity, there are (in general) only local inertial frames, and then the straight lines are called geodesics.
Light would move on straight lines due to symmetry even without fermats principle. The fermats principle is usefull when there is no symmetry and the actual paths must be computed
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OPE of three operators In the process of thinking about this question, I realized that I don't understand something very fundamental about operator product expansions.
Consider a product of 3 local operators in a 2d CFT:
$$ X(x) Y(y) Z(z) = \sum_{n=-N}^{\infty} A_n(x) Z(z) (x-y)^n, $$
where we have substituted $X(x) Y(y)$ for the $XY$ OPE. This expression contains the singular terms for $x = y$.
Now because by definition of OPE $A_n(x)$ is a local operator at $x$, we can use the $A_n Z$ OPE again:
$$ X(x) Y(y) Z(z) = \sum_{n=-N}^{\infty} \sum_{m=-M}^{\infty} B_{nm} (x) (x-y)^n (x-z)^m. $$
This expression contains the singular terms for $x = y$ and $x = z$.
Question: where did the $y = z$ singular terms go?
This is likely related to the convergence of the series, but I wasn't able to formulate a convincing argument.
| Let me redo the calculation, while explicitly writing OPE coefficients. Let $(A_n(z))_n$ be a basis of operators at $z$. We use the two OPEs
$$
Y(y)Z(z) = \sum_n c_n(y,z) A_n(z)
$$
and
$$
X(x)A_n(z) = \sum_m d_{m,n}(x,z) A_m(z)
$$
We end up with the result
$$
X(x)Y(y)Z(z) = \sum_{m,n} c_n(y,z)d_{m,n}(x,z) A_m(z)
$$
where the coefficient of $A_m(z)$ is $\sum_n c_n(y,z)d_{m,n}(x,z)$. This coefficient is an infinite sum, and it can very well be singular as $x\to y$, although this is not manifest. For instance,
$$
\frac{1}{x-y} = \sum_{n=0}^\infty (y-z)^n (x-z)^{-n-1}
$$
You can recover a similar result in your calculation by distinguishing more clearly the operator basis from the OPE coefficients. Your operators $B_{m,n}(x)$ should not all be linearly independent, and you should rewrite them in terms of a basis of operators.
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Is motion smooth? It's obvious that for every particle velocity is smooth i.e it cannot undergo sudden finite change in its position in infinitisiminal time.
Similarly any particle's velocity cannot undergo a change instantaneously (Infinite acceleration can't happen, intuitively).
Does this pattern apply to higher time derivatives of position like jerk? If yes then till how much higher derivative? 10th? 100? Infinite?
| Typically these higher derivatives are assumed to be smooth.
The key question will be what causes a discontinuity in the n-th derivative. If you focus on classical mechanics, the forces on an object boil down to the positions of particles in the system, which are continuous. This means there would need to be a discontinuity (such as a divide by zero) in the equations of motion in order to have a non-smooth higher derivative.
When you push towards quantum mechanics, the terms get really murky quickly, because position ceases to be a single observable number. But if you stick to classical mechanics, we find things stay nice and smooth.
Now that being said, when modeling real systems, we very often assume instantaneous changes in velocity or higher derivatives. This is because, in many cases, we can get away with ignoring the precise acceleration or jerk function and treat it as-if it were a simple discontinuous system. A straight forward example of this is a billiard ball collision. For most intents and purposes, this collision is "instantaneous" and the velocity of the balls changes in a discontinuous manner. However, if we look closer, with a slow motion camera, we find that the collision is not actually instantaneous -- position and its derivatives smoothly change over time. In fact, if you look hard enough, you can even see the ripple as the effects of the impact race across the surface of the ball. But, for the purposes of determining the result of a trick shot, these fine details are immaterial, and calculations assuming an instantaneous change in velocity are used.
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2 windows - will I see the reflections?
I have a question regarding photons nature. Let's say I have a single source of light - regular bulb and the observer - in the same room.
The observer looks through a glass window (normal glass window-nothing special about it) and sees his reflection, but some of the light is passing through the window.
Now I put a second window - as presented on the attached photo. Will the second window reflect some of the light back to the observer, or it will pass 100% of the light forward?
This may be pretty basic but this question sparked a discussion and there where no definitive answer.
| It will reflect some of the light back, depending on the thickness of the glass. The partial reflection of light from the two glass windows varies from 0%-16% (that is, on average 8% of the light is reflected back by both windows combined). Definitely read QED by Feynman for more detail.
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Movement of bubbles in a cup of tea Why is it that when I try to scoop bubbles out from the surface of a cup of tea, they always slip off the spoon even if some of the tea does not?
| For contrast, consider what happens when you scoop dirt with a shovel. As you lift the shovel, a pile of dirt stays on the shovel, and an empty space opens up below the shovel as you lift it. You can lift the pile of dirt all the way up out of the hole, above the level of the surrounding ground, and the dirt still stays piled up on top of the shovel, leaving a hole underneath. That's good, because otherwise trying to dig a hole with a shovel would be a very frustrating experience.
Tea is different. As you lift the spoon through the liquid, the liquid above the spoon does not remain in a pile on top of the spoon, and it does not leave an empty space underneath the spoon. Instead, the liquid above the spoon runs off to the sides to keep the surface of the liquid at a uniform height across the width of the teacup, and liquid from around the sides of the spoon flows underneath to fill in what would have been the empty space. Bubbles float on top of the liquid, so they follow the flow of the surface-liquid that would have been lifted up in a pile (if it were dirt) as it spills off the edges of the spoon to keep the level uniform.
Some tea remains in the spoon, but only enough to fill the spoon up to the height of the spoon's edges. By the time you've lifted the spoon from even a slightly lower depth, the liquid that was above that height has already flowed off to the side, carrying the bubbles with it... unless you're able to keep the bubble so perfectly centered in the spoon that it can't decide which way to go, like balancing a pencil on its tip. I haven't tried this, but I'll bet it's very difficult.
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Can a wire having a $610$-$670$ THz (frequency of blue light) AC frequency supply, generate blue light? We know that when we give alternating current across a wire then it will generate an electromagnetic wave which propagates outward.
But if we have a supply which can generate 610 to 670 terahertz of alternating current supply then does the wire generate blue light?
| Even at tens of gigaherts, one does not carry current "in" a conductor -- it is carried along the outside (Google "skin effect").
There are transmission lines for high-frequency RF that basically launch an RF wave along a single naked wire, and catch it at the other end -- think of a coax without the outer shield. If you take this analogy and pursue it into absurdity and beyond, then if you take a really well polished wire, and really carefully launch blue light along its length, then as long as the wire doesn't bend too suddenly, the light -- or some portion of it -- will be refracted and "stick*" to the wire.
I think you could achieve a setup in a lab that involved people looking at a blue-glowing end of a carefully-maintained copper wire or gold wire and going "oooh!". I doubt there is much potential for practical use here.
* Imprecise language used on purpose -- I'd have to do a lot of work to do the math on this one!
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Does rotation happen throughout the whole x axis of an object simultaneously? For example, if I draw a line on the side of a pencil top to bottom, then snap one end of it as in launching it due to the pressure of my fingers. Anyways, if I record the pencil launch in slow motion (perhaps it’s my phone that has to do with it) but it will focus on where the line was, and it appears that only some of the actual line is there, or out of focus. So that leads to the question, does rotation happen simultaneously down the pencil axis? Maybe I’m completely clueless and I’m missing something but figured I’d ask.
| In general for a rigid body you have the combined motion of translation of the center of mass and rotation about the center of mass at any instant.
The COM moves with $\boldsymbol{v}_C$ and the body rotates (simultaneously) with $\boldsymbol{\omega}$.
So the motion of any other point P, located at $\boldsymbol{r}$ relative to the center of mass is
$$ \boldsymbol{v}_P = \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{r} $$
The particulars on how much translation vs. rotation happens depend on the situation. This is the area of kinematics, where constraints on rigid bodies only allow a certain subset of the motions that a body would normally have.
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How much energy is needed to make fire? I'm so curious about fire. So I searched a lot in the internet. And now, I knew that fire is some kind of chain reaction and combustion energy make the other molecule hot and the other molecule makes other chemical reaction and so on...
Then does the first given energy make this reaction start and maintain chemical reaction(fire)? if so, if a long time passes, fire goes out? and how much energy is needed to make this chain reaction start?
to sum it up,
*
*when cigarette lighter make candlelight start how much energy is needed?
*if a long time passes, this candlelight goes out?
(Assuming that oxygen and material(will be burned) are supplied)
*how much energy go out with light or heat?
(I'm not good at english, so please understand me)
| The amount of energy needed to start a fire depends on the combustibility of the substrate. The energy supplied needs to be enough to heat a small quantity of the material to its ignition temperature. If it is something highly inflammable like methane,hydrogen or aviation spirit,a tiny, almost invisible spark is enough to do the trick.a spark of static electricity for instance. For something less combustible,like wood or coal,a lot more energy is required to heat it to ignition temperature & maintain that temperature till it is well ablaze.
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Difference continous - discrete symmetry I am trying to understand the difference between the two types of symmetries.Wiki
Wikipedia says that
*
*Translation in time : $t \rightarrow t + a$
is a $\textbf{continuous}$ symmetry, for any real $t,a$
but
*Time reversal: $t \rightarrow -t$ is a $\textbf{discrete}$ symmetry.
But if we choose $a=- 2t$ we get the same transformation - can somebody explain to me why this is no contradiction?
| When we say a symmetry is continuous, it is shorthand for saying that the group of symmetries is continuous.
For time translations, the group consists of all translations for any a, where $a$ is a continuous parameter. There are an infinite number of these group elements, and there are elements of the group which are infinitely close together.
For time reversal, there are only two group elements and they are not close together in any sense.
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Unorthodox way of solving Einstein field equations Usually when we solve field equations, we start with a stress energy tensor and then solve for the Einstein tensor and then eventually the metric. What if we specify a desired geometry first? That is, write down a metric and then solve for the resulting stress energy tensor?
| You can certainly go from the metric to the energy-momentum tensor, but then you’re not “solving” anything. There are no differential equations to be solved if you’re doing that. It’s just a straightforward, although often tedious, computation (of the Einstein curvature tensor, which is proportional to the energy-momentum tensor) that requires nothing more than differentiation and algebra.
It’s not a particularly useful thing to do. Trying lots of metrics and seeing what density and flow of energy and momentum they correspond to doesn’t really give you insight. It’s generally the energy-momentum tensor that is simple, and the metric that’s complicated, so you need to start with the former and solve for the latter.
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Why does a helium balloon rise? This may be silly question, but why does a helium ballon rise? I know it rises because helium is less dense than air. But what about the material of the ballon. It is made up of rubber/latex which is quite denser than air. An empty ballon with no air in it falls, so why does a helium filled balloon rise?
| Lots of good answers here, but none address the main issue with your questions. The balloon doesn't rise due to any direct mechanism. Instead, the denser air around the balloon pushes itself down, with the net effect pushing the balloon up.
This is similar to the misstatement that 'heat rises'. Again, it doesn't. Instead, cold falls, pushing the heat up.
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How can system of charges be static? I am a beginner in electrostatics.
I don't understand the below:
When we have a set of charges, those charges exert a force on the test charge but what I don't understand is how are these set of charges at rest?
Isn't this against the Coulomb's law?
Won't each charge affect each other causing them to move and not letting them to be in rest?
| Remember that for charges to flow, we require a potential difference, and a low resistance path(not an insulator). So, charges can't flow in case of insulators, when no free charges are available, even if there is a difference of potential.
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Temperature relation to Archimedes principle I have a probably very simple problem which I cant solve. I am weighing animals in seawater by placing them on a small PVC arm that is fixed on a balance and reaches into a beaker of seawater. On the arm, I place an organism and measure the specific weight of the animal. Now, the weight of the animal changes a lot between days which cannot be explained by physiological processes. I believe this might be due to the water temperature which I also collect alongside. Can anybody tell me how to correct for water temperature if I dont know the volume of the organism?
Sample data:
| It is highly unlikely the differences you are seeing are due to temperature unless this is occuring outdoors.
As you can see on this page, the density of water changes less than 1% between -20 and +20 C. If you are indoors, it is highly unlikely it's changing outside +15 to +25, which would be much less than 1%.
Even outdoors, even day/night changes would be unlikely to have a measurable effect, and a full winter/summer change would only be 2 or 3%.
What sort of differences are you actually measuring?
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Why does the Schrödinger equation work so well for the hydrogen atom despite the relativistic boundary at the nucleus? I have been taught that the boundary conditions are just as important as the differential equation itself when solving real, physical problems.
When the Schrödinger equation is applied to the idealized hydrogen atom it is separable and boundary conditions are applied to the radial component. I am worried about the $r=0$ boundary near the nucleus. Near the proton, the electron's kinetic energy will be relativistic and looking at the Schrödinger equation itself for how this boundary should behave seems dangerous because its kinetic energy term is only a non-relativistic approximation.
Is there any physical intuition, or any math, that I can look at that should make me comfortable with the boundary condition in this region?
| The boundary conditions that pick up the hydrogen wave functions are the "constraints" placed on the wave function solutions. Remember that the observable is the probability distribution from the $Ψ^*Ψ$, not a particular location. Please read the link. The solutions are within the quantum mechanic postulates after all.
There isn't any relativistic boundary condition, because there are no orbits, only probability distributions.
Thus the solutions do not have a singularity at r=0, and in general there is a small probability of finding the electron at the origin, if the quantum numbers allow for an interaction, as with electron capture in nuclei. For the hydrogen atom there is not enough energy for a neutron to appear.
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If entropy decreases for cold systems, isn't the heat death of the universe a state of low entropy? Entropy is a consequence of heat. The heat death of the universe results in an approach to absolute zero temperature. Does this mean the end of the universe is low entropy?
| First of all, I agree with @anna v answer. I just want to expand (no pun intended) on the last paragraph.
Entropy is related to the dispersion of energy. In that sense, it is related to the universe gradually moving to its most probable state on order that all the energy is equally dispersed throughout the universe. At the present time energy is concentrated in particular areas (you might some of them "hot spots") but as the theory goes it will eventually spread out all over the universe (whatever all over means) because that is the most probable state.
Now as the temperature of the universe goes down there is less energy available to do work. A the present time we take advantage of temperature differences and the fact that heat flows from high to low temperature by taking some of that heat from the high temperature body and doing work in a cycle, while necessarily rejecting some of it to the lower body due to the second law. The energy rejected to the low temperature body is no longer available to do work in the current environment. We would need to take our heat engine and operate it between the lower temperature body and yet another even lower temperature body. Each time we do this we have less of the original heat energy available for work since our engine can never be 100% efficient. What's more we need to continue seeking lower temperature bodies to reject heat to.
Eventually as we approach absolute zero we will no longer be able to find a lower temperature body. In other words, the temperatures will be equal and we will no longer be able to do work.
Hope this plus anna v's answer helps.
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Is plane altitude limited by engine power and if so does air density cause this? I notice that, for example, human-powered flight operates at low altitudes. This might of course be due to safety but I wonder if in fact the delta in air pressure is greater at lower altitudes and this prevents low-powered aircraft from reaching higher altitudes?
| Human-powered flight is only barely possible because humans cannot generate large amounts of power. Human-powered planes therefore fly almost exclusively in what is called ground effect, where the plane is no more than about one wingspan off the ground. When flying this close to the ground, the plane is partly supported by a "bubble" of air that is caught between the underside of the wing and the ground, which reduces the amount of drag experienced by the plane and enhances the lifting power of its wings. This in turn reduces the power requirement to keep the plane in the air.
Regarding small planes with low-powered (piston) engines, they are limited in how high they can fly because as they climb they move into air which is less dense and which therefore furnishes less oxygen for the engine to burn with its fuel. This causes the power output of the engine to decrease with increasing altitude, and at some point the engine's power output is so diminished that the plane can climb no further.
In small planes, this maximum altitude point can be increased by putting a bigger engine in the plane, reducing the plane's weight, or putting a compressor on the engine's intake to furnish it with denser air at high altitudes so it can deliver more power.
In large planes with jet engines, those engines are specifically designed to deliver high power and best economy at high altitudes.
| {
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Why choosing for prime numbers eliminates vibration? I have read that the spokes of a car wheel are usually five because, besides other substantial reasons, five being a prime number helps to reduce vibrations.
The same also happens with the numbers of turbine blades and the way a microwave grill is spaced. Prime numbers are always preferred.
| In trying to answer this question I came across a lot of interesting phenomena related to primes. This is not a very detailed answer but will hopefully I can share the intuition and feel of the concepts involved.
The phenomena we are dealing with is resonance.
In any machine, there are several parts. Each part has some resonant frequency(a natural frequency). Now a noticeable vibration occurs when the magnitude of this oscillation increases. How will that happen? If you have two parts whose resonant frequencies are the same or multiples of each other, when every one part vibrates it sort of hits the other and makes the other part vibrate with a larger amplitude. Therefore slowly the amplitude of oscillation in both parts will increase due to a feedback mechanism and eventually such a large vibration can damage the machine.
The same goes for a wheel. Actually you cannot just look at spokes. There will also be some mechanism underneath which is holding on to the wheel with certain number of extensions. For example for the axle of the car to have good grip on the wheels it must have some elements extending from the axle to lock with the wheel. Now let the number of extensions be n1 and number of spokes be n2. If n1 and n2 are multiples of each other, due to the feedback mechanism discussed, large vibrations could be caused. Hence in general n1 and n2 need to be co-prime and in most cases the numbers 3,5,7 are used. As far as I understand it, co-prime numbers reduce the vibrations because they cancel each other out.
I got an idea for this answer from two phenomena. One is described here - https://www.reddit.com/r/askscience/comments/1aulwq/why_are_frequencies_in_hz_which_are_prime_numbers/
The second object was a pedestal fan. Have you ever wondered why the number of blades in a fan and the number of spokes in it's casing are not the same? The same concept applies. That is why for a 3 blade fan, a 5 spoke casing is used even though it costs more money to make such a casing than a 3 spoke casing.
| {
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Why don't we use Ampere's law to find the magnetic field due to a wire of finite length at its perpendicular bisector? I know that finite length doesn't have symmetry and thus it's hard to apply maths here but take the case of magnetic field of a wire of finite length at a distance $r$ from axis of the wire exactly at perpendicular bisector of wire.
The magnetic field is symmetric here, but still, Ampere's circuital law doesn't apply here. Why is it so?
| we take circular amphere loop for infinite wire because field due to the part parallel to that wire is neglected .but when you talking about a finite wire(whose other part of given current carrying circuit) loop we have to consider the field due to other part of the circuit hence by applying the amphere circuital law you will not get the correct value of magnetic field
| {
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If a RG fixed point (FP) is CFT, do all theories flowing into FP CFTs? Suppose that a RG (renormalization group) fixed point of some RG trajectory (or flow) is a CFT. Then do theories in this trajectory have to be CFTs as well?
| No. The points along the RG trajectory represent the theory’s behavior at some scale. The most baby example of this is a massive free field $\mathcal{L} = (\partial \phi)^2 + m^2 \phi^2$. This can be viewed as a perturbation of the Gaussian fixed point by the relevant operator $\phi^2$. In the deep UV the mass is negligible and the theory approaches massless free fields. In the deep IR the mass is effectively infinite and we get the trivial conformal theory, with no low-lying excitations. The spectrum is gapped above the vacuum. For generic points along the flow, the theory is of course non-conformal.
Families of CFTs correspond to manifolds of fixed points in the coupling constant space.
| {
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Neutral lithium ionization energy for removing an electron from the $1s$ orbital How much energy does it take to singly ionize a neutral lithium atom by removing an electron from the $1s$ orbital?
| You can obtain this directly from the energy-levels part of the NIST Atomic Spectra Database.
*
*In Spectrum, put Li I for the energy levels of neutral lithium, or Li II for the energy levels of the first cation.
*Choose whatever Level units you're most comfortable with.
Click Retrieve Data to get the results.
The ionization energy you want is not listed explicitly, and you need to get it as the sum of two factors:
*
*The ionization energy from Li I to Li II, which is listed in the spectrum of Li I, at the entry Li II ($1s^2$ $^1S_0$) at the left-hand column.
*The excitation energy from the $1s^2$ of the cation to the relevant $1s\,2s$ of the cation, which are listed in the Li II spectrum.
Note that there are two such states, with a small $\sim 1\:\rm eV$ energy difference between them, depending on how the two remaining spins are aligned. (Though, frankly, given how ridiculously hard it is to excite the cation to anything other than the ground state, this difference may well be negligible.)
| {
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If the pressure inside and outside a balloon balance, then why does air leave when it pops? Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.
But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?
| But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?
You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.
Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.
Hope this helps.
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Why charge induction is limited to some amount? Polarization occurs when a charge is placed next to a conducting object. When we ground the side of the object opposite the charge, that side becomes neutral.
But this neutralization disturbs the previous equilibrium. Thus, more polarization should be induced by the charge on the end of the object next to the charge and so more and more polarization should induced. Why doesn't this happen in the real world?
| Let me help you clear your concept first. Consider $-q$ charge placed next to the conductor. Let the side facing the conductor be $A$ and the opposite side be $B$. The $-q$ charge induces a positive surface charge density ($\sigma_A=q/\text{area}$) on surface $A$. Therefore, a negative surface charge density ($\sigma_B=-q/\text{area}$) is induced on surface $B$. When we ground the conductor the electrons will flow to the ground from surface $B$ while the positive charge at surface $A$ remains held there due to the attractive force of the $-q$ charge placed next to the conductor (hence the equilibrium is not disturbed yet).
When the grounding wire is removed the conductor gains a permanent positive surface charge density . Therefore polarization, or charge by induction, is stable.
| {
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Could $\nu_e+p\rightarrow e^-+\pi^++p$ occur via the weak interaction? If so, how would be the Feynman diagram? I checked if the reaction conservate baryon number, charge and leptonic number and it seems like it does. But I tried to draw the Feynman diagram and I don't understand how the proton interact, so maybe the reaction can't occur? Or is something about the Feynman diagrams that I still don't understand?
I'm new to Feynman diagrams, so any insight will be appreciated.
| No problem.
The neutrino emits a virtual $W^+$ and thereby turns into an electron.
The $W^+$ is absorbed by the $d$ quark in the proton, which turns it into a $u$ quark.
One of the many virtual gluons around in the proton splits to become a $d$ and a $\overline d$.
The 4 quarks and the antiquark arrange themselves into $p$ and a $\pi^+$.
I don't have a drawing packing handy to show this but hopefully that's enough information for you to do it.
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Localization of $4f$ in rare earth and $3d$ electrons in transition metals Is there a justification for why are the $4f$ electrons strongly localized about the nucleus in rare-earth atoms but the $3d$ electrons in transition metals extend further out from the nucleus? I have hit upon this question while reading about magnetism. Here is a reference from Blundell's Magnetism in Condensed Matter Sec.4.2.2 second paragraph.
A slightly different quote can be found in Kittel's Solid State Physics book which says,
The difference in the behaviour of the rare earth and the iron group salts is that the $4f$ shell responsible for paramagnetism in the rare earth ions lies deep inside the ions, within the $5s$ and $5p$ shells, whereas the iron group ions in the $3d$ shell responsible for paramagnetism is the outermost shell.
| The crystal field splittings are much larger in transition-metal salts (mostly due to covalency, the ligand-field effect), of the order of electronvolts. In rare-earth salts, crystal field splittings are two orders of magnitude smaller.
| {
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Does the second-order correction to degenerate perturbation theory vanish? Consider a degenerate two-state system with states denoted by $|1\rangle$ and $|2\rangle$. If we apply a perturbation $H^\prime$, the first order correction to the energy is obtained by choosing two linear combinations of $|1\rangle$ and $|2\rangle$ that diagonalizes $H^\prime$. So can we say that the second order correction always vanish in this case because $H^\prime_{12}$ vanishes? I am disturbed by the denominator which blows up.
| For degenerate levels the first order correction is obtained by the exact diagionalization of the Hamiltonian for the degenerate states. If all the states are degenerate, belonging to the same energy eigenvalue, then this is equivalent to the exact diagonalization of the Hamiltonian - no perturbation theory is needed (It would be necessary, if we have other states or multiple degenerate energies.)
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Conservation of Angular momentum or Work = 0 , which is valid?
In the figure, the block on the smooth table is set into motion in a circular orbit of radius "r" around the Center hole. The hanging mass is identical to the mass on the table and remains in equilibrium. Neglect friction. The string connecting the two blocks is massless and inextensible.
The question then asks about Angular momentum if the hanging block is pulled down.
I couldn't understand whether angular momentum would be conserved or not.
Because the force(Tension) is always acting perpendicular to the motion of the block , so the work done on the block should be zero and thus the velocity should remain constant and as r decreases, the angular momentum should change as $L=mvr$.
But on the other hand,since the force is central (r and F act in same direction) , the torque about the centre is zero and thus angular momentum should be conserved
What is it that I am missing here
|
Because the force(Tension) is always acting perpendicular to the motion of the block , so the work done on the block should be zero...
This is not true. As you pull the hanging block down the tension force does do work on the spinning block. The spinning block moves inward, hence it has some motion along the direction of the tension force. This means the work is non-zero.
You are correct in saying that angular momentum about the center point is conserved, since there is no net torque about the center point.
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How thick would the atmosphere have to be to block all the sunlight? this is a question I haven't been able to find an answer to anywhere.
How thick would our atmosphere have to be for the surface of the planet to be completely dark? Is that even possible?
Thank you!
| From a Wikipedia article on the Solar Constant:
At most about 75% of the solar energy actually reaches the earth's surface
You can roughly estimate that if the atmosphere were $n$ times as thick, the fraction would be $(0.75)^{n-1}$ compared with what we get. For example, if the atmosphere were 10 times as thick, sunlight would be reduced to 8%; at 20 times, to 0.4%; at 30 times, 0.02%. It never gets completely dark, but at 100 times you get less than one trillionth of the sunlight we get.
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Rotation as an example of symmetry in classical mechanics I modified the question because it was confused.
On my book there is this mathematical definition of symmetry transformation:
"The equations of motion have a symmetry, if the solutions of the equations transformed by the symmetry are still solutions of the equations of motion, namely, there is a symmetry if the transformed equations of motions have the same form of the original".
I don't understand the meaning of this sentence, do you think is it a good (and easy) mathematical definition of symmetry transformation? Anyway what "equal in form" means?
Then, i know rotation of an isolated system is a symmetry, can you make an easy example of an isolated system in which if we apply a rotation the mathematical definition of symmetry apply?
Can you make an example of non symmetric transformation and show me why the mathematical definition doesn't apply?
| Suppose you keep track of the total energy in the system at hand and it remains unchanged with time. This is Conservation of energy. However you transform the time coordinate, the energy remains constant. Symmetry is in the system if some transformation leaves certain elements of the system unchanged. This includes translations in time.
Now suppose you consider the energy as it changes with position. It's important to consider the energy measurements in the SAME coordinate system. Coordinates refer to a real, singular point. You need to compare the change in energy at different points, not the same point at different coordinates. Keeping this in mind, we know that momentum remains constant in time if the space derivative of the Hamiltonian is 0 everywhere in the system.
This applies to every conjugate momentum. So If the energy is constant under rotations, then angular momentum is conserved.
Look up Hamiltonian Mechanics and conjugate coordinates for more information.
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Time dilation and twin paradox - can this be explained in plain English? If one twin goes with an extremely fast rocket to Alpha Centauri and back, time is going slower for him, so when he comes back, the twin that stayed on the Earth is older.
But... if we look from the relative position of the rocket, the Earth rushed 4 light years away and then came back -- so the twin on the rocket should be older.
When the rocket is at cruising speed, neither accelerating nor decelerating, aren't Earth and the Rocket equally "correct"?
So, does that mean that acceleration and deceleration make the twin in the rocket younger?
But then acceleration and deceleration cause time dilation and not speed?
I know there are multiple questions already there about twins, but I can't understand the answers...
Can this be explained for someone with no physics background, in plain 5th-grade level English without manifolds, integrals, homotopy, etc?
|
So, does that mean that acceleration and deceleration make the twin in the rocket younger?
The twin on the rocket changes inertial frames. The twin on Earth does not. This is why the situation is not symmetric. If you want to not invoke acceleration, then do the following. At the time the twin would (instantaneously) turn around, instead just have a third twin (triplet? I guess it can just be some other person) that is traveling towards Earth pass the rocket twin and synchronize clocks. Then follow the third person back to Earth.
The conclusion is still the same that the rocket trip takes less time, and this shows that it isn't necessarily the acceleration, but rather the change in reference frames associated with an acceleration.
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Which particle mediates the Aharonov-Bohm effect? BACKGROUND
The Aharonov-Bohm (AB) effect induces phase shifts between the two paths that an electron could take around an enclosed magnetic field. In radial coordinates, assume that the magnetic field is localized around the origin and that the two paths traced by the electron form two complementary half-circles at radius some R. Assume further that the magnetic field is initially switched off.
QUESTION
At the moment the magnetic field is switched on, which particle travels outward from the origin towards the electrons' path so as to mediate the phase shift? And at what speed? Clearly, such a particle can't be a disturbance of the electromagnetic field since the magnetic field is restricted to the origin and its vicinity.
| Useful physics exercise here would be to study the problem of the solenoid with a current increasing with time. The field is growing inside the solenoid, and so is the field energy there. Where is the energy coming from? Answer is: work done by the current against an electric field. The electric field exists not only at the wire but also throughout space within the region accessible at the speed of light since the current started to change. It's not hard to calculate: you have non-zero $d{\bf B}/dt$ so non-zero curl ${\bf E}$. Integrate this over the interior of a circle of whatever radius you like and convert to line integral:
$$
\oint {\bf E} \cdot d{\bf l} = -\frac{d}{dt} \int {\bf B} \cdot d {\bf S}
$$
The symmetry tells you ${\bf E}$ is in loops around the solenoid, so
$$
E = -\frac{1}{2\pi r} \frac{d \phi}{d t}
$$
where $\phi$ is the flux in the solenoid.
So, to answer your question: the influence of the change at the solenoid is carried by this field, and therefore it is mediated by photons (whether real or virtual).
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Why isn't my calculation that we should be able to see the sun well beyond the observable universe valid? I recently read an interesting article that states that a human being can perceive a flash of as few as 5 or so photons, and the human eye itself can perceive even a single photon. The brain will filter this out, however.
I wanted to calculate how far away you'd have to be standing from our sun for not a single one of its photons to be hitting your pupil over a given second.
The first thing I did was assume that the sun emits $10^{45}$ photons per second, because, well, that's the only number I could find through internet research.
The next step is to assume that the average angle between photons emitted from the sun is pretty much the same, and is equal to $3.6 × 10^{-43}$ degrees.
The next step is to assume that the average human pupil diameter is 0.005 meters, and then draw a triangle like so:
The length of the white line through the center of the triangle equals the distance at which two photons from the sun would be further apart than your pupil is wide, meaning not even one photon should hit your eye.
I broke the triangle into two pieces and solved for the white line by using the law of sines, and my final result is ridiculous.
$3.97887×10^{41} $ meters is the length of the white line. For reference, that's over $10^{14}$ times the diameter of the observable universe.
My conclusion says that no matter how far you get from the sun within our observable universe, not only should some of the photons be hitting your pupil, but it should be more than enough for you to visually perceive.
But if I was right, I'd probably see a lot more stars from very far away every night when I looked up at the sky. Why is my calculation inconsistent with what I see?
| The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.
So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by
$$I = \frac{N}{4\pi r^2}.$$
This comes from spreading out the photons over the surface of a sphere surrounding the sun.
The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye:
$$n = IA_\text{eye} = \frac{N}{4\pi r^2}A_\text{eye}.$$
You are looking for the distance beyond which you would see less than one photon per second:
$$n = \frac{N}{4\pi r^2}A_\text{eye} \lt 1$$
Solving for $r$ gives
$$r > \sqrt\frac{NA_\text{eye}}{4\pi}$$
Plugging in your numbers gives
$$r > \sqrt{\frac{(10^{45})\pi(0.005\,\textrm{m}/2)^2}{4\pi}} = 4\cdot10^{19} \,\textrm{m} \approx 4000\,\textrm{light-years}$$
This distance is still well within our own galaxy.
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Fourier optics - Transfer function of free space In all consulted literature, the transfer function of the free space is given as follows:
$$\exp(-i k_z d) = \exp(-i2 \pi d \sqrt{1/\lambda^2 -\nu_x^2-\nu_y^2})$$
When referring to this source, they derive the transfer function from the following equation:
$$H(\nu_x,\nu_y) = \frac{f_{in}(x,y)}{f_{out}(x,y)}$$
I'm wondering why they do it this way (note: I've already seen this in other sources). I thought the transfer function is defined in the frequency domain rather than in the time domain (frequency and space for the fourier optics respectively).
| The source is specifically referring to the input and output being a plane wave. The key point on that slide is that complex exponentials in the form $\exp(i 2\pi\nu_x x)$ are eigenfunctions of linear, shift-invariant optical systems. Thus, a complex exponential signal goes into the system, and another complex exponential with the same frequency goes out. A plane wave has the same form as this complex exponential. As a plane wave propagates through a linear system, it remains a plane wave. It can only be amplified/attenuated and phase-delayed. New spatial frequencies (plane wave components) cannot appear in the output.
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Temperature of vacuum chamber on Earth Does the temperature of a vacuum chamber drop if left for 24 hours or more, since some in space in quite cold somewhere and quite hot how does vacuum become cold near to absolute zero.
| @AlonsoPerezLona already explains it, I just want to clarify even further in terms of thermodynamics.
It's all about thermodynamic equilibrium. Everything exchanges heat through vacuum via thermal radiation, so if you have a vacuum chamber with walls at different temperatures, these walls will eventually equilibrate. And if you put something in a chamber, it will equilibrate to the same temperature, too.
For a vacuum chamber at equilibrium, you can say that the vacuum itself has a temperature. It's not actually empty, it's full of thermal radiation (and photons are just as valid particles as everything else). Away from all stars, the outer space has temperature of about 2.7K (cosmic background), and if you are near a star, you will feel equilibrate at some kind of average, so that outgoing radiation will match the received radiation - for our Earth, this equilibrium sets the average temperature somewhere around 10-15°C.
This is the reason a good vacuum flask has mirror walls. Vacuum doesn't stop heat exchange (only stops exchange through thermal conduction), but mirror walls slow down the radiation exchange. Eventually, anything in a vacuum flask will reach the average temperature of its walls.
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Deriving Planck's Constant from Wien's Displacement Law So I'm reading an introductory book on Quantum Theory (David Park, 3rd ed.) and I am having trouble with the following question:
"According to Wien's displacement law, the wavelength $\lambda_m$ at which blackbody radiation at temperature T has its maximum intensity is given roughly by $\lambda_mT \simeq $ 3 mm K. Assuming that the quantum energy at this temperature is of the order of kT where k is Boltzmann's constant, estimate the value of Planck's constant."
My attempt at a solution is as follows:
$\lambda_mT \simeq$ 3 mm K $,\quad$ $\lambda = \frac c \nu$ $\quad \rightarrow\quad$ $\frac {cT} \nu = 3\cdot10^{-3} mK$ $\cdot (\frac k c )\quad \rightarrow \quad$ $\frac {kT} \nu = 1.38\cdot10^{-34}Js$
$(\frac 1 2 mv^2)_{max} = kT_{max} = h\nu - e\phi$
$\frac {kT_{max}} \nu + \frac {e\phi} \nu =h$
$1.38\cdot10^{-34} + \frac {e\phi} \nu = h$
At this point I get stuck, the orders of magnitude and units seem to be right but I'm not sure what to invoke/what is given in the question that can solve/eliminate this last term to get h, or if I'm even on the right track.
| You do not need to convert the electromagnetic energy into the energy of a particle. You just posit that a photon with wavelength $\lambda_{m}$ has energy roughly $kT$; then you get that $h\approx ckT/\lambda_{m}\approx\times10^{-34}$ kg$\cdot$m$^{2}\cdot$s$^{-1}$, which is the correct order of magnitude. (Note that you only have one significant figure accuracy in the quantities you are given.)
If you want better than order of magnitude precision, the numerical calculation of the precise constant of proportionality between $kT$ and $hc/\lambda$ is outlined on the Wikipedia page for Wein's Law.
| {
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Percentage of Neutron Stars that leave the Galaxy When a neutron star(NS) is born the supernova kick gives the newborn NS a certain boost in a supposed isotropic direction and, depending on how strong the boost is and on the position of the NS at birth time, its kinetic energy might exceed its gravitational potential one and the NS will definitely leave the galaxy.
For what I found in literature the absolute value of the kick velocity follows a Maxwellian distribution with a mode value around 400 km/s.
With this in hand, together with a certain model for the galactic potential and initial spatial distribution, isn't it trivial to find the percentage of NS that eventually leave the galaxy?
I ask this because I cannot find anything about it in literature.
| I couldn't tell you exactly what percentage of neutron stars leave the galaxy, neither could anybody else, but it must be a very tiny percentage indeed. After a supernova explosion, most neutron stars stay where they are. If you look at supernova remnants, there is usually a neutron star at the dead centre. To send the neutron star flying off like an outsize cannonball, the supernova would have to be asymmetrical, but very few are. It does sometimes happen, but I would think the supernova would have to be on the outskirts of the galaxy for the neutron star to leave the galaxy altogether.
| {
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Why is the drag force proportional to $v^2$ and defined with a factor of $1/2$? $$Drag = \frac{1}{2}C_d \rho Av^2$$
I understand that the strength of the drag depends on the density of the fluid the body passes through, the reference area of the body, the drag coefficient, and the velocity of the object.
I don't, however, understand the 1/2 and the $v^2$ in the equation.
| The drag force is doing negative work on the object that it is decelerating. By the work/kinetic-energy theorem, the work done is equal to the change in kinetic energy that the object experiences. Since kinetic energy is defined as $E_k = 1/2 mv^2$, you can expect the "1/2" and the $v^2$ terms to show up in the equation.
| {
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In relativistic QFT, is it ever possible that the bare mass be finite and equal to the physical mass? In renormalization, one follows the philosophy that the bare mass is unobservable and could be infinite, and the physical mass comes from the pole of the two-point function. Is it possible that in any case the bare mass is same as the physical mass? Do we have an immediate example (perhaps in some extensions of the Standard Model or more mundane)?
| Yes, it is possible.
One possible example is the non-interacting theory, for example, a free scalar (bosonic) field with action
$$ S[\phi] = \int d^4 x \left( \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{m^2}{2} \phi^2 \right). $$
If you're looking for examples of interacting theories, Yang-Mills theory for any compact group $G$ has massless gluons in both the bare and renormalized actions. This is an example of the mass term (or rather the absence of thereof) protected by the symmetry.
| {
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Why can't we take space as universal frame of reference? Suppose we have a ball filled half with water in space with nothing else around (nothing else in the whole space except the ball) and suddenly it accelerates for time t. obviously, there would be movement in water which will tell us that the ball underwent motion. But since we have nothing to compare the motion with how can we say that it was in motion? can we say that it was at point A inertially (in space) and then in point B (again in the empty space)?
| I don't know if you came with this thought experiment by yourself but if you did you are crazy smart. This is actually very close to the argument made by Newton, usually called Newton's Bucket. He believed that absolute space was a real physical scenario where physical processes were to happen. And something similar to that is his main argument for it's existence.
The truth is that today we don't consider this to be empirical evidence for that idea necessarily. To join a concept of relative space been the only one in existence with the results of this thought experiment you can just suppose that the objects of the rest of the universe are somehow mysteriously influencing the water communicating it what its movement relative to other objects is. In the extreme case were only your ball filled with water exists the believe is that water would actually not slosh at all. This is calles Mach's principle.
Today with the Einsteinian view we think in another way. Space is nothing more than a conceptual construct, an abstract aid for calculations, but space-time is something physically real and something entirely different.
| {
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How to know if a sine wave must be described as a sine function or as a complex exponential? In problems, sine waves (electromagnetic, acoustic, ...) are often described as $\psi(\mathbf{x},t) = \psi_0\sin(\mathbf{k}\cdot\mathbf{x} - \omega t)$.
However, they sometimes need to be described as $\psi(\mathbf{x},t) = \psi_0 e^{j(\mathbf{k}\cdot\mathbf{x} - \omega t)}$.
Is there a way to guess the required description of a given problem ?
| Answered by Lith in comments.
The exponential notation is introduced to make calculations easier by using the complex plane, but you are free to use whichever formulation you prefer, they are equivalent (as long as at the end of the problem you recover real function).
| {
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Peskin and Schroeder Section 7.1 Mass Shift I'm slowly reading my way through Peskin and Schroeder. Near the end of section 7.1 they compare the mass shift of the electron from QFT to the classical value, both of which are divergent but in different ways.
The calculation from QFT gives:
$$\delta m = \frac{3\alpha}{4\pi}m_0\log\left(\frac{\Lambda^2}{m^2_0} \right)$$
Which diverges logarithmically as $\Lambda\to\infty$.
Versus the classical expression which diverges linearly, $\alpha\Lambda$.
The bit I don't understand is the argument they use after to explain why the divergence should be logarithmic:
"$\delta m$ must vanish when $m_0=0$. The mass shift must therefore be
proportional to $m_0$, and so by dimensional analysis, it can depend
only logarithmically on $\Lambda$".
(Taken from the paragraph immediately following the $\alpha\Lambda$ result.)
The understand the first part of this statement, its the dimensional analysis I don't get. Why is it logarithmic specifically and not some other dimensionless function of $\Lambda$?
| Fundamentally, it has to do with 't Hooft's technical naturalness argument: since the electron mass term breaks the chiral symmetry, the quantum correction to the electron mass has to be proportional to the weakly broken mass scale $m_0$, thus linear in $m_0$ (with logarithmic divergent factor).
A counter example is the quantum mass correction to a bosonic field, which is of the order $\Lambda^2$. Why? The root cause is that the bosonic mass does NOT break the chiral symmetry, thus it's NOT protected by 't Hooft's technical naturalness principle. The Higgs mass issue is exhibit A, hence the vexing hierarchy/naturalness problem that keeps physicists up at night.
| {
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Nuclear Physics Question How could I calculate the chance of a proton actually joining the nucleus and 2 protons and 3 and so on? Assuming a large number of protons are fired at a substance?
| What you're looking for is the absorption cross section $\sigma$ of the target. This quantity has units of area, and is typically measured in barns (1 barn = 1 b = 10-28 m2.) There are many such cross-sections one can define, for any particle interaction you might care to name. If you have a beam of protons with a flux of $f$ (i.e., $f$ protons per area per time), and you send this beam at a target for a time $\Delta t$, then the expected number of absorptions is
$$
\langle N \rangle = f \sigma \Delta t.
$$
Calculating these cross sections from first principles is basically impossible (but then, that's true of most calculations in nuclear physics.) Instead, they must be experimentally measured, or looked up in the literature. Note that the cross section usually depends on the energy of the incoming protons; it also depends on the target, so once the target has absorbed one proton, the cross-section to absorb a second proton will be different since the target has changed.
Finally, note that proton capture is relatively rare compared to the more common neutron capture, since the proton and the target are both positively charged and tend to repel each other. Cross sections for scattering of protons off of nuclei can be defined in much the same way. There may not be a huge amount of data out there for proton capture cross sections, compared to either proton scattering or neutron capture.
| {
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Torque and angular acceleration in elliptical orbit I am stuck in a supposedly simple aspect. Consider the Sun-Earth system. The torque is zero and angular momentum is conserved. $L = I\omega$ is constant, but since $I$ changes, $\omega$ should change as well. That means there is a non-zero angular acceleration.
Now consider $\tau = I\alpha$ which should be zero. Since $I$ can't be zero, angular acceleration must be zero.
I can't explain the apparent contradiction. Your answers would be much appreciated.
| The problem here is that your system is not a rigid body. We can think of it as a point particle (the Earth) rotating about another point particle (the Sun). The system is not rigid because the distance between these two point particles changes over time. Therefore, we need to be more careful with how we apply these rotational analogs of Newton's laws.
Typically we have:
$$\tau=\frac{\text dL}{\text dt}$$
and, for constant $I$ this means:
$$\tau=\frac{\text d}{\text dt}\left(I\omega\right)=I\alpha$$
However, as you have pointed out, $I$ is changing, therefore we really have:
$$\tau=\frac{\text d}{\text dt}\left(I\omega\right)=\frac{\text dI}{\text dt}\omega+I\alpha$$
Now I won't go into any specific calculations for elliptical orbits (you can do this if you want), but you can see how if the particle is moving inward then $I$ would be decreasing (negative derivative) and the particle would be speeding up (positive $\alpha$), and if the particle is moving away then $I$ would be increasing (positive derivative) and the particle would be slowing down (negative $\alpha$).
The above qualitative argument shows that the two terms in this equation will have opposite sign, and this shows how it could be that these two terms will always cancel out in the orbit to give a net torque of $0$
| {
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Gaussian path integral is equivalent to saddle-point? If we have a path integral involving many fields,
$$Z = \int \mathcal D \phi_1 \cdots \mathcal D \phi_n \exp(-S[\phi_1,\ldots, \phi_n]),$$
and $\phi_n$ occurs only quadratically-- i.e. the $\mathcal D \phi_n$ integral is Gaussian-- I've been told that integrating over $\phi_n$ is equivalent to solving for $\phi_n$'s equation of motion
$$\phi_n= f(\phi_1,\ldots, \phi_{n-1})$$
using Euler-Lagrange and plugging in. Up to normalization. Can one show in general why this is true?
| The gaussian integral
$$
\int dx\,e^{-\frac12 a x^2 + bx + c} = \sqrt{\frac{2\pi}{a}}\, e^{c+b^2/(2a)}\,,
$$
is similar to its path integral counterpart, which is
$$
\int \mathcal{D}\phi\,e^{-\frac12\phi \cdot A\cdot \phi + \phi\cdot b + C} \propto \exp\left(C + \frac{1}{2} \, b\cdot A^{-1}\cdot b\right)\,.
$$
By the dot I mean $a\cdot b \equiv \int a(x)\, b(x)$, $a\cdot B \cdot c \equiv \int a(x)\, B(x,y)\, c(y)$. Moreover $A^{-1}$ satisfies
$$
\int A(x,y)\cdot A^{-1}(y,z) = \delta(x-z)\,.
$$
The equations of motion for $\phi$ are
$$
-A\cdot \phi + b = 0\qquad \Longrightarrow\qquad\phi = A^{-1}\cdot b\,.
$$
Replacing this on the action yields the same result
$$
-\frac12\phi \cdot A\cdot \phi + \phi\cdot b + C \quad\to\quad -\frac12\,b \cdot A^{-1} \cdot A\cdot A^{-1}\cdot b + b\cdot A^{-1}\cdot b + C = \frac12\,b\cdot A^{-1}\cdot b + C\,.
$$
If the dot notation is confusing I suggest to expand everything in integrals. The operator $A$ usually is just $(\square_x + m^2) \delta(x-y)$ and $A^{-1}$ is $G_F(x-y)$, the Feynman propagator.
Edit: As a comment pointed out, this does not take into account the $(\det A)^{-1/2}$. If $A$ is a constant operator, this does not pose any problem in perturbative computations because we only need the partition function modulo overall factors.
On the other hand, if $A$ is a function of the remaining fields $A(\phi_1,\ldots,\phi_{n-1})$, it will not pass through the subsequent integrals. The way this is normally handled is by exponentiating it as
$$
(\det A)^{-1/2} = e^{- \frac12 \mathrm{Tr}\log A}\,,
$$
(with a suitable regularization procedure) and this typically yields a non-local action
| {
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Rotation of a rigid body within another orbiting body I would like to understand the physical behaviour of the following rotating sytstem.
There are 3 rigid bodies (1 is blue, 2 are red). We assume that there is no friction between the blu body and the red bodies, i.e. no friction along the dashed line.
The blue body is rotated clockwise with constant angular velocity with respect to its center of mass (point O in the figure).
What kind of motion do the red cylinders exhibit with respect to
*
*The lab's frame?
*A frame rotating as the blue body?
| Lab Frame
You have posited the system is began with a rotational rate $\omega$. In this state, only a radial force is required to satisfy the constraints of the blue object. With only a radial force and an initial tangential velocity, the motion of the CM of the red objects is uniform circular motion.
The red objects would not change their orientation in the lab frame, due to the lack of friction and due to the initial conditions. For instance, if you painted an arrow on each red object, and made both arrows parallel, they would remain parallel through the rotational motion.
Rotating Frame
In a frame attached to the blue object, the motion of the red objects would be uniform rotational motion about their centers. Again, this falls from the fact that the system was began in an initial state of uniform circular motion. If we take the same parallel arrows as before, then from the rotating frame the arrows rotate in the opposite direction but at the same rate $\omega$.
| {
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Idealization of Eletric Field at a point According to Jackson "Classical Electrodynamics", in the first chapter about electrostatics:
[...] point charges or electric fields at a point must be viewed as
mathematical constructs that permit a description of the phenomena at
the macroscopic level, but that may fail to have meaning
microscopically.
I think I've understood why point charges represent an idealization, but why also the electric field at a point fails to have meaning microscopically?
Moreover, what does it mean in practice that in electrostatics we are only interested in macroscopic phenomena? Does this also affect electrodynamics?
|
why also the electric field at a point fails to have meaning microscopically?
You have no problem with a point particle being an idealization, right? Now, remember
$$
∇.E = ρ
$$
If you put garbage in the right hand side, do not expect anything better out of the left hand side.
what does it mean in practice that in electrostatics we are only interested in macroscopic phenomena?
If you knew the position of a charged particle with absolute precision (assuming for now that it is still a point), then, by the uncertainty principle, you would know very little about about its velocity. Since the velocity is basically unknown (and because moving charges produce magnetic fields), you cannot have pure electrostatics at the microscopic level where quantum mechanics prevails. In the classical regime, where your charged body is made out of many charged particles, electrostatics will appear because the tiny magnetic fields average out.
By the way, if someone tells you that 'virtual photons are not real particles', ask back 'what is particle?' They might try to handwave something about virtual being temporary. Ask them then 'what about resonances?'
| {
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How does a heated lid in a thermal cycler prevent evaporation? A thermal cycler is a chemistry lab device that increases or decreases the temperature of the material inside it. The lid of the device is heated to prevent condensation and evaporation of the mixture inside.
I understand the condensation bit, as a hot lid would keep the gaseous particles gaseous. But I don't understand how a hot lid can prevent evaporation of the mix. Is there a physics law that states liquids cannot evaporate if the temperature of the gas above them is higher than the liquid? Or is it about pressure? Or both?
Edit: Adding some more info since this was migrated to physics. This site provides a decent overview on thermal cyclers. The following paragraph that explains the basics is from there:
The basic idea of a thermal cycler is that it provides a thermally
controlled environment for PCR samples. A thermal cycler usually
contains a heating block with holes or depressions in it that receive
sample tubes (though other types of sample vessels are now possible
also; see below). For the PCR reactions to work properly, the block
must change temperature at specific times, and spend specific
durations of time at specific temperatures. The researcher programs
the temperature cycling information into the thermal cycler either by
computer or via a console on the instrument, or uses a preprogrammed
routine built into the machine.
The main idea is that the machine cycles the temperatures of its contents, the temperatures usually range from around 50 to around 95 C.
| You were wise to be suspicious, because the answer is that thermal cyclers do not reduce evaporation relative to a similarly sealed chamber.
| {
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Why is it that when a chalk board gets cleaned, the area that used to have chalk is the cleanest? Why is it that when you erase a chalk board, the area where the chalk used to be becomes the cleanest? By that I mean that when you erase a chalk drawing, the board gets smeared with chalk dust, but the area where the drawing used to be has less dust on it than the rest of the board.
For example: In the first picture below I draw a simple chalk smiley face. Here the face is noticeable because it is the area with the most chalk. For the second picture, I erase it. You can still make out the picture, but notice that you recognize it because it is now the area with the least chalk.
I would expect that if chalk was stuck to a certain region of a chalk board, then after erasing it, some chalk residue would remain, but instead it seems like the opposite happens. I don't have a good answer for this problem.
| My guess is that the chalk you have just applied binds to the existing chalk particles underneath. Then, when you swipe it, you are drag these particles along with the big lump of chalk that makes up your drawing. This is possible thanks to the internal cohesion of the material.
In the other areas of the board, you have reached a sort of steady-state with the current state of your eraser. This time there is a more direct contact with the surface of the eraser because there is no big lump of chalk that has been applied between it and the board. But the eraser is not clean and always leaves some amount of dust behind. This dust is not applied with the same pressure as with the chalk itself and is also too dispersed to form a cohesive lump, so that you are not able to drag all of it as a unit.
| {
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Wouldn't gravitons themselves theoretically be affected by gravity? (I wrote this using the assumption that the graviton exists, which I know is not necessarily true; this is asked from a theoretical standpoint)
It is my understanding that General Relativity states all massive and/or energetic particles must be affected by the curvature of spacetime; this stipulation aligns with all known elementary particles. I understand that the graviton, being a particle that reliably interacts with photons, is going to be virtually impossible to detect experimentally. It is almost certainly massless and of a remarkably low energy, but it must possess energy nonetheless. As such, shouldn't gravitons themselves be affected by gravity, creating a paradox in which (increasingly virtual ?) gravitons must interact with other gravitons? Also, assuming that our conception that massless particles always travel at the speed of light holds true for gravitons, wouldn't gravitons themselves not be able to escape from a black hole, annulling the gravitational force of the black hole in the process?
| Yes, gravitons must necessarily respond to each other via gravity. This means that in addition to communicating forces between objects that bend spacetime, the force carrier particles bend spacetime themselves.
But for very weak gravitational fields, the math gets easier to deal with because in this simplified case, the effect of the gravitons themselves on the curvature of spacetime can be ignored.
| {
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How did Einstein know the speed of light was constant? I often hear the story of how Einstein came up to the conclusion that time would slow down the faster you move, because the speed of light has to remain the same.
My question is, how did Einstein know that measuring the speed of light wouldn't be affected by the speed at which you are moving. Was this common knowledge already before Einstein published his paper on special relativity? If not, what led him to that conclusion?
| When using the term 'the speed of light' it is sometimes necessary to make the distinction between its one-way speed and its two-way speed.
The "one-way" speed of light, from a source to a detector, cannot be measured independently of a convention as to how to synchronize the clocks at the source and the detector. What can however be experimentally measured is the round-trip speed (or "two-way" speed of light) from the source to the detector and back again.
Albert Einstein chose a synchronization convention see Einstein synchronization that made the one-way speed equal to the two-way speed. The constancy of the one-way speed in any given inertial frame is the basis of his special theory of relativity.
Experiments that attempted to directly probe the one-way speed of light independent of synchronization have been proposed, but none has succeeded in doing so.
The paper A Comparison between Lorentz's Ether Theory and Special Relativity in the Light of the Experiments of Trouton and Noble by Prof. Michel Janssen provides comprehensive historical and theoretical analysis of development of the theory.
Some thought and references on conventionality of distant simultaneity can be found here
| {
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What are change of frame and change of coordinates? What's the difference between a change of frame and a change of coordinates?
I feel like both are transformations on the coordinates but change of frame changes also the vectors.
| It depends what you mean with 'change'. If 'change' is 'rate of change' i.e. the operator $\frac{d}{dt}$ acting on a vector $\vec v$, then the following is the case:
Any vector that is the element of a vector space can be split into its basis vectors $\vec e_i$ and its scalar components $v^i$, so that $\vec v = \sum_{i=1}^{n=3} v^i \vec e_i \equiv v^i \vec e_i$, where the last definition just uses the summation convention for indices.
If you now take $\frac{d}{dt}\vec v$ then according to the product rule you get
$$\frac{d}{dt}\vec v = \left(\frac{d}{dt}v^i\right) \vec e_i + v^i\left(\frac{d}{dt}\vec e_i\right).$$
Now $\left(\frac{d}{dt}v^i\right)$ is what I would call a change of coordinates, and $\left(\frac{d}{dt}\vec e_i\right)$ a change of frame. The change of coordinates is just the change of the scalar numbers denoting some position in the given coordinate system $\vec e_i$. Should the coordinate system be accelerated, or, you just wish to change the 'grid' with which the $v^i$ are computed, then you need to change the basis vectors accordingly.
Mathematically there is no difference between a change of coordinates at a given time, or an accelerated coordinate system w.r.t time, they give you only different transformation Jacobians.
On another note, the change in basis vectors can impact our understanding of physics fundamentally. Simply changing from cartesian to spherical coordinates in Newtons equation of motion (at a given time) already introduces a term that behaves like a centrifugal force, although we only changed our 'viewpoint'.
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Is there a way to see the sun's outer layers from earth outside a solar eclipse using polarization? ¿Is there a way to filter sky light using light polarization?
in https://en.wikipedia.org/wiki/Coronagraph#Invention it says:
High Altitude Observatory's Mark IV Coronagraph on top of Mauna Loa, use polarization to distinguish sky brightness from the image of the corona: both coronal light and sky brightness are scattered sunlight and have similar spectral properties, but the coronal light is Thomson-scattered at nearly a right angle and therefore undergoes scattering polarization, while the superimposed light from the sky near the Sun is scattered at only a glancing angle and hence remains nearly unpolarized.
¿How does it work?, i tried to watch to the day sky through a polarizing filter and i didn't see it much more dark.
| The SOHO satellite uses polarization analysis to study direction of coronal magnetic fields. I guess this could also be done from Earth.
| {
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How can I understand if an object stay (zero velocity) or moving with constant velocity (zero acceleration) I thought a scenario like; lets say I am looking an object and there is nothing except this object. Is there a way to understand that if this object is stay on its position or if object moving with a constant speed and also I am moving as same constant speed with this object ? (consider there is not any friction etc.)
| You cannot tell the difference. In fact many would say there is no difference. If you're really in a universe where there's nothing except the object like you say, there would be no difference between just sitting still and moving at a constant speed. There are no stars to see whizzing by, no atmosphere to make you feel the wind, nothing to make the light so you can measure redshift with special relativity.
If you're curious, this is one of the fundamental assumptions of special relativity.
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In a circuit having one resistor why do the electrons lose all their potential energy across that resistor and not do so if there are many resistors In a simple circuit which consists of a battery and one resistor, why do electrons lose all their potential energy across this one resistor regardless of the magnitude of the voltage or the resistance? Why is the voltage drop of this one resistor equal the voltage of the battery?
If there is more than just one resistor in series then in the first resistor happens a voltage drop which is not equal to the voltage of the battery. The electrons don't lose all their potential energy across the first voltage. it seems that the resistor knows that there is another resistor after it so it doesn't eat all the electrons' potential and leaves some potential for the next one. Why does this happen?
| The voltage drop across an ohmic resistor is given by Ohm's law:
$$V=IR$$
Therefore, for a set resistance the potential drop is determined by the current. So really, your question about adding a resistor becomes Why does adding a resistor to the circuit decrease the current flowing through the circuit?
For resistors in series, the total resistance of the circuit increases as you add more resistors. This makes sense if you thinking of adding resistors in series as just adding a longer section of the circuit that resists the flow of current$^*$ More resistance means less current, which means less potential drop across each resistor.
$^*$ Note that for resistors in parallel the argument is not the same. It's not true in general that adding a resistor to a circuit increases the overall resistance in the circuit. This answer is only concerned with your specific series example.
| {
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Does throwing a penny at a train stop the train? If I stand in front of a train and throw a penny at it, the penny will bounce back at me.
For the penny to reverse its direction, at some point its velocity must go to zero. This is the point it hits the train. Two objects in contact have the same velocity, so the train must come to a stop for the penny to change its direction.
I assume I'm getting some principles wrong. Is it because I assumed a perfectly rigid body, when in practice the train actually deforms ever so slightly?
|
Two objects in contact have the same velocity
No, not always. Take two balls and make one roll faster than the other and make them collide. They do collide with different velocities!
But, in your case, if the coin hits the train and exerts a force capable of accelerating it in the opposite direction of its movement until its velocity becomes zero then yeah, the coin will stop the train.
| {
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Electromagnetic waves according to Maxwell If a variable Electric field creates a variable magnetic field and VICE VERSA (according to Maxwell's equations), then why don't we enter a loop where E vector and B vector keep creating one another until they reach infinite magnitudes?
| This can not happen if only by energy conservation. The Maxwell equations may seem to indicate otherwise but electric and magnetic fields do not generate one another. They are mutually dependent quantities and it is this fact that is expressed in two of the four equations.
| {
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Zero mass Kerr metric When mass in Kerr metric is put to zero we have $$ds^{2}=-dt^{2}+\frac{r^{2}+a^{2}\cos^{2}\theta}{r^{2}+a^{2}}dr^{2}+\left(r^{2}+a^{2}\cos^{2}\theta\right)d\theta^{2}+\left(r^{2}+a^{2}\right)\sin^{2}\theta d\phi^{2},$$
where $a$ is a constant. This is a flat metric. What exactly is the coordinate transformation that changes this into the usual Minkowski spacetime metric form $$ds^{2}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}?$$
| As mentioned in @Umaxo's comment,
according to Boyer-Lindquist coordinates - Line element:
The coordinate transformation from Boyer–Lindquist coordinates
$r,\theta,\phi$ to Cartesian coordinates $x,y,z$ is given
(for $m\to 0$} by:
$$\begin{align}
x &= \sqrt{r^2+a^2}\sin\theta\cos\phi \\
y &= \sqrt{r^2+a^2}\sin\theta\sin\phi \\
z &= r\cos\theta
\end{align}$$
Proving that this is the desired transformation
is a straight-forward but very tedious task.
First calculate the differentials of the above:
$$\begin{align}
dx &= \frac{r}{\sqrt{r^2+a^2}}\sin\theta\cos\phi\ dr \\
&+ \sqrt{r^2+a^2}\cos\theta\cos\phi\ d\theta \\
&- \sqrt{r^2+a^2}\sin\theta\sin\phi\ d\phi \\
dy &= \frac{r}{\sqrt{r^2+a^2}}\sin\theta\sin\phi\ dr \\
&+ \sqrt{r^2+a^2}\cos\theta\sin\phi\ d\theta \\
&+ \sqrt{r^2+a^2}\sin\theta\cos\phi\ d\phi \\
dz &= \cos\theta\ dr
- r\sin\theta\ d\theta
\end{align}$$
Then insert these differentials into the Minkowski metric:
$$\begin{align}
ds^2 &= -dt^2+dx^2+dy^2+dz^2 \\
&\text{... omitting the lengthy algebra here,
exploiting $\cos^2\alpha+\sin^2\alpha=1$ several times} \\
&= -dt^2
+\frac{r^2+a^2\cos^2\theta}{r^2+a^2}dr^2
+\left(r^2+a^2\cos^2\theta\right)d\theta^2
+\left(r^2+a^2\right)\sin^2\theta\ d\phi^2
\end{align}$$
which is the Kerr-metric for zero mass $M$, angular momentum $J$,
and charge $Q$.
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Definition of "specific gravity" I've learnt that a specific quantity is an extensive quantity divided by the mass. How does the definition of specific gravity fit into this scheme?
| That's one definition of the term "specific quantity"; but the qualifier "specific" is used in other ways for other terms.
For example, Wikipedia's page actually gives plenty of examples of other terms using specific that don't rely on mass.
Basically, the definition of "specific quantity" that you use would not include the term "specific gravity", and that's perfectly fine, specific gravity doesn't have to be a "specific quantity", you just have to know that not every use of "specific" means per unit mass.
In this case, specific gravity is a term for the relative density of something compared to water.
| {
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Is $X\otimes X$ not the simultaneous position operator? I had thought that $X\otimes X$ would be the operator on $H_1\otimes H_2$ to simultaneously measure the x-positions of two particles.
But there seems to be something wrong with this -- for a given eigenvalue $z$, there is an entire subspace $\mathrm{Span}\left(|x\rangle\otimes|z/x\rangle\right)$ associated with it, so we don't get precise positions from measuring it, just the product of positions $x_1x_2$.
If so, what's the right operator representing "simultaneous measurement" of the x-positions? Is that even possible -- to have "vector eigenvalues"? Or do we just need a spacefilling curve or something?
| It's easier to think about one particle first. In a one-dimensional case, $\hat{x}$ is the position operator. When we move to three dimensions, the position operator must have vector eigenvalues, because position is vector-valued. This is achieved by using a vector operator,
$$\hat{\mathbf{x}} \equiv \begin{pmatrix} \hat{x} \\ \hat{y} \\ \hat{z} \end{pmatrix}$$
which is a vector of operators. There is of course no difference when you have two particles. In this case their positions are described by six parameters, so you need a six-element vector,
$$\hat{\mathbf{X}} \equiv \begin{pmatrix} \hat{x}_1 \\ \hat{y}_1 \\ \hat{z}_1 \\ \hat{x}_2 \\ \hat{y}_2 \\ \hat{z}_2 \end{pmatrix}.$$
The operator $\hat{x}_1 \otimes \hat{x}_2$ doesn't represent a simultaneous measurement of $\hat{x}_1$ and $\hat{x}_2$, it represents measuring the single quantity $x_1 x_2$.
| {
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Do objects besides strings, ropes, and rods have tension? Why do we define tension only in strings and ropes and rods and such? Shouldn't every object experience tension force? Like when you pull a paper from opposite sides, it gets taut, and experiences what seems like a state of tension. If every object does experience tension, can you define tension?
| Tension is not defined only for strings.
However, the unique thing about ideal strings is that they can ONLY experience tension, whereas rigid bodies can experience tensions and compression. Ideal strings would collapse.
Hope this helps
| {
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If I stood next to a piece of metal heated to a million degrees, but in a perfect vacuum, would I feel hot? A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.
| The black body answers are fine, but I would like to point out that no one has accounted for the amount of material present. If you had a metal gas with 100 atoms obeying a Maxwell-Boltzman distribution at the stated temperature, you would feel nothing.
| {
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What is a mass moment? I am currently reading through a document Finding Moments of Inertia from MIT, page 4, and I am a little confused as to one of the concepts that they use.
In this document, there is mention of a mass moment. Could someone possibly define this for me please? I can't find anything too clear on the Internet.
Is this synonymous with the first moment of mass?
| Mass moment is slightly different from moment of inertia. It is moment of inertia x total Mass
| {
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Trouble with Math in Physics I am a current high school student and I am very interested in physics, especially particle physics (that stuff is super cool!). Unfortunately, my school only teaches classical physics, so I have to continue my study at home. I've read several books and watched videos and online lectures on quantum mechanics and have gotten a basic overview of the big ideas, but when I try to dig even just a little bit deeper the math immediately gets too confusing to handle. Beyond basic summaries and oversimplifications, what other resources can I use to continue studying physics but avoid getting frustrated by math that I don't understand?
EDIT
Based on feedback so far, I think the question I should really be asking is "What are some resources that can help teach me the prerequisite math I need to know for quantum mechanics?"
| Now is not the time to worry about quantum mechanics. There's a reason it's not taught at high school: it's not a simple topic, and you need prerequisite mathematics (especially calculus & linear algebra) that takes time to learn.
If you must attempt it anyway then in my experience the most common QM undergraduate textbook is David Griffith's Introduction to Quantum Mechanics. However, as mentioned, you will have trouble understanding this book without the prerequisite mathematics.
I suggest not feeling bad about it though, because when I did undergraduate studies, the program left QM to 2nd year, with the first year spent making sure everyone had the necessary mathematics.
| {
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Intuition of Maxwell's Equations Is there an intuitive explanation for Maxwell's equations?
I know they are axioms but is there a logical understanding of why instead of mathematical. Both forms don't explicate the scientific reasoning behind them to me.
I would appreciate a non- or minimally mathematical approach to them.
| They’re not axioms: They’re experimental results.
Coulomb, Faraday, etc did a lot of experimental work to observe and systematize the underlying phenomena. Maxwell then reformulated them (though not in the modern form) and added the displacement current term which itself was later confirmed experimentally.
So the “why” historically comes down to “because people observed this”
Today, the underlying “why” is “because these emerge from the quantum field theory of QED” and even fro electroweak interactions. But that’s a very long leap to make intuitive.
So if you want an intuitive understanding of Maxwell’s equations themselves, the best place to look is at the experiments that underlies them. It’s a lot to put in an Answer, because it’s a lot of physics.
For example, forces on small charges led to Coulomb’s law, which led to the idea of an electric potential hence electric field and Gauss’ law.
| {
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Do air conditioner power ratings violate the 2nd law of thermodynamics? I just got a new AC rated at 6000 BTU and wanted to determine its power consumption. Some research on AC conventions quickly reveals that 6000 BTU really means 6000 BTU/h, where BTU is a measure of energy (British thermal unit). This is supposedly the rate at which heat is pumped from inside to outside. By the second law of thermodynamics, the power consumed to run the device must be even greater.
A direct conversion gives 6000 BTU/hr = 1758 W, so we should expect a power consumption greater than 1758 W. But I was surprised to see on the side panel of the unit (in the electricity specs) a voltage of 115 V and current of 4.66 A, suggesting a power of
$$115\text{ V} \times4.66\text{ A} \approx536\text{ W}.$$
At first I thought I must be misenterpreting the side panel, so I checked the official specs on the manufacturer's website. They boast a higher current of 6.5 A, and claim the electrical "rated input" of the device is 700 W. Still way too small. How is this possible?
P.S., all the sources I could find about this via googling seemed to contain similarly impossible numbers, without mentioning this seeming violation of the second law.
| The industry rates air conditioner power consumption based on an average which takes into account the fact that once the air conditioner has got your room to the desired temperature, the compressor (which draws most of the power required to run the unit) is cycling on and off every few minutes, so the overall duty cycle or per cent time the compressor is actually running is typically around ~25%.
However, the heat transfer rate in BTU's assumes a 100% duty cycle, which is a confusing way of rating the performance of an air conditioner.
| {
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Why do jet engines sound louder on the ground than inside the aircraft? Everyone is familiar with the whirring sound of jet engines when seeing an aircraft taking off from a nearby airport. It is distinctly very loud on the ground and one can hear it even when the airplane is miles away.
Although one can hear a 'white noise' like sound when inside an airplane, the engines don't sound very loud in spite of being just meters away from them. I understand that the cabin is well insulated from the outside, but I would expect to hear a similar whirring sound of the engines.
So what is the phenomenon that makes jet engines sound louder on earth compared to inside the aircraft cabin?
| There is a howling 500 - 600 mph gale blowing outside the aircraft, the like of which you have never experienced on the ground, and this carries away most of the sound before it can enter the fuselage. Some sound is also reflected by the fuselage. A few passenger aircraft like the Caravelle, VC10 and Tristar have engines at the rear, which also helps to reduce sound levels in the cabin.
| {
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Understanding Shankar's standard deviation: Is $\langle\psi|\Omega|\psi\rangle$ shorthand for $\langle\psi|\Omega|\psi\rangle I$? On page 128 of Shankar's book, he defines the standard deviation as follows:
$$\Delta\Omega=[\langle\psi|(\Omega-\langle\psi|\Omega|\psi\rangle)^2|\psi\rangle]^{\frac{1}{2}} $$
Now, this equation is undefined as is, because $\langle\psi|\Omega|\psi\rangle$ is a scalar and $\Omega$ is an operator. Is $\langle\psi|\Omega|\psi\rangle$ shorthand for $\langle\psi|\Omega|\psi\rangle I$, or is it something different?
| So if you want to explicitly combine $(\Omega-\langle\psi|\Omega|\psi\rangle)^2$ into a single operator without actually using it on a state, then you would need to put the identity operator in for the math to make sense.
However, as pointed out in the comments, multiplying a state by a scalar is a perfectly valid operation. Therefore, you could easily apply the distributive law for the operator/scalar terms and then be perfectly fine with not even thinking about the identity operator. i.e.
$$\langle(\Omega-\langle\Omega\rangle)^2\rangle=\langle \Omega^2\rangle+\langle\langle\Omega\rangle^2\rangle-2\langle\langle\Omega\rangle\Omega\rangle=\langle \Omega^2\rangle-\langle\Omega\rangle^2$$
where I have condensed the notation to avoid clutter of $|\psi\rangle$ terms.
In my opinion, there is nothing wrong with putting in the identity operator, so if that is what makes the most sense to you then go ahead and think of it like that.
| {
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How can we see the microstructure of steel samples at room temperature? When we intend to see the microstructure of a steel sample at a temperature say 950 $^\circ$C they say we quench it in order to freeze the microstructure. However, if we quench it are we not going to have Martensite formation? Again if we anneal it are we going to see the microstructure at room temperature or at 950$^\circ$C. What is the way out? How do I see the microstructure that I wish to see (the ones at elevated temperatures) at room temperature?
| By choosing the correct quench rate for the alloy at hand, you will indeed freeze into the metal the microstructure it had at the temperature at which you began the quench. Different quench rates will yield different microstructures, with the slowest quench rate producing room temperature microstructure.
In all cases, the microstructure is studied in the lab after the quench with a metallographic microscope which uses reflected light, confocal with the microscope's optical path, and sometimes polarizing filters.
Before inspecting the quenched part with the microscope, it must first be polished to mirror smoothness and then subjected to a selective grain boundary etch to resolve the microstructure. These processes are carried out in an electrochemical etch bath.
| {
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Momentum operator in QM - scalar or vector? The momentum operator for one spatial dimension is $-i \hbar\frac{\mathrm d}{\mathrm dx}$ (which isn't a vector operator) but for 3 spatial dimensions is $-i\hbar\nabla$ which is a vector operator. So is it a vector or a scalar operator?
| $-i\hbar \frac{d}{dx}$, a scalar, is the position space representation of $\hat{p}_x$, the $x$ component of the momentum operator, a scalar. The momentum operator itself, $\hat{\textbf{p}}$, is a vector operator. The position space representation of $\hat{\textbf{p}}$ would be $-i\hbar \nabla$, a vector.
Again, the momentum operator is a vector operator. The components of the momentum operator are scalars operators.
| {
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Delayed eraser experiment (Kim experiment) Will I receive an interference pattern on D0 detector https://en.wikipedia.org/wiki/Delayed-choice_quantum_eraser#The_experiment_of_Kim_et_al._(1999) if D1, D2, D3, D4 exist in the set up of the experiment, but they are thousands of light years far? Simply, to set up the experiment so that I detect all of the signal photons on D0 before I detect any of idler photons!
| The interference patterns in the delayed-choice quantum eraser experiment are only ever observed via post-selection: using the notation from Wikipedia, the $D_0$ detector will detect a formless blob, which will only resolve into complementary interference patterns once you post-select on detections by $D_1$ and $D_2$ (as opposed to detections by $D_3$ and $D_4$).
As such, the interference patterns can be argued to be "implicit" in the data you've taken, but they can only be revealed once you receive the data from what was measured on the other side. Since you specified that the two sides of the experiments are thousands of light-years apart, this means that you will need to wait for several thousand years after measuring on $D_0$ before you can tease apart the constituent interference patterns in your data.
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Is there 100% pure white? Is it possible to have an object 100% pure white without sky blue or sun color tinting the pure whiteness of the photons reflecting/deflecting off an object? Are there any lights that can produce pure white photons (RGB)? And can we see that the color is white or is our eyes going to trick us into thinking the white is a different color? Can scientist produce what can be seen macroscopically as "pure white".
| I would think that a light source which has an emission spectre $f(\omega)$ that looks something like
$$
f(\omega)=\begin{cases}
\infty & \text{if infrared} < \omega < \text{ultraviolet}\\ 0 & \text{else}
\end{cases}
$$
could be considered pure white, but then, as others have pointed out, "pure white" has to be defined.
| {
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Is the only absolute difference between types of light frequency? Probably a bad question but for some reason, it seems too simple in my head that anyone at home could theoretically create anything from radio waves to gamma waves by generating electrical signals at different frequencies.
Say I had a electronic frequency generator that was able to produce a signal at any frequency, and for illustrative purposes, say there was a diode hooked up this generator that could receive its signals.
If it created a signal at $10^{12}$ Hz, the diode would give off infrared radiation.
If I increased the signal to $10^{20}$ Hz, the diode would give off gamma radiation.
I’m using this example just to emphasize my question, is frequency the absolute and only differentiator in types of light on the electromagnetic spectrum?
| Your question is a little confused at the end, but I think the answer to what you're trying to ask is "yes". Names like infrared and gamma apply here to ranges that have been divided up for historical and practical reasons, but they do not denote something other then an electromagnetic wave within certain frequency ranges. If you had the hypothetical device that you mentioned, then you could create waves of any of the types that you mentioned. (Although I know of no such single device that covers such a range.)
Also note that there can be more than one name for a particular range. For example "radio" waves have different bands and those bands go by different names by country and by science / engineering discipline. For example, K Band or X Band.
The "electromagnetic spectrum" would cover all frequencies by definition, so your device would not create something "outside" of the spectrum. It might, I suppose, create a wave in a frequency range that has no conventional name.
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In QFT, are forces made out of multiple fields? I’ve been reading about 1,5 books about quantum physics and I’ve also watched a few YouTube videos. In one book, I learnt that there are fields, such as the electromagnetic field, which carries forces between particles (vibrations in the field) via virtual particles. But in another book and on YouTube, they instead say that every elementary particle (quarks, photons, electrons etc.) have their own fields. But what I can’t understand is what the relation of the fields I read about that carries forces to the ones with the particles? Are these forces made up of interactions between multiple fields?
| There is no clear notion of what exactly a "force" is in quantum field theory (obligatory xkcd).
Every field has particles associated with it, and we tend to associate those particles/fields with "forces" whose interactions with the rest are such that they have a major influence on their behaviour even when they are not directly involved (e.g. there are no real photons involved in the process that gives the Coulomb force in its classical limit, and no real W/Z-bosons involved in the process that turns protons into neutrons in $\beta^+$-decay).
Massless particles that couple to others in the same way the photon does always produce a classical force like electromagnetism (see this question for the technical background), while massive particles with such a coupling produce a classical force that decays exponentially (cf. Yukawa potential, and see this answer of mine for the residual nuclear force mediated by pions as an example). But even this is oversimplified, since e.g. gluons are massless yet produce no classical force at all since no free color charges can occur due to confinement.
| {
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What is the difference between position, displacement, and distance traveled? Suppose the question is somewhat like this:
If $v=8-4t$ and the position at time $t= 0\ \rm s$ is $2\ \rm m$, find the distance traveled, displacement, and final position at $t=3\ \rm s$
Since $\text dx/\text dt=v=8-4t$, then $\text dx=(8-4t)\text dt$. After integrating we find $x(t)-2=8t-2t^2$, and substituting the value of $t=3\ \rm s$ we get $x(3)=8\ \rm m$.
Is the answer that I found displacement, position or distance?
It can't be distance. I am sure of this. But is it position or displacement?
|
So the 8m is displacement or distance?
The 8m is the position of the particle. The displacement is 6 m.
When you integrated the velocity function to get position you included a constant of integration (2 m) to account for the initial position of the particle, that is, its position at time t=0, which is given as 2 meters. If you omitted the constant of integration then you would be calculating the change in position, i.e. the displacement, and not the position. In this case the displacement would be 6 m.
Position can be considered as a point in a coordinate system relative to the origin of the coordinate system. The diagram below shows the position $s$ of the particle in a one dimensional coordinate system, relative to the origin of the system, s=0. In effect, when the particle's position is 2 m the timer is started, i.e. t=0. Displacement is then distance between two positions of the coordinate system as a function of elapsed time (3 s), which in this case 6 m.
Hope this helps.
| {
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Circuit - solve Kirchhoff's laws with determinant? Here is an exercise from my textbook.
]1
At first I solve it by using Kirchhoff's laws directly and using complex impendance:
$$U_{in}=\frac{1}{Cs}(i_1-i_2)+i_1R_1$$
$$0=R_2i_2+\frac{1}{Cs}i_2+\frac{1}{Cs}(i_2-i_1)$$
By solving above equations, both $I_1$ and $I_2$ become function of $U_{in}$. Then I sub them into $U_o=i_1R_1+\frac{1}{Cs}i_2$ to get the relationship between $U_o$ and $U_{in}$. This method works and lead to the answer provided in the book.
However I found a solution looks quite different in solution manual with determinant:
Let,
$$\Delta=\begin{vmatrix}
G_2+Cs & -Cs & -G_2\\
-Cs & G_1 + 2Cs & -Cs \\
-G_2 & -Cs & Cs + G_2
\end{vmatrix}$$
then,
$$V_j=\dfrac {\Delta_{1j}} \Delta I_1~~~~~~or~~~~~~
\dfrac {V_3}{V_1}=\dfrac{\Delta_{13}I_1/\Delta}{\Delta_{11}I_1/\Delta}$$
Therefore the transfer function is
$$\dfrac {V_3}{V_1}=\dfrac{\Delta_{13}}{\Delta_{11}}=\dfrac{\begin{vmatrix}-Cs & 2Cs+G_1\\-G_2 & -Cs\end{vmatrix}}
{\begin{vmatrix}2Cs+G_1 & -Cs\\-Cs & Cs+G_2\end{vmatrix}}$$
$$=\dfrac{C^2 R_1 R_2 s^2+ 2CR_1 s+1}{C^2 R_1 R_2 s^2+(2R_1+R_2)Cs+1}$$
So far I am quite confused on what does the first determinant stands for and could this determinant be written down directly without any draft? (If so, I think it will be faster and easier to solve problems like this and I really want to learn about it.) Also, why do we need to use the Minor of that determinant?
| As has been pointed out, $\Delta$ is the determinant of the nodal admittance matrix $\mathbb{Y}$. It's also clear that $\mathbf{V}$ is the vector of node voltages. What wasn't clear to me is that $\mathbf{I}$ is the vector of injected currents rather than the currents the OP has drawn in red.
That is, $I_1$ is the (phasor) current injected to node 1, i.e., the current into the input port of the network.
Since node 2 (the junction of the two capacitors and $R_1$) is an internal node, the current $I_2$ injected is zero.
Assuming the output port is terminated with an open circuit, the current $I_3$ injected is also zero.
Thus, the only independent variable is $I_1$.
Solving for $\mathbf{V}$ yields
$$\mathbf{V} = \mathbb{Y}^{-1}\mathbf{I} = \frac{\mathbb{C}^T}{\Delta}\mathbf{I}$$
where $\mathbb{C}$ is the cofactor matrix. Since $I_2 = I_3 = 0$, it follows that
$$V_j = \frac{C_{1j}}{\Delta}I_1$$
| {
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What are estimated magnetic properties of liquid metallic hydrogen? I understand that liquid metallic hydrogen isn't easy to produce, or keep it stable on Earth, but can be liquid metallic hydrogen magnetised? Does it have magnetic properties at all?
| There is no reason to expect metallic hydrogen to be any more magnetic than say metallic sodium, that is, patamagnetic. One caveat is that at sufficiently high pressure and low temperature it may become superconducting.
Apparently, rotating liquid sodium will generate a magnetic field as this experiment aims to prove : https://m.youtube.com/watch?v=OsPyQ4Sv-DA&feature=youtu.be
| {
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Does stacking two magnets together increase the magnetic strength? Just wondering if adding two (or more) identical magnets together increases their magnetic strength (in Tesla) as I am doing a physics write-up on the Lorentz force (Fleming's left-hand rule) as I tried stacking magnets together but found that the force produced is still the same. Wondering if I'm doing anything wrong.
| Assuming that the field from one magnet does not effect the magnetization of the other, then the resultant field at each point will be the vector sum of the two separate fields. That in tern depends on the position and orientation of the two (as described by Andrew Steane).
| {
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Motion between two particles in a relative manner Suppose a particle A is travelling in east direction with velocity of x m/s and another particle B is travelling with velocity y m/s in the west direction. Why does the the particle B appears to move towards A with a velocity of x+y and not just y m/s?
| The top diagram shows the velocities in the lab frame. Particle $A$ is moving east at speed $x$ and particle $B$ is moving west at speed $y$. I'm taking the east direction to be positive, so the velocity of $A$ is positive and the velocity of $B$ is negative.
To find the velocity of $B$ relative to $A$ we have to transform to the frame where $A$ is stationary, and we do this by adding the velocity $-x$ to everything as shown in the middle diagram. Then the velocity of particle $A$ is $v_A = x + (-x) = 0$, so this makes particle $A$ stationary as we require.
And as the bottom diagram shows, when we add $-x$ to the velocity of article B we get $v_B = -y + (-x) = -(x + y)$. That's why the velocity of $B$ relative to $A$ is $-(x+y)$.
| {
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A best definition of proper acceleration In this link of Wikipedia https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity) there is a definition of proper acceleration:
The proper acceleration to a particle is defined as the acceleration that a particle "feels" while accelerating from one inertial reference system to another.
$$\alpha=\frac{1}{(1-u^2/c^2)^{3/2}}\frac{du}{dt}.$$
May I ask, please, if there is a better alternative definition to this or possibly some documentation in English clearer than that of Wikipedia?
| As far as I know, the definition of the proper acceleration was the derivative of the proper velocity with respect to proper time, $$\vec{\alpha}=\frac{\mathrm{d} \vec{w}}{\mathrm{d} \tau}=\frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} \tau^2}.$$ It is the spatial part of the four acceleration, and in Wikipedia in English it says that the equation that you cite is valid iff the proper acceleration is directed parallel to the line of motion. The relation between proper acceleration and ordinary acceleration is instead $$\vec{\alpha}=\frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \gamma^2+\frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \frac{\mathrm{d} \gamma}{\mathrm{d} \tau}$$ by the chain rule. You can identify $\vec{a}=\frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2}$ as the ordinary acceleration.
| {
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Why is "a very silent room" louder than "leaf noise" / "breathing noise"? I recently was at night in my bed room and used the Schallmessung Android App (on a Samsung S7). It showed 20 dB while laying on some furniture. I could not hear anything; to me, the room was completely silent.
Now I looked at some examples of noise levels on the German Wikipedia. Two of them confused me a lot:
*
*Sehr ruhiges Zimmer (very silent room): 20-30dB
*Blätterrauschen, ruhiges Atmen (leaf noise, breathing noise): 10dB
Why is a "very silent room" (I held my breath, no AC, at night, windows closed) where I can't hear anything louder than "leaf noise"? How accurate can I expect the results of the app to be?
| They are most likely talking about just the "leaf noise" by itself. If you were in your silent room with additional "leaf noise" then your app would obviously record a higher decibel reading (assuming it is sensitive enough).
| {
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Boundary condition for partial reflection I want to solve a wave equation for the wave $\psi(x,t)$.
One boundary is moving, therefore I impose the velocity
$$v(x=0)=v_a\cos(\omega t)$$
the other boundary is fixed, but reflecting. If the reflection is total the proper boundary condition is
$$ \frac{\partial \psi}{\partial x}\bigg|_{x=L}=0 $$
Which is the right boundary condition if the reflection is partial? Therefore, my boundary has a reflection coefficient $R$ and an absorbing coefficient $A$ (no transmission), so that $R+A=1$.
| Consider a boundary condition of the form
$$
\alpha \psi + \beta \frac{\partial \psi}{\partial x} + \gamma \frac{\partial \psi}{\partial t} =0
$$
on the boundary, where $\alpha$, $\beta$, and $\gamma$ are real coefficients. Standard Dirichlet boundary conditions correspond to $\beta = \gamma = 0, \alpha \neq 0$, while Neumann boundary conditions correspond to $\alpha = \gamma = 0, \beta \neq 0$.
If we impose this condition at $x = 0$ for an incoming wave of the form $\psi(x,t) = e^{i(kx - \omega t)}$, then the wave solution for $\psi$ will be
$$
\psi(x,t) = e^{i(kx - \omega t)} + A e^{i(-kx - \omega t)}
$$
where $A$ is the (complex) amplitude of the reflected wave. Some algebra then reveals that we must have
$$
A = \frac{i (k \beta - \omega \gamma) + \alpha}{i (k \beta + \omega \gamma) - \alpha},
$$
and so
$$
R = |A|^2 = \frac{(k \beta - \omega \gamma)^2 + \alpha^2}{(k \beta + \omega \gamma)^2 + \alpha^2}.
$$
By appropriate choices of $\alpha$, $\beta$, and $\gamma$, one can "tune" the reflection coefficient to be anywhere between 0 and 1.
A few notes on this expression:
*
*This reflection coefficient will be frequency-dependent unless the medium is non-dispersive and $\alpha = 0$. In particular, if the reflection coefficient is frequency-dependent, this means that a pulse sent towards the wall will not be reflected with the same shape. If you want pulses to retain their shape on reflection, use $\alpha = 0$.
*For a given value of $R$, the coefficients $\alpha$, $\beta$, and $\gamma$ are not uniquely determined. This is for two reasons. First, we can always divide or multiply all three coefficients by the same number to get the same boundary condition; you can always use this freedom to set one (non-zero) coefficient equal to 1 if you like. Second, the reflection coefficient alone doesn't tell you everything about the reflected wave; the reflected wave can also be phase-shifted.
*If $\gamma = 0$, we have $R = 1$ identically. This makes some sense; in such a case, the equations have time-reversal symmetry, whereas absorption involves an inherent "arrow of time".
| {
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Can the flow be irrotational if the viscous forces act on fluid? I tried to answer the question only using the definitions and the Navier-Stokes equation:
$$\rho \frac{Dv}{Dt} = -\nabla P +\rho g -\mu[\nabla \times(\nabla \times v)] $$
In my opinion if the vorticity is zero, then the fluid is irrotational, regardless of presence of the viscous forces, thus $\mu$ can have a non-zero value which implies the existence of viskeuze forces, while the $\nabla \times v = 0$.
| First, unlike the other answers, I believe your equations are correct thanks to the identity
$$\nabla \times \nabla \times \mathbf v = \nabla (\nabla \cdot \mathbf v)-\nabla^2 \mathbf v$$
Irrotational flow means $\nabla \times \mathbf v=0$ so in fact as you note correctly Navier-Stokes predicts viscosity plays no role in irrotational flow.
The question of whether the flow is irrotational is answered using the vorticity equation, which predicts the evolution of the vorticity $\mathbf \omega = \nabla \times \mathbf v$:
$$\frac{D\mathbf \omega}{Dt}= (\mathbf \omega \cdot \nabla) \mathbf u+\nu \nabla^2 \mathbf \omega$$
In many situations the flow is irrotational if and only if $\mathbf \omega =0$ (but there are exceptions). Therefore we see that if $\nu=0$ (no viscosity) and the vorticity is initially zero, then $\mathbf \omega =0$ at all times (this is because $\mathbf \omega =0$ is a solution and the solution is unique). However note that if $\nu \neq 0$ then in general vorticity will develop even if there was none initially. Of course you can still impose the requirement that the flow is irrotational if you so wish.
As a final remark, I'd like to point out that even though in many cases one may assume that the flow is irrotational, then one may run into issues: for instance you need viscosity to impose the no-slip condition on a boundary. In these situations one assumes that viscosity is relevant near the boundary and the flow is irrotational elsewhere.
| {
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Can a macrostate of a system be defined upon distributional quantities of the energies? Is it valid to define for a system, a macrostate based upon statistical quantities of the particle energy distributions? Eg. that a macrostate is based upon the variance of the kinetic energies of the particles? Would another potentially valid macrostate value be the skew of the distribution of the energies? Those macrostate values are to be used in order to calculate the entropy changes in a system over time as the number (size) of microstates that satisfy the macrostate.
| Mathematically, the answer is yes. Whether this makes physical sense is a different story.
Let's do the math first. We have a collection of particles $i=1,2,\cdots$ with energies $E_i$ and all we know is their variance. By the maximum entropy principle we seek the probability distribution $p_i$ that maximizes the entropy functional,
$$
-\sum_i p_i \log p_i
$$
under the constraints
$$
\sum_i p_i = 1,\quad
\sum_i p_i E_i^2 = \sigma^2
$$
This variational problem has a known solution: $p_i$ is a Gaussian function with variance $\sigma^2$ regardless of its mean (see wikipedia; on the same web page there is also the case of a distribution with known variance and skewness).
Does it make physical sense? A physical macrostate is a macroscopic state in equilibrium that can be prepared so as to satisfy a number of specifications. For example, a $(U,V,N)$ system is prepared by putting $N$ particles in a rigid insulated box with volume $V$ (we may have to play with the temperature to hit the exact value of $U$ we want). Can we prepare a system of particles such that the variance or skewness of the distribution is exactly what we want, independently of the mean of the distribution? I think it's safe to say that answer is no.
| {
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Huygens-principle Huygens principle states that every point on the wave front acts as a source. If it is true, then why can't a single source (let's say a bulb) illuminate a whole big room? Why is it dark after some distance from the bulb? According to him it should continue to infinite.
Where am I wrong?
|
Why is it dark after some distance from the bulb? According to him it
should continue to infinite.
That's a great observation! I must have seen Huygen 100 times and never thought of this.
The basic answer is that Huygen's principle is mostly concerning the phase of the waves, in a way. He's talking about how the wave front propagates. This is what you see in the typical diagram consisting of arcs emanating from a point source.
The amplitude of the wave, which corresponds to the illumination in this case, is a function of its spread. This is actually baked into the principle, although you'd never know that from common explanations or diagrams.
A better illustration would have the arcs grow increasingly light as they move away from the source.
| {
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Normalization constant of a planar wave As we know for the plane waves ( $ae^{i k x}+b e^{-i k x}$), the normalization constant can be easily obtained from the integral $\int^{x_{2}}_{x_{1}}\psi^{*}\psi dx=1$ by the relation $|a|^{2}+|b|^{2}=1$. But what happens if the parameter $k$ is imaginary, i.e. $k=i \kappa$ where $\kappa$ is real. Do we have the same relation for the normalization?
| Using your parameterization, the wave is $ae^{-\kappa x}+be^{\kappa x}$. Note that this particular wavefunction blows up at $x=+\infty,-\infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$. If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize the wave.
Remember that $k=\sqrt{2m(E-V)}/\hbar$, so that it will be imaginary in regions where $E<V$. In particular consider a wave incident in $x<0$ on a step potential barrier of height $V_{0}$ for all $x>0$. If $E<V_{0}$, it will have the form $ae^{-\kappa x}$ at $x>0$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.
| {
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Are the electrons' orbitals the same for all atoms? Are the electronic orbitals of an atom always quantified in the same way (i.e. the same energy required to reach the next level), or does each atom have its own values for each level?
If the quantification is universal, then the creation of photons (due to the deexcitation of the electrons) at the wavelength / color corresponding to the transition should be more abundant in the universe than all the other frequency. Except one detects no more photon of a given wavelength than of another.
So where is my reasoning error?
| The diagram you posted shows the electron energy levels for the hydrogen atom only. Other atoms will have different energy levels. Also note that the hydrogen atom itself has an infinite amount of energy levels which the diagram does not really show.
This information is very useful. The energy levels for each atom are unique, and the corresponding frequencies of light emitted by the electrons can serve as a signature of the atoms presence. This gave rise to the field of spectroscopy and it is useful in many applications. For instance, the frequencies of the light coming from other planets is used to determine the composition of the atmosphere.
| {
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Why are you doing no work when carrying a body through a horizontal distance? Work done on an object is equal to $$FD\cos(\mathrm{angle}).$$
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction,
| Work-energy theorem would be quite helpful to you.
$$dW_{net}=dK_{system}$$
As speed is constant,$$dK_{system}=0$$
$$dW_{hand}+dW_{resistance}=0$$
It doesn't imply that work done by your hand is zero but the net work done by your hand as well as resistive forces is zero.
| {
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Entropy vs energy graphs How to interpret temperature from entropy vs energy graphs?..
And what do the different behaviour of the graph signify and how should I interpret them?(provided volume of the system and number of particles are kept constant)
| As pointed out in the comments (or in any good text book), the temperature is given by
$$\frac1T=\left(\frac{\partial S}{\partial U}\right)_{V,N}$$
where $T$ is the temperature, $S$ is the entropy, $U$ is the internal energy, $V$ is the volume, and $N$ is the number of particles.
You stated the graphs already assume constant $V$ and $N$, so all you need to do is look at the slope of your graph. The slope determines the temperature.
This is all you need to interpret your "different graphs".
| {
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Is kinetic energy always conserved in an elastic collision/impact? While working out through some problems I encountered this problem :
A ball moving with a velocity $v$ hits a massive wall moving towards the ball with a velocity $u$. An elastic impact lasts for a time $\Delta t$
Now I have to answer whether the Kinetic energy of ball increases or remains same after collision.
In the theory books which I read, it is mentioned that Kinetic energy is conserved before and after in an elastic collision.
So that way for the above question Kinetic energy should be conserved.
But the answer given is that Kinetic energy increases.
So my question is how is it possible for Kinetic energy to increase after an elastic impact ? Is it because of the time interval $\Delta t$?
| While @Rumplestillskin has basically answered your question, here is an intuitive explanation. The total kinetic energy of the wall and the ball together is preserved after the ball bounces off. Now, since the wall was moving towards the ball, however small the ball is, it had slowed the wall a bit. So part of wall's energy has passed to the ball, thus ball's energy has increased. Imagine that the wall is actually a tennis racket - it is relatively massive, and hitting the incoming ball. Naturally, the ball returns faster than it came in when you hit it.
| {
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Complex Scalar Fields and Killing Vectors In a stationary and axisymmetric spacetime, there are two Killing vectors, say $\zeta^\mu$ and $\xi^\mu$, one timelike and one space like.
I understand that for a real scalar field, $\phi$, that obeys the symmetry of this spacetime one can say
$$\zeta^\mu \nabla_\mu \phi = \xi^\mu \nabla_\mu \phi = 0 $$
and therefore since the gradient of $\phi$ is orthogonal to both Killing vectors, $\nabla_\mu \phi \nabla^\mu \phi \geq 0$.
Now, my question is in regards to how this logic translates across if $\phi$ becomes a complex scalar field, where $\Phi = \phi_1 + i\phi_2$.
$\nabla_\mu \Phi \nabla^\mu \Phi \geq 0$ no longer makes sense to me, since the gradient is complex, however is it a valid comment to say that the real components of the complex field must be spacelike or null, if the complex field is to still obey the spacetime symmetries, i.e. $\nabla_\mu \phi_i \nabla^\mu \phi_i \geq 0$
| You're right to conclude $\nabla_\mu\phi_j\nabla^\mu\phi_j\ge0$ for $j\in\{1,\,2\}$, so $\nabla_\mu\Phi^\ast\nabla^\mu\Phi=\sum_{j=1}^2\nabla_\mu\phi_j\nabla^\mu\phi_j\ge0$.
| {
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Total force on upper block in two block system If a block $m$ is placed on another block $M$ and a force $F$ is applied on bolck $M$. Then how many forces are acting on block $m$.(Friction is non zero)
The image is taken from this site.
Is pseudo force acting on block $m$ or not?
| No pseudo-force ever acts on anything. A pseudo-force is - as the name indicates - imaginary and non-existing. It is just a "feeling" or "illusion" of a force. So don't worry about pseudo-forces; just stick to inertial reference frames (references that aren't accelerating), and you never have to worry about them.
In your case, if the bottom block accelerates because of that force $F$, then friction - and only friction - acts on the upper block in the horizontal direction. This friction pulls the upper block along with the lower block.
Vertically, there is a normal force as well as gravity, of course. They cancel each other out, so only the friction causes its acceleration.
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How does the introduction of the charm quark suppress FCNC? I did some reading on the GIM mechanism today, and simply fail to understand how it works. I understand how the CKM-matrix can be used to do the basic calculation of the probability of, say, observing an up quark after a strange quark's decay over the weak interaction. However, I don't understand how the CKM matrix is applied to these Feynman diagrams:
I see that the vertices for the weak interactions of the quarks are labeled with the corresponding matrix elements, but how do I take these two Feynman diagrams and infer that they cancel each other out?
| Well, the two interfering amplitude diagrams do not quite cancel each other out: they almost cancel each other out. That is to say in the notional limit that the mass of the u and the mass of the c were identical, the two diagrams would be identical except for the minus sign of the Cabbibo (CKM for 3 generations) matrix influence on the vertices, which you reassure us you are comfortable with.
To the extent the masses of the two quarks in the internal lines differ, the effect of them on the respective propagators differ, and so the respective results of the loops differ. In fact, good SM books compute the nonvanishing, but vastly suppressed amplitude. It is a function of $m_c/m_u$ which goes to 0 as that ratio goes to 1. Something like $\propto g^4 \frac{m_c^2}{M_W^2} ( 1-m_u^2/m_c^2)$.
So you might sensibly object that the term in the parenthesis is much closer to 1 than it is to 0. But, look at the factor multiplying it, $\alpha^2 m_c^2/M_W^2$, and how small it is: do this. (Still, if that were a part of your puzzlement, perversely, the introduction of c actually increases the $\Delta S=1$ rate instead of suppressing it! Historically, the rate was used to bound the mass of the then hypothetical c from above!)
When the third generation is introduced, the 3x3 unitary analog matrix (CKM) performs the same function. The quark mass ratios are bigger, but the couplings are suppressed at the vertices, so, if I recall, the effect of neglecting the 3rd generation is not dramatic.
*
*Extra credit hypothetical: In an imaginary world where the masses of u and c differ by one part per million, but are huge, say half the mass of the W, would you have this strangeness-changing neutral current amplitude be suppressed, or not?
| {
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"timestamp": "2023-03-29T00:00:00",
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Symmetry reason why magnetic dipole transitions are suppressed In the theory of light-matter interaction, electric dipole transitions between two atomic states of same parity are forbidden. This is because the Hamiltonian conserves parity. Is there a symmetry reason why magnetic dipole transitions happen but their amplitudes are extremely suppressed?
| This comes from the heirarchy of length scales in the atomic problem. Very briefly, the EM-atom transition matrix element has a term like
$$\int e^{i\vec{k} \cdot \vec{x} }\psi'(\vec{x})\left[\vec{A_0} \cdot (-i\hbar \vec{\nabla} )\right]\psi(\vec{x}) d^3x $$
A standard approximation to this is the dipole approximation, in which we say that $$e^{i\vec{k} \cdot \vec{x} }\approx 1 \text{ .}$$
This is justified by the fact that $\psi$ typically has a term like $e^{-r/a_0}$, so the spatial integral is over a volume about the size of the atom. This is usually much smaller than the wavelength of light considered, so that the EM field is basically constant spatially over the atom. It turns out that when you take this approximation you end up with an electric dipole coupling, because the integral reduces to something like
$$\hat{\epsilon}\cdot\int \psi'(\vec{x})\left[ e\vec{x}\right]\psi(\vec{x}) d^3x $$
where $\hat{\epsilon}$ is the photon polarization.
Taking the next order in the expansion of $e^{i\vec{k} \cdot \vec{x} }$ gives the magnetic dipole transition, as well as the electric quadrupole. Lots of books give up to the dipole approximation, like Sakurai and Shankar, but I'm not sure of a good treatment of the higher terms.
For atoms, one can say from dimensional arguments that the ratio of a generic magnetic to electric dipole matrix element is $(Z\alpha)^2$, with $Z$ the atomic mass (which determines the typical atom size) and $\alpha$ the fine structure constant (1).
Regarding Ben's point, for the first nucleus I found searching ($^{12}$C), it looks like $r/\lambda\approx 0.01.$ So it might naively appear that the dipole approximation is not too bad, but this is still two orders of magnitude larger than a typical value for atoms.
(1): Foot, Atomic physics.
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Why do all fields in a QFT transform like *irreducible* representations of some group? Emphasis is on the irreducible. I get what's special about them. But is there some principle that I'm missing, that says it can only be irreducible representations? Or is it just 'more beautiful' and usually the first thing people tried?
Whenever I'm reading about some GUT ($SU(5)$, $SO(10)$, you name it) people usually consider some irreducible rep as a candidate field. Also, the SM Lagrangian is constructed in this way. (Here, experimental evidence of course suggests it.)
| This is only semantics. A reducible representation $\mathbf R$ of the symmetry group can be decomposed into a direct sum $\mathbf R_1 \oplus \cdots \oplus \mathbf R_N$ of irreducible representations. A field that transforms as $\mathbf R$ is the same thing as $N$ fields, which transform as $\mathbf R_1, \dots, \mathbf R_N$. When talking about fundamental fields, we can therefore assume that they transform as irreducible representations of the symmetry group.
| {
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"timestamp": "2023-03-29T00:00:00",
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Basic band structure reading In band diagrams (e.g. GaAs below), symmetry points are sometimes indicated with a numerical index. I first thought it was the band index in the Bloch function (so two electrons with the same wave vector but different band index would have different energies), but the same point appears at different energies...
What is this index really corresponding to ?
| Those are the labels of irreducible representations corresponding to the wave-function on that symmetry point, derived via group theory. They classify how wave-function at that symmetry point evolves under crystal symmetry operations, similar as the classification of odd and even function $f(x)$ by whether it changes sign under inversion operation of $x$. Here is a pretty nice introduction to this topic of group theory and band structure (if you have access to it):
https://link.springer.com/content/pdf/10.1007%2F978-3-642-00710-1_2.pdf
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