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Different frequencies working together How do the different waves of EM spectrum present in the environment not interfere with each other? If they do, how does everything work properly?
The radio waves of mobile phones and wi-fi work together. Why don't they collide with each other, since they are physically present?
| They do interfere with each other, but the outcome is not a problem so long as the device that's receiving or using the signals is tuned to pick out the relevant frequency.
For instance, resonant circuits will only be driven properly in a narrow range of frequencies around the resonant frequency. A mixture of waves with different frequencies will effectively be filtered by such a circuit to pick out the signal of interest.
Where you do get problems is broadband interference or noise that covers a wide range of frequencies, because obviously some of this will occur over the narrow range of frequencies that a device uses.
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Is there a commonly used unit of measure (other than temperature units) that is not absolute? I live in a country where we use Degree-Celsius(°C) to measure the temperature.
Sometimes from one day to the other, the temperature rises from 10°C to 20°C and I hear people say, "Wow! Today is twice as hot as yesterday!".
I try to explain that today is not two times hotter than yesterday because Celsius(°C) is not an absolute unit of measure, that is 0° does not mean the absence of temperature.
If I convert Celsius(°C) to Fahrenheit(°F) or Kelvin(K) it gets clear, but I wish I could provide another example of non-absolute unit of measure to clarify things.
In short, do you know any other unit of measurement (except for temperature ones) where 0 does not mean the absence of the physical phenomenon that is being measured?
| Gravitational potential energy is another physical quantity that is measured relative to an arbitrary starting point.
In many derivations, the zero point for the potential energy of a mass is taken to be a point an infinite distance away, and all the potential energy is negative.
A hydro electric engineer would take the bottom of his dam for the zero point of potential energy for the water he has going through his generators, so the higher the water level above the dam, the more energy he has available.
His colleague downstream will take the bottom his second dam as a convenient reference point for the energy available at that dam.
Or the entire system could use sea level as a zero point, and rely on height difference at each dam for their calculations. It's all relative...
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How can the thrust due to radiation pressure be amplified in photonic laser thruster? The thrust is amplified due to repeated bouncing of photons between two mirrors as shown in the diagram in this: Why does repeated bouncing of photons produce amplified thrust when the answer in 'Mirror problem of radiation pressure' indicates that the radiation pressure will only be doubled?
| There is a difference between "multiple photons" and "repeated photons". If one photon keeps bouncing back and forth, it can carry a little bit of impulse from one mirror to the other on each round trip. When you have multiple photons, each of them can do this. So your "amplification" comest from either
*
*Waiting a long time, so the photon makes lots of round trips
*Adding lots of photons.
In either case, the total impulse imparted scales with number of (photon-round trips).
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Current in a fluorescent tube that is not in a circuit In Walter Lewin's 8.02 Electricity and Magnetism course, he places a fluorescent tube pointing radially outwards from a large Van de Graaff (VDG) generator. Due to the VDG's E-field, this causes a large potential difference between both ends of the tube. He then says that this high voltage causes a current in the tube, which creates light.
Subsequently, he then touches the end of the tube furthest the VDG, receives a shock, and by grounding this end, allows more current to pass, thus creating more light.
If the tube is not in a circuit, that must mean all the charges that were moved by the potential difference are accumulating in one end of the tube, causing the tube to act like a capacitor and eventually stopping current flow. Obviously this is not what is happening because the tube creates light continuously.
So what exactly is the path of the current, and why should the tube light up more if Dr. Lewin grounds one end through his body?
EDIT: He also does this experiment with a neon flash tube, and it creates light as well
| When the VDG sphere is 'charged' to say $300kV$ or greater, the moisture in the air around it is able to conduct a very small $<100\mu A$ electric current, which is enough to dimly light the fluorescent tube.
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How were the ratios of distances between planets and the Sun first calculated? I was reading some literature and I found that long before the actual distances between other planets and Earth or distance between Sun and Earth were known, physicists had calculated the ratios between these distances. Can anybody tell me the technique used at that time to measure these ratio? This must have been done before 1650.
| The relative distances of the earth, sun and moon were determined by Aristarchus. See my summary here. By measuring the size of the earth (as e.g. Eratosthenes did) these can be turned into absolute distances.
Once heliocentrism was introduced the planetary distances could be determined as follows:
Distance from Venus (or Mercury) to the sun: continually measure the angle VES; when it is at a maximum the angle EVS will be right, and we know ES so we can find VS. (Since Venus and Mercury move much faster than the earth, the earth can be considered stationary for the purposes of this demonstration.)
Distance from an outer planet P to the sun. Note when P is in opposition, i.e., when SEP is a straight line. Then wait for the earth and planet to move until the angle SE'P' becomes a right angle. Since we know the orbital times of E and P we know the angles ESE' and PSP' (assuming the orbits to be circles centred at the sun). The angle P'SE' follows, and we already know angle SE'P' and length ES so we can compute SP'.
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The interference pattern of a wave along a line Why the answer is B, but not C?
Is it true that the pattern cannot go beyond $A_0$ and $-A_0$ at the same instant?
| The integral of both initial waves must be ~0 (it is cancelled out) and adding those two together yields a 0 integral too. So the resulting wave-function must be symmetric. Also, energy conservation implies that the amplitude must be $\sqrt 2 A_0$.
So answer C seems to be correct.
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What is dimension and how many types of dimensions are there in the universe? What is dimension and how many types of dimensions are there in the universe? I mean how many total dimensions are there? I have only heard about 2d and 3d. Other than these two, are there any other dimensions. If yes than please explain each of them clearly.
| 'What is meant by "dimension"?'
As the question is asked above, I think we are talking about the nature of reality. In our physical world, 3 dimensions are required to differentiate any and all points that any physical object can occupy.
If our universe was static that would be sufficient but the universe and its contents are not static so a 4th dimension is necessary to describe/observe events: time.
Time exists at every point in space. If an object exists at a point in space it is also defined by a point in time to complete the description. Time is not an observable dimension which opens the idea of other dimensions which are not observable but still exist at every point in our 3 dimensional space.
If one believes that there is a spiritual world then it may very well account for other dimensions but that crosses into philosophical discussions. Other dimensions may be the domain of the non-physical forces we do observe, such as gravitational, electro-magnetic, weak and strong.
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Pressure in the grand canonical ensemble when momentum integration limits depends upon volume When one does not want to consider the thermodynamic limit, it is possible in some systems to consider a dependance of the volume on the integration limits of the momentum. For example:
$$\mathcal{Z} = \sum\limits_{N} \int d^{3N}r \int\limits_{0}^{f(V)}d^{3N}p \, e^{-\beta(\mathcal{H} - \mu N)} $$ where $f(V)$ is an arbitrary function of the volume. You can think of an ideal gas in which the particles has less momentum near the walls than in the center.
If one wants to calculate the pressure, there are two ways, that in this example, does not yield the same results if one doesn't take the thermodynamic limit. They are, $pV= k_B T \ln \mathcal Z$ and $p = -\frac{\partial (-k_B T\ln\mathcal Z)}{\partial V}\Big|_{T,\mu}$. Why does this happens? Which one is wrong? Is it because the identification of $-k_B T\ln \mathcal Z$ with the grand potencial is neessary to use the first one, and it cannot be done without taking the thermodynamic limit so mean quantities are equal to most probable?
Any help will be much appreciated.
| You are right, $pV= k_B T \ln \mathcal Z$ is only guaranteed in some systems (namely extensive ones). If you suspect your system is not extensive, for example if it has physics that depend on surface area and not just volume, then you can't use this expression and you have to use the partial derivative you named.
But careful, in this case your system's pressure is probably not just a single number. If you change the system boundary in one place you will see a different pressure than making a change in another place. In other words in extensive systems there is not just volume, but also shape dependence.
For example take a fluid system in a gravity well, not quite the same as your system but it definitely it is non-extensive since the particle's energy is position-dependent. Then you find that pressure is different on the top and the bottom.
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What is antimatter? Can you give a visual example of what is antimatter?
With the re-opening of Large Haldron Collider scheduled in Mar 2015, I'm reading that they smash two particles together to try to re-create particles that might have been there are the beginning of the Big Bang, and this includes antimatter?
Is antimatter something we can see, or it is some invisible field. I'm trying to get my arms around this concept.
| Paul Dirac in the 1930's described a vacuum as not empty space but as an infinite sea of negative energy particles.
His famous 'Dirac Equation' has a negative energy solution for the electron, he interpreted this solution by saying that since Pauli's Exclusion Principle states that 'no two electrons (fermions with half-spin) can occupy the same quantum states', all of the negative energy states below the ground state predicted by his equation must already be occupied by particles, say for examples electrons.
Now, if you imagine a gamma ray photon transforming via pair production into an electron and a positron, you can think of the positron as a hole in this infinite sea of negative energy which is what Physicists detect as being an antimatter particle, with the same mass, spin etc as its matter counterpart, the electron, but opposite sign on its charge.
Below is a visual representation of this;
I wouldn't take this way of thinking of antimatter too literally, however I find it to be a useful way of picturing the essence of antimatter.
Hope this helps.
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Who is doing the normalization of wave function in the time evolution of wave function? In the Schrödinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations simultaneously. Why not? I know it does not yield, but I am always baffled, who is doing the normalization of the wave function? Observers, the system, the measuring process, God?
| I think it's a very good question.
As a specific case for example for the $\psi$ of a particle, we say that $\int||\psi_t(x)|| = 1$, and what does it mean? it means that we have a particle. it means it can be found in a time in some space. and how do we say that?
I think it's just a logical reasoning and it's according to what we have persevered of nature from the beginning up to now that: if we have a particle it is (must BE) in some space-time. So the probability of finding it in all space and time (universe under which we do experiment) should be equal to 1.
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Why do things here on Earth fall down? I want to have an answer with that question above for my physics lesson. I really don't have an idea about it, so, I ask help from you guys and hope that someone can help me with it.
| In Newtonian mechanics:
Things fall due to the gravitational force of Earth. This force between 2 masses $M$ and $M'$ is given by: $$F = G\frac{M \cdot M'}{r^2}$$.
This force is a central conservative force due to mutual attraction.
The rise of General Relativity:
Albert Einstein,in 1915, looked at the gravity through new eyes. For him, the fact that all objects fall toward the earth with $g$,whatever their size,implied that this must be in some truly profound way a geometric and kinematic result,not a dynamic one. He regarded it as being on a par with Galileo's law of inertia,which was accurate in straight-line motion.
Building on these ideas,Einstein developed the theory that
Objects which are at rest tend to remain at rest, and objects which are moving tend to move along geodesic paths - that is to say,the most economical way of getting from one point to other, with uniform motion,unless some force acts on them.
Objects fall simply because massive objects like Earth modifies the geometry locally so that the shortest straight lines become geodesics. The state of affairs in the vicinity of a massive object is,in this view,to be interpreted not in terms of a gravitational field of a force but in terms of a CURVATURE OF SPACE.
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How do electromagnetic waves travel in a vacuum? This is perhaps a total newbie question, and I will try to formulate it the best I can, so here it goes. How does an electromagnetic wave travel through for example, the vacuum of space?
I usually see that waves are explained using analogies with water, pieces of rope, the strings of a guitar, etc, but it seems to me that all those waves need a medium to propagate. In fact, from my point of view, in those examples the wave as a "thing" does not exist, it's just the medium that moves (involuntary reference to The Matrix, sorry).
But in space there is no medium, so how does a wave travel? Are there free particles of some sort in this "vacuum" or something? I believe the existence of "ether" was discarded by Michelson and Morley, so supposedly there isn't a medium for the wave to travel through.
Moreover, I've seen other answers that describe light as a perturbation of the electromagnetic field, but isn't the existence of the field, potential until disturbed? How can it travel through something it does not exist until it's disturbed by the traveling light in the first place? (this last sentence is probably a big misconception by me).
| Since, electro magnetic waves have electric and magnetic vector. Due to this EM waves show electric and magnetic field. An electric and magnetic field have no need a medium to show thier effect. Hence in the presence of electric and magnetic field vector which vibrate perpendeculer to each other and get pertervation EM waves travels in vacuum.
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Electromagnetic waves in vacuum If there is no source then also there is electromagnetic waves described by Maxwell equation. how if there is no source then existence of EM waves. What gives energy to this EM waves. Is it vacuum fluctuation or something else?
| The fact that you can solve Maxwell's equations in vacuo means that the vacuum can propagate electromagnetic waves. It cannot generate them, since as you point out there are no sources (charges and currents). The waves were generated somewhere else, where there was a current or a charge moving around, and you are just looking at a region where there's "nothing"*.
The energy of those waves is the energy stored in the oscillating electric and magnetic fields, and it comes from the source. E.g. for dipole radiation (see link later) an accelerating charge slows down and the lost kinetic energy goes into creating EM waves.
Remember that the Maxwell's equations are differential equations, i.e. they require boundary conditions. One of the things to have to fix is the amplitude of a wave $E_0$. You can set it to $0$ and have no electromagnetic waves whatsoever. But if you indeed have a wave propagating, then it means that $E_0 \neq 0$ and its value must be computed with information about how this wave was generated. For example, in the case of dipole radiation, you can compute the pre-factor.
*: Let me just stress that this is in the context of classical electrodyanmics. If you were to venture into the quantum world, then you would have particle-antiparticle pairs and their respective electric fields popping out of nowhere.
| {
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Are we comoving observers of space expansion? In cosmology:
A comoving observer is the only observer that will perceive the
universe, including the cosmic microwave background radiation, to be
isotropic. (Wikipedia)
According to this definition, is Earth considered as a comoving reference frame, or are we supposed to have a "peculiar velocity"?
What is the current precision for measuring if a frame is comoving or not, and for measuring its peculiar velocity? Or: From which speed (with respect to Earth) a frame would be considered as peculiar?
| We have a small peculiar velocity with respect to the comoving frame, this can be seen as a dipole in the CMB data. (CMB gets doppler shifted)
This dipole (and the monopole) is usually subtracted before doing further analysis of the CMB. I think (but I am not sure about this) that measuring the CMB-dipole is the best and easiest way to find earths peculiar velocity with respect to the comoving frame.
There is no sharp division between an object with a peculiar velocity and one without, the question is, how large is the peculia velocity compared to what scales we are talking about.
The numerical value for the peculiar velocity is:
(369 $\pm$ 0,9) km/s.
You can find it here: https://arxiv.org/abs/1303.5087
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Sign of the totally anti-symmetric Levi-Civita tensor $\varepsilon^{\mu_1 \ldots}$ when raising indices I am confused with the sign we get when we want to raise or lower all indices of the totally anti-symmetric tensor of any rank. Take the metric to be mostly plus ($-+\ldots+$). Then is it
$$\varepsilon^{ijk}=\varepsilon_{ijk}$$
or
$$\varepsilon^{ijk}=-\varepsilon_{ijk}?$$
so I am confused to as which one is true.
And if we consider higher rank does something change? For example
$$\varepsilon^{ijkl}=\varepsilon_{ijkl}$$
or
$$\varepsilon^{ijkl}=\varepsilon_{ijkl}?$$
| To not confusing by these things you must know exactly what are the Levi-Civita symbols $\varepsilon$ and what are the Levi-Civita tensor $\epsilon$. Then you also need to know that the Hodge dual has no any confusion when they had defined with Levi-Civita tensor.
$$
\star H^{\mu \nu} = \frac{1}{2}\epsilon^{\mu \nu}{}_{ \rho \sigma} H^{\rho \sigma}\Bigg( = \frac{1}{2}\epsilon^{\mu \nu \rho \sigma} H_{\rho \sigma}=-\frac{1} {2\sqrt{-g}}\varepsilon^{\mu \nu \rho \sigma} H_{\rho \sigma}\Bigg)
$$
and so
$$
\star H_{\mu \nu} = \frac{1}{2}\epsilon_{\mu \nu}{}^{ \rho \sigma} H_{\rho \sigma}\Bigg( = \frac{1}{2}\epsilon_{\mu \nu \rho \sigma} H^{\rho \sigma}=-\sqrt{-g}\varepsilon_{\mu \nu \rho \sigma} H^{\rho \sigma}\Bigg)
$$
| {
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Are there new physics scenarios that predict low lying hadrons? There is a significant ongoing experimental effort to search for new hadrons with masses in the GeV range. This is used to find the spectra of QCD bound states, with a particular emphasis on finding exotic resonances such as the tetraquark. To my knowledge, they have not found any state whose mass is in contradiction with the theoretical prediction using lattice QCD thus far (though e.g., there are a few tetraquark candidates, such objects are not in conflict with lattice predictions).
These searches are clearly very important as they confirm our understanding of QCD and in particular, they verify the validity of lattice QCD, which can subsequently be used to study new phenomena. But my question is, are there any mainstream new physics scenarios which predict a deviation in the spectra of QCD and could be found at for example, LHCb?
EDIT: I'm interested in changes to the low energy ($\sim$ GeV) bound state spectrum measured by these experiments
| Some popular supersymmetric extensions of the Standard Model predict masses of the squarks (superpartners of the quarks) that could be detected with the LHC.
If you mean any mainstream model that predicts a different mass spectrum of the quarks than observed at LHC? I don't know of such a model and if there would be any I would say it conflicts experimental data and won't become mainstream for that reason anyway.
| {
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Defining Reference Directions for Voltage and Power (sign convention)
My professor decided to use the above reference directions when calculating power in circuits. He says that when power > 0, power is consumed. When p < 0, power is generated.
This definition is counter intuitive to what I would have assigned--I would have said negative means power is consumed and positive power means power is generated.
So my question is this: Are these sign conventions actually what is happening physically? In other words, when you look at the diagrams above, what determines (or what are the formal conditions based on the parameters above) that a system is generating or consuming power?
Do we understand it like this: On the left diagram, charge is flowing from "high" to "low" potential so charge is being used. And on the right, charge is flowing in its "opposite" direction so power is being generated?
| Consider a resistor. Label one terminal "+" and the other one "-".
When current flows in to the + terminal (like in your left-hand diagram), Ohm's law tells us that the voltage at the + terminal is higher than at the - terminal and the value is given by $V=IR$.
And we know that in this situation, electrical power is consumed by the resistor and turned into thermal power.
Conversely, when a battery turns chemical energy into electrial energy and delivers it to the rest of the circuit, current flows out of the more positive terminal of the battery.
Of course this is just a convention. We could have defined current with the opposite sign (for example, we could have defined current in the direction of negative charge flow), and then we'd have the opposite result: When negative charge flows from + to - in a resistor, the voltage would be negative, giving a negative product of voltage and "current", but the resistor would still be consuming electrical power.
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Do particles always flow from high to low pressure? In a recent question, it was stated that
particles in high pressure air always flow to lower pressure.
In a pipe with a constriction, fluid flows from from low to high pressure after the constriction. (From here.)
How are these concepts related?
| The statement "fluid flows from region of high pressure to region of low pressure" in the original question may be corrected as "velocity of fluid increases from region of high pressure to region of pressure and decreases from region of low pressure to region of high pressure." Comments solicited.
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Closed system, gas change in time I'm trying to find out how to determine how a pressure of gas changes in time, but I can't find a way.
I have a system with constant volume $V$ (closed system, exchanges heat with environment through walls), the system is heated to temperature $T_0$ and then left to cool down on its own (heat source is removed after reaching $T_0$) until steady temperature $T_1$ is reached.
Is a way to determine the time dependent function for change in gas pressure inside the system (for example gas in some container)?
Assuming an ideal gas is okay.
At first I thought maybe differentiating $pV = nRT$ as $dp/dt = nR/V * dT/dt$, but I don't know where to go next.
Then I thought that maybe finding how heat is transferred through the walls to the environment and through the gas and using that to find $dT/dt$ (and eventually $dp/dt$) but I don't know how to go about approaching the problem that way.
| I am not sure how much it will help, but normally temperature drops in proportion to the difference in temperature - and as you point out pressure is directly proportional to temperature - so the temperature will drop by exponential decay and the pressure will also drop in this way as well.
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If wave speed is dependent on medium only, then how to reconcile $v\propto f$? I have read and learnt in many places that velocity of a wave depends only on the medium through which it travels. It is clear from this that the velocity of a wave doesn't depend on the frequency of the wave because both the sound of a roaring lion and crying baby reaches our ear with the same speed. But we also know that $\text{speed} = \frac{\text{distance}}{\text{time}}\ \Rightarrow\ v=\frac{\lambda}{T}\ \Rightarrow\ v=\lambda f=\text{wavelength}\times\text{frequency}$. In this derivation, velocity is found to be dependent on frequency.
Can anyone please explain this contradiction? Is there any fault in my perception of the concept?
| The equation is correct, but you incorrectly stated that wavelength "depends" on frequency.
The two (wavelength/frequency) are both part of the same property of a medium. As the wavelength increases in a given medium, the frequency decreases. If the frequency increases, the wavelength will decrease inversely.
Wavelength and frequency are inversely proportional to each other, but neigher "depends" on the other...because...they both depend on the "medium" for their values.
Hope this helps.
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Magnetic field between two concentric inductors Suppose we have two inductors with $n_1$ and $n_2$ turns, of radii $R_1$ and $R_2$ $(R_1 > R_2)$ respectively and length $l$. They are aligned concentrically and the smaller inductor is wholly inside the bigger one (i.e. the smaller one doesn't stick out). The currents flowing are $I_1, I_2$ respectively (they might be of different signs to indicate different direction of the current).
The magnetic field inside both indutors (inside the smaller one) is of course homogeneous and equals $B = B_{1_{in}} + B_{2_{in}} = \frac{\mu\mu_0}l(I_1n_1 + I_2n_2)$. But why is the field between these inductors $(r \in (R_2, R_1))$ homogeneous as well and what's its induction value equal?
| A Note About A Certain Simplifiation
When we say that the field between these inductors is homogeneous, it's actually a simplification. These inductors have an effectively zero magnetic field outside of them because the field produced from one segment of wire effectively cancels out the field produced from a segment on the other side. This simplification lets us say: outside of an inductor, there is no magnetic field from that inductor.
This simplification breaks down as you get close to the solenoid/inductor. To be more specific, when the distance from one side of the inductor to the other diverges from zero, this simplification no longer holds. In practice, you rarely do anything where this simplification doesn't apply.
The Answer
Given the simplification that the magnetic field produced from a solenoid is zero when outside the coils of that solenoid, you can say that there is no magnetic field from the interior inductor in that space. Therefore it is as if that interior solenoid is not there. Assigning the variables to the larger, exterior solenoid with a subscript 1, you get $B = B_{1_{in}} = \frac{\mu\mu_0}l{I_1n_1}$ when $R_2<r<R_1$.
| {
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Is differential geometry used in solid state? I'm an undergraduate in physics interested in a career in solid state. While I know that any additional math is helpful--I am on time constraints, and can only take a few supplemental classes.
That said, is differential geometry used much in solid state physics? I'm aware of things like Fermi Surfaces, but wonder if much diff. geo. techniques are actually used.
| On the very theoretical side of solid state physics is the holographic AdS/CFT correspondence which links strongly coupled condensed matter systems to gravitational theories. Recent work has been done on describing things like phase transitions in this theory. For example models of superconductivity in the gravity dual are promising in describing difficult condensed matter systems in terms of easier differential geometry.
See for example the final chapters of http://arxiv.org/abs/1409.3575
| {
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How quickly is motion transferred in a solid object? Just for example: assume an iron bar one foot in length. If you push on one end, the entire bar will move. This seems instantaneous. but actually, from my understanding, the atoms all push against each other in a very fast "wave" - making the entire bar move.
Now, say the bar is 2 light-years long. We are at one end of the bar, and rotate it 90 degrees. Clarification: rotation is the same as if you held a pencil horizontally by the eraser, then turned it vertical. This seems to mean the other end of the bar stays where it started until our motion travels through the entire bar. How long would that take?
| This is actually a really interesting question, but the answer is simpler than you might think. The pressure wave propagates along the bar at the speed of sound in whatever material the bar is made of. This is because sound is nothing more than pressure waves, and so the speed of sound is by definition exactly what you're looking for.
For example, the speed of sound in iron is around 5130 m/s (from this site), so any disturbances would propagate along the bar at that speed.
| {
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Would a rotating magnet emit photons? If so, what causes the torque that gradually slows the rotation? If a magnet is rotating around an axis perpendicular to the axis north-south axis of the magnet (which I assume to be cylindrically symmetric), in space (so no-gravity/freefall or friction), should it still slow down because it emits electromagnetic radiation/photons?
I would think so, due to conservation of energy. The power output of the oscillating magnetic field should mean a decrease of the rotational energy of the magnet.
But what causes the torque that gradually slows the magnet's rotation? One way of looking at it would be conservation of (angular)momentum and the fact that photons have momentum. But how would you express the torque in terms of electromagnetism/Maxwell's equations?
| If you move a magnet, you create an electrical field around the magnet perpendicular to its direction of motion. As the magnet accelerates, the electrical field evolves, and generates a magnetic field. I don't know the specifics, but I'm guessing that if you spin the magnet, this magnetic field will act counter to the spin of the magnet, and slow it down.
| {
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How to determine angular frequency of this system? I am self studying on harmonic motion and springs. One of the problems is:
Two identical objects of mass m are placed at either end of a spring of spring constant k and the whole system is placed on a horizontal frictionless surface. At what angular frequency w does the system oscillate?
The answer is sqrt(2k/m), but I don't understand how this answer was obtained. I got sqrt(k/m) because that is the angular frequency of a regular spring attached to the wall system and this system seems to have the identical forces as this system.
| The trick here is to realize that both masses are moving, and the point on the spring that is stationary is midway (in general, at the center of mass). Once you know which part of the spring is not moving you can fix it, and look at just one mass. You now have a shorter spring - length $\ell/2$ - and thus a higher frequency.
O/\/\/\/\/\/\O
^
center of mass
Equivalent to
O/\/\/\||||||||
^
fix the spring to a solid object here
| {
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Is the shell theorem only an approximation? I've read the shell theorem during gravitation lectures, i.e. I know it states that the net gravitational field inside a 3D spherical shell or a uniform 2D ring is zero.
Now, assume a thin spherical shell. If I put a particle inside the shell, so that it was infinitesimally close to one of the regions of the shell, shouldn't the particle move towards the shell and touch the portion of the shell it was closest to? (Since as the distance goes to zero, the magnitude of the field between the particle and that portion of the shell should be very high, when compared to the field from other regions.)
But in the same case if I apply the shell theorem, the particle shouldn't move at all! Since it states the net gravitational field inside the shell is zero.
Can anybody explain this difference, or if there isn't any, how am I wrong?
| If you put a particle very close to the border, the force from matter very close to it will be very strong, as you say. But that is only a small portion of the shell; all the rest is pulling the other way, towards the center. The shell theorem guarantees that these forces cancel exactly.
| {
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If an object falls - regarding air resistance - does the Potential or Kinetic energy get converted into thermal energy as it is falling? I read a paragraph on the transfer of potential energy to kinetic energy and heat from this website:
Even if air resistance slows down the ball, the potential energy is
the same (Mb x g x H). But if air resistance is in the way, not all of
the potential energy can be converted to kinetic energy. Some of the
energy has to be used to push the air molecules out of the way. When
that happens, the energy of the air molecules is increased. The air is
actually "heated" up by the falling ball.
This text indirectly mentions that the Potential Energy gets converted into heat. However my common sense (for lack of a better term) makes me think that the Potential Energy gets converted into Kinetic Energy which in turn gets converted into Thermal energy. Would someone be able to enlighten me on this please?
Also, as a side question, it isn't called Heat Energy right? A lot of websites seem to be saying that, but heat is just the transfer of Thermal Energy...
| The friction with the air indeed slows down the falling ball. The friction is minimal if the ball stays in place (i.e. there may be only some friction due to air currents.) But if the ball moves the friction is bigger because as the ball moves, it pushes away the air molecules to make room. The friction opposes the ball movement. So, indeed, the heating takes from the kinetic energy through friction, not from the potential energy. The meaning of the text, as I understand, it that, bottom line, the kinetic energy acquired by the ball during its fall is not equal to the total potential energy $M_b gH$.
| {
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Is there a physical interpretation of a tensor as a vector with additional qualities? What is a tensor? has been asked before, with the most highly up-voted answer defining a tensor of rank $k$ as a vector of a tensor of rank $k-1$.
But if a scalar is defined as a physical quantity with a magnitude, a vector as a physical quantity with both a magnitude and direction, can a tensor be defined in a similar way as a vector with additional qualities?
| I think a safer way, at base level, of thinking of a tensor, is to think of it as a function that takes in $n$ vectors, and spits out a number, and in each of it's arguments, satisfies:
$$T(a{\vec v} + b {\vec w}) = aT({\vec v})+ bT({\vec w})$$
where $a,b$ are numbers, and ${\vec v}, {\vec w}$ are vectors, and the other arguments of $T$ are suppressed. There is certainly additional detail to get into (about one-forms, gradients, and coordinate transformation), but this is the base idea.
| {
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What position of the center of gravity can make the front wheels of the car lift off the ground? I have a question regarding the position of center of gravity required to just lift off the front wheel of a vehicle
Consider a vehicle of mass $m$ having a center of gravity at height $h = 0.5m$ from the ground. The coefficient of friction between the tire and the ground is 1. Assume that the engine supplies just enough torque to utilize all the friction force without causing the wheels to spin.
My question is where should the center of gravity of the vehicle should be located in relation to the rear wheels to make the wheels just lift off the ground. I have solved the question as shown, and I get a 'NEGATIVE' value for L2 meaning that COG should be 0.5m behind the rear wheel, but the solution says that it should be 0.5 m infront of the rear wheels.
Can someone help me out on this!
Since, my handwriting is not clear, I am writing the equations here too
Equilibrium in vertical direction
(1) $N_1 + N_2 = mg$
Equation of motion in horizontal direction
(2) $F_{tr} = ma$
Here $F_{tr}$ is the traction force on the rear tire which propels the vehicle. Also,
(3) $F_{tr} =\mu N_2$
The balance of the torque on the rear tire $ \to $ the net moment on the rear tire about the contact point is zero
(4) $mgl_2 + 0.5 \ ma = 0$.
Now, since the vehicle must just lift off the ground
(5) $N_1 = 0; \ N_2 = mg$.
Using the equations (2), (3), and (5),
(6) $a = \mu g$.
Now using the equation (6) in the equation (4),
(7) $l_2 = -0.5 \ \mu = -0.5$
Now, the final value of $l_2$ is negative, which means that it is opposite to the assumed direction. So the center of mass should be 0.5 m behind the rear wheels.
The only difference between my method and the solution manual which I am referring to is that they have considered inertia force(pseudo force) on the vehicle, and thus, they get the answers $l_2 = 0.5$, which means 0.5 m in-front of the rear wheels
| We often use the sum of torques about an axis to find the case of zero rotation by finding where the sum is zero. The reason this works is that the case of no rotation must also be a case of no rotational acceleration and therefore no change in angular momentum.
$$\tau_1 + \tau_2 + .... = I \frac{\Delta L}{t} = 0 $$
But in your case the car is accelerating. The angular momentum of your system is
$$ L = d \times v$$
The offset distance between the center of mass of the car and the axis is fixed ($0.5m$), but the velocity of the car is changing. That means the angular momentum is also changing and there is a net torque. You cannot solve for the case where net torque is zero.
In the frame of the car, the axis is no longer accelerating, but you have a fictitious force on the center of mass to account for.
| {
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How much additional light does Earth receive from the Sun due to Earth's gravitational field? I was reading about how gravity affects light, and that got me wondering how much additional light is collected by the Sun due to the Earth's gravitational field.
Is it a significant amount of light (>1% of total light)? Is it significant enough to be considered when estimating the surface temperature of a planet?
| Earth's gravitational field causes Earth to retain a gaseous atmosphere, which both absorbs light itself and refracts light towards the surface. Estimating the altitude of the optically thick part of the atmosphere as somewhere between 6 km and 60 km, this atmosphere effectively increases the cross-sectional area of the Earth for interacting with sunlight by between 0.1% and 1%; the lower end is a better estimate. Not all of this atmospherically captured sunlight is absorbed by the Earth, but the same is true for directly incident light as well.
So, interaction with Earth's gravitationally-bound atmosphere increases insolation by something like 0.1%, subject to local, daily, and seasonal fluctuations due to things like clouds.
Atmospheric refraction of sunlight in the ideal case bends the light by about half a degree, or 1800 seconds of arc. In the same ideal configuration, the general-relativistic deflection of light by the sun is 1.75 seconds of arc. Scaling the GR deflection by $M/R$ for mass $M$ and radius $R$ to about 0.6 milliarcseconds, I get that the atmospheric refraction is about three million times larger than the general-relativistic refraction.
So general relativistic refraction of light is a parts-per-billion corrections to Earth's insolation. Not important.
| {
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General relativity: is curvature of spacetime really required or just a convenient representation? I'm not really far into the general theory of relativity but already have an important question: are there formulations that can do without spacetime curvature and describe the general theory of relativity/all associated gravitational effects in global cartesian coordinates? Idea: Einstein chose spacetime curvature so that one had not to build gravitational effects into the rest of physics equations like Maxwell's equations. I assume that spacetime curvature provides a convenient way to apply physics laws not related to gravity unmodified locally because at small distances we may use special relativity as an approximation.
Please correct me everywhere I am wrong.
| In traditional GR, the possibility that spacetime can be curved is a fundamental requirement and moreover gravity is a fundamental force. However, in entropic gravity, it's argued that gravity is not fundamental and is an entropic effect and hence curvature is also not fundamental but an emergent property.
| {
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Conservation of angular momentum in a free rod When a collision is elastic and no external torque acts on a system, angular momentum is conserved
I found this example and checked the results:
A ball (m = 1 Kg , v = p =+22 m/s, Lm = +11, Ke = 242 J) hits the tip of a rod (M = 10Kg , length = 1m, $I = 10*1^2/12$ = 5/6 ) in an elastic collision.
If the rod is pivoted, the ball bounces back with v, p = -11.846 m/s , L = -5.923, (Ke = 70.16) and the rod rotates with $\omega$ = 20.3 , L is conserved : Lr = (20.3 *5 /6) = 16.923 and Ke = 70.166 + 171.834 = 242 J
the rod translates with v = 3.3846m/s , p = 33.846 , (Ke = 57.2J) and rotates about its CoM with $\omega$ = 16.58 (Ke = 114.556).
If the rod is not fixed to a pivot, in order to conserve linear momentum the rod must translate with p = (11.846 + 22) = 33.846 (v = 3.3846, Ke = 57.28) and the energy of the rotating rod becomes Ke = 114.5 and $\omega = \sqrt(2E/I)$ = 16.58
angular momentum L was +11
after the impact we have
Lm = -5.923
Lr = 16.58 ( $\omega * I$ ) 5/6 = 13.82
13.82 - 5.92 = +7.9
It seems that angular momentum is not conserved. Is there a case in which also L is conserved?
In that case the bouncing speed must necessarily be different from 11.8 m/s, if that case exists, can you explain why the bouncing speed is different, whereas the masses are the same?
| Yes, angular momentum is conserved if you do the problem correctly.
If you assume the ball bounces back along exactly the same path it followed before the collision, there are three degrees of freedom: the velocity of the ball, the velocity of the rod, and the rotational rate of the rod. There are three constraints: conservation of linear momentum in the direction of motion of the ball, conservation of angular momentum, and conservation of kinetic energy. You seem to understand this.
From there, it's unclear what your approach was. How did you get the particular numbers you cite? There are infinitely-many ways to conserve linear momentum and kinetic energy. Conserving just those two applies two constraints to a problem with three degrees of freedom, so there is an entire one-dimensional manifold of solutions. IE you could give the ball any velocity you like (up to a maximum), then choose the rod's velocity and rotation rate to fit the two constraints. If you simply pick one of these solutions at random, then it is very unlikely to conserve angular momentum. You must use all three constraints to solve the problem.
You should do this an confirm that the correct figures are
$$v_{ball} = -\frac{66}{7} m/s$$
$$v_{rod} = \frac{22}{7} m/s$$
$$\omega = \frac{132}{7}s^{-1}$$
| {
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How to observe Floquet state? The Schrodinger equation is
$$i\hbar\partial_t\psi(t)=H(t)\psi(t).$$
Now, given that the situation that the Hamiltonian is periodically driven, i.e., $H(t+T)=H(t)$, then the equation can be solved by the Floquet ansatz,
$$\psi(t)=e^{-i\varepsilon_n t/\hbar}u_n(t),$$
where $\varepsilon_n$ is called the quasienergy, and the Floquet state $u_n(t)$ satisfies the periodic condition $u_n(t+T)=u_n(t)$.
This can be written in a compact form by defining the Floquet Hamiltonian as
\begin{align}
H_F(t) & =H(t)-i\partial_t\\
H_F(t)u_n(t) & =\varepsilon_n u_n(t).
\end{align}
Then if we treat time $t$ as a new degree of freedom, the $H_F$ is then defined in the composite Hilbert space $\mathcal{R}\otimes\mathcal{T}$. Formally, $H_F$ is analogous to the Hamiltonian for stationary states of the "time"-independent form and all mathematical conclusion of the stationary theory can be obtained.
My question is, although we can formally define the Floquet state $u_n(t)$ and the quasienergy $\varepsilon_n$, how can we detect them? I think that only the real wave function $\psi$ describes the real physics, so how can we draw the information of the Floquet state $u_n(t)$ from $\psi$?
| They have been observed in the following experiment:
Observation of Floquet-Bloch States on the Surface of a Topological Insulator.
YH Wang, H Steinberg, P Jarillo-Herrero and N. Gedik. Science 342, 453 (2013), arXiv:1310.7563.
In this paper, there is a periodic laser pulse hitting the sample which gives a time-periodic Hamiltonian. They then used ARPES (angle-resolved photoemission) to see the band structure repeated in energy steps, $\epsilon_n=\hbar \omega_n$. Without the time-dependent laser pulse, however, the "replica bands" are not visible. Of course the experiment is not perfect, which manifests itself in the fact that there are not infinitely many "replica bands" but at least this is strong evidence that under perfect experimental conditions, a perfect Floquet state could exist.
| {
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Can a conservative field produce a torque? I am asking whether the following Lagrangian for a point moving in a conservative field, can be correct :
$L(r, v, \omega) = \frac {mv^2}{2} + \frac {I \omega^2}{2} - U(r)$.
$r$ is the distance between the equipotential surface on which the movement begins and the equipotential surface on which the movement ends, $v = \text d r/ \text d t$, $\omega$ is angular velocity of rotation around some fix point in space (see example on the bottom of the text), $I = m\rho^2$, where $\vec \rho$ is the vector connecting the fix point in the space with the current position of the moving point (see the example).
What I am not sure on, is the presence of the term $I \omega^2/2$. I think that $\omega $ can vary only if the potential energy can produce a torque ($\vec F \ \text x \ \vec \rho$), and in that case $U$ should also depend on a variable $\theta$, indicating the angle between the vector $\vec \rho$ and a fix axis in the rotation plane.
But, if there is a torque, if $U$ depends not only on the distance between equipotential surfaces, but also on an angle $\theta$, is this anymore a conservative field?
I know that in a conservative field the mechanical work doesn't depend on the path followed by the point, but on the distance between equipotential surfaces, however that doesn't help me in my question.
(As a simple example, one can think that the field is produced by an electric charge uniformly distributed on an ellipsoid. Then $d$ is the distance to the surface of the ellipsoid measured perpendicularly on the equipotential surfaces, and given a point $P$ in the field, $\vec \rho$ is the vector from the center of the ellipsoid to the pint $P$.)
| The answer to the title of this question, "Can a conservative field produce a torque?" is yes. For example, a non-uniform gravity field (e.g., the Earth's gravity field) results in a gravity gradient torque on an object with a non-spherical mass distribution. This torque is sometimes problematic for artificial satellites, other times something those satellites can take advantage of as a stabilizing influence.
| {
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Advection operator How are exactly $u_j\partial_ju_i$ and $u_i\partial_j u_i$ related?
And what is their relation to ($\boldsymbol{u}\cdot\nabla)\boldsymbol{u}$ and $\boldsymbol{u}\cdot(\nabla\boldsymbol{u})$ ?
I ask this because:
$$[\mathbf{u}\cdot(\nabla\mathbf{u})]_{i}=u_{j}\partial_{i}u_{j}=u_{x}\partial_{i}u_{x}+u_{y}\partial_{i}u_{y}$$
$$[(\mathbf{u}\cdot\nabla)\mathbf{u}]_i=u_{j}\partial_{j}u_{i}=u_{x}\partial_{x}u_{i}+u_{y}\partial_{y}u_{i}$$
from this it would seem they are different, but:
$$[(\mathbf{u}\cdot\nabla)\mathbf{u}]=(u_{x}\partial_{x}+u_{y}\partial_{y})\left(\begin{array}{c}
u_{x}\\
u_{y}
\end{array}\right)$$
$$[\mathbf{u}\cdot(\nabla\mathbf{u})]=\left(\begin{array}{c}
u_{x}\\
u_{y}
\end{array}\right)\left(\begin{array}{cc}
\partial_{x}u_{x} & \partial_{x}u_{y}\\
\partial_{y}u_{x} & \partial_{y}u_{y}
\end{array}\right)=\left(\begin{array}{cc}
\partial_{x}u_{x} & \partial_{y}u_{x}\\
\partial_{x}u_{y} & \partial_{y}u_{y}
\end{array}\right)\left(\begin{array}{c}
u_{x}\\
u_{y}
\end{array}\right)=\left(\begin{array}{c}
u_{x}\partial_{x}u_{x}+u_{y}\partial_{y}u_{x}\\
u_{x}\partial_{x}u_{y}+u_{y}\partial_{y}u_{y}
\end{array}\right)
$$
from this it would seem that they are the same. I am quite suspicious about my definition of $\nabla\boldsymbol{u}$. Could someone clarify this?
| Your math is correct, $\left(\mathbf u\cdot\nabla\right)\mathbf u\equiv\mathbf u\cdot\left(\nabla\mathbf u\right)$. This should make sense because the commutative property holds for dot products.
Personally, I prefer to view $\mathbf u\cdot\nabla$ as an operator that acts on something (in this case, a vector, but it could be a scalar or higher tensor as well). Thus, I would use $\left(\mathbf u\cdot\nabla\right)\mathbf u$ over the other way. This also has the added benefit of viewing the index-form more clearly:
$$
\left(\mathbf u\cdot\nabla\right)\mathbf u\equiv u_i\partial_iu_j\tag{1}
$$
In this form, you can see that the indices of $\mathbf u$ and $\nabla$ must be the same due to the dot product.
The confusion, it seems is from your alignment of indices. Equation (1) is not equivalent to the 2nd way you write it:
$$
u_i\partial_iu_j\not\equiv u_j\partial_iu_j
$$
The latter term, $u_j\partial_iu_j$, is actually a column vector times the matrix $\partial_iu_j$ (making a row vector) while the first term, $u_i\partial_iu_j$, is a row vector times a matrix (making a column vector, which is what you actually want as a result).
| {
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How does a supersonic flight speedometer work? I'm sure today they can use GPS and radar, but I was pondering the queation when I saw a film clip of a vintage analog dial labeled in mach number. I'm supposing that the usual way of measuring the pressure drop of the air flow would not work in this case. So what does?
| The pitot tube measures stagnation (i.e. dynamic) air pressure.
There is also a static port that measures the actual air pressure.
This is enough information.
First of all, the pilot cares about "knots indicated air speed" (KIAS).
That is not true speed over the ground.
(A "knot" is one nautical mile per hour.
A nautical mile is one arc-second of latitude on the earth, or about 1.15 statute miles.)
Rather it is the speed the wings care about, for aerodynamic behavior.
If a plane is flying at constant KIAS, its speed over the ground is higher at higher altitudes because the air is thinner.
The static air pressure determines what altitude the altimeter reads.
(Actually, there is an adjustment for meteorological air pressure, so it can tell the actual height above mean sea level. The pilot needs to stay above towers, mountains, etc.)
The speed of sound decreases in the thinner air at higher altitude.
The airspeed indicator takes that into account, so it knows the plane's current Mach number, which is what you asked.
GPS is of no value in these calculations. It can only give speed over the ground and, in modern versions, altitude.
It knows nothing about wind and atmospheric pressure.
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Need of small charge in definition of electric field? Why do we need infinitesimally small charge in definition of electric field? Since the test charge cannot exert force on itself, F on test charge will not change whatever the value of test charge q is. So, F/q will be same for any value of test charge. Then why do we need this limit of infinitesimally small charge?
| Ideally, test charge should not affect the charge distribution of the source. An infinitesimal charge will ensure, for example, that the electric field it produces does not redistribute charges on any conductors in your system. A large test charge would polarize nearby objects, thus affecting the field you're trying to measure in the first place.
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Why is an airplane able to increase thrust without moving? I was just watching some documentaries and saw planes building up power in the turbines without moving. I thought about it and remembered, that this happens before every take off.
So, why is this possible? A planes thrust isn't related to the ground, but to the air, so brakes would just increase friction but won't reliably prevent the plane from starting.
So, how is the thrust compensated? I just don't get the clue.
| Brakes, m'boy, brakes.
Big planes, not surprisingly, have big brakes. And they are easily capable of locking the wheels up. The engine thrust would have to exceed the static friction breakaway force for that plane's weight before it started sliding down the runway.
Aircraft carriers work a bit differently - when your available thrust exceeds your gross weight (and the aircraft can therefore accelerate straight up) the brakes aren't enough. They use a hold-down bar that is engineered to break in two once the catapault engages. The pieces are then discarded. Rather labor-intensive, but simple and reliable.
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Is there a mathematical explanation for why there occur bound states if the effective potential falls below zero? Usually in physics textbooks, if the effective potential of the radial schroedinger equation
$$-\frac{d^2}{dr^2}u(r) + \frac{\ell(\ell+1)}{r^2}u(r) + V(r)u(r) = E u(r)$$
falls below zero in some subinterval $I\subset(0,\infty)$, a (physical) argument is made for the existence of bound states.
Is there actually a way to mathematically proof the existence of bound states in this case?
| There need not be any bound state if $V(r)$ is chosen appropriately. Even a potential $V(r)=-V_0$ for $r<R$, $V(r)=0$ for $r>R$ will have no bound states if the potential is sufficiently shallow: see wikipedia.
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How does "pushing-start" a dead-battery manual car work? A few days ago the battery of my car went (almost) dead. As it is a manual car, my father once told me that the way to get it going without jumper cables was to push it or let it roll down a hill, sink the clutch, shift to 2nd gear and then let go the clutch.
After the joy of being able to 'revive' the car, I got to wonder of the reason for why this works. All I could think of was Electromagnetic Induction, however I couldn't find anything on the web to support this. I'm sure that there might be all sorts of engineering details, so I'm only looking for the physical principles involved and the basic process that make this work.
| The purpose of the starter motor (using the battery) is to get the engine moving so that the combustion cycle can be initiated. The combustion cycle is self-sustaining, but it relies on the pistons moving to compress the air/fuel mixture, so you have to get them moving in the first place somehow. By putting the car into gear and rolling it down the hill, the motion of the wheels will transfer to the engine and start the pistons moving, in the same way as the starter motor would.
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Sending information faster than light If I could ever send my friend any information faster than light it would violate causality. If he just guesses the information and acts on it before he could ever receive it, everything is fine. What is different here?
I can understand that nothing can ever move faster than light but I can't understand why causality would be violated if something did. Or does it really have to? Since wormholes are mathematically sound, is it only a question of traversing faster than light? Is it ok if I find a way of transferring information faster than light as long as I don't move anything faster than light?
| The equations of special relativity imply that a hypothetical superluminal signal would arrive at the receiver before it was transmitted from the source. Since the effect precedes the cause, the "law of causality" would be violated. Therefore superluminal signals are not considered possible -- if the special theory is correct.
If you could send a superluminal signal, special relativity would be falsified. Any attempts to use the equations of special relativity would give absurd/impossible results, such as the violation of causality, and particles with imaginary masses.
Recently, superluminal neutrinos were thought to have been detected at the large hadron collider. The implication was not that the neutrinos had violated causality, or had been sent backwards in time, the implication was simply that special relativity had been falsified.
Note that quantum nonlocality implies that superluminal causal connections do exist, and that entangled particles "influence" one another superluminally. However this does not violate special relativity because the quantum state information transmitted between entangled particles cannot be used to send a superluminal signal.
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Which modes are excited when a drum is struck? I've been searching quite extensively for an answer to this question but I cannot find anything definitive. The most I can see is that "one or several" modes become excited, but this is never parameterized by any relevant factors. I imagine it depends on the location of the strike and the force applied. I do not need an exact formula, but I would appreciate some intuition here.
I have seen several videos now of the individual modes of a circular membrane being excited by oscillators at targeted frequencies. But when the drum is actually struck, does the membrane vibrate in some superposition of these modes? Are all possible modes - all infinitely many - excited simultaneously, just with all higher modes having infinitesimal amplitude? Or is there some finite fixed limit of the number of excited modes depending on say the force of the impact? (Further, I imagine if you strike it hard enough, the material will break, so if this is the case then some higher modes would maybe never become excited, but perhaps we can gloss over such details.)
| The functions which describe the vibrating modes of a circular drum are the Bessel functions of the first kind $J_n$, they form a complete set so you can express any function as a series of Bessel functions (Fourier-Bessel Series), they also satisfy an orthogonality relation, then knowing the function $F$ which approximates the applied force you can calculate the coefficients in the series projecting $F$ onto $J_n$. The magnitude of the coefficients will tell you which modes are more easily excited.
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If the solar system is a non-inertial frame, why can Newton's Laws predict motion? Since there is no object in the universe that doesn't move, and the solar system likely accelerates through space, how did Newton's Laws work so well? Didn't he assume that the sun is the acceleration-less center of the universe? Shouldn't there be many psuedo-forces to account for planetary motion?
| There are two main reasons it is practical to ignore the pseudo forces due to the rotation of the earth/sun about the galaxy. First, the accelerations are pretty small, and second, they are pretty uniform.
The sun moves around galactic center at about 800,000 kilometers per hour, but it takes around 250 million years to complete a single orbit of galactic center.
Using $v=2\pi r/T$ we get $r=vT/2\pi$.
So for a circle $a=v^2/r=v2\pi/T\approx 2\times 10^{-10} m/s^2$ which is pretty small.
The other factor is that the acceleration is pretty uniform. Tidal forces fall off like $1/r^3$ instead of $1/r^2$ so they are even smaller for large distances.
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Absolute (as opposed to relative) concept of inertial frame In mechanics there is a relative concept of "inertial frame": frame A is inertial with respect to frame B if A moves uniformly with respect B. That concept is easy to understand.
There also seems to be an absolute concept of "inertial frame". I keep reading things like "A is an inertial frame", without specifying with respect to which other frame B. Every time I read that kind of statements I get stuck. I cannot see how A can be "inertial". I can only see how it can be "inertial with respect to B".
Related to this, I keep reading things like "the solar system is accelerating" or "an object is moving" (for example here and here). Those statements I simply can't understand, unless they specify with respect to what frame (or object) that movement is defined.
I suspect my inability to understand the absolute concept of inertial frame is related to my inability to understand the statement "an object is moving". I only keep wondering: "with respect to what?".
So, my question is: can you really say "A is inertial" or "B is moving" in an absolute sense? (i.e without adding "with respect to C"). If so, how is that interpreted?
| An inertial frame is one with respect to which Newtons second and first laws are valid.There is no ideal inertial frame in the universe although the heliocentric refrence frame fixed with the center of the sun can be regarded as an inertial frame with a high degree of accuracy. If we assume the heliocentric frame as an inertial one then all other frames moving with constant velocity with respect to it are also inertial.
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What is a reasonably accurate but simple model of the Milky Way's gravitational field? I am putting together a toy program which shows how stars move around in the galaxy.
To run the simulation I need to know strength of the Milky Way's gravitational field at any location in it. I'm looking for a model (e.g. a collection of uniformly dense planes/rods) rather than a database of potentials.
Where can I get such a model?
I could simply construct an infinite plane of uniform density, but is that good enough? This is only a toy so I'm looking for something which preserves integrity of the overall shape and statistics of the galaxy, rather than worrying about the specific location of any particular star.
| Deriving the galactic mass from rotation has the following chart (on the right) for the enclosed mass as a function of radius
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Why do we call a white led with high color temperature "cool"? one can buy LED bulbs with defined color temperature.
why cool white = many kelvins (= high temperature?)
why warm white = few kelvins (= low temperature?)
| Any and every object naturally gives off light that is to a good approximation that of a blackbody. The amount of each wavelength given off is dependent on the object's temperature.
For cool enough objects, most of this emitted light is too far into the infrared to be seen by our eyes. Once something gets to be hot enough, a significant part of its emission will be visible. Our eyes take in all the visible wavelengths and interpret their sum (weighted by how much of each wavelength is emitted) as a single color. As you vary the temperature, this average color varies.
That linked article contains a diagram to this effect:
As you can see, low (but still rather hot) temperatures lead to red, then the color of blackbody light shifts through orange and yellow to white and whitish-blue. When it comes to blackbodies, red is cooler and blue is hotter. This applies to any object that is glowing because of its heat, whether it be a piece of hot metal or a star. (Yes, if a star and a newly-forged sword have the same exact color, that is because they have the same exact temperature; composition and solid/gas/plasma phase don't matter at all.)
So why do we think of blue as "cool"? This is just a coincidence of evolution/environmental exposure. Much of the visible light we see on a day-to-day basis is not pure blackbody emission; it comes from more complicated physics. Water tends to reflect the sky, which tends to be blue because blues are scattered by particles in the air better than reds. And natural fires tend not to be hot enough to make anything glow white-hot, much less blueish-white hot.
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Metric tensor in SRT I just read on this webpage that we have (click me) $g_{\alpha \beta} = g_{\alpha}^{\beta} = g^{\alpha \beta}.$
Now, although I understand that the first and the last one are equal, I don't think that the term in the middle is the same as the other two, cause we should have $(g_{\alpha} ^{\beta}) = (g_{\alpha \alpha'})(g^{\alpha' \beta})$. This should be equal to the identity matrix.
What am I doing wrong?
| That statement is nonsense.
While it is true, that, in flat space, the components of $g^{\mu\nu}$ and $g_{\mu\nu}$ are exactly the same, the equation $g^{\mu\nu} = g_{\mu\nu}$ is not a valid equation - the indices don't match.
As you correctly observe
$$ {g_\mu}^\nu = g_{\mu\rho}g^{\rho\nu} = {\delta_\mu}^\nu$$
since $g_{\mu\nu}$ is the inverse matrix of $g^{\mu\nu}$.
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Is $∣1 \rangle$ an abuse of notation? In introductory quantum mechanics it is always said that $∣ \rangle$ is nothing but a notation. For example, we can denote the state $\vec \psi$ as $∣\psi \rangle$. In other words, the little arrow has transformed into a ket.
But when you look up material online, it seems that the usage of the bra-ket is much more free. Example of this usage: http://physics.gu.se/~klavs/FYP310/braket.pdf pg 17
A harmonic oscillator with precisely three quanta of vibrations is
described as $|3\rangle$., where it is understood that in this case we
are looking at a harmonic oscillator with some given frequency ω, say.
Because the state is specified with respect to the energy we can
easily find the energy by application of the Hamiltonian operator on
this state, H$|3\rangle$. = (3 + 1/2)$\omega h/2\pi |3 \rangle$.
What is the meaning of 3 in this case? Is 3 a vector? A scalar? If we treat the ket symbol as a vector, then $\vec 3$ is something that does not make sense.
Can someone clarify what it means for a scalar to be in a ket?
| What they're saying is that $|3\rangle$ represents the third energy eigenstate of the oscillator. So, it replaces something like $\psi_3$.
Writing $|3\rangle$ requires context - you would have to explain that you were going to number the nth energy eigenstate of the harmonic oscillator as $|n\rangle$ before using that notation. It's not an abuse of notation, it's just not very self-descriptive.
You could use this notation too - the nth energy eigenstate of the harmonic oscillator is $|N_{energy}^{harm.~osc.} = 3\rangle$, but it would be pretty tedious to write.
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Superpositions: forcing particles into a state According to a MinutePhysics video ("How to See Light") physicists won a Nobel Prize for showing that creating a certain superposition state for a particle and observing the particle state after it interacted with light, may allow us to prove the presence of a photon without directly observing it.
If this is true, it must mean that we are capable of create certain superposition states for particles or creating particles that are already in a superposition.
If this is possible, is it possible to force a particle currently in a superposition into a state of our choosing?
Example:
Two spin 1/2 fermions are entangled and both in superpositions. Detecting either of them in the same way will result in one being measured as spin up and the other spin down. However, if you could force the first particle to become spin up, the other will instantly become spin down. This would allow faster-than-light communication in cases where the particles are far enough, because changing the state of one particle would affect the state of the second particle.
By creating 32 pairs of entangled particles one could transfer a 32-bit integer across an unlimited distance without the constraints of data transfer speed.
| The latest study on this subject that I could find, "All-Optical Control of a Single Spin Employing Coherent Dark States," uses 2 lasers to keep the particle in a coherent "dark state," which I understand to be a superposition where spin direction has not been determined, and hence, not measured. My question is, could that unmeasured particle be somehow split into 2 entangled particles (or however entangled particles are created) so that the original, still controlled with lasers, would maintain that superpositional state, forcing one that gets moved to maintain that state, where they could be separated by a long distance, where the original one would then be released from the superposition, signaling the other as it also is released?
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Do we know where newly formed quark-antiquark pairs come from in the process of hadronization? The only explanations I have found are very vague, such as "spontaneously created from the vacuum" and because "it is more energetically favorable".
| Here is a Feynman diagram of hadronization, i.e. parton showers
For this argument I am using the diagram as input of a quark antiquark pair from an incoming hadron interacting by a gluon and scattering as an off mass shell quark- antiquark.( in the figure the input is QED)
the curls are gluons, the light blue and purple and brown arrows are quarks.In this diagram, real particles, i.e. on mass shell, are the jets on the right coming from the paired quark-antiquarks. All the rest are within the interaction region and are virtual, off mass shell.
The answer in this context to the question:
Do we know where newly formed quark-antiquark pairs come from in the process of hadronization?
They come from the quark gluon sea within the interaction region. This link might help. Strong interactions make sure that quarks cannot become on mass shell in the lab.
Fig. 3: A somewhat more accurate picture of a proton, filled with gluons (g) and quarks (u,d,s for up, down and strange) and antiquarks (same letters but with an overline bar.) These particles are whizzing around at speeds that are a significant fraction of the speed of light. The number of gluons and quark-antiquark pairs is enormously understated, for reasons of clarity. (If you look carefully, you'll see there are two more up quarks than up antiquarks, and one more down quark than down antiquark; that EXCESS of two up quarks and one down quark is what leads to the shorthand: "a proton is made from two up quarks and one down quark.")
Excess energy in proton proton scattering carried by the two scattered quarks in the first figure raise the energy of the virtual gluons and quark pairs in the sea, the energy ending up in jets of hadrons.The neutral in color pairs of quarks and triplets can take the energy away from the interaction region, being on mass shell.
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Basic quantum entanglement question Please consider commenting on this basic quantum entanglement question or point me to articles that may enhance my knowledge.
Does quantum entanglement only occur in pairs, or can three or more particles become entangled?
| Entanglement with more than two particles is allowed. Mathematically, you could write, for example,
$$|\psi\rangle = a|111\rangle + b|000\rangle.$$
Here, if you measure particle 1 to be in the "0" state, then you know immediately what the entire wave function is after measurement.
One applications is quantum error correction.
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What is the speed of sound in space? Given that space is not a perfect vacuum, what is the speed of sound therein? Google was not very helpful in this regard, as the only answer I found was $300\,{\rm km}\,{\rm s}^{-1}$, from Astronomy Cafe, which is not a source I'd be willing to cite.
| Given the low density of gas, the speed of sound would be a direct function of the temperature of the gas ie the speed of the molecules/atoms. Since this varies from about 2.7K to millions of degrees near some stars, the speed of sound can change quite a bit.
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Why would two protons repel? I understand that two protons would repel due to them both being positively charged, however, wouldn't the strong force act on the two protons pulling them together? Would this mean that in this case the electromagnetic repulsive force is greater than the strong force? If so why? If not why would they repel?
| The electromagnetic repulsion between two protons is a long-range force, depending on $1/r^2$, where $r$ is the separation of the two protons.
The electromagnetic repulsion between two protons is not the reason that they do not stick together; if they are forced together (or can tunnel through the Coulomb barrier) then short-range strong nuclear forces are much stronger than the electromagnetic force over separations $<1.7\times10^{-15}\ m$, yet they are unable to make a bound state consisting of two protons.
The reason for this is that although the nuclear interaction is symmetric to the isospin of the nucleons (i.e. to first order it does not depend on whether the nucleons are protons or neutrons) it does depend on the spins of the two particles. The attractive nature of the force is only strong enough to bind the two nucleons if they have aligned spins (as in the bound state of the deuteron which has a neutron and proton with aligned spins and total angular momentum 1). If the two nucleons were identical however, i.e. a p+p or n+n interaction, then a bound state state with aligned spins would be forbidden by the Pauli exclusion principle.
The deeper reasons behind this spin dependence will need an answer from someone with a much better understanding of these issues.
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What does the statement "the laws of physics are invariant" mean? In the first paragraph of Wikipedia's article on special relativity, it states one of the assumptions of special relativity is
the laws of physics are invariant (i.e., identical) in all inertial systems (non-accelerating frames of reference)
What does this mean? I have seen this phrase several times, but it seems very vague. Unlike saying the speed of light is constant, this phrase doesn't specify what laws are invariant or even what it means to be invariant/identical.
My Question
Can someone clarify the meaning of this statement?
(I obviously know what an inertial frame is)
| "the laws of physics are invariant..."
means that they wouldn't vary or change.
Meaning that any experiment done in one inertial frame would give the same result as the same experiment done in another inertial frame.
It could be any experiment at all, for example, seeing how the momentum of a ball changed if a given force were applied, measuring the angle of the maxima in a diffraction experiment, timing the oscillation of a mass-spring system etc...
Every experiment, if set up the same, would give the same result. Hence two experimenters would deduce the same laws of physics from their experiments.
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Tensor product of two different Pauli matrices $\sigma_2\otimes\eta_1 $ I'm solving problem 3.D in H. Georgi Lie Algebra etc for fun where one is to compute the matrix elements of the direct product $\sigma_2\otimes\eta_1$ where $[\sigma_2]_{ij}\text{ and }[\eta_1]_{xy}$ are two different Pauli matrices in two different two dimensional spaces.
Defining the basis in our four dimensional tensor product space
$$\tag{1}\left|1\right\rangle = \left|i=1\right\rangle\left|x=1\right\rangle\\
\left|2\right\rangle = \left|i=1\right\rangle\left|x=2\right\rangle\\
\left|3\right\rangle = \left|i=2\right\rangle\left|x=1\right\rangle\\
\left|4\right\rangle = \left|i=2\right\rangle\left|x=2\right\rangle$$
Now we know that when we multiply representations, the generators add in the sense of
$$\tag{2}[J_a^{1\otimes2}(g)]_{jyix} = [J_a^1]_{ji}\delta_{yx} +\delta_{ji}[J_a^2]_{yx}, $$ where the $J$s are the generators corresponding to the different representations $D_1$ and $D_2$ ($g$ stands for the group elements).
Using all of this I find that in the basis of $(1)$ the matrix representation of the tensor product is given by
$$\tag{3}\sigma_2\otimes\eta_1 = \begin{pmatrix}
0 & \mathbf{1} & -i & 0 \\
1 & 0 & 0 & -i \\
i & 0 & 0 & 1 \\
0 & i & 1 & 0 \end{pmatrix}$$
(The bold $\mathbf{1}$ is just notation, see below!)
I am not asking you to redo the calculations for me but does $(3)$ make sense?
Appendix.
My calculations were done in the following fashion [using equation $(2)$]:
$$\tag{4}\langle 1| \sigma_2\otimes \eta_1 |1\rangle =
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=1\rangle
\\
=
[\sigma_2]_{11}\delta_{11}+\delta_{11}[\eta_1]_{11}
\\
= 0.$$
Similarly for eg
$$\tag{5}
\langle 1| \sigma_2\otimes \eta_1 |2\rangle =
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=2\rangle
\\
=
[\sigma_2]_{11}\delta_{12}+\delta_{11}[\eta_1]_{12}
\\
= 1.
$$
This is how the bold $\mathbf{1}$ was obtained.
So are my calculations $(4), (5)$ totally wrong?
The Pauli matrices
$$\begin{align}
\sigma_1 &=
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix} \\
\sigma_2 &=
\begin{pmatrix}
0&-i\\
i&0
\end{pmatrix} \\
\sigma_3 &=
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix} \,.
\end{align}
$$
| Each Pauli matrix has two non-zero elements. Therefore, direct product of Pauli matrices will have four non-zero elements. Your answer, unfortunately, has eight.
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Is $ds^2$ just a number or is it actually a quantity squared? I originally thought $ds^2$ was the square of some number we call the spacetime interval. I thought this because Taylor and Wheeler treat it like the square of a quantity in their book Spacetime Physics. But I have also heard $ds^2$ its just a notational device of some sort and doesn't actually represent the square of anything. It is just a number and that the square sign is simply conventional.
Which is true?
| As already mentioned by others, $\mathrm ds^2$ is used as suggestive notation for the metric tensor
$$
g = \sum_{\mu,\nu}\mathrm g_{\mu\nu} \, dx^\mu\otimes\mathrm dx^\nu
$$
In case of a positive definite metric and given a curve $\gamma:[0,T]\to M$, it has a precise meaning in terms of either the length function
$$
s_\gamma(t) = \int_0^t \sqrt{g(\dot\gamma(\lambda),\dot\gamma(\lambda))}\;\mathrm d\lambda
$$
with derivative
$$
\mathrm ds_\gamma = \sqrt{g(\dot\gamma,\dot\gamma)}
$$
or equivalently in terms of the pullback
$$
\gamma^*g = \mathrm ds\otimes\mathrm ds
$$
where $s$ denotes the induced normal coordinate on the interval.
| {
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Is the expression $S=K \log(\Psi)$ appearing in Schrödinger's first paper well defined? I am currently reading Schrödinger's papers and happen to have some questions that maybe some expert in the field could clarify for me. Like what happens with
$$S = K \log(\Psi)$$
when $\Psi<0$. If I understood correctly, in his paper Schrödinger assumed that both $\Psi$ and $S$ were real, something that could not be true for a wave function. Thanks in advance for any help with this issue.
Note: I am aware that you can get around this issue by defining
$$S = \frac{K}{2}\:\log(\Psi^2)$$ which reduces to the original expression when $\Psi>0$ and behaves correctly when $\Psi<0$. I wonder if this may be the answer. A different possibility would be to make $S$ a complex number.
| If you consider the complex logarithm, then $\log z=\log|z|+i\arg z$. (plus integer multiples of $2\pi i$ for different branches.)
Now if $z\in \mathrm{R}$ and $z<0$, then $\log z = \log(-z) + i \pi$. So the action $S$ has just changed by an additive constant. However, adding something to your action does not change the physics described by it, since the Euler-Lagrange equations of motion are the same.
Therefore I think, writing $S=K\log\psi$ is perfectly valid and as you stated yourself, you can always rewrite it as $S=\frac{K}{2}\log\psi^2$.
| {
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Is the method of images applicable to gravity? It is well known that the method of images is a useful tool for solving electrostatics problems. I was wondering why this technique is not applied when considering newtonian gravity?
Obviously there is no "negative mass" to correspond to a negative charge in electromagnetism, but surely could the unphysical nature of negative mass be ignored and considered a mathematical trick to solve a given problem?
The classic example for the method of images is the point charge near an infinite conducting plane, is there a way to apply a similar method to calculate the gravitational field between a point mass and an infinite thin plane?
Some preliminary research online has resulted in no resources on this idea so any references for/against this would be great.
| The method of images works on the electrostatic case because the axis of symmetry of the mirror charges induces an equipotential line that is equivalent to the infinite conductor surface. In gravitational physics, there are no known instances of a physical surface that is at the same potential in the gravitational field.
| {
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What physical evidence is there that subatomic particles pop in and out of existence? What physical evidence shows that subatomic particles pop in and out of existence?
| $$\sin(x) = x-\frac{x^3}{3!} + \text{trigonometric fluctuations}$$
Above you can see why I don't like the language of "quantum fluctuations" -- what people mean by them is just "terms in perturbation series we can't make classical sense of". Similarly the phrase
... particles pop in and out of existence...
is a yet another naive attempt of describing quantum effects in a classical language. And there is no classical analogy that would reflect the quantum description of the world in a total accuracy.
On the other hand I cannot say that this language is wrong -- it is formally correct. Problem is that it just puts the cart before the horse, making a lot of unnecessary confusion.
To sum up my answer: your question is wrong, since you are asking for "evidence" for a popular naive description of quantum phenomena in a classical language.
What you should actually ask about is the experimental evidence for quantum mechanics and quantum field theory. And the experimental evidence for quantum description of the world is made of plethora of famous, not-so-famous and not-at-all-famous experiments. There are even mundane devices that exploit intricacies of quantum mechanics for the benefit of human beings. (I'm pretty sure that you can find those without my help.)
| {
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Why do most formulas in physics have integer and rational exponents? I mean, why is $F=ma$? Why not $m^{0.123}$, $a^{1.43}$ or some random non-integers or irrational?
I hope you understand that my question isn't limited just to force, energy, velocity, etc.; it also extends to the area of a square, circle, etc. and all other formulas.
I think the whole thing starts with direct proportionality.
Most of them tell about the area of a circle, $A = πr^2$, where π is 3.14159..... an irrational number! It's not about the constant. I am talking about the power of a physical quantity.
I know why it has pi. It's because we chose the constant for area square to be one.
If we chose it to be 1 for the circle then the area of a square will have a constant, 1/pi.
I've edited the question to 'rational exponents' since all are giving me examples of decimal non-integers.
| I think everyone has missed the obvious answer: because the equations of physics simply use math to model the way the universe works. Put some fractional exponent in your F = ma, and the answers come out wrong.
Now if you're asking why the universe happens to be that way... Well, I don't know, but I think it's more of a question for philosophy than physics.
| {
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Why is Bose-Einstein condensation a phase transition? Bosons may succumb to a Bose-Einstein condensation at a certain critical temperature $T_c$, thus entering the BEC phase.
The only thing I know about the BEC is that since we are talking about bosons it is possible for all of them to occupy the same ground state. As the ground state is the energetically lowest one it makes sense that given low enough temperature most of the bosons will occupy the ground state.
Why however is it a phase transition? For the transition to the BEC to be a proper phase transition we need an order parameter that suddenly becomes non-zero when crossing into a certain realm of temperatures. Also we need a broken symmetry below the critical temperature. I have no idea what either of those are in the BEC case!
Related: How can one experimentally show that the magnons in a ferromagnet have formed a Bose-Einstein condensate? I guess probing the solid with neutrons would show a big increase in intensity for low-energy magnons. However how can we for sure tell that we are in a BEC state and not just at low temperatures?
| This is a very good question. It turns out that the phase transition occurs precisely when the chemical potential becomes equal to zero (assuming that the ground state energy is at zero). The order parameter in the BEC is the "macroscopic wave function" or rather the square root of the single-particle reduced density matrix. The broken symmetry is usually said to be gauge symmetry. There is a very good discussion of these points in Quantum Liquids by Leggett where he also briefly discusses the difficulty in thinking about gauge symmetry breaking. Though this is the mainstream scientific opinion, Leggett disagrees.
I'm sorry I cannot be of more help with regard to your question on magnons.
| {
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Could dark energy just be particles with negative mass? The title speaks for itself.
Dark matter: We see extra attractive force, and we posit that there are particles which create such a force, and use the measure of that force to guess their locations.
Dark Energy: We see extra repulsive force.
Only thing is, dark energy is uniform. So I suppose the stuff would have to be (at least somewhat) uniformly distributed throughout the universe. How uniform do we know it to be? Could the "stuff" be somehow a part of empty space itself?
| Matter and dark matter are also evenly distributed throughout the observable universe, at least on the largest scales. What makes dark energy different isn't that it is uniformly distributed, but that it has a constant density. The amount of dark energy per cubic meter of universe is the same regardless of the total volume of the universe. If the universe is twice as big, there is twice as much dark energy, so the universe expands twice as fast. Hence the accelerated expansion. This must be a property of space itself. If these were particles they would simply dilute away as volume increased.
| {
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How can we define the energy stored in a (conservative) force field? I have come to know from my textbook that energy is stored in the E-field of a capacitor, in the B-field of an inductor and so on. Take the example of an inductor. The derivation bewilders me completely. From Kirchhoff's Loop Rule, we take the the voltage drop along the inductor, multiply by current then integrate it wrt time to get the energy stored in an inductor. They say that the energy is stored in the B-field of inductor.
Analogically lets take the same derivation for a freely falling body opposed by drag (ohmic resistor) accelerating downwards due to g (inductor). We can find the work done by G-field and say that the gravitational potential energy of the body changes by this much.
Can we say that this much energy is being stored in the gravitational-field?
In the same way the term
$$\frac{1}{2} L \,i^2$$
represents the energy change of the charge flowing per unit time through the inductor.
How does this relate with the energy stored in the inductor or in the B-field?
So my question, how can we define the energy stored in a force field, or at least visualise it, and why is it needed to consider that this is being stored in the field?
| The energy stored in a field is the energy required to create it. In your case of the inductor there is no field when no EMF is applied. When we apply an EMF a current flows and does work, and the work goes into creating the field.
When we talk about the energy of e.g. a charge in an electrostatic field, we normally assume the charge is small enough that its effect on the field is negligable. However any charge, no matter how small, will affect the overall electric field and this changes the energy stored in the field.
| {
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What information do $|\psi(0)\rangle$ and $|\psi(t)\rangle$ represent? I am starting to feel comfortable with the role of the unitary operator in quantum mechanics. For instance, one of the equations I have seen is
\begin{equation}
|\psi(t)\rangle = U(t) |\psi(0)\rangle
\end{equation}
I understand what a unitary operator is in that
\begin{equation}
U^*U=UU^*=I
\end{equation}
I also understand that if we have a vector space containing our wave function $|\psi(0)\rangle$, then the operator maps to $|\psi(t)\rangle$.
My question:
What information do $|\psi(0)\rangle$ and $|\psi(t)\rangle$ represent?
If the answer is probability amplitude, then perhaps someone can clarify what exactly that is. As far as I understand, the $\|\psi\|^2$ represents the probability density. But that the amplitude is more fundamental for some reason.
I ultimately hope to understand why we are using unitary operators in the first equation, but I think I need to figure out this question first before tackling that.
| The Schrödinger equation is an ordinary differential equation of first order and its solution requires one initial condition which is your $\psi(0)$. Imagine you would like to do a quantum experiment. the first step is to prepare your system in a known state ( eg $|\uparrow\rangle$ for a qubit). This state corresponds to your $\psi(0)$. Then you want to do your experiment which is described by the Hamiltonian and a corresponding unitary time evolution $U(t)$. Your system which was prepared initially in the state $\psi(0)$ is after a certain time t in the state $\psi(t)=U(t)\psi(0)$.
Basically you need the unitary operator to describe the time evolution of a system.
| {
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Why is the bottom part of a candle flame blue? What’s the explanation behind the bottom part of a candle flame being blue? I googled hard in vain. I read this. I don’t understand how it’s explained by the emission of excited molecular radicals in the flame. I read that a radical is a molecule or atom which has one unpaired electron. That made me more confused. I want a more detailed, clearer explanation.
| The red, orange, yellow, and white parts of a candle flame results from glowing soot. The color in this part of the flame is indicative of the temperature. The spectrum in this part of the flame is fairly close to that of a black body.
The blue part of the candle flame at the bottom of the flame results from chemiluminescence. Chemiluminescence is not black body radiation. The spectrum of that blue part of the flame has narrow peaks. It is nothing like the nice smooth curve of a black body. The color of that lowermost part of the flame is not indicative of temperature. The blue light is instead a byproduct of the chemical reactions taking place in that part of the flame, which is why the spectrum is so peaky.
| {
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Spacelike to timelike four vectors First at all, let me just say that I'm not a Physicist, I study mathematics. So, I have this question. If you have a spacelike four vector, is there any transformation that could change it to be a timelike four vector? I mean, I know that every Lorentz Transformation (LT) preserves this properties (timelike $\rightarrow$ timelike, spacelike $\rightarrow$ spacelike, etc.), but I was thinking in another frame $S'$, different from the former $S$, where a spacelike four-vector (in $S$) will be timelike (in $S'$). If it is possible to have this other frame then, the way to relate events between frames is not a LT? or I'm missing something?
| Yes, there are transformations that take timelike vectors into spacelike vectors and viceversa. Consider $(t,x,y,z) \mapsto (x,t,y,z)$. You could event throw a Wick rotation, $(t,x,y,z) \mapsto (i\ t,x,y,z)$, as a transformation that takes timelike vectors and returns spacelike vectors.
Now, these transformation do not correspond to coordinate transformations between physical observers. This last fact is, as mentioned in a comment, a fundamental law of Physics.
| {
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Closure Relation For Operator With Degenerate Spectrum Suppose we have an observable represented by a Hermitian operator, $A$. Further, $A$ has at least one eigenvalue that is degenerate. For concreteness lets say $A |\alpha_i> = \alpha | \alpha_i>$ for $i = 1,2, \ldots, n$. When writing the closure relation must we include all the degenerate eigenkets in the sum or is it only necessary to take one?
| You must include all of the eigenkets.
To show this, I must create notation for all of the non-degenerate eigenkets. Let's call them $\left|\beta_i\right\rangle$, each of which has a unique eigenvalue $\beta_i$, and $i=1,2,3...,m$. Now, I'll prove by contradiction that leaving out even one of the degenerate eigenkets contradicts the closure relation:
Assume the operator $P$, defined as $\sum_{i=1}^{n-1} \left|\alpha_i\right\rangle \left\langle\alpha_i\right| + \sum_{i=1}^{m} \left|\beta_i\right\rangle \left\langle\beta_i\right|$ is equal to $1$, the identity operator. Then, by definition of the identity, we require $P\left|\alpha_n\right\rangle = \left|\alpha_n\right\rangle$. However, since the eigenkets are orthogonal, $P\left|\alpha_n\right\rangle = \sum_{i=1}^{n-1} \left|\alpha_i\right\rangle \left\langle\alpha_i\right|\left|\alpha_n\right\rangle + \sum_{i=1}^{m} \left|\beta_i\right\rangle \left\langle\beta_i\right|\left|\alpha_n\right\rangle = \left|0\right\rangle$, the null ket. This is a contradiction, hence we must include every degenerate eigenket (in this case $\left|\alpha_n\right\rangle$, the one we left out) in our closure relation. QED
| {
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$\mathrm{\rho^0}$ meson decay via the weak interaction? Of course, the $\mathrm{\rho^0}$ meson can decay in $\mathrm{\pi^{+}\ \pi^{-}}$ through the strong interaction. Using Feynman diagrams, I cannot understand why the same decay couldn't happen through the weak interaction. I attach the diagram I've drawn.
Strong decay:
Weak decay:
| Note that rho meson has no strange quark i.e. its strangeness quantum number is zero. Similarly the decay products positive as well as negative pion has no strange quark ie. Strangeness quantum number is zero for decay products too. Thus, strangeness is conserved in this decay process of neutral rho meson. Now note that it is the rule of nature that strangeness is conserved in strong interaction/decay but NOT in weak interaction/decay. Thus this decay of neutral rho meson can proceed only via strong process i.e. via gluon and can never happen through weak process i.e can never happen through W boson, a mediator of weak interaction/decay.
| {
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Cosmology: what is a quantity that is called "$h$" in regard to angular size of a galaxy? I am trying to solve a Cosmology problem, but a certain quantity $h$ appears in it, of which I do not know the definition (I have never seen it mentioned anywhere before). So I thought maybe someone here could tell me what this $h$ is?
The problem goes as follows:
If $a$ is the scale factor in a FLRW-universe and $1+z=1/a$ is the redshift, then the luminosity distance of a far away object that is emitting light is given by $d_L=a_0r(1+z)$ (where $a_0$ is the scale factor today and can be scaled to $a_0=1$). Knowing the luminosity distance, we can get the angular diameter distance $d_A=l/\theta=d_L/(1+z)^2$, where $l$ is the proper size of the source and $\theta$ the apparent size. Knowing $l$, what is the minimum angular size of the radiating object? Be sure to first express your finding in terms of $h$, and then use $h=0.7$ to get a numerical value.
So, the problem asks to find the minimum angular size (which is $\theta$ I assume), and I can do that. But I have no idea what $h$ is supposed to be! Can someone clarify? Thanks for any suggestion!
| This is explained in the Wikipedia article on Hubble's Law: http://en.wikipedia.org/wiki/Hubble%27s_law
In particular,
"Dimensionless Hubble parameter
Instead of working with Hubble's constant, a common practice is to introduce the dimensionless Hubble parameter, usually denoted by h, and to write the Hubble's parameter $H_0$ as 100 h km s −1 Mpc−1, all the uncertainty relative of the value of $H_0$ being then relegated on h."
| {
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Why doesn't the speed of the wind have an effect on the apparent frequency? A boy is standing in front of stationary train. The train blows a horn of $400Hz$ frequency . If the wind is blowing from train to boy at speed at $30m/s$, the apparent frequency of sound heard by the boy will be?
The answer: The frequency remains the same at $400Hz$
MY QUESTION:
Why doesn't the speed of the wind have an effect on the apparent frequency?
| The Doppler Shift formula $f'=f(\frac{v \pm v_{obs}}{v \mp v_{source}})$ only works if the wind (or the medium that the sound is moving in) is constant. Therefore, if the wind is moving at a constant speed, change the reference frame so that the wind is stationary.
In your case, change the reference frame so that it is moving from the train to the boy at $30m/s$. This way, the wind is stationary in the new reference frame. Then, calculate the speed of both the observer and source in the new reference frame: $f'=f(\frac{v \pm v_{obs}}{v \mp v_{source}})=f(\frac{343m/s + 30m/s}{343+ 30m/s})=f=400Hz$
Note: If the windspeed changes, the frequency will be different as the reference frame will be changing all the time.
| {
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Bounded operator - definition? As mentioned also in Bounded and Unbounded Operator, an operator $A$ is said to be bounded, if
$$\|Af\|\leq k \|f\|,$$ where the constant $k$ does not depend on the choice of $f$ (let us consider a map to the same Banach space).
However, in a mathematical physics text I came across a definition: a symmetric operator $B$ is said to be bounded from below if there $\exists$ a constant $c$ such that $$\langle\psi,B\psi\rangle\geq c\|\psi\|^2$$ for all $\psi$ in the domain of $B$.
Both definitions are logical (in the second one we can imagine $B$ being the Hamiltonian, than the system energy is bounded from below and hence the system is stable).
The only think that bothers me is when we rewrite the first definition into a similar form to the second one (we assume the norm comes from an inner product), namely:
$$\langle Af, Af\rangle \leq k\|f\|^2,$$ we get something quite different on the left-hand side, so the same words (bounded operator) refer to different things. Any hints how I can clarify this to myself?
| TL;DR: The property bounded, bounded from above, and bounded from below are different things, cf. Wikipedia.
In detail, consider a densely defined symmetric linear operator $A:D\subseteq H \to H$ in a complex Hilbert space $H$. Let $$\langle A \rangle_{\psi}~:=~ \frac{\langle \psi, A\psi\rangle}{||\psi||^2}$$ for $\psi\in D\backslash\{0\}$.
It follows that $\langle A \rangle_{\psi}\in\mathbb{R}$ is real.
*
*That $A$ is bounded from below means that
$$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\geq ~C. $$
*That $A$ is bounded from above means that
$$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\leq ~C. $$
*That $A$ is bounded means that
$$\exists C\geq 0~ \forall \psi\in D\backslash\{0\}: ~~ \frac{||A \psi||}{||\psi||}~\leq ~C, $$
which is equivalent to $A^{\dagger}A$ $(=A^2)$ being bounded from above,
which in turn is equivalent to $A$ being bounded from both above and below.
| {
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Diffusion current and resistance of a diode In a diode, there is a resistance in the p and n region, and electric field is required to move the charges in forward bias conditions. My question is: when there is diffusion current is the electric field helping the diffusion current because of the resistance? Thus can we say this is also a drift current?
| There seems to be some confusion about carrier motion on your part. Carrier diffusion occurs all the time - in a field-free region the net will eventually go to zero in steady state (but carriers are still moving and diffusing around). Drift current is the result of the (slight) bias in charge carrier motion caused by an applied electric field. No field, no drift current, just diffusion. With field, both drift and diffusion, but the drift is the portion caused by the presence of the field. Resistance is caused by carrier scattering processes, and impacts both drift and diffusion equally.
The separation of current into drift and diffusion is physicists imposing a model for carrier motion where we want separate it into those two terms. It isn't like you can paint some electrons red, and say they are only drifting, while those blue ones over there are only diffusing - it is all the same electrons executing a complicated motion that is a combination of random diffusion (field or no field) and drift under the applied field.
Resistance is then the scattering (and resulting momentum scrambling) of the electrons. The more times they scatter and forget where they are going, the harder it is to make them eventually go in the direction the field is pointing.
| {
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Eigenstates into which a system can be projected after a measurement I'm currently reading Dirac's Principles of Quantum Mechanics, on page 36, he says:
Another assumption we make connected to the physical interpretation of the theory is that, if a certain real dynamical variable $\xi$ is measured with the system in a particular state, the states into which the system may jump on account of the measurement are such that the original system is dependent on them.
On what physical basis can we make this assumption and why is it reasonable?
| This phenomenon is called the collapse of the wave function. It is one of the tenets of the Copenhagen interpretation of quantum mechanics.
The eigenstates $|\xi_i\rangle$ of the $\Xi$ operator form a complete set. From linear algebra we have
$$I=\sum_i|\xi_i\rangle\langle \xi_i|$$
where $I$ is the identity operator. We apply this to the state vector $|\psi\rangle$:
$$|\psi\rangle=\sum_i|\xi_i\rangle\langle \xi_i|\psi\rangle$$
We have now expressed the state in terms of the $\Xi$ eigenstates. When we measure $\Xi$ and get $\xi_j$, we project the state vector onto the eigenstate using the projection operator $\mathbb{P}_j=|\xi_j\rangle\langle\xi_j|$. So after measurement we get
$$\psi\longrightarrow N\mathbb{P}_j|\psi\rangle=N\langle\xi_j|\psi\rangle|\xi_j\rangle$$
where $N$ is the new normalization constant.
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How could a cord withstand a force greater than its breaking strength? How could a 100 N object be lowered from a roof using a cord with a breaking strength of 80 N without breaking the cord?
My attempt to answer this question is that we could use a counter weight. But I don't really understand the concept behind counterweights so I hope someone can clear that up for me and if there is a better answer I'll love to know it.
| Use a ramp, an incline of 53° will work.
Otherwise You need to double up the cord.
The third option is to just carry the 10 kg object down the stairs.
| {
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What is the correct way to treat operators that has "time" in QM? I don't know if this question has already been resolved but considering that $i\hbar\partial_t$ is the energy operator, and $\partial^2_t$ is the waves operator (or helmholtz), I can't accept that $t$ itself isn't an operator
What is the argument here that says $t$ is not an operator?
| To say that something is a (linear) operator you have to specify the space where it acts. You may say that, for example, wavefunctions of quantum mechanics are maps: $t\to \psi(t)$ that are continuous in $t$ with values in $L^2(\mathbb{R}^d)$. If we restrict to compact time intervals $[0,T]$, we may denote the space of these maps by $C^0([0,T],L^2(\mathbb{R}^d))$.
On $C^0([0,T],L^2(\mathbb{R}^d))$, with norm $\lVert \psi(\cdot)\rVert_{C^0}=\sup_{t\in [0,T]}\lVert\psi(t)\rVert_{L^2}$, both $t$ and $-i\partial_t$ are densely defined linear operators [actually the multiplication by $t$ is bounded, with norm $T$; the derivation has domain $C^1([0,T],L^2(\mathbb{R}^d))$].
On $C^0(\mathbb{R},L^2)$, also the multiplication by $t$ is unbounded, and since every wavefunction of QM satisfies, by means of Schrödinger equation $\lVert\psi(t)\rVert_{L^2}=\lVert\psi(t_0)\rVert_{L^2}=k$ for any $t,t_0\in \mathbb{R}$ (where $k\geq 0$, usually $k=1$); we see that every nonzero wavefunction is outside the domain of definition of $t$, as a multiplication operator on $C^0(\mathbb{R},L^2)$ [because $\sup_{\mathbb{R}}\lvert t\rvert \lVert\psi(t)\rVert_{L^2}=k\sup_{\mathbb{R}}\lvert t\rvert=+\infty$].
| {
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What is basically the difference between static pressure and dynamic pressure? What is basically the difference between static pressure and dynamic pressure?
While studying Bernoulli's theorem, I came before these terms. The law says:
When the fluid flows through a small area, its pressure energy decreases & kinetic energy increases and vice versa.
Now that's wierd as I know due to having KE, ie. having momentum, one can impart pressure. Then why distinction ? What is then pressure energy??
In order to understand that I went to wikipedia & quora & others; there I found fluid exerts two pressure: Static & dynamic. But really nothing could be understood more than that. What are they actually?
| Dynamic pressure is not really pressure at all. It represents the amount that the pressure would increase if all the fluid's kinetic energy per unit volume could be converted to pressure by blocking the flow (say with a pitot tube). On the other hand, static pressure is another name for just "plain old pressure" exerted by parcels of fluid on adjacent parcels of fluid, or on the walls of the flow channel. For incompressible flow, it is, more precisely, the isotropic part of the stress tensor.
| {
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Non-zero potential measured on the surface of an insulator After driving a screw into an alumina firebrick and applying a 5kV potential, I measured a ~4.5kV potential using a high voltage probe at other locations on the brick surface.
Firebrick is generally considered an insulator at room temperature. Using a mega-ohmmeter, we measured an ~2GigaOhm resistance along the length of the brick.
Treating the firebrick as a large resistor, we electrically grounded a second screw -- driven in to the opposite end of the first screw. In this case, no current flow was measured.
What would explain a high resistance element's small potential drop that isn't due to current flow?
| You present both a static and a quasi-static case here. Let's consider them one at a time:
1) Static case: You applied a large potential to the screw relative to ground, which charged the screw. The firebrick is a good insulator, but not perfect, so the charge on the screw should polarize the material slightly, creating a slight electric field inside the firebrick. Potential is the integral of the field, so you see a small potential drop from the screw to other positions on the brick.
2) Quasi-static case: Again, the firebrick has high resistivity, but not infinite. This means the conductivity is low, but not zero. In this case the potential difference between the two screws is large, so the line integral of E between the screws is large. Applying Ohm's law, J = c E where c is the conductivity. Thus there is actually a very very tiny current flow between the screws (too small to be measured by your meter), and hence the voltage drop from the first screw to other locations.
| {
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Calculate the approximate number of conduction electrons So i have the following problem: A cube of gold 0.1 meters on an edge, calculate the approximate number of conduction electrons whose energies lie in the range from 4.0 ev to 4.025 ev.
But I'm not clear on how to start. could someone offer any help?
| I wasn't going to answer since I don't know exactly what your professor had in mind, but since I've been prodded by Sofia here is my suggestion. However don't treat this as gospel as I may have completely the wrong idea of the question.
I would guess that you're supposed to treat the cube as an infinite 3D potential well aka particle in a box. In that case the system has discrete energy levels given by:
$$ E_{ijk} = \frac{\hbar^2}{2m}k_{ijk}^2 \tag{1} $$
where:
$$ k_{ijk}^2 = \frac{\pi^2}{\ell^2}(i^2 + j^2 + k^2) $$
where in this case $m$ is the mass of an electron and $\ell$ is the size of the box.
The density of gold is 19.3 g/cc and the atomic weight is 197, so the mass of the gold cube is 19.3 kg and dividing by the atomic weight (in kg) gives us the number of moles, 97.7. Multiplying by Avagadro's number gives the number of atoms in our cube as about $5.9 \times 10^{25}$. We'll assume that each atom contributes one electron to the conduction band.
In principle you could start counting up the energy states starting at $111$ and counting upwards until you reach 4eV. However this isn't a practical way to do the problem. You need to knoiw the expression for the density of states. I probably learned how to derive this for the particle in the box, but I have long since forgotten the details so I just Googled to find:
$$ g(E) = \frac{\pi \sqrt{E}}{2E_{111}^{3/2}} \tag{2} $$
The density of states tells us the number of states between two energies $E_a$ and $E_b$ is:
$$ N_{ab} = \int_{E_a}^{E_b} g(E)dE $$
So the question requires you to do this integration with $E_a = 4$eV and $E_b = 4.025$eV. The integral is straightforward and gives:
$$ N_{ab} = \left[ \frac{\pi}{3} \left(\frac{E}{E_{111}}\right)^{3/2}\right]_{E_a}^{E_b} \tag{3} $$
As a sanity check let's put $E_a = 0$ and $E_b = 4$eV and see how this compares with the number of electrons in the cube. I get $E_{111} = 1.13 \times 10^{-16}$ eV, and putting this into equation (3) I get the number of states between zero and 4ev to be about $7 \times 10^{24}$. Reassuringly, this is comparable to our estimate of $5.9 \times 10^{25}$ electrons in the conduction band.
NB remember we can fit two electrons into each state, one spin up and one spin down.
And we're basically done. If we set $E_a = 4$eV and $E_b = 4.025$eV in equation (3) we get the answer:
$$ N \approx 6.6 \times 10^{22} $$
And the number of electrns in the energy range is just twice this.
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When does Pauli's exclusion principle kick in? Imagine that I prepare a fermion in the $\left|\uparrow \right\rangle$ state and a second one far away in the $\left|\downarrow \right\rangle$ state and set them in a path for collision.
According to Pauli's exclusion principle, the composite wave function must be anti-symmetric. Does the wave function become anti-symmetric as they collide or was it like this from the start? Can one predict if the composite wave function will correspond to a singlet or a triplet state from the moment we prepare the separate fermions?
| To say you have put them into states, you have implicitly considered them as part of a single system (independent of how "far away" they are from each other). The wavefunction is anti-symmetric by definition, so they will behave corresponding to the Pauli Exclusion Principle "from the start".
| {
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Understanding magnetic force on charged particle if we put any charged motionless or static particle in the constant magnetic field, then why does it don't feel a magnetic force?
Mechanism by which electric and magnetic fields interrelate
I have read the above article which suggested that the magnetic field is the relativistic effect of the electric field then why does the static charge particle does not feel the magnetic force and the magnetic field is also an one type of electric field, and any static charged particle always feel the electric force in electric field.then what really happens there?
| The Lorentz force law is
$$\mathbf{F}=q[\mathbf{E}+\mathbf{v}\times\mathbf{B}]$$
Only a moving particle experiences a magnetic force, but the electric force is always felt.
| {
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Are electrons miniature black holes? For something to be a blackhole, it must have gravity and the radius must be smaller than the schwarzschild radius for its mass.
-Electrons have gravity
-Electron are theoretically believed to be infinitely small points
Since it has gravity it is capable of being a black hole. Since its radius is infinitely small, it must have a schwarzschild radius and thus be a black hole.
| Electrons are very close to the energy of self-capacitance of a quantum of charge.
The size of the electron is very close to $r_e$, the energy supposed if one tries to charge a sphere of that radius with a single electronic charge, ie $mc^2 = e^2/4\pi\epsilon r_e$.
| {
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What is the meaning of the negative sign in $W = -\Delta U$? What is the meaning of the negative sign in $W=-\Delta U$?
As far as I understand, $W=-\Delta U=-(U_\mathrm f-U_\mathrm i)=U_\mathrm i-U_\mathrm f$.
While $U_\mathrm i$ is the initial potential energy (before applying the work), and $U_\mathrm f$ is the final potential energy.
But that doesn't work out when calculating the work done to bring an object from the face of the Earth to a height $h$ above the sea level:
$$W=-\frac{GM_\mathrm Em}{R_\mathrm E}-\left(-\frac{GM_\mathrm Em}{R_\mathrm E+h}\right)=GM_\mathrm Em\cdot\left(\frac1{R_\mathrm E+h}-\frac1{R_\mathrm E}\right)\lt0$$
The result is negative, but a work that is done against a force field should be positive.
That negative sign always confuses me.
| $U$ in your equation in potential energy, and $W$ is internal work. That is, the work done by forces within the system. The system in question here comprises the object and the earth, and the internal force is gravity. The work that you do to lift an object is external to the system, and does not appear in your formula.
In your scenario, lifting an object "up", the direction of the force is down, and the direction of the displacement is up. Thus the work is negative, and the change in potential energy is positive, as expected.
The external force of your hand presents a common source of confusion. It does play an important role, of course: It provides external work, bringing energy from outside the object-earth system to within the system.
| {
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Solution to Schrödinger equation I'm trying to solve the Schrödinger equation for a given potential. With some assumptions I end up with:
$$\frac{\hbar^2}{2M}\frac{d^2u(r)}{dr^2} = - \left(E - V(r)\right)u(r)$$
Since it's a square well potential I'm looking at, I have for the first region ($r \leq r_0$) that $V(r) = V_0$. So if I plug that into the above equation and solve that differential equation, I end up with:
$$u(r) = A\sin(kr) + B\cos(kr),$$
where $k = \sqrt{2M(E+V_0)}/\hbar$. I can then make some conditions and stuff to reduce it even further, but that's not my problem.
The next region ($r > r_0$) I get that $V(r) = 0$, which mean I have to solve:
$$\frac{\hbar^2}{2M}\frac{d^2u(r)}{dr^2} = - E\,u(r)$$
In my books, and websites I've seen, the solution to this is the same as above (With different constants) and the expontential functions instead of cosine and sine. And this is what I don't understand, why is that ? Whenever I try to solve that on my computer, I still get cosine and sine, but no exponential functions - which I think makes sense, since the only difference is some constant for the potential.
So what am I missing ? Am I solving it the wrong way, or is there some trick I don't know ?
| First, the $k$ in the first region is incorrect. Check your signs. You should have
$$k_{I} = \sqrt{2M(E-V_o)}/\hbar.$$
Second, your solution to the second region will have sine and cosine solutions with
$$ k_{II} = \sqrt{2ME}/\hbar.$$
Real exponential solutions will occur in each region if $E<V_o$ in region one and/or $E<0$ in region II.
Look at the DE; if it has the form
$$ f'' = -af$$
where $f$ is a differentiable function and a is a positive constant, the solution is sines and cosines.
If the form is $$f''=a f $$, the solution will be exponentials.
| {
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How eddy current brakes function Take the following example:
where a rectangular sheet of metal is entering a constant magnetic field at $v \dfrac{m}{s}$. Due to Faraday's law of induction + Lenz's law, we can state that an eddy current will be generated to oppose the increase of magnetic flux through the sheet of metal, so as to produce a magnetic field coming out of the page (represented by the red dots). Intuitively, I believe that this induced magnetic field should act as a 'brake' on the metal plate, as Lenz's law implies that the induced current should always in some way act against the motion, but I don't see how to calculate this 'retarding' force that would act to reduce the plate's speed?
| I had a fundamental misunderstanding of eddy currents. I believed that eddy currents were formed simply in the part of the metal that was already submerged in the magnetic field, but in reality it is actually something like
(source: boredofstudies.org)
this, where only half the eddy current is actually in the field. If this is the case, then you can just use $F = qv*B = IL*B$, probably with some integration, and you can find the force. So the retarding force is just a variation on the lorentz force.
| {
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Flux linkage of 2 coils in series I have a question about magnetic circuits. I am an engineering student, therefore I will neglect some minor errors.
There is a simple magnetic circuit with 2 coils ($C_1$ & $C_2$), with $N_1$ & $N_2$ turns and $i_1$ & $i_2$ currents. Coils are in series and connected with 1 magnetic core of "O" shape (we may assume some reluctance in form of air gap = $R_c$). The flux linkage of $C_1$ is as follows:
$$
\lambda_1 = N_1 \Phi= N_1 \left(\frac{N_1 i_1}{R_c} + \frac{N_2 i_2}{R_c}\right)
$$
so in the end I got something like
$$
\lambda_1=N_1^2i_1\left(\cdots\right)+N_1N_2i_i\left(\cdots\right)
$$
The $(\cdots)$ are some constants regarding air gap and core geometry.
I know that magnetic circuit resembles electric circuits. If there would be similar circuit with 2 batteries in series and 1 resistor, the overall voltage would be $V = V_1 + V_2$
In above mentioned magnetic circuit the overall flux linkage would be
\begin{align}
\lambda_{tot}&=\lambda_1+\lambda_2\\
&=N_1^2i_1\left(\cdots\right)+N_2^2i_2\left(\cdots\right)+2N_1N_2i_2\left(\cdots\right)
\end{align}
My questions are then
*
*Why is there the term $2N_1N_2 i_2(...)$ in the last part?
*Why is the electric circuit $V = V_1 + V_2$ and no $V_{12}$?
|
Why is there the term $2N_1N_2i_2$(...) in the last part?
Two coupled inductors are not two two-terminal circuit elements but, rather, one two-port network, e.g.
where
$$M = k\sqrt{L_1L_2} $$
thus, the reason for the $N_1N_2$ terms. Note that as the coupling $k$ goes to zero, we recover the two independent inductors.
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Can an object appropriately isolated from its surroundings become colder than its surroundings? Consider a sealed box, well-insulated on all sides, except for the lid which is transparent to infrared. An object is placed inside the box and the box is evacuated (purpose being to thermally isolate the contents of the box from its surroundings). The box is placed outdoors (in an everyday atmosphere) on a clear night. Let's assume that at the start of this experiment, the box and its contents are in thermal equilibrium with its surroundings. The object inside the box will radiate infrared according to its temperature, which should escape through the lid of the box. With nothing but clear dark sky above, I assume there is nothing to radiate appreciable heat back into the box and maintain the object's temperature.
Question: will the object cool below the ambient temperature outside the box?
| Yes, this works. It's called radiative cooling. This phenomena has been known for a long time, considering the ancient Egyptians used to make ice this way.
Ideally, something open only to a clear sky would "see" the temperature of space, which is the microwave background radiation. In practise there is enough stuff in our atmosphere that radiates so that it won't get anywhere near that cold.
Try it some time. You can do this experiment yourself fairly easily. Hollow out a bowl-shaped depression in a block of styrofoam, then paint it black. fill the bowl with water, but leave a little room at top, then cover the top with plastic wrap. There will still be some conduction thru the air between the bottom of the plastic wrap and the top of the water, but this is still a more effecient setup than the ancient Egyptians had. It doesn't have to be perfect to clearly show that the method works. Place the contraption outside on a clear night open to the sky with as few other objects around as possible. It won't take long for the temperture of the water to go below the air temperature.
| {
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Proof that a traceless strain tensor is pure shear deformation How can i proove that the traceless part of linear strain tensor $e$ in the Euler description:
$$e_{i,j}={ 1 \over 2 } \left({ \partial u_i \over \partial x_j}+{ \partial u_j \over \partial x_i} \right)$$
is alway a pure shear deformation i.e. it does conserve volume. In case it is not clear this is the Euler strain tensor with the assumption $ { \partial u_i \over \partial x_j} << 1$, which means the part $\sum \limits_k{ \partial u_k \over \partial x_j} { \partial u_k \over \partial x_i}$ is neglected.
This apperently always has to hold when decomposed into traceless part $ e^t$ and a generally not traceless part $e^l$:
$$u= \underbrace{{ 1 \over 3 } Tr[e] \mathbb{I}}_{e^l} +\left( \underbrace{e -{ 1 \over 3 } Tr[e] \mathbb{I}}_{e^t} \right) $$
Even just the direction of how this can be done would be appreciated.
EDIT NOTE: the notation in my script is confusing so i changed it
EDIT UPDATE:
One hint might be that if we look at small local deformations the strain tensor can probably be approximated as equal in all directions in first order. So we can write $e= \mathbb{I} \epsilon+...$. The first term must be nonzero for small deformations, and so the trace in that order does not vanish, but ist there a better more general Argument? This seems very hand waving.
| this website may help you I fonund this on Google books
https://books.google.com/books?id=yNXVBAAAQBAJ&pg=PA234&lpg=PA234&dq=Proof+that+a+traceless+strain+tensor+is+pure+shear+deformation&source=bl&ots=RmwzNuwndU&sig=BeNMc5oRLSxrUWBOfD_bpSrIWG0&hl=en&sa=X&ved=0CCwQ6AEwAmoVChMImYGa0uHmyAIVliuICh35UAiy#v=onepage&q=Proof%20that%20a%20traceless%20strain%20tensor%20is%20pure%20shear%20deformation&f=false
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If a black body is a perfect absorber, why does it emit anything? I'm trying to start understanding quantum mechanics, and the first thing I've come across that needs to be understood are black bodies. But I've hit a roadblock at the very first paragraphs. :( According to Wikipedia:
A black body (also, blackbody) is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.
OK, that's nice. It's an object that absorbs (takes in itself and stores/annihilates forever) any electromagnetic radiation that happens to hit it. An object that always looks totally black, no matter under what light you view it. Good. But then it follows with:
A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation.
Say what? Which part of "absorbs" does this go with? How can it absorb anything if it just spits it right back out, even if modified? That's not a black body, that's a pretty white body if you ask me. Or a colored one, depending on how it transforms the incoming waves.
What am I missing here?
| In simple words
Black body radiation means a body - independent of its color - that absorbs all the wavelength falling on it in the form of energy and does not reflect any of that wavelength, but instead it radiates what it absorbed with different wavelength.
That is why a star is considered a black body, since it just radiates and doesn't reflect.
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My physics teacher gave us this equation $v= -3 +3t$ She asked us if the body was accelerating or slowing down, and I immediately said that it was accelerating (because the $a=3>0$). Then she said that I was wrong because the direction of the acceleration vector was the opposite of the direction of initial speed($v_0=-3$). I do not understand why it slows down, because with the passage of time the body moves faster. Can someone give me an explanation?
| The sense of acceleration has nothing to do with the sense of velocity. Bodies always have negative acceleration due to gravity regardless if they are going up or down. What is important is the convention as to which direction a positive displacement occurs.
In your case all you know that the acceleration vector is in the same direction as a positive displacement ($a=+3$) and your answer is correct.
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Missing a factor of $\sqrt{\frac{\hbar}{m}}$ in a QFT Practice Problem. Can someone explain why? I am doing problem 2.3 on page 27 of Quantum Field Theory for the Gifted Amateur.
Use eqns 2.46 and 2.62 to show that
\begin{equation}
\hat{x}_j = \frac{1}{\sqrt{N}} \left(\frac{\hbar}{m}\right) \sum_k \frac{1}{\sqrt{2\omega_k}} [\hat{a}_k e^{\text{i}kja} +\hat{a}^\dagger_k e^{-\text{i}kja}]
\end{equation}
Here's the two equations:
\begin{align}
\hat{x}_j &= \frac{1}{\sqrt{N}} \sum_k \tilde{x}_k e^{\text{i}kja}\\
\hat{x}_k &= \sqrt{\frac{\hbar}{2m\omega_k}}\left(\hat{a}_k + \hat{a}^\dagger_{-k} \right)
\end{align}
This seem pretty straightforward. I just substituted (2) into (1), distributed, then adjusted the value of the $e$ exponent by making the indices on the creation operators positive and the exponent negative:
\begin{align}
\hat{x}_j &= \frac{1}{\sqrt{N}} \sqrt{\frac{\hbar}{m}}\sum_k\frac{1}{\sqrt{2\omega_k}}\left(\hat{a}_k + \hat{a}^\dagger_{-k} \right) e^{\text{i}kja}\\
\hat{x}_j &= \frac{1}{\sqrt{N}} \sqrt{\frac{\hbar}{m}} \sum_k \frac{1}{\sqrt{2\omega_k}} [\hat{a}_k e^{\text{i}kja} +\hat{a}^\dagger_{-k} e^{\text{i}kja}]\\
\hat{x}_j &= \frac{1}{\sqrt{N}} \sqrt{\frac{\hbar}{m}} \sum_k \frac{1}{\sqrt{2\omega_k}} [\hat{a}_k e^{\text{i}kja} +\hat{a}^\dagger_k e^{-\text{i}kja}]\\
\end{align}
But as you can see, this doesn't match what they say I should end up with. I am missing a factor of $\sqrt{\frac{\hbar}{m}}$.
My Question:
Why am I missing this factor of $\sqrt{\frac{\hbar}{m}}$?
Intuitively, the fact that everything else matches up seems like a good sign. But it should work out then and it doesn't. So either I'm missing something or the problem as stated is written wrong.
| Dimensional analysis shows the book's answer is wrong. Let's work it out. I should obviously get length because this is a position operator. Since neither $e^{\text{i}jka}$ nor the creation annihilation operators have units, I can ignore those terms.
This reduces to
\begin{equation}
\sqrt{\frac{1}{m}s\frac{m \cdot L^2}{s^2}}\sqrt{s} = \sqrt{\frac{L^2}{s}}\sqrt{s} = L \frac{\sqrt{s}}{\sqrt{s}} = L
\end{equation}
So the answer I gave was right by the above.
I also realized I could look this up by going to their errata page. Next time I have a problem like this, that's what I'll do first instead of posting to SE. But good to learn!
| {
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Neutrino Reaction: Is the Following Reaction Allowed? Is the following reaction allowed and why?
$$
\nu_e \to e^- + \mu^+ + \nu_{\mu}
$$
I would say it is allowed since individual lepton number and charge are conserved.
| Also : Lepton number conservation law, respectively Barion number conservation law, are laws which explain if any reaction can occur or not. Every particle has its own leptonic (barionic) number, and for the reaction to occur, the sum of those numbers in the right side of the equation MUST be equal to the sum of leptonic numbers in the left side of the equation. So if you have any doubt in any of these reactions, just look at the leptonic respectively barionic number of the particle.
| {
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Why aren't all black holes the same "size"? The center of a black hole is a singularity. By definition, a singularity has infinite density. So how can a black hole with a different mass or density be described?
| Although we don't have a quantum theory of gravity, we think we have some reliable knowledge about the properties of black holes from general relativity.
One thing we think we know is the so-called "No-hair conjecture", which says that black holes can be described by just three numbers: mass, charge, and angular momentum (i.e. how much they are spinning).
Black holes with different mass differ by the size of their event horizon (the point of no return); for a common black hole solution in general relativity (Schwarzschild black hole), the relationship is linear:
$$
R \propto M
$$
So that's the answer to your question: even if black holes of all masses all contain a singularity, heavier black holes have bigger event horizons.
| {
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What kinds of behavioural anomalies can a zero-field-cooled (ZFC) / field-cooled (FC) split indicate? If a material shows a spiltting in the ZFC and FC curves, is it necessarily superparamagnetic, or could there be any other reason for the irreversibility?
I have heard spin glasses also show ZFC-FC split; but whatever magnetisation measurements I have observed till now (from potential superconductors, to low dimensional magnets, to optically active materials, to thermoelectrics, to multiferroics, to rare earth - transition metal pyrochlores, to molecular magnets, and some strongly correlated systems - I just wish to clarify the range of systems I have seen measurements of so that there shouldn't be any bias in the category of materials) have always shown a split between the ZFC and FC curves. Would this imply that ALL these materials have a tendency to show superparamagnetism? (Because it has been confirmed that not all of them show spin glass behaviour).
Also a superparamagnetic material would not show a straight line M-H behaviour at temperatures like 2 and 10 K, from whatever I understand of the system. But most of these do. So what other kind of systems show ZFC FC split?
P.S. - I have also seen materials which show a superparamagnetic - like M-H curve at 2K (not a straight line, and no area inside of the hysteresis like curve) but no FC-ZFC split. What properties can lead to such a measurement?
| A zero-field-cooled/field-cooled split in the magnetic susceptibility vs. temperature doesn't have to be superparamagnetism. In the case of superconductors, if we apply a field to the material and cool past T$_c$, some flux can be trapped inside, but if we cool first and then apply field, that flux will be shielded away, resulting in greater diamagnetism.
| {
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Time dilation as an effect of energy density Has any relation been observed or postulated to exist between the energy-density (or the surrounding space) of an object and time dilation?
i.e. Higher energy density==>Slower rate of time?
| In both GR and SR, the passage of time is dependent upon the energy state - in the special theory, the passage of time logged by two clocks in relative motion depends upon the kinetic energy $(v^2/c^2)$ difference whereas in the general theory, the passage of time depends upon the gravitational potential $(2GM/rc^2)$. This in turn, is simply the escape velocity (the kinetic energy required to extricate a mass from the gravitational well). So the bottom line is, time dilation in both SR and GR can be expressed by the same factors
| {
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How to calculate the increased pressure drop due to a restriction? Assume I have a straight pipe with diameter D1 and a volumetric flow, Q flowing through it. The inlet pressure is P1 and the outlet pressure is P2.
Now, assume I've added a restriction in the middle of that pipe, which takes the form of another pipe with a smaller diameter D2 and a length L.
Presumably, the new outlet pressure P'2 would drop due to the increased restriction in the pipe. How can I calculate that drop in outlet pressure P'2?
| In an ideal fluid, assuming the diameter of the pipe after the contraction is the same as the diameter of the pipe before the contraction, $P_2^\prime = P_2$. There is no effect of the contraction downstream from the contraction itself.
If we consider an inviscid, isentropic, incompressible flow, the total pressure in the flow is constant along streamlines (and since all streamlines originate from the same source, it is also constant throughout the flow). This means $P_0 = P + 1/2 \rho U^2 = \text{const}$. We also know that $\rho_1 U_1 A_1 = \rho_2 U_2 A_2$ for any positions 1 and 2 where $\rho$ is the density, $U$ is the velocity and $A$ is the area of the pipe.
So, we know density is the same because it is incompressible. We know $A_{inlet} = A_{outlet}$ because the pipe is the same diameter away from the contraction. Therefore, the velocities are the same $U_1 = U_2$. And since the total pressure is constant, the values of $P_2$ and $P_2^\prime$ must be the same also.
This is probably counter-intuitive. You would expect something to be different. And in a real fluid, it very well likely would be. Turbulence, separation, friction, heating, all kinds of losses will add up to have an influence. But in an ideal world where none of those things happen, there is no effect of the contraction away from the contraction.
| {
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Weinberg's spontaneous broken symmetries Steven Weinberg in his second volume of QFT's book (in section about spontaneously broken symmetries, in subsection about Goldstone bosons) writes following:
if we have linear transformation of fields
$$
\tag 1 \varphi^{m} \to \varphi^{m} + i\varepsilon \sum_n t^{mn}\varphi_{n},
$$
under which classical action and path integral measure is invariant, then quantum effective action $\Gamma [\varphi ]$ is also invariant under $(1)$. Then he says that let's consider only case when theory is translational-invariant and fields are constant in space-time. The next calculations are based on this assumption.
But did he discuss only this case? Many results about Goldstone bosons which are obtained in case of constant fields be generalized in an arbitrary case? If no, why Weinberg assumes only this case?
Edit. It seems that the answer is following. We consider the vacuum state of the theory, i.e., stationary point of corresponding action. Since theory is translation invariant, stationary point doesn't depend on coordinates.
| He considers this case, because for this case, the effective action $\Gamma[\phi] = -\mathcal{V}_3TV(\phi)$, such that the $<H>_{\Omega}/\mathcal{V}_3 = V(\phi)$. Basically so that he could use the results of section 16.3.
| {
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Length contraction alongside acceleration Let's have a ship, a target and a ship traveler which we use as point of view. Assuming no other object are observed and we are so far from any other object that gravity distortion are negligible, I'm confused of the effect of length contraction during acceleration (usual length contraction seems quite clear).
If the ship is currently $5ly$ away from target and distance between them does not change (ship is at rest), then the ship begin to travel to target at speed $0.6c$, so the distance between the ship and the target by length contraction becomes:
$$
l = l_0 \sqrt{1-\frac{v^2}{c^2}}
$$
Measuring the speed in "speed of light" unit where $c=1$:
$$
l = 5 \sqrt{1-\frac{(0.6)^2}{1}} = 5 \sqrt{0.64} = 5 * 0.8 = 4
$$
So the distance is reduced by $1ly$ at that speed. However how fast can we reach that speed? If we reach it in a matter of minutes, let's say $1$ minute, then from the point of view of the ship traveler, the distance between his ship and the object was reduced by $1ly$ in $1$ minute. This exceeds the speed of light, which seems incorrect.
What is the expected behavior of the observed distance object from the point of view of the ship traveler, while he is accelerating from zero speed to near-light speed? How length contraction applies, but still keeping the speed of the change in distance less than the speed of light?
|
This exceeds the speed of light, which seems incorrect.
During acceleration, the speed of light may seemingly be exceedet from the viewpoint of the accelerated observer. This is why one talks about "uniform relative velocity" when talking about inertial frames, in which the speed of light may not be exceedet. When you accelerate, you change your reference frame, so you are no longer in an inertial frame. Nevertheless, from the viewpoint of every inertial frame, the speed of light is of course never exceedet, since the rulers and clocks relative to which they measure the distance by time do not contract or dilate in their inertial frame. Also see Ruslans answear in this thread.
| {
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Mass, energy, and entropy. I have a seemingly simple question about the relation between these three that for some reason doesn't make sense to me. If entropy is the disorder of a system, then a low entropy state is one of higher energy. As we know, mass is energy. From here we must say that the more mass something has, the lower its entropy because the mass can be converted to energy. Why then are black holes, the most massive things known, considered to be of such high entropy?
| Drop a glass on the floor -> Entropy increases.
Drop another glass on the floor -> Entropy increases again.
From above we conclude that a big pile of broken glass contains more entropy than a small pile of broken glass. We also happen to know that a big pile of broken glass has larger mass than a small pile of broken glass.
Drop a glass on the floor -> Entropy increases.
Hammer the pieces of glass to smaller pieces -> Entropy increases again.
From above we conclude that breaking a solid to very small pieces causes a large increase of entropy. We also happen to know that breaking a solid to very small pieces increases the internal energy of the (former) solid.
| {
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Why is a hexagon such a stable shape for materials? A hexagonal lattice is famously the shape of graphene, the source of the 2010 Nobel prize. The shape also shows up in beehives and in the basalt columns of Giant's Causeway in County Antrim.
"Causeway-code poet-4". Licensed under CC BY-SA 2.0 via Wikimedia Commons
What makes hexagons so stable?
| Hypothesis: Considering transverse stress and strain, the hexagonal unit completely tiles the space within the structure without gaps and combined with this property, comes closest to a emulating a circle which most symmetrically distributes stress and load.
| {
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Conceptual question on Gauss's Law
*
*By my understanding, the electric field in the surface integral expression for Gauss's Law represents the total electric field and any point on a closed Gaussian surface. However, when we employ Gauss's Law to find the electric field within a uniform charge distribution, we ignore any charges outside the Gaussian surface and look only at the charge enclosed. My question is - if the electric field expression in Gauss's Law represents the total E field at a point, why do we only consider enclosed charges and not those outside a defined surface?
For example, consider a charge distribution of uniform charge density $ρ$, total charge $Q$, and radius $R$. When asked to find $\vec E$ at radius $ r<R$, we can solve by the following.
$$\int \vec E \cdot dA = \frac{Q_{enclosed}}{ε_0} $$
where $$ Q_{enclosed} = ρV = \frac{r^3}{R^3}Q $$
$$E = \frac{k_eQ}{R^3}r $$
This solution would have been the same regardless of whether charges external to the Gaussian surface of radius $r$ had existed. In other words, external charges seemingly do not affect the magnitude of $ \vec E$ in this case. How can this be possible?
*I am trying to determine the direction of electric field within a uniformly charged, insulated sphere of radius $R$ and total positive charge $Q$
My solution:
knowing that $V = -\int \vec E \cdot dR $, I find the expression for potential as a function of radius $ r$ and define electric field direction as that opposite to increasing potential.
$$V(r) = V(R) - \int_R^r \vec E \cdot dR$$ where, as previously determined, $ E$ for $ r<R$ is $\frac{k_eQ}{r^3}r$ .Thus: $$ V(r) = \frac{k_eQ}{R} - \frac{k_eQ}{2R^3}(r^2-R^2)$$
As $r$ increases, $V(r)$decreases and therefore $\vec E$ is directed radially outward.
My question - is there an easier method to visualize why?
| Gauss law is sometimes used to find the electric field, but this is usually confined to situations where there is a strong symmetry that allows you to conclude that the electric field you're looking for is precisely the one in Gauss law, and moreover it reduces to something easy to evaluate, like the product of the magnitude of $\mathbf E$ with a surface. Apart from these cases, Gauss law simply relates the flux of the total electric field through a closed surface and the total charge contained within it. For (essentially) geometrical reasons, outside charges do not contribute to the flux, but they indeed contribute to the total electric field.
| {
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