Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Have there been attempts to create heavier antimatter particles? This is a follow-on from previous question:
It seems as though all experiments concerning antimatter have only been conducted with antimatter protons - as they are obtainable through decay.
Have any heavier elements been attempted? eg. has there been anti-helium or anti-carbon created?
Also, when an anti-hydrogen comes into contact with a heavier element (say carbon), are both atoms totally annihilated, or does the heavier element ( of matter) simply reduce by one proton and release the energy?
Thanks
| Antihelium has been observed: http://www.theguardian.com/science/2011/apr/24/antihelium-antimatter-brookhaven (sounds like just the nucleus, not a neutral atom).
The AEgIS experiment http://aegis.web.cern.ch/aegis/research.html plans to make neutral antihydrogen atoms and measure their gravitational acceleration.
Also, when an anti-hydrogen comes into contact with a heavier element (say carbon), are both atoms totally annihilated, or does the heavier element ( of matter) simply reduce by one proton and release the energy?
It wouldn't be possible, e.g., for both atoms to annihilate and produce nothing but gamma rays, because that would violate conservation of lepton number and baryon number. However, the annihilation of the antiproton with a proton will produce an energy of ~1 GeV, which is much more than the binding energy of the carbon nucleus. Therefore the carbon nucleus will likely fly apart as individual neutrons and protons.
In science fiction, antimatter is often portrayed as a way of storing large amounts of energy to run a spaceship. This kind of makes sense, because the energy density is orders of magnitude higher than, e.g., uranium used to fuel a fission reactor. However, it's not necessarily easy to extract the energy for useful purposes:
Antimatter Propulsion System
Would matter-antimatter annihilation create a fireball or not?
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Are critical exponents below and above the critical point always same? The scaling relations don't distinguish the the critical exponents below and above the critical value. In the mean field level, I understand these critical exponents are same whatever one approaches the phase transition from the order phase or disorder phase. However, beyond the mean field treatment, are they always same?
Are there examples where they are different below and above the critical value? Or are there some theoretical arguments that the must be same? Can anyone give me some hints or good reference?
| Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly above or below the critical temperature.
An obvious exception to this is the scaling of the order parameter,
$$ m \sim (-\tau)^{\beta} \, .$$
Above the critical temperature, $\tau > 0$, we have $m=0$ and there is no critical exponent.
Edit (15 sept 2015): I recently read this paper. They show, using Renormalisation Group (RG) methods, that critical exponents can be different above and below a phase transition if there is a dangerously irrelevant operator involved. An anisotropic term is added. This term breaks a continuous symmetry and forces the order parameter to take one of $n$ countable values. They consider $n=6$. The value of $n$ depends on the way the symmetry is broken and is not important.
Basically, below $T_c$, the RG flow becomes able to double past the previously attractive Infra-Red (IR) RG fixed point where it terminates in the symmetric case. See the figure in the paper. The closer you are to the phase transition, the closer you come to this fixed point and the larger the susceptibility gets. There is an additive contribution to the critical exponent of the susceptibility because of this. When $T>T_c$, the IR attractive fixed point is Gaussian and its attractiveness is not affected by the anisotropy. Then the critical exponent is not affected.
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Explain why we use this formula to calculate total resistance Why do we use this formula to find the total resistance? Lets say we have three resistors in a parallel circuit
$$R_t = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}$$
Where does it come from?
|
Why do we use this formula to find the total resistance?
Because parallel connected conductances sum just as series connected resistances.
Ohm's law is
$$V_R = RI_R $$
The dual of Ohm's law is
$$I_G = GV_G $$
Since, for parallel connected circuit elements, the voltage across is identical, it follows that, for parallel connected conductances
$$I_{total} = V\left(G_1 + G_2 + G_3 + \;... \right)$$
But
$$G = \frac{1}{R} $$
thus
$$I_{total} = V\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \;...\right) $$
or
$$\frac{V}{I_{total}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \;...} = R_{eq}$$
| {
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Why is force a localized vector and not a free vector?
A vector which is drawn parallel to a given vector through a specified point unlike free vector in space is called a localised vector. The effect of a force acting on a body depends not only on the magnitude & direction but also on its point of application & line of action.
What is localized vector and how is force a localized vector? Can anyone clarify me the book's reasoning?
Another property of the force is that the point of application of a given force acting on a rigid body may be transferred to any other point on the line of action without altering the effect of the force.
What is this all about? I couldn't understand this property infact the whole para. What does the book talk about force in this para? Plz help clarifying the meanings.
| Vectors are considered free (of location) and are the same if they have the same magnitude (length) and orientation. So a vector A in one coordinate system with origin O is the same as a vector A in another coordinate system with origin O* even if O* is moving (translational and/or rotational motion) with respect to O. However, the meaning of the vector depends on the origin. If r is a vector describing the location of a particle with respect to O, the same vector r with respect to O* is not the location of the particle even though both vectors are equal. The location of the particle with respect to O* is r* = r - h where h is the vector from O to O*. similarly, r* is the same vector in O and O* but only in O* is it the position of the particle with respect to O*. If O* is rotating with respect to O, although the vector A is the same in both coordinate systems, its Cartesian components (magnitudes in x, y, z directions) differ- e.g., rx != rx*; also dA/dt is not the same in both coordinate systems. See Mechanics, Keith Symon; chapters 3, 7, and 11.
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Pulling on a weakened rope - where will it tear? Let's say I have a rope of 10m length and it is weakened in 3 spots:
at 2.5m, at 5m and at 7.5m. Weakened means that if enough tension is applied it will tear at these points (all points are equally weakened).
This rope is connected to an immovable wall. There is a person on the other side pulling it.
My question is where will the rope tear?
Follow up question: If two people are fighting over a piece of weakened rope, what is the optimal strategy in applying the pulling force, assuming that the rope is weakened at several points and the winner is the one who has the largest piece of rope at the end?
| I would say that not in the middle, either one of the other points. This is my argument:
The string when pulled from both extremes (the same in both cases you ask) it will vibrate in its eigen-frequencies, shown in the picture below taken from Wikipedia. This oscillation will add strain to the points of rope.
But as can be seen, while the point in the middle receives strain contribution from half of the eigen-frequencies (i.e. is displaced and therefore stretched), the points in $1/4$ and $3/4$ will receive contributions from 3 out of every 4 eigen-frequency. Therefore the center point should resist a bit more than the others. Due to symmetry, it is hard to predict which one of the other would break easier.
| {
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Benefits of rear spoiler in cars What would be the benefits of rear spoilers in cars, like this one:
| The spoiler spoil the air flow at the rear side. When the car is cruising, from the frame of reference of the car, air flows around it from front to back. If the air flow is smooth and if there is a large velocity at the rear side then there will be pressure drop. Thus, there is a low pressure at the rear side and higher pressure on the front side, creating a drag.
| {
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Why in PN junctions the octet rule is dominating electrostatic repulsion? The depletion region in PN junctions is created by charges from the N part diffusing into the P part, thus completing an octet of covalent bonds in the P part. This shift however leaves positive ions in the N region and negative ions in the P region, which in turn resist this shift of charges.
What physical principle makes the "octet rule" dominate the electrostatic repulsive force, allowing the depletion region to form in the first place?
| The electron in the $n$ semiconductor and the hole in the $p$ type semiconductor are delocalised and not bound to any particular atom, so arguments based on completing octets aren't useful.
This diagram shows roughly how the depletion layer forms:
In the $n$ type semiconductor the doping creates donor states in the band gap, and electrons from these states are thermally promoted to the conduction band where they are delocalised. In the $p$ type semiconductor the doping creates acceptor states and electrons are thermally promoted into these states to leave holes in the valence band, and again these holes are delocalised. An electron in the conduction band on the $n$ side can lower its energy by recombining with a hole on the $p$ side.
The electron lowers its energy by about the band gap, but its energy increases because it has to do work crossing the depletion layer against the potential across the depletion layer. The charge separation and depletion layer thickness increases until these two energies are equal.
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How does water help extinguish fire? How does water extinguish fire? Heat energy from the fire is transferred to the water, isn't that how it works? How does water deprive oxygen and stop combustion? How is the specific heat of water connected to this? If we use hot water instead cold water, does that make a difference?
| To sustain fire, it is true that you need the tri-factor of oxygen,fuel, and heat.
However extinguishing fire through the use of water, is different than one would think.
Indeed, water "sucks" energy in order to change its phase, and thus reduces the heat factor,
but the real crux lies in the water expansion properties.
Water is heavier than hot air, and as such sinks down into the fire.
It is at the base of the fire that most of the work is done.
The water is heated up, granted, this takes up some heat, but most important of all, it turns into vapor. by the virtue of this process water expands by a factor of 300 (or maybe 3000? I can't remember that factoid from my fire-fighter days), and serves to actually create an upward moving "blanket" that separates the oxygen from the fire source.
The reasons water is to be avoided when dealing with electric fires or chemical fires are numerous:
1. In electric fires, the main concern is, that there is live wires in the vicinity, and using water might actually increase the affected area, and create additional victims/problems.
2. In chemical fires, a lot of burning agents/accelerators are lighter than water (oils for example), thus using water might actually increase the perimeter by "convecting" flammables out of containment, and into an oxygen-rich environment.
3. Some chemical agents, when burned and introduced to vapor might become airborne in forms of aerosol, while these usually do not burn as such, they can be hazardous on their own as toxins etc.
So to sum it up - the prime usage of water as a fire extinguisher is due to their ability to "inflate & separate" more than any other property they have.
There are other/ preferable methods of fire-fighting and extinguishing, that are used today, Helon gas, extinguishing foams etc. each is targeting a different (sometimes even 2) aspect of the fire-triangle.
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How to make a rotating linearly polarized (not circular polarization) beam from a single beam? One way to make a linearly polarized beam rotating at frequency $\Delta f\approx10\mbox{MHz}$ is by combining two circularly polarized beams, one left-handed and one right-handed, and where one beam is at a frequency $f$ and the other at $f\pm\Delta f$. Is there another way to do this using only a single beam passed through some active optical device (eg electro-optic-modulator)? I'm interested in the mid-infrared (10.6um).
| A Pockel cell followed by a quarter waveplate will do this. The Pockel cell acts as an electrically controlled waveplate which will give an elliptical state. The waveplate will then convert this to a rotated linear state. Producing a 10MHz driving signal at the several kilovolts required might be a challenge but not impossible.
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What happens to waves when they hit smaller apertures than their wavelenghts? I was wondering this for quite a long time now. Let's say you have a water wave (like ripples, not the ones you see during tsunamis) with wavelength 10 m. Imagine you put a boundary with an opening of 1 m. Will diffraction be observed? According to my research, no. But, then, what would be seen?
| I cannot speak for how to match with a mechanical resonator but you can match arbitrarily well any EM radiator that is small relative to its wavelength at one (1) frequency. This means that the radiator (loop or dipole, monopole, etc.) will absorb all radiation coming from directions in which it can radiate as a source without any reflection. The rest is diffracted. All elementary dipoles, be they electric or magnetic, are limited by bandwidth and not the ability of radiating if matched properly by some finite set of inductors and capacitors, and because of reciprocity this property also holds in receiver as well as transmitting antennas.
| {
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What is the physical meaning of electric potential, potential difference, and voltage? When resembling the electricity flow through a wire to people walking through a street: electrons are people, current is the number of people, resistance is the barriers on the way.
But what is the electric potential in this example?
What is the physical meaning of 2 V potential difference? And what is changed in the flow of electrons when the potential is 1 or 2 V?
| In my opinion electrostatic potential at any point in an electric field can be viewed as
Suppose a positive charge makes an electric field
In that Field if any positive or negative test charge is present
Let us assume positive test charge
Then this positive test charge will tend to move away from Source charge due to force of repulsion
And if it does not move. Definitely it should have enough energy available with it to counter this force of repulsion otherwise it would have moved away.
So Electric potential at any point in an Electric field may be defined as energy required to be possessed by a unit charge to hold its position countering the force due to E on it.
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How is antimatter made? How is antimatter made in laboratory? Can anyone explain, at the particle level, specifically how anti-protons and anti-electrons are made?
| Creating anti-protons is straightforward in principle because any high energy collision produces a shower of protons, antiprotons and various types of pions. The pions decay in a few nanoseconds, so you just have to wait for the pions to decay then separate the antiprotons from the protons.
At Fermilab a 120GeV proton beam was collided with a nickel target to produce the protons and anti-protons. The proton/antiproton mixture was first passed through a lithium lens to produce a collimated beam, and subsequent magnetic separation produced separate beams of protons and antiprotons.
Although making antiprotons is easy, if by making antimatter you mean making neutral antimatter like antihydrogen that's much, much harder. The energies of antiprotons created from the nickel target are far higher than the ionisation energy of (anti)hydrogen, so the antiprotons need to be cooled and trapped then reacted with positrons to create antihydrogen.
The first significant amounts of antihydrogen to be made were created by the Alpha group at CERN. There's a nice video that describes their experiment here. They hold the antiprotons in a magnetic trap then adjust the field to allow them to come into contact with the positrons.
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How much entropy is produced in evaporating water due to irreversible evaporation towards equilibrium (humidity=100%)? $dS=dH/T_{boil}$ for the increase in entropy by changing a phase at saturation ($T=373\text K$ for $p=1\text {atm}$ for water). However, water also obviously evaporates below boiling point when equilibrium is not reached (below 100% relative humidity).
dS can also be interpreted using chemical potentials ($u$). The difference in the chemical of $u(p,T,N)$ between the 2 phases not at equilibrium is what drives the phase change. However, the increase in entropy when computed comes to
$$ds=(u_{liquid}-u_{gas})\times N/T$$
where $N$ is the number of moles converted. But should the temperature $T$ be the "average" temperature of the system (i.e. the ground level atmospheric conditions for say a lake that evaporates water), or should $T$ be the temperature at which the water boils?
I ask in multiple interests, but motivated to ask due to the temperature discrepancy. Other ideas that can quantify entropy produced by non-equilibrium thermodynamic state is also appreciated.
Thanks
| Specific entropies ($\mathrm{kJ\,kg^{-1}\,K^{-1}}$):
\begin{array}{lrll}
& \mathrm{^\circ C} & \text{liquid} & \text{vapor} \\
\text{Triple point} & 0.01 & 0 & 9.155 \\
\text{Normal boiling point} & 100. & 1.307 & 7.355 \\
\text{Critical point} & 374.15 & 4.430 & 4.430 \\
\end{array}
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Reconciling Units in Classical System Analogies: Why Does Torque Have Units of Energy? In classical physics we often cast an analogy between translational and rotational systems
Force < > Torque
Energy < > Rotational Energy
Momentum < > Angular Momentum
and considering SI units we have [Force] = N, [Torque] = N-m, [Energy] = [Rotational Energy] = N-m (Joules), [Momentum] = N-sec and [Angular Momentum] = N-m-sec.
Physically this analogy seems to make sense, but if you ponder the units in a simplistic way, questions come up like:
Why does torque, which is an analogy of force have the same units as energy, but force does not?
and
If there are differences in units between the analogy for force and torque, why not also a difference between energy and rotational energy?
Is there a simple way to reconcile these questions, or do you have to step outside classical physics?
| Torque and work are different physical quantities,so it makes sense to use different units. Since torque is a vector and work is a scalar, one idea would be to use "$\mathrm{N\times m}$" for torque, instead of "$\mathrm{N \cdot m}$". $\mathrm{N\times m}$ would be consistent with torque, or cross product, while work is dot product! Moreover, $\mathrm{N\cdot m = J}$. But it is never easy to change units, as history teaches us.
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What make us fall when we move on ice? This is a common phenomenon and most of the people have encountered this while moving on ice:falling down! Why does this occur?
According to me, it is due to lack of static friction. But why will it be absent here? What is the cause? I have heard that there is a thin layer of water above the ice . But why,unlike other solids, is this thin layer present? Will it not cool down to ice?
| Walking on ice can can be hazardous if you're wearing a nice pair of dress shoes with slick soles. The backward-sloping phase transition between water and ice cannot explain this phenomenon, and it certainly cannot explain why a hockey puck slides so easily on ice. Nor can frictional heating.
That ice is slippery is instead a boundary layer effect. That the surface of ice was somehow wetted was first proposed by Michael Faraday, but scientists forgot that hypothesis for a while. Scientists didn't have the tools to study this phenomenon until the 1960s, and it remains an area of active research to this day.
The lack of ice molecules above means the water molecules near the boundary layer can't quite organize themselves into a nice crystalline structure. Instead, those boundary layer molecules form a quasi-liquid layer at the ice/air boundary. The thickness of this layer is quite temperature sensitive. This means ice and snow stop being slippery when the temperature gets too cold. Pressure obviously plays a role, too; your dress shoes would make rather lousy ice skates. How pressure plays a role is still a bit up in the air.
The Physics Today article on the subject by Robert Rosenberg,
"Why is ice slippery?" Physics Today 58.12 (2005): 50, is highly readable and very interesting. Read it for more info. And don't be afraid to follow the links in my google scholar search.
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Is it possible to improve the efficacy of light bulbs with a semipermeable mirror? (The greenhouse bulb) Let us imagine that there exists a material that reflects infrared radiation, but is transparent for visible light. Could we take an incandescent bulb and add this material to the inner surface of the glass hull? Then most of the infrared radiation isn't lost, but reflected onto the wire again. So we would need much less energy to heat up the wire to a sufficient temperature. The result would be an efficient light bulb with a broad, gapless spectrum in the visible range.
Is this possible like that? Have I done a fundamental mistake? Or do the required materials not exist?
| Many light bulbs already do this. See for example this article. I was in the lighting technology business at one point. At that time it was done in some tungsten-halogen incandescents. I don't know current state of things.
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Why does gravity enact force? Gravity would cause two objects in a vacuum to move toward each other. I understand that gravity is a force that exists as a product of energy's original conversion into mass and the continuing change in form of energy and mass. I also understand that the potential energy in the two objects changes to kinetic energy as a result of gravity. My real question is: what causes the form of energy to transform? Why do the objects' masses cause them to move?
| Generically, if I have a particle which has potential energy $\phi(x,y,z)$, then the force on that particle will be given by ${\vec F} = - {\vec \nabla}\phi$. So, generically, the motion of particles will "try" to minimze the potential energy. In particular, the only points where the particle will not move will be those points where $\nabla \phi = 0$, or, in simple terms, the potential energy doesn't change under small displacements.
Furthermore, these points will only be stable stationary points if they also represent a minimum of the potential.
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Failure of the Steady State Theory I was reading a journal of astronomy and came to the most famous opponent of Big Bang theory:
The Steady State Theory:
The 20th - century theory was proposed by Hoyle,Gold and Bondie. The theory is based on the Perfect Cosmological Principle which states that universe has no change in its appearance and is homogeneous. It is isotropic in nature. When an old star dies,new star replaces it. So everything remains the same . According to the theory, the universe has neither any beginning nor any end. Universe was and will always the same through the whole time.
Then I was surprised when the journal wrote that this theory has no scientific existence now. It is obsolete today. While reading this, I have found nothing wrong. Isn't the universe isotropic? Why did the Steady State theory fail?
| The following passage has been extracted from the book Parallel worlds:
Finally, in Nature magazine in 1965, Hoyle officially conceded defeat,
citing the microwave background and helium abundance as reasons
for abandoning his steady state theory.
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Why do vapour cones form around jet fighters? Apparently this phenomenon has nothing to do with jets breaking the sound barrier and has something to do with the Prandtl-Glauert singularity as described on Wikipedia. But, the Wikipedia article isn't very detailed and it doesn't explain why the cone arises.
Is there a reason why a "cone", of all the possible shapes, forms around the jets?
| We have to understand there are two phenomena occurring. One is what creates the vapor. The other is what shapes the vapor into a cone. The vapor is created in this case by the localized pressure dropping below the dew point in the air immediately around the structure moving through it. The cone is formed by the high and low pressure waves propagating from the structure as it moves through air.
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How fast would someone have to run to travel vertically up a wall? I am currently doing a physics project on the effects of so-called 'super-speed'. I was wondering how fast you would have to run to vertically travel up a wall? That is, to negate the force of gravity. Is it even possible? Help would be appreciated!
| We can take three steps or so against a wall, what we do is not walking or running, we kick the wall about one meter (our leg) before the collision, change the rotation of velocity, thus throwing our body upside and taking advantage of friction.
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Why are ammeters used in series and voltmeters in parallel? As I'm reading a course on electricity, this one says that an ammeter should be branched inline and not on a bridge. Can someone explain to me physically why we branch a amperemeter inline and a voltmeter on a branch?
| The ammeter measures the current flowing through itself. If you want to measure the current flowing through another component, then you must make the current through the ammeter equal to the current through the component. If you wire it in series, that's true. If you wired it in parallel, the current would be unevenly divided between the component and the ammeter.
Similarly: The voltmeter measures the voltage across itself. If you want to measure the voltage across some other component, you must make the two voltages equal. If you wire it in parallel, that's true. If you wired it in series, the voltage would be unevenly divided between the component and the voltmeter.
| {
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How is Kirchhoff's voltage law understood in the water flow analogy? I met the Kirchhoff circuit laws in the past, but now I'm trying to associate them with a practical representation to be sure to understand them.
Let's start with the Kirchhoff current law: If I say that the electrons are like water going though a pipe, it can clearly be understood that the water flow will be divided when meeting a junction.
The problem is how to understand (practically and physically) the voltage law - what analogy can I use to understand it?
| The Kirchoff Voltage law states that the sum of emfs in a circuit is equal to the total potential drop in the circuit.
So for a simple example, where you a 6V cell, for example, and 2 resistors in series.
The 6V cell can be seen as a place where the water is given potential energy - if we imagine a ramp, it would be the water being pumped up to the top of the ramp.
The water then flows along a straight path (wires assuming 0 resistance) until it meets a resistor. A resistor can be seen as another ramp, however since there are 2 resistors, (assume they are the same resistance, however this doesn't really matter) all the potential cannot be dropped across one resistor so this ramp is smaller than the ramp up from the 6V battery. As water falls down this ramp, it loses potential energy - this is 'the same' as the drop in potential difference across a resistor.
You may ask - but the water speeds up when it falls down the ramp? well for this we should assume that there is no gain in kinetic energy and all gained kinetic energy is dissipated falling down the ramp - this can be analogous to current remaining constant in a series circuit.
| {
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Poisson brackets of the Kepler Problem For the hamiltonian of a particle of unit mass in a kepler potential:
$$H = \frac{1}{2}\mathbf{p} \cdot \mathbf{p} - \frac{\mu}{r}$$
The angular momentum vector is given by: $\mathbf{L} = \mathbf{r} \times \mathbf{p}$
I know and can show that the poisson brackets of $\mathbf{r} \cdot \mathbf{r}$, $\mathbf{r} \cdot \mathbf{p}$ and $\mathbf{p} \cdot \mathbf{p}$ with any component of the angular momentum vector vanish algebraically, but what is the geometric reasoning behind this? I am trying to develop a better intuition about this. Could someone explain? Thanks!
| Let me give a further comment (not exactly an answer)
The quantity $\mathbf{r} \cdot \mathbf{r}$ represents the magnitude of the radius as such it does not change under rotation (poisson commutator with ang. momenutm $L$)
The quantity $\mathbf{p} \cdot \mathbf{p}$ represents the magnitude of (linear) momentum, as such it also does not change under rotation
The quantity $\mathbf{r} \cdot \mathbf{p}$ represents the magnitude of action, as such it also does not change under rotation
| {
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Clausius statement of the 2nd Law I'm slightly messed up with the Clausius statement of the 2nd Law.
I've seen at least two versions, which seem to be conceptually different.
a) It is impossible to transfer heat from a colder body to a hotter body without any other effect.
b) It is impossible to transfer heat from a cold (thermal) reservoir to a hot (thermal) reservoir without any other effect.
I would like to use the Clausius statement to rule out heat transfer from a colder body to a hotter body, where both have finite thermal capacity. No work put in. If we suppose this transfer could happen, then the hotter body would become hotter and the colder body would became colder. So, there actually is some other effect (temperature changes). How does a) rule out this procces? Obviously b) does the job.
| The statement
without any other effect.
means without external work being acted upon the system. Actually nobody uses the word "effect" and most textbooks use the correct terminology of external work.
Reservoirs are bodies too, therefore 1) directly applies to 2).
| {
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Tesla Coils - Is there a risk that the discharge can create x-rays? I've built a Tesla coil that stands about 3 ft tall and uses a spark gap as the interrupter for the primary circuit. Judging by the size of the streamers it's reaching at least a million volts.
Someone once told me that you have to be careful with Tesla coils because they can create x-rays. I had been skeptical, but then read about how x-rays can be produced by unwinding scotch tape. So now I am somewhat concerned.
So are harmful x-rays a risk with Tesla Coil operation, and if so how can I easily test my system to see if it's safe?
| I defer to:
https://webhome.phy.duke.edu/~rgb/Class/safety.html
They do list X-ray generation, however the primary risks are ozone (which destroys organic compounds; your body is primarily composed of organic compounds...) with insufficient ventilation and electric shock.
X-rays are typically generated using 20kV to 60kV. A Tesla coil can definitely reach that range of voltage.
They will also fry electronics, including pacemakers.
| {
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Influence of applied voltage to an electron of a metal I would like to ask what would happen to the potential well of an electron being trapped in a metal? If I apply a voltage trying to accelerate the electron out of the potential well. Would It make the potential well shallower? Or would it rather shift the fermi level to a lower fermi energy level?
| First of all I think we are thinking about the free electrons in the metal rather than the ones which are bound to individual atoms.
Generally to apply a voltage we attach connections to each end of the metal and apply a potential difference. One way of thinking about what happens when we apply a voltage is that the fermi level gets slightly tipped from one end of the metal to the other so that it is slightly raised at one end and lowered at the other and electrons flow because of this.
Another way to think about this is using the free electron model where electrons fill the 3D space of the metal. This has to be solved quantum mechanically and is like the particle in a box calculation except, of course, because electrons are femions they each have to have separate levels and fill up to the fermi level. Now we often think of levels in the particle in the box as having different energies, but in the 3D box the electrons are confined to we can also think of the electrons as having different wave vectors $k$ ($E$ is proportional to $k^2$. The wave vector $k$ has direction and points anywhere in 3D. If we fill up to the fermi level we will fill up all states up to some maximum $k$ value. We can plot these states in 3D phase space where the axes are $k_x$, $k_y$ and $k_z$ and we would get a filled sphere centred on the origin - we can think about the electrons in this sphere travelling in all different directions with a net current flow of zero because the sphere is centred on the origin....
.... Now applying a voltage to the metal will slightly shift this sphere in $k$ space so that it is no longer centred on the origin. This will meant that there is a net flow of electrons.
If you want to think about this without using metal contacts then UV photons can knock an electron out of the metal by the photoelectric effect.
Alternatively in Field Emission Microscopy a high voltage near a sharp metal tip 'pulls' electrons out of the metal - this raises the fermi level at the tip to the vacuum level and electrons can escape.
| {
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Why doesn't a star's core cool down when it expands as a red giant? When a star starts to run out of hydrogen to fuse, it begins to collapse due to gravity until the central core temperature rises to $10^8~\text{K}$
Then due the force generated by the fusion of helium, the star expands again and becomes a red giant.
So, why doesn't the expansion cool the core?
| First, a star does not become a red giant when helium fusion begins, instead it becomes a red giant earlier when an inert degenerate core of helium forms and a shell of hydrogen begins fusion. When shell hydrogen fusion begins, the star expands to be a red giant.
The core is degenerate (sustained from collapse by electron degeneracy pressure) and therefore cannot cool by expansion, as explained here:
http://burro.astr.cwru.edu/Academics/Astr221/LifeCycle/redgiant.html
Later, helium fusion begins at which point the star is a horizonal branch star, rather than merely a red giant. At that point the core can cool by expansion, as explained in the reference.
| {
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Renormalization in non-relativistic quantum mechanics I read many articles about renormalization in the Internet, but as I currently don't know much of QFT (currently just studying classical field theory and QM), and as all this looks quite interesting, I'd like to still get some bit of understanding and feeling of it — in the context of non-relativistic quantum mechanics.
So, my question is: are there any (hopefully simple) examples of quantum mechanical problems, applying perturbation theory to which will give divergent series, which can then be regularized by renormalization procedure? What are they, what does this process of renormalizing them look like?
| http://www.roma1.infn.it/~amelino/appunti1.pdf
Here, page 16: "Aside on perturbative renormalizability".
You can find a quite simple but enlightening example of renormalization applied to a non-relativistic theory where you also have the exact energy spectrum and eigenfunctions to be compared to.
| {
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Can action be unbounded from below? While solving the problem in this question, I found cases where the numerical optimization failed, suspecting unboundedness of the function being minimized. The function approximates the action of the system in question.
I decided that this result could be explained by an unbounded from below action. But I'm still in doubt because it may be my implementation problem.
So, the question is: do there really exist such physical systems with finite number of degrees of freedom, where the action is unbounded from below, given fixed values for $q(t_1)$ and $q(t_2)$? If yes, how can one decide whether a given system is of such type?
| Example: Consider an action functional
$$\tag{1} S[q]~=~\int_{t_i}^{t_f} \! dt ~L, \qquad
L~=~\frac{1}{2}m\dot{q}^2-V(q),$$
with Dirichlet boundary conditions (BC)
$$\tag{2} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f,$$
where the potential $V$ has a repulsive pole
$$\tag{3} V(q_0)~=~+\infty$$
at $q=q_0$. Then it is possible to choose a virtual $C^1$-curve $\gamma:[t_i,t_f]\to \mathbb{R}$ that satisfies the BC (2) and sits for a while at the pole $q_0$, so that the action functional
$$\tag{4} S[\gamma]~=~-\infty$$
is unbounded from below.
In particular, if the exists a unique stationary path $q_{\rm cl}$ [which satisfies the Euler-Lagrange equation and the BC (2)], it cannot minimize the action functional $S$.
| {
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single-mode approximation in spinor F=1 BEC with dipolar interactions How can a single-mode approximation be justified in spinor F=1 BEC with dipole-dipole interactions? Or maybe this kind of approximation will never take place and condensate components are always separated. What do You think?
Good literature on the topic is welcome.
| In the field of multi-component condensates, the single mode approximation (SMA) means that different dipole states are assumed to share the same spatial wave function. Thus, there are no dipolar textures.
SMA is well justified when the inter-component (e.g., spin-dependent or dipole) interactions are much weaker than the interactions independent of the components.
My guess is that the OP is concerned about the applicability of the SMA for ultracold atoms due to their experience with solid-state physics, where
even a weak dipolar interaction spontaneously induces spatially varying dipole moments.
Indeed, the SMA is not always applicable for BECs. My favorite reference on the topic is cond-mat/0606099 , they give phase diagrams, explicitly showing where the SMA is applicable, and when it breaks down. More references can be found in a recent paper arXiv:1306.0398 , and also in papers citing the first paper.
| {
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Why does Earth behave like Natural Bar Magnet? What is the reason for the Earth to behave like a bar magnet and have poles (North and South poles)?
| The Earth's magnetic field is caused by eddy currents in the liquid parts of the planet's interior. We believe the field is not due to a permanent magnet because: (1) Its direction and strength change over time, and (2) the planet's interior is hotter than the Curie temperature of its elements, and so a permanent magnet would not retain its magnetism.
However, saying the field is due to eddy currents is not sufficient. Simply swirling around a conductive fluid does not produce a magnetic field. There has to be some source of electric current to make the whole process work, and we don't know what that source is. Several possibilities have been proposed. Examples include: (1) Gravitationally separated elements produce a voltage difference, as in a battery; (2) Gravitationally separated elements combined with the temperature differential between the Earth's core and its mantle creates a thermocouple junction that produces voltage, and therefore current; and (3) Flowing electrons produced by Beta decay of radioactive material in the core. However, none of these explanations are accepted as proven, and some are thought to be highly unlikely. So the source of the electric current remains a mystery.
| {
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Making a plastic scintillator glow? I have a plastic scintillator I had purchased off eBay, which is believed to be BC408
http://www.phys.ufl.edu/courses/phy4803L/group_I/muon/bicron_bc400-416.pdf
Upon shining a small UV LED (10-40mA, maybe ~400nm wavelength) on it, it seems to glow mostly from the refraction of the plastic or how it was milled and the UV beam itself, rather than appearing to fluoresce, such as the beam just bouncing off imperfections and passing through the bar. I had also tried to point it at an old CRT screen as a test because I believe they generate near-xray levels (or is that kept behind leaded glass?) and saw nothing interesting.
How can I cause the scintillator to glow/observe flashes with objects one can source around the house or easily? Would I need a higher energy UV at less energy to make the fluorescent effect more dominant than the light hitting it and reflecting off itself?
| You can get original Fiestaware pieces online or at some antique stores. The uranium glaze is a decent low-level x-ray and gamma emitter that is safe to handle.
My orange salt-shaker reads a couple hundred Bequerel on a good Geiger tube held a few centimeters away.
You can also order nano-curies of a number of isotopes without any paperwork (in the US, no idea about other countries). Google for "needle source"
I would not recommend disassembling a smoke detector or otherwise trying to get access to radionucleides used in consumer products because you don't know how they are contained and the last thing you want is to ingest or inhale a alpha emitter.
| {
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Distribution of gravitational force on a non-rotating oblate spheroid Suppose a person is standing on a non-rotating$^1$ oblate spheroid of uniform density. He first stands on one of the poles, then on the equator. In which case is the gravitational force greater?
In Case 1, the distance between the centre of the person and the spheroid is lesser than in Case 2, but in Case 2, there is lesser 'distribution' of force, i.e., more force is concentrated in one direction.
Summary
By applying parallelogram of forces, Newton's inverse square law of gravity, and any other rule, in which case will the Gravitational force be greater? Or will they both be equal?
--
$^1$ The attentive reader may ponder how come the spheroid is oblate in the first place if it is non-rotating and there is hence no centrifugal force to stretch it? Well, it is a hypothetical question. Please accept the slightly unrealistic premise.
| Neglecting rotation, it seems to me that if one takes the case where the diameter at the equator is very much larger than that at the pole, we can consider it to be a disk.
Standing on the edge, all the mass is directly below you and all of it contributes to attracting you to the centre. Standing at the centre of the disk, the overwhelming bulk of the disks mass is at the periphery, and the gravitational effect you experience from it is only the component which points to the centre which will be tiny. Most of the effect will be at 90 degrees to this and balance out.
I would expect one to to be heavier at the equator than at the pole.
| {
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Have you ever seen in your car a mosquito? it's not capable of going 60 mph. How come it can still keep up with the car as if it weren't trying For instance, a fly is staying at the same spot, then your car suddenly moves forward at a rate of 60 mph. The mosquito IS staying at the same spot but for some reason it moves along with the car as if nothing happened.
| The air inside the car moves transiently as the car speed up (is being pushed by the rear of the car, sits, etc), if this happens slow enough this will occur without much turbulence or wind-like movements. The mosquito is pushed by the air (assuming he is flying), so in his frame of reference, relative to the inside of the car, he is still moving slow. Only a person in the street outside the car will measure a mosquito speed of 60 mph
| {
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How to represent a laser pulse in quantum optics Every quantum optics textbook that I've found says something like "a coherent state represents the output of a laser", but a coherent state is a static thing (aside from the oscillating phase of the complex parameter); how do you represent a laser pulse propagating in space and time? I'm guessing you need to take some kind of sum over k-modes (as a single k-mode is presumably completely delocalised?)...
Also, any references you can suggest which discuss this in detail would be greatly appreciated!
| A coherent state is a monochromatic sinusoidal field. The electric field pulse is inherently not monochromatic, but instead has a spectrum of frequencies which are superimposed on top of one another so as to all add constructively once every pulse repetition time. Therefore, to represent the field pulse in quantum optics you actually need to bring in more modes: the full state is a tensor product of the state of each mode, each being itself in a coherent state.
| {
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How to know if a vehicle is moving without any external source of information? The situation is the following:
I'm inside a vehicle (plane or a car, it doesn't matter) and I need to know if the vehicle is moving at a constant speed BUT I cannot perceive any external change like visual changes, vibration, etc.
How can I know if the vehicle is moving? Do I really can know?
Additional question Can I know my speed?
| Does humidity and similar cues count in the answer or not?
For example if moving continously after a while certain environmental parameters will change. Which cannot be screened in the vehicle.
For example humidity, temperature etc..
Assuming the vehicle can actually screen all those parameters. Then one can use non-inertial cues: For example rotations, turns, acceleratios etc..
Assuming no non-inertial cues are present, then as already stated, the principle of relativity (for inertial frames) makes the uniform motion and no-motion equivalent.
| {
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Bose Enhancement Factor How may one explain the fact that the probability of a boson transferring to a state with an occupation number n is 'enhanced' by a factor of (1+n), compared to the classical case? (In the classical case, the probability is supposed to be independent of the occupation of the final state.)
| A simple way to see this is by considering the fact that the probability to transition from a particular initial state to a particular final state is the same as for the inverse process where one considers the transition from the final to the initial state. This is because the square of the modulus of the matrix element is the same for both cases.
This means that instead of considering the transition to a state where there are already n bosons present and one will be added, we can just as well consider the transition rate from the final state containing n+1 bosons to the initial state where one of these bosons will go somewhere else. Obviously, this will have to be proportional to the number of bosons in the final state, so the rate will be proportional to n+1.
Note that in a classical setting you can also get to this result, but then "classical" refers to treating the field of the bosons in a classical way (take e.g. the electromagnetic field, in that case you can derive the formula for stimulated emission, in a full quantum mechanical treatment you get the classical result by ignoring certain commutators which then yields a factor of n instead of n+1).
Instead, if the classical limit is taken where the volume of the system is effectively sent to infinity so that the density of states goes to zero, then the occupation numbers go to zero as well and then n+1 becomes 1.
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Analytical solution of Liouville's equation for classic harmonic oscillator I'm interested in the analytical solution of the simple PDE:
$$\frac{\partial f}{\partial t} - m\omega^2x\frac{\partial f}{\partial p}+ \frac{p}{m} \frac{\partial f}{\partial x} ~=~ 0.\tag{1}$$
With: $$f(x,p;t\!=\!0)~=~f_0(x,p) \quad \mbox{arbitrary smooth},\tag{2}$$ $$x(t)~=~x_0 \cos(\omega t) + \frac{p_0}{m\omega}\sin(\omega t),\tag{3} $$ $$p(t)~=~p_0 \cos(\omega t) - m\omega x_0 \sin(\omega t).\tag{4}$$
And $x_0, p_0$ constants.
| Hints:
*
*OP's eq. (1) is the equation for a constant of motion $\frac{df}{dt}=\{f,H\}_{PB}+\frac{\partial f}{\partial t}=0$ of a harmonic oscillator $H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2$.
*Let us assume for simplicity that $m\omega=1$, and leave it to the reader to generalize to arbitrary $m$ and $\omega$.
*Complexify $z=x+ip\in\mathbb{C}$. Then the solutions (3) and (4) read $z(t)=e^{i\omega t}z_0$.
*The solution $f(z,t)$ to eq. (1) with initial condition (2) is then $f(z,t)=f_0(e^{-i\omega t}z)$.
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Why does the Copenhagen interpretation assert randomness if this cannot be tested? Why does the Copenhagen interpretation of QM assert that random events occur if such a claim cannot ever be proven or disproven?
A related question:
How to tell if QM is really random?
Edit On second thought it could certainly be disproven. It couldn't, to my understanding, be shown that randomness as Copenhagen asserts is provable in any sense of statistical significance. Is this true? If it is, why does Copenhagen assert randomness instead of asserting that no non-disprovable deterministic rules are known?
Edit 2:
I've created a proper place for arguing about what an interpretation is:
What is an interpretation of quantum mechanics?
| We do experiments and come up with distributions of the variables that can be measured.
There exists a very deterministic theory, Quantum Mechanics, that predicts accurately probability distributions for our measurements and the theory is continuously validated.
Probabilities exist in the classical physics regime and mean exactly the same thing as the probabilities measured and fitted with quantum mechanical solutions.
Assuming randomness is on the lines of KISS (keep it simple stupid). If there were not all these people trying to find a classical framework underneath the quantum mechanical probabilities there would be no need for the assertion of randomness, it would be self evident, as it is for the throws of dice.
So it is the theories that assert the existence of deterministic underlayers for quantum mechanics that, by contrast, force an "assertion" of randomness. If quantum mechanical probabilities could be shown to be biased (the way a dice can be biased by cleverly weighting it) then new standard models and theories would be developed. At the moment as far as I know deterministic underlayer theories have problems with Lorenz invariance, which is another well validated theory of particle physics.
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How does a single charge produce magnetic field? I have studied in Introduction to electrodynamics (Griffiths) that magnetic field is actually due to effects of relativity
unequal Lorentz contraction of the positive charge and negative lines, a current- carrying wire that is electrically neutral in one inertial system will be charged in another.
(this was ,'as explained in Griffiths.)
But here (MIT website, chapter 9, http://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ )
I have seen,
a single charge also produces magnetic field
but how this is possible? How can I imagine contraction of single charge (not a line of charge).
| A single charge cannot produce a static magnetic field. It can produce a time-dependent magnetic field. It is also a relativity effect because both electric and magnetic fields are changed with change of the reference frame.
In a moving reference frame (or when the charge moves) one observes not only electric, but also a magnetic field $\mathbf{H}'\propto\mathbf{V}\times\mathbf{E}$
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Semiclassical quantization of bouncing ball Consider an elastically bouncing ball of mass $m$ and energy $E$. This has a triangular potential
$$ V(x)~=~\left\{\begin{array}{ll} mgx & \text{if } x>0, \\
\infty & \text{if } x<0, \end{array}\right. $$
where the $x$-axis points upwards. Let $\hbar = m = g = 1$, so that the maximum height reached is $E$.
The classical frequency of motion is $$\omega = \frac{\pi}{\sqrt{2 E}}.$$
I can recall there is a quantization rule that says that the spacing between energy levels equals the classical frequency (in the semiclassical regime). This would imply that
$$
E_{n+1} - E_n \propto E_n^ {-1/2}
$$
so
$$
E_n \propto n^{-2}.
$$
However, the area in phase space enclosed by the orbit is proportional to $E^{3/2}$. This area must be an integer times $2\pi$, which gives a different quantization condition
$$
E_n \propto n^{2/3}.
$$
What went wrong? I am pretty sure the second result is correct, but why is the first result wrong?
| I think your error is in assuming that $E_{n+1} - E_{n}$ is proportional to $n$. At least, I assume you assumed it; it's the only way I can see that you could go from the statement
$$
E_{n+1} - E_n \propto E_n^{-1/2}
$$
to the statement
$$
E_n \propto n^{-2}.
$$
Really, what the first proportionality above implies is that
$$
\frac{dE}{dn} \propto \frac{1}{\sqrt{E}}
$$
in the limit of large $n$; and if you solve this equation, you get
$$
\sqrt{E} \, dE \propto dn \quad \Rightarrow \quad E^{3/2} \propto n \quad \Rightarrow \quad E \propto n^{2/3}
$$
as expected.
| {
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How noisy are photon detectors? I have a single photon detector and $N$ photons per second arrive at the detector. Then something happens and the number of incoming photons per second changes by the factor of $\alpha$. So now $(1+\alpha) N$ photons per second arrive at the detector.
What are currently the best photon detectors to measure the smallest possible change $\alpha$?
Of course I know that this depends on the noise and also on $N$.
Say I have $N=1000$ photons per second and I want to measure a change of 1%, hence I want to measure $\alpha= 0.01$. Is that possible? If not, is there a limit in principle?
| As a practical matter we generally use these devices in cases where the photon arrivals are random. That is the mean rate may be known, but the actual arrivals are distributed according to a exponential time law rather than being periodic.
In those case, counting statistics usually dominate the uncertainty (in low background detectors shot-noise can generally be made a small effect).
So the question becomes one of how long you can integrate the before and after the change.
I'm going to use $R$ for the actual rate, and $r$ for the measured rate and use a prime to denote "after the change" while unprimed symbols come before the change.
To put some numbers in it, the expected number of counts in time $t$ is $R t$, and the standard deviation in the number observed is $\sqrt{R t}$. That makes the standard deviation in the measured rate $\Delta r = \sqrt{R/t}$. We have the measurement before the change:
$$ r = R t \pm \sqrt{\frac{R}{t}} \,.$$
Taking $R' = R(1 + \alpha)$ the measurement after the change is
$$ r = R' t \pm \sqrt{\frac{R'}{t'}} \,.$$
Now because we assume that $\alpha \ll 1$ it makes sense to use our time evenly so I set $t' = t$, and form the difference between the measured rates:
$$
\begin{align*}
d &= r' - r\\
&= \left( R' \pm \sqrt{\frac{R'}{t}}\right) -
\left( R \pm \sqrt{\frac{R}{t}}\right) \\
&= \alpha R \pm \sqrt{\frac{R'}{t} + \frac{R}{t}}\\
&= \alpha R \pm \sqrt{ \frac{R(2+\alpha)} {t} } \,.
\end{align*}
$$
This is only statistically significant when the main terms is several times the error term
$$
\begin{align*}
\alpha R &= \text{(a few)} \times \sqrt{ \frac{R(2+\alpha)} {t} } \\
\alpha^2 R^2 &= 10 \times \frac{R(2+\alpha)} {t} \\
t &= 10 \left( \frac{2 + \alpha}{\alpha^2} \right)\frac{1}{R} \\
&\approx \frac{20}{\alpha^2 R}\,.
\end{align*}
$$
In other words, is $\alpha$ is small and $R$ is anything but very large you have a long wait in front of you. Using Andreas' numbers ($R = 1000 \,\mathrm{Bq}$ and $\alpha = 0.01$)--which represent a fairly desirable arrangement--we get $t = 200\,\mathrm{s}$ of counting on either side of the change to get a 3ish sigma measurement of a one percent change in the rate.
This is a specific case of the general rule that measuring small changes is hard.
| {
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How does air pressure affect the speed of sound? How does air pressure affect the speed of sound in relation to its kinetic theory etc?
I have tried searching this but i have not found a suitable answer as other websites simply related to air pressure waves.
| To first order, the speed of sound is not affected by pressure. Pressure waves can be shown to fulfill the D'Alembert wave equation $(c_S^2\,\nabla^2 - \partial_t^2)\psi=0$ where the wavespeed $c_S$ is given by:
$$c_S = \sqrt{\frac{K}{\rho}}$$
where $K$ is the bulk modulus of the medium in question and $\rho$ its density. Now, for an ideal gas, the bulk modulus $K$ is in most conditions proportional to the pressure; if the compression is adiabatic (good approximation for high frequency sound, as there is little time for heat to shuttle back and forth in the gas), then $K=\gamma\,P$, where $\gamma$ is the Heat Capacity Ratio or Adiabatic Index. However, from the ideal gas law $P\,V=n\,R\,T$ we have:
$$\rho = \frac{n\,M}{V} = \frac{P\,M}{R\,T}$$
where $M$ is the mean molar mass of the gas in question in kilograms. Thus the pressures cancel out in the speed of sound:
$$c_S = \sqrt{\frac{\gamma\,R\,T}{M}}$$
Thus we see that the speed is also weakly affected by the humidity - more water in the air lowers the mean molecular mass. If we put $M=0.029$, $T=300K$ and $\gamma = 1.4$ for air, we get $c_S=347{\rm m\,s^{-1}}$.
| {
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Why would different metals glow red at different temperatures? According to everything I've been taught about incandescence and black-body radiation, and some quick Googling to confirm I'm not crazy, just about everything, regardless of composition, should start glowing red at about the same temperature- 798K, the Draper point, where sufficient power in the black-body radiation curve crosses into the visible spectrum to be visible.
I have just been informed by a metallurgist friend, however, that different metals in his experience begin to glow red at wildly different temperatures; typically, just below their melting points. For example, apparently aluminum glows red at much lower temperatures than steel.
My hypothesis so far: The metals in question are far from perfect black bodies (reasonable, since most metals are shiny), and differing levels of emissivity in the low end of the visible spectrum require different temperatures to raise total emission in that range to visible levels.
This, however, does not explain why there should be any connection between glow-point and melting point.
Am I close to correct? Is there another better explanation? Or is my friend simply crazy?
| Different metals would to glow at different temperatures because of their different abilities to hold on to electrons. Some metals hold their electrons very weakly and some hold it very tightly. The temperature at which it glows is dependent on the strength of this force. Things glow either because of absorption or emission spectrum. The lesser the force with which the nucleus holds the electron, the lesser is the energy required to make it glow. Actually, to get a more satisfying answer, you must post this question in chemistry stack exchange.
Edit: The ability to remain solid is dependent on this force too, hence the connection between melting point and glowing red.
| {
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Have we proven that higher dimensions exist? Have we proven higher dimensions exist?
| I think dimension means the number of independent coordinates required for describing or specifying the position and configuration in space of a dynamical system. As example when we consider the Minkowski's four-dimension world we consider the position coordinate is homogeneous to time coordinate. So an event is represented by $(x,y,z,\tau) $ where $\tau=ict$. And in this space $s^2=x^2+y^2+z^2+{\tau}^2=x^2+y^2+z^2-c^2t^2$., which remain invariant under Lorentz transformation.
As far as you say about the graphical representation of these fore coordinates, we plot the time in imaginary axes.
So, yes. Higher dimension exist.
| {
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What's the difference between frequency domain and time domain spectra?
If I have a mechanical oscillator and want to observe the dynamical behavior of the oscillator, is there any additional information to observe it in time domain and frequency domain? Normally, we observe the frequency domain spectra (power spectral density) as the information of oscillator. In fact, I'm solving a dynamical behavior of two coupled mechanical oscillator, like the picture above. While someone told me that I could get different information from time domain than frequency domain. In my opinion, the difference between time domain and frequency domain is just the transform of Fourier. So what's the difference
| A signal spectrum can be decomposed in its component frequencies in two ways:
*
*Time domain - For example with wavelets (you will see the different frequencies along the Y axis and the increasing time in the X axis)
*Frequency Domain - For example with the Fast Fourier transformation or multitaper transformation where you will find the frequency power in the Y range and the frequency of time in the X axis. Usually the time frequency has as its maximum range half the length of the total time and is given in percentage of the range.
| {
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What caused the expansion of the universe to slow down after the inflationary epoch? As everyone knows, when the big bang happened, the universe expanded at an unbelievable rate and this was called the inflationary epoch (or more popularly cosmological inflation) which lasted for about $10^-$$^3$$^4$ seconds. But after that, the rate of expansion suddenly reduced and is now gradually increasing. Can someone tell me what is the probable reason or the resistive force that caused expansion to slow down greatly after the big bang?
The question is based on the simple fact that expansion could not have slowed down if there was no resistive force acting or more simply inertia of motion.
| If you take and measure the accelerated expansion of space occurring between a random pair of galaxies, say around 20 to 30 million light years away from us, then multiply that acceleration by 13.8 billion years, the age of the universe. Now, compare the answer to the distance they are currently apart. The numbers don't match. The two galaxies don't actually merge onto a single point, IE big bang. They would still be over 2 million light years apart, even after the space between them accelerated between 9% to 11% in the last few years. Remember, if we were to go backwards in time then the acceleration would decelerate, making the distance they potentially traveled in 13.8 billion years much less. The two galaxies still would have never occupied the same spot. Hence no big bang. Besides, cosmological inflation actually violates the 3rd law of motion, once in motion always in motion unless acted upon by an outside force. If inflation happened it would not stop. Galaxies would not have formed if rapid inflation occurred right after a big bang. If the amount of energy and mass in the singularity was unable to prevent the big bang from happening then it wouldn't be enough to ever slow down the theorized inflation. There has to be a better explanation. Future observations by the infrared telescope James Webb should uncover the solutions to every unexplained mystery, even how and when the universe began. Go JWST.
| {
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Nature of Cooper pairs Some people say it is bound state, some say it is not. Which is more accurate?
Problem is that I read in some books, including Ziman, that Cooper pairs are bound states but my teacher says that it is not true and that Bardeen had to explain it many times even to his peers.Now, i know that it has something to do with the resonance in scattering cross section...but oversimplifications with hand-waving about phonons that mediate interaction creating an attractive force and a bound state, like some kind of electron-electron molecule just make me angry. I know it is phonon mediated but it is not that simple, right?
| As Cooper wrote in his paper (1956), one Cooper pair is a bound-state. The point is : it's a bound-state if you consider 2-electrons on top of the Fermi sea with an attractive potential. In the many-body case, better not to think in term of bound state. A Cooper pair is a fictitious particle which has no clear meaning, although the main characteristics of superconductivity (bosonic electrodynamics of charge 2e) suggests that such a picture might be fruitful.
In field theory, a Cooper pair is just a non-trivial correlation between fermionic creation operators in the ground state : the anomalous Green-Gork'ov functions $F^{\dagger}\sim\left\langle c^{\dagger}c^{\dagger}\right\rangle $.
Following some comments from @tparker : Cooper pair has nothing to do with any kind of motion. "Cooper pair" simply means that the correlation function made from two creation operator (say $\left\langle c^{\dagger}c^{\dagger}\right\rangle $ with the average procedure done on the ground state) is non-vanishing, that's all. For historical/conventional superconductivity, the correlations appeared between two time-reversal electronic states, the so-called $s$-wave superconductivity. It makes the superconducting state robust to some interaction, in particular it gives to the condensate an immunity with respect to disorder (the Anderson's theorem). But this can be seen as extra properties on top of the correlations.
| {
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Everyone calls Electromagnetic Induced Transparencyan interference phenomenon, but is it also an interference phenomenon in classical systems? Electromagnetically induced transparency is a hot topic in physics. However I'm curious about its mechanics in physics. Physicists think that it's a phenomenon of interference from transition of two states which is explained by the atomic levels of the system. Could it also be interference in a classical system? If it could, what would be the mechanics of their interference? Or what would interfere with what?
The paper below describes a classical system which displays EIT in a classical system which may help you save your time in understanding.
Garrido Alzar, C. L., Martinez, M. A. G. & Nussenzveig, P. Classical analog of electromagnetically induced transparency. American Journal of Physics 70, 37 (2002). (arXiv link)
| EIT and the mechanical analogy you posted are examples of coupled resonator systems. In the RLC circuit, the system includes two RLC circuits that are coupled. The EIT system consists of a pair of two-level systems that couple to a single shared level. An RLC circuit and a two-level system both exhibit resonance effects: the energy transferred into the system depends on the frequency of the drive. The interference occurs between the power transferred by the probe field (applied voltage in the RLC circuit) and the power transferred from the pump field (the right-hand loop in the RLC circuit). These two halves of the system try to drive the system and compete to the point of canceling out at zero detuning.
| {
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Is a photon really massless? If a photon travels at a speed of light and its massless then it must have no energy but this is not the case as we see in photo electric effect. Also help me to know what are photons made of, how are they created?.
| When people claim that a photon is massless, they mean that a photon has zero rest mass. In special relativity, the formula for the energy of a particle with mass $m$ possessing a momentum $p$ is
$$ E = \sqrt{p^2c^2 + m^2c^4}$$
If we set $m = 0$ for a photon, we'll end up with
$$E = pc$$
Here the momentum of a photon is described by quantum mechanics to be
$$p = \frac{h}{\lambda} $$
with $\lambda$ being the wavelength of light.
So it can have an energy, while still being massless.
| {
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Meaning of Einstein's equation $E=mc^2$? Meaning of Einstein's equation $E=mc^2$? How can a $1\,\mathrm g$ mass possess energy equal to $9\times10^{13}\,\mathrm J$? What does it actually mean?
| The rest energy of mass...is really a vibratory energy, caused by gravitons colliding with the protons and nuetrons mass, as the gravitons flow thru them at the speed of light, in a vibratroy fashion. The total energy of gravitons contained in the mass, is transferred to the mass (proton), for example. I can send you a paper that explains this. Ed Wilke
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Laplace equation vs Gauss theorem Two concentric conducting spherical shells of radii $a_1$ and $a_2$ ($a_2>a_1$), are charged to potentials $φ_1$ and $φ_2$, respectively. Determine the electric potential and field in the region between the shells. Also determine the charge on the inner shell..?
Why we need Laplace's equation to solve this problem?
Why I can't just solve it by Gauss law, as the charge on first shell will be $4π
ε oa_1φ_1$ (as potential $φ1=q/4π ε oa_1$ and also since the shells are conducting, charges will distribute on the surface of the shells uniformly, so there will be spherical symmetry)?
| Consider that the electrostatic potential $\varphi$ isn't directly observable. The potential $\varphi+C$ where $C$ is a constant gives the same electric field, and so the same physics. Because changing $\varphi$ by a constant should give the same physics, you cannot conclude that the charge on the shell is $4\pi\epsilon_0 a_1 \varphi_1$. The charge is a physical, observable quantity, but the value of the potential at a point is not.
To fix this constant you need some boundary condition, for example that $\varphi$ is $0$ at infinity. This boundary condition is the one used when we say that the potential from a charge is proportional to $q/r$. But actually, any $\varphi$ of the form $$\varphi = \frac{q}{4\pi \epsilon_0 r} + C \tag{1}$$
will do. This form of the potential can be found by finding the electric field with Gauss's law and integrating, remembering to add the integration constant. With the potential in the from (1) your boundary conditions are \begin{align}
\varphi_1 = \frac{q}{4\pi\epsilon_0 a_1} + C \\
\varphi_2 = \frac{q}{4\pi\epsilon_0 a_2} + C \\
\end{align}
Here the unknowns are the charge $q$ and the integration constant $C$, and the two equations for two unknowns can be solved.
You can derive an equivalent system of equations by applying Gauss's law, more similar to your original suggestion. By the usual argument of spherical symmetry, the electric field must be $\mathbf E = \frac{q}{4\pi \epsilon_0 r^2}\hat r$. Integrating along any path from $r = a_1$ to $r = a_2$, we have $$\int_{a_1}^{a_2} \mathbf E \cdot d\mathbf r = \varphi_1 - \varphi_2 = \frac{q}{4\pi\epsilon_0}(-\frac{1}{a_2} + \frac{1}{a_1})$$
which is an equation for $q$. Using that the general potential is of the form (1), we obtain also the constant $C$.
| {
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Structure of white light? White light is a mixture of different wavelengths.
If so what will be the structure of a beam of white light ? Is there a separation between different colours ? what does it actually mean ?
Does a beam of white light shows any spacing between different wavelengths ?
| The general idea is that if you had empty space, with only visible white light moving through it, and you put a line of electrons a few thousand nanometers long, they would start accelerating up and down, maybe in a pattern like this:
where each tick of the horizontal axis is something like hundreds of nanometers and the vertical axis is small.
This looks like it has no rhyme or reason to it, but it is actually the sum of a lot of different things. I take the function for red light (say it's of the shape $\cos(x+\varphi_1)$ and it propagates to the right at the speed of light), and I add it to the functions of higher and higher frequencies ($\cos(2x+\varphi_2)$ and $\cos(3 x+\varphi_3)$ and so on). The variable $\varphi$ are just arbitrary constants.
In an idealized detector, even though you've summed them, an individual frequency will activate an individual sensor. It doesn't matter what else you add into the function - if it's not precisely the frequency you want, the sensor won't get activated.
So, when all of your sensors are being activated, you can be sure you're looking at a sum of these ideal frequencies. And that's what white light is like.
(I'm trying very hard here to NOT write about Fourier transforms, and ignore difficulties with human rods and cones, and ignore the difficulties of "pure frequencies" when the source of light has finite length in time, and also to ignore the difficulty of what the spectrum of white light "actually" looks like and accurately reproduce how much the frequencies taper off!)
(The graphic was generated with the mathematica code Plot[Sum[Cos[n t + RandomReal[2 Pi]], {n, 1, 40}], {t, 0, 2 Pi}, Evaluated -> True])
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Is it possible to create a parachute large enough to stop all velocity? This idea came to me while playing Kerbal Space Program. I noticed that the larger my parachute was, the slower my rocket would fall back down to Kerbin. I would like to know if it is possible to create a parachute so large in the real world that it might stop all velocity, essentially making whatever is attached to it float in mid-air. Common sense is telling me "no," but I could always be wrong, and I would love some explanation behind whether or not it is possible.
| It would be possible in theory, but only in a very side-thinking way: if you make a parachute so large it encapsulates the whole Earth, it will in effect act as a balloon and not fall down, due to the internal pressure of the atmosphere.
This wouldn't work in practice for obvious reasons, but maybe in Kerbal you might be able to do something like it..
| {
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Does the conductivity of a wire in a vacuum decrease over time? Does the conductivity of a wire in a vacuum decrease over time, say over the period of years or decades? In other words: Does current degrade a wire, making it less conductive? If so, by how much, and why does this occur? Does it have something to do with the 2nd Law of Thermodynamics?
(I'm not looking interested in mechanical effects of currents on wire degradation, but thermodynamic effects.)
| The line itself does not change much over years. What changes and therefore needs maintenance on power transmission lines is insulators, connectors and spacers. Insulators get dirty or simply break, connectors work loose due to thermal expansion and contraction, mechanical stresses and oxidation, and spacers can be damaged by wear due to these same physical forces.
Edit due to change in question:
The original question asked about degradation of power transmission lines over time. The edited version asks a completely different question about conductivity over time in a vacuum. This edit adds a an answer to that new version of the question:
In short, physically, yes, there is an effect on the conductor over time. It's called electromigration and is due to the migration of ions within the conductor. This effect is noticeable at micron-scale and below, (e.g. within integrated circuits), however with larger conductors electromigration effects are largely irrelevant within meaningful human timespans.
| {
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How do radio waves reach receivers without being canceled out by interference?
When I think of waves traveling through a medium, I tend to think of the double slit experiment or waves in a pond. In those cases, waves are canceled out by destructive or constructive interference.
I'm curious as to how the EM wave can be sent out through an antenna without being canceled out by other EM waves from nearby antennas (or attenuated when it travels through some attenuating medium)? How is it possible that I am able to hear the sounds from the radio knowing that it was transmitted by a radio tower miles and miles away?
Or rather, if the magnitude of the EM wave is not of concern, then what exactly is the information which is sent and what information is received and decoded by the receiving antenna?
| What makes you think they don't?
When I was a wee lad, my physics teacher showed a video to me of exactly that. A man was driving down a road listening to a radio station (BBC if memory serves). At regular intervals, the music would get quieter and cut out.
The issue was that the road in question was nearly along the line between two radio transmitters, which setup a standing wave between them.
This is typically a rare occurrence, as governments have setup agencies to govern usage of radio frequencies, in hopes that this wouldn't happen.
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Will a person experience micro or zero gravity inside an aeroplane in free fall Let us say that a person is inside a stationary aeroplane being held by a stationary helicopter in the sky. Now, the cable which supports the aeroplane breaks. Thus the aeroplane is now not bound by anything.
1) Will the person inside the aeroplane experience micro or zero gravity when the aeroplane starts falling?
2) Will he, at any point in time, be pushed to the ceiling of the aeroplane?
3) What will be the answer to the above questions, in case the above scenario happened in vacuum?
| 1) will feel zero gravity. The confusion could be that some (including the NASA website) call microgravity to free fall apparent gravity, others to case where gravity is very small, such as in an asteroid.
2) it depends, if it is on the seat he will now only feel the force of the cushion trying to gain its original shape. That could give him a little push up, but not too strong.
| {
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Is the world we are living in discretized? I do not know how to use professional words to ask my question, so I will try to use a layman language. Please bear with me for a moment.
A ROUGH GUESS
The world our eyes are seeing every moment is a picture reflected in our eyes. I guess our eyes are like cameras, that are taking pictures "continually". I suppose there is a frequency in this picture taking. Say it is 1/10000s, the time it takes a picture, let's assume it is negligible. Something like that.
THE QUESTION
My question is, if we take a picture at 0/10000s, 1/10000s, 2/10000s, etc. How do I know that between the time 1/10000s and 2/10000s, the world exists?
So now:
*
*If my guess is wrong, then what is the real picture? What is
happening in reality?
*If my guess is correct, how do we know the
world exists continually? How would you use experimental methods to
prove it? Or there might be theoretically, many worlds in our time gaps coexisting with ours?
EDIT
I feel I still have not got a satisfied answer for my first question. Could anyone explain to me: is our vision equipment (i.e., our eyes) functioning continually or discretely?
| Fortunately for experiments in physics we have better proxies than the accuracies of our five senses. We have detectors and computers and ....
With these tools a theory of how the universe is made has been developed, from elementary particles with the theory of quantum mechanics building up the observables around us, to the astrophysical models that fit the observations with newtonian mechanics and general relativity.
The theories have been validated over and over again with experiments. Discretization of space time would particularly involve violations of Lorenz invariance, and this has not been seen. Within the limits of experimental measurements discreteness of space time is no go as a proposition, though there are important people working similar lines , as G 't Hooft.
| {
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Why is the energy of particles in accelerators much higher than the energy of the particles they are trying to find? I have been wondering. In the LHC, or other particle accelerators for that matter, they are colliding particles with energies above TeV. The LHC is going to be 14 TeV or something like that the next time they are turning it on, right ?
So, when they were searching for the Higgs, it turned out it had a mass around 125 GeV, if I'm not mistaken. So basically my question is: Why is so high colliding energy needed to detect masses that are way below that ?
The only thing I've heard of is, that the particle might have the TeV energies, but the quarks inside does not, do to something I can't remember. But is that really it, or is there a better reason I can't seem to figure out ?
| The proton is not fundamental. It is made up of quarks and gluons. It is these constituents that are colliding in the LHC to produce, in your example, a Higgs boson. The quarks and gluons only carry a fraction of the energy of the proton.
In addition, the colliding gluons or quarks in general do not have the same momentum. Therefore some of the energy will be "wasted" boosting the produced particles. Finally, sometimes the particle (eg Higgs) will be produced in association with other particles which will carry some of the energy of the interaction.
| {
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Are there eight or four independent solutions of the Dirac equation? I edited the question as a result of the discussion in the comments. Originally my question was how to interpret the four discarded solutions. Now I'm making a step back and hope that someone can clarify in what sense it is sensible to discard four of the eight original solutions of the Dirac equation.
From making the ansatz ${\mathrm{e}}^{+ipx}$ and ${\mathrm{e}}^{-ipx}$, with $E=\pm \sqrt{ (\vec p)^2 +m^2} $ we get eight solutions of the Dirac equation. $u_1, u_2, u_3 , u_4$ and $v_1,v_2,v_3,v_4$.
Conventionally the four solutions ($u_3 , u_4,v_3,v_4$.) following from $E=- \sqrt{ (\vec p)^2 +m^2}$ are said to be linearly dependent of the remaining four solutions with $E=+\sqrt{ (\vec p)^2 +m^2}$ two ($u_1,u_2$) are commonly interpreted as particle and two ($v_1,v_2$) as antiparticle solutions.
Nevertheless, in order to be able to construct chirality eigenstates we need the other four solutions and I'm unsure in how far we can then say that four of the eight solutions are really linearly dependent. A chiral eigenstate must always be of the form $ \psi_L= \begin{pmatrix} f \\ -f \end{pmatrix} $ for some two component object $f$. In order to construct such an object we need all eight solutions. For example $\psi_L= u_1 - u_3$, as can be seen from the explicit form of the solutions recited below.
In addition, I'm unable to see that the eight solutions are really linearly dependent, because for me this means that we can find numbers $a,b,c,d,e,f,g,h \neq 0$, such that $a u_1 + b u_2 + c u_3 +d u_4 + e v_1 + f v_2 + g v_3 + h v_4=0$. As pointed out in the comments, this can be done, but only for one point in time. Is this really enough? In what sense is then for example the basis used in the Fourier expansion $\sum_n (a_n e^{in x} + b_n e^{-in x}) $ linearly independent? With the same reasoning we could find numbers for one $x$ to show that all these $e^{in x}$ and $e^{-in x}$ are linearly dependent...
The explicit solutions
This is derived for example here
Two solutions follow from the ansatz ${\mathrm{e}}^{-ipx}$ with $E=+ \sqrt{ (\vec p)^2 +m^2}$ and two with $E=- \sqrt{ (\vec p)^2 +m^2}$ .
In the rest frame the solutions are
$$ E=+ \sqrt{ (\vec p)^2 +m^2} \rightarrow u_1 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad u_2 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} $$
$$ E=- \sqrt{ (\vec p)^2 +m^2} \rightarrow u_3 = \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad u_4 = \begin{pmatrix} 0 \\ 0 \\0 \\ 1 \end{pmatrix} {\mathrm{e}}^{-imt} $$
Analogous four solutions from the ansatz ${\mathrm{e}}^{+ipx}$, we get four solutions.
$$ E=+ \sqrt{ (\vec p)^2 +m^2} \rightarrow v_1 = \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} \qquad v_2 = \begin{pmatrix} 0 \\0 \\0 \\ 1 \end{pmatrix} {\mathrm{e}}^{imt} $$
$$ E=- \sqrt{ (\vec p)^2 +m^2} \rightarrow v_3 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} \qquad v_4 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} $$
Examples for chiral eigenstate are, with some two component object $f$
$$\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_1 - u_3 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} - \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad \text{ or } \qquad \psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_2 -u_4 $$
$$\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = v_1 - v_3 \qquad \text{ or } \qquad \psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_2 - u_4 $$
And similar for $\Psi_R = \begin{pmatrix} f \\ f \end{pmatrix}$.
Are four of the eight solutions really dependent? If yes, how can this be shown explicitly ? Any source, book, pdf would be awesome. Is it possible to interpret the solutions $(u_3,u_4,v_3,v_4)$ that can be discarded for many applications, but that are needed in order to create chirality eigenstates?
| OP's solutions $u_3$, $u_4$, $v_3$, $v_4$ don't solve the Dirac equation. For example, let's try $u_3$:
$$
\begin{align}
0&=(i \gamma^\mu\partial_\mu -m) \psi, \qquad\text{with}\enspace \psi= u_3 = (0,0,1,0)^T e^{-imt}\\
&=\big[i\big(\gamma^0\frac{\partial}{\partial t}+\vec\gamma\cdot\nabla
\big)-m\big]u_3\\
&=[i\gamma^0(-i m)-m]u_3\\
&=\textstyle\Big[\pmatrix{m\\&m\\&&-m\\&&&-m}-\pmatrix{m\\&m\\&&m\\&&&m}\Big]\pmatrix{0\\0\\1\\0}e^{-imt}\\
&=\pmatrix{0\\&0\\&&-2m\\&&&-2m}\pmatrix{0\\0\\1\\0}e^{-imt}\qquad\text{(false)}
\end{align}
$$
It is because these solutions don't solve the Dirac equation that there are only four degrees of freedom. It is said that the Dirac equation projects out four physical solutions out of a possible total of eight degrees of freedom.
| {
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What is "trivial" about the trivial topological superconducting phase? Once more I am stuck on my favorite word: "trivial". I am reading a bunch of stuff about topological superconductors at the moment and people keep talking about having to distinguish between the topologically "trivial" and "nontrivial" phases.
To have the easiest possible concrete example, consider the one dimensional p-wave superconductor:
$H=H_{metal} + H_{pairing}\\
H_{metal}= \sum_p c_p^\dagger\left(\frac{p^2}{2m}-\mu\right)c_p\\
H_{pairing}= p\frac{1}{2}\left(\Delta c_p^\dagger c_{-p}^\dagger + \Delta^* c_{-p}c_{p} \right)$
which has the following energy spectrum
$E_{\pm}=\pm \sqrt{\left(\frac{p^2}{2m}-\mu\right)^2+|\Delta|^2p^2}$
The book I am reading (Topological Insulators and Topological Superconductors, by
B. Andrei Bernevig and Taylor L. Hughes), refers to the $\mu<0, \; \Delta=0$ case as the trivial insulating limit (page 198, bottom).
I think I do understand why the $\mu<0$ and $\mu>0$ are topologically distinct. However I can't deduce why the $\mu<0, \; \Delta=0$ case
*
*is said to be trivial. What is the trivial referring to?
*is the insulating limit. I thought $H_{metal}$ describes a 1D metal of spinless fermions.
I am most grateful for any clarifying answers.
| "Trivial" is used because the equation reduces to something a bit simpler. Take the equation
$E_{\pm}=\pm \sqrt{\left(\frac{p^2}{2m}-\mu\right)^2+|\Delta|^2p^2}$
Setting $\Delta$ to $0$ gets rid of the last term, and making $\mu<0$ means that the first term will reduce to
$$E_1=\pm \left(\frac{p^2}{2m}-\mu\right)$$
which is pretty simple. That's what triviality boils down to. From Wikipedia,
In mathematics, the adjective trivial is frequently used for objects (for example, groups or topological spaces) that have a very simple structure.
That's certainly the case here.
I can't answer your second question, because I'm not too familiar with the subject.
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What is the effect of water 'climbing' over a cup via a wet string? I have noticed (with tea bags, etc) that if you get the whole string wet, and you have one end inside the cup and the other end below the level of the liquid (outside), the water drains down the string outside the cup at a slow pace... Seeming to defy gravity and move its way up the string until it gets to the cup edge...
What is this effect called, and do we observe it in nature at a larger scale somewhere?
| This effect is called Capillary Action. Yes we do in fact observe it in nature in a large scale: How do you think plants are able to "suck up"1 water through its roots and send it to the leaves? One of the major forces responsible for it is capillary action.
Here, have a quote from the article mentioned above:
Wicking is the absorption of a liquid by a material in the manner of a candle wick. Paper towels absorb liquid through capillary action, allowing a fluid to be transferred from a surface to the towel. The small pores of a sponge act as small capillaries, causing it to absorb a large amount of fluid. Some textile fabrics are said to use capillary action to "wick" sweat away from the skin. These are often referred to as wicking fabrics, after the capillary properties of candle and lamp wicks.
This is an interesting phenomenon as it relies on the tendency of adhesion of water, i.e. water molecules tend to cling to other substances or surfaces, and its surface tension. The narrower the tube or capillary is (like the ultra-fine tubes of Xylem vessels or the gaps in the thread), the more surface area the water gets to cling to and pull itself up, defying gravity.
1Plants can't really "suck up" water, they provide the pathway for the water to pull itself up.
| {
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Triple integral $\iiint_{\mathbb{R}^3} d^{3}q ~\delta^{3}(\vec{q})\frac{(\vec{p}\cdot\vec{q})^2}{q^{2}} $ involving Dirac Delta function
I am trying find $$\iiint_{\mathbb{R}^3} d^{3}q ~\delta^{3}(\vec{q})\frac{(\vec{p}\cdot\vec{q})^2}{q^{2}},$$ where $\vec{p}$ is some fixed vector.
The answer should be $\frac{p^2}{3}$. Below is my attempt, which seems to lead to the wrong answer $\frac{p^2}{2}$.
Attempt: Let's align $q_{z}$ with $\vec{p}$, so we measure $\theta$ wrt $\vec{p}$. Since there is no $\phi$ dependence so I can write $$\delta^{3}(\vec{q})=\frac{\delta(q)\delta(\theta)}{2\pi q^{2}\sin(\theta)}.$$
Therefore I have
$$p^{2}\int_{0}^{\infty} dq \delta(q)\hspace{1mm}\int_{-\pi}^{\pi}d\theta\hspace{1mm} \delta(\theta)\cos^2\theta .$$
I understand $$\int_{0}^{\infty}\delta(q)dq = \frac{1}{2},$$ if we treat $\delta(q)$ as a limiting case of a symmetric Gaussian distribution. While the $\theta$ integral is $1$. So my answer to my question is $\frac{p^2}{2}$. Which is different from the correct answer $\frac{p^2}{3}$.
So my questions are:
*
*What went wrong in my derivation?
*How do you derive and justify the answer $\frac{p^2}{3}$ from first principles?
| Hints:
*
*In mathematics, a distribution is usually only defined wrt. smooth testfunctions. However the function ${\bf q}\mapsto({\bf q}\cdot{\bf p})^2/q^2$ is not continuous at the origin ${\bf q}={\bf 0}$. Nevertheless, we can e.g. try to evaluate the triple integral using the following representation of the 3D Dirac delta distribution
$$\tag{1} \delta^3({\bf q})~=~ \lim_{\varepsilon\to 0^+} \frac{1}{4\pi} \frac{3\varepsilon}{(q^2+\varepsilon)^{\frac{5}{2}}}, \qquad q~:=~|{\bf q}|,$$
where it is implicitly understood that the limit $\lim_{\varepsilon\to 0^+}$ should be taken after the triple integration.
*For given $\varepsilon>0$, the integrand is integrable on $\mathbb{R}^3$. And it is bounded at the origin ${\bf q}={\bf 0}$, so we can use spherical coordinates. As OP mentions, in spherical coordinates with ${\bf p}$ along the $z$-axis, we have
$$\tag{2}\frac{({\bf q}\cdot{\bf p})^2}{q^2}~=~p^2\cos^2\theta.$$
*Substitute $q\to \sqrt{\varepsilon}q$ in the triple integral. The $\varepsilon$-dependence disappears. Perform the triple integral.
| {
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If direction of torque is upwards(or downwards), why does the body rotate perpendicular to the direction? We know that torque is given by $$\vec{\tau} = \vec{r} \times \vec{F}.$$ Its direction is given by the right-hand rule which says that torque acts perpendicular to the plane where force applied and position vector do exist. But if the direction of the torque is upward, why does the body rotate in the plane perpendicular to the direction?
For clarification, if one applies a force $\vec{F}$ on the door at a point which is $\vec{r}$ away from the hinge, the torque $\vec{\tau}$ is given by the above relation. But in which direction does it act? It acts perpendicular to the force i.e. along the length of the door. But why does the door rotate perpendicular to the direction? What does the direction of the torque imply if the body doesn't move in that direction?
|
Now, torque τ⃗ is given by the above relation. But in which
direction does it act?
Torque, like angular momentum is a pseudovector and not a vector. It is a conventional way of showing the direction (anti/clock -wise) ot the rotation. It is devised in such a way that you can apply the right-hand rule.
| {
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Why aren't there spherical galaxies? According to the Wikipedia page on Galaxy Types, there are four main kinds of galaxies:
*
*Spirals - as the name implies, these look like huge spinning spirals with curved "arms" branching out
*Ellipticals - look like a big disk of stars and other matter
*Lenticulars - those that are somewhere in between the above two
*Irregulars - galaxies that lack any sort of defined shape or form; pretty much everything else
Now, from what I can tell, these all appear to be 2D, that is, each galaxy's shape appears to be confined within some sort of invisible plane. But why couldn't a galaxy take a more 3D form?
So why aren't there spherical galaxies (ie: the stars and other objects are distributed within a 3D sphere, more or less even across all axes)? Or if there are, why aren't they more common?
| It is due to the combined effect of rotation and "dissipation". A rotating cloud of gas consists of particles which interact strongly with each other (colliding physically) on relatively short timescales can radiate away some of their energy and momentum by emitting photons. For both of these reasons, a dense cloud of rotating gas will collapse to form a rotating disk. But there are some star systems that do remain pretty spherical, they called globular clusters.
On the other hand, if the gas in a cloud forms stars very quickly, so that the particles in it are stars rather than atoms, then these stellar "particles" do not interact strongly on short timescales (for instance the time between direct collisions for a star in a globular cluster is > $10^{10}$ years, and globular clusters are pretty much spherical)
can not radiate away their energy and momentum by emitting photons; they can emit gravitational radiation, but that's not as effective
For these reasons, a spherical cluster of stars will remain spherical for very long periods of time; much longer than the current age of the universe.
| {
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Fluid dynamics - immiscible liquids Question:
A U-tube of uniform cross-section is partially filled with a liquid I. Another liquid-II, which does not mix with liquid I is poured into one side. It is found that the liquid levels of the two sides of the tube are the same while the level of liquid-I has risen by 2cm. If the specific gravity of liquid-I is 1.1, what will be the specific gravity of liquid-II?
My logic:
Shouldn't the specific gravity be the same? Otherwise the two levels would never become the same. There is the same height of liquid 2 as the height that liquid 1 has risen by. So both will exert the same pressure. But obviously my logic is flawed as the answer is 1.12
| The specific gravities would be the same if the levels of the two side were the same after liquid-II was added.
When I try this, my logic seems to be flawed too. I don't get 1.12.
The level on side II has not changed. The level on side I has risen 2 cm. So 2 cm of liquid-II were added.
Consider the horizontal plane 2 cm below the top of side II. Below this level on both sides there is liquid-I
Above this level on side 2, there is 2 cm of liquid-II. Above this on side I, there is 4 cm of liquid-1. That should tell you the answer.
| {
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All geodesics are inextendable? I think the title is true, because geodesics has a tangent vector with a constant length parametrized by an affine parameter.
Probably, it is easier to think about timelike or spacelike geodesics. In this case, its affine parameter measures the length of the curve. It is difficult to image such kind of curve has a future endpoint (or past endpoint).
Is it correct?
| The property you are referring to is called geodesic completeness. It is an important concept in the study of singularities in general relativity. There are somewhat trivial examples of geodesic incompleteness where you are just "missing" part of the spacetime, i.e. you could have Minkowski space with a point removed. In these cases you usually consider extending the spacetime ("isometrically embedding" it in a bigger one, without the "hole").
The real interesting thing is when your spacetime has singularities. For example, the maximally extended Schwarzschild black hole spacetime has a singularity at $r=0$, and geodesics run into it within finite affine parameter. These geodesics cannot be extended.
The proof of the singularity theorems of Penrose and Hawking basically amounts to proving that a spacetime is geodesically incomplete. Hence, geodesic incompleteness can be taken as the definition of a spacetime with a singularity.
| {
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Why do the high frequency waves have the most number of modes? While reading the Wikipedia page of Ultraviolet Catastrophe, I came across how Rayleigh and Jeans applied the equipartition theorem. They told that each mode must have same energy. Now as the number of modes are greatest in small wavelengths or large frequency, energy radiated will be infinite.
What is mode actually? And why do the large frequencies have the most modes? Please help me explaining these.
I need a math-free explanation.
| The "modes" in this case refer to the standing waves that can exist in a cavity. A very nice diagram / explanation is given at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html
To summarize: if you consider a cavity of dimension $L$, the modes that fit inside the cavity have wave numbers $n_1$, $n_2$, $n_3$ such that
$$n_1^2 + n_2^2 + n_3^3 = \frac{4L^2}{\lambda^2}$$
It then goes on to show that the number of these modes (represented as dots on the surface of a sphere) increases as the radius of the sphere increases (which it does with the inverse of the wavelength).
See the link for all the details.
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Why are unilateral Laplace transforms suitable for causal systems and bilateral aren't? https://en.wikipedia.org/wiki/Two-sided_Laplace_transform#Causality
The above section says that bilateral transforms will not necessarily make sense for causal systems. In the course of advanced engineering mathematics our instructor told us that's because the unilateral version is integrated over time $t=0$ to $t= ∞$.
But that's more like a memory aid than a reason. There's no reason why, when it comes to real systems, I can integrate from time $t=0$ to $t= ∞$, but can't for time $t=-∞$ to $t= ∞$. The relation this has to causality isn't obvious. I was not able to find a proof or a more solid grounding for this using Google.
| Usually, what you are taking the Laplace transform of, is the system response, $h(t)$, e.g. system's response to a delta input, $\delta(t)$. If system's delta response is nonzero at $t<0$ it would mean the system had been anticipating your delta input at $t=0$ and already started responding, which is non-causal. Therefore a causal system must have $h(t)$ such that $h(t)=0$ for $t<0$. If you are using a unilateral Laplace transform it automatically makes sure that you are at least ignoring the non-causal part ($t<0$) of $h(t)$. So really, you should make sure that you have modeled your system correctly, so that it is causal at the first place and not rely on the kind of Laplace transform you are using. For example, $h(t)=\exp(-at)u(t)$ is a causal response, where $u(t)$ is Heaviside's step function, but $h(t)=\exp(-at)u(t+1)$ is not.
| {
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Water behaviour under theoretical near-infinite pressure conditions I've asked a similar question here but the answer given shows the behaviour of water under general conditions.
I'd like to know what the behaviour of water is like as pressures increase towards infinity without being able to escape it's confinement.. i.e. a ball of water at the core of a galactic mass.. maybe this question is more for theoretical physics since we can't really measure or experiment?
| Yes, the question is theoretical and so the response. Under enough pressure water will become a solid, regardless of temperature. That is, as far as it is still water. If pressure is high enough, the atoms will collapse and form neutron-degenerate matter (theorized to exist in the cores of neutron stars). I am not sure if there could be an intermediate mixed phase in between water and "neutronium" in which only one of the atoms collapese first (either H or O) and the other at a larger pressure.
| {
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Does the lagrangian contain all the information about the representations of the fields in QFT? Given the Lagrangian density of a theory, are the representations on which the various fields transform uniquely determined?
For example, given the Lagrangian for a real scalar field
$$ \mathscr{L} = \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi - \frac{1}{2} m^2 \varphi^2 \tag{1}$$
with $(+,-,-,-)$ Minkowski sign convention,
is $\varphi$ somehow constrained to be a scalar, by the sole fact that it appears in this particular form in the Lagrangian?
As another example: consider the Lagrangian
$$ \mathscr{L}_{1} = -\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu + \frac{1}{2} m^2 A_\mu A^\mu,\tag{2}$$
which can also be cast in the form
$$ \mathscr{L}_{1} = \left( \frac{1}{2} \partial_\mu A^i \partial^\mu A^i - \frac{1}{2} m^2 A^i A^i \right) - \left( \frac{1}{2} \partial_\mu A^0 \partial^\mu A^0 - \frac{1}{2} m^2 A^0 A^0 \right). \tag{3}$$
I've heard$^{[1]}$ that this is the Lagrangian for four massive scalar fields and not that for a massive spin-1 field. Why is that? I understand that it produces a Klein-Gordon equation for each component of the field:
$$ ( \square + m^2 ) A^\mu = 0, \tag{4}$$
but why does this prevent me from considering $A^\mu$ a spin-1 massive field?
[1]: From Matthew D. Schwartz's Quantum Field Theory and the Standard Model, p.114:
A natural guess for the Lagrangian for a massive spin-1 field is
$$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,\tag{8.17}$$
where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are
$$ ( \square + m^2) A_\mu = 0,\tag{8.18}$$
which has four propagating modes. In fact, this Lagrangian is not the Lagrangian for a amassive spin-1 field, but the Lagrangian for four massive scalar fields, $A_0, A_1, A_2$ and $A_3$. That is, we have reduced $4 = 1 \oplus 1 \oplus 1 \oplus 1$, which is not what we wanted.
| Field $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ with a given spin and mass (i.e. field which transforms under irrep of the Poincare group) must satisfy some determined conditions called irreducibility conditions:
$$
\tag 1 \hat{W}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = -m^{2}\frac{n + m}{2}\left(\frac{n + m}{2} + 1\right)\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}},
$$
$$
\tag 2 \hat{P}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = m^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}.
$$
Here $\hat{W}$ is Pauli-Lubanski operator and $\hat{P}$ is translation operator. Representation with equal quantity $\frac{n + m}{2}$ are equivalent.
As for details, look here for the information about indices and Lorentz representations and here for fields as representations of Poincare group.
If you construct lagrangian which leads to $(1), (2)$, you will uniquely determine transformation properties of field with a given mass and spin.
| {
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How can we show that pressure is exerted sideways too? We can show that pressure upward and downward in a fluid is caused by weight of fluid column or volume. A simple derivation of this:
$$\text{Pressure}=\frac{\text{Force}}{\text{Area}}=\frac{\text{Weight of the fluid column}}{\text{Area}}$$
$$~~~~~=\frac{\text{mg}}{\text{A}}=\frac{\rho Vg}{\text{A}}=\frac{\rho g\times Ah}{\text{A}}=\rho g h$$
I am able to show that pressure in a fluid at a height is $\rho gh$ upwards and downwards, but how can I show that it acts sideways too?
I found a pretty reasonable answer here. So is this the correct reason?
Fluids are made of a large number of very small particles, much too
small to see. These particles are in constant, rapid motion. They bump
into one another. They bump into the walls of any container that holds
them. They bump into objects in the fluid.
As the particles of a fluid bump into an object in the fluid, they
apply forces to the object. The forces, acting over the object’s surface,
exert pressure on the object. When the pressure in a fluid increases, the
particles bump together more frequently. This increases the pressure
on objects in the fluid.
The pressure a fluid exerts on an object in the fluid is applied in
all directions. That is because the particles that make up the fluid can
move in any direction. These particles exert forces as they bump into
objects in the fluid.
| It is not true that the pression of a fluid is due to gravity. Do you think fluids inside a bottle in the space station have no preassure? The right answer is the one you have quoted. The first link only refers to buoyancy, that does require gravity.
| {
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Best way to heat something in aluminum foil? Let's say we have a wet piece of paper, wrapped in aluminum foil, that we need to heat up in the fastest and most energy efficient way possible (no flamethrower).
What would that be?
Details regarding the methods would be highly appreciated.
| The microwave answer given above is good, especially if you have only one paper wrapped in foil because it would transfer a large fraction of the energy produced to the sample. If you have many of these (for example as a step on an assembly line) then immerse it in a hot medium. This would provide really efficient transfer of heat energy for each sample but at higher overhead. Since you are targeting 250 C, if you want to use water then it has to be in vapor form. You can heat the vapor with microwave energy and improve your energy transfer efficiency over a single unit in dry air. Also, at 250 C the water inside the foil will pressurize the foil if it is sealed, and explode it unless it can hold about 40 atmospheres. If unsealed then it will vaporize and you will have your sample immersed in a vapor anyway. :-)
| {
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Why does the light reflecting off of ocean water sometimes appear 'smoother'? Looking out the window at some water in the Harbour - I noticed that some parts of the water appear 'smoother' than others.
My question is: Why does the light reflecting off of ocean water sometimes appear 'smoother'?
| Two potential reasons among others: the wind is not uniform, if it blows larger on one sector it will move the water surface stronger. Second, different smal surface waves that come from different directions can interfere either negatively or positively, depending on their relative phases, over some small region, which again wouls result in inhomoheneities in the observed disturbances
| {
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Gibbs free energy intuition What is Gibbs free energy? As my book explains:
Gibbs energy is the energy of a system available for work.
So, what does it want to tell? Why is it free? Energy means ability to do work. What is so special about this energy? Can anyone simply explain?
I just want a math-free intuition.
| First, you have system with some energy, named $U$ by physicists. You think you have all the information you need to characterize the system but then some guy comes near and says:
"Whoa, that's bad, the volume of your system can change."
You say:
"No problem, we just add here $pV$. Our new energy is $H=U+pV$."
"But hey," they say, "your temperature can change by external heat, you have to count that also."
"No problem, we subtract $TS$ from our energy, and rename it. $G=U+pV-TS$"
There you are.
People name it "available" energy, because if your system increases in size, it creates work: $V$ increases, so $U$ decreases. But energy is always bounded from below. When you drain everything you can, your system goes to $G=0$.
You cannot drain all $U$ because you would need an external system with $T=0$.
| {
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Must Matter Particles Have A Hard Edge? It's my understanding that electrons are particles, and it's also my understanding that their location while orbiting an atom cannot be determined precisely and must be determined by statistics and probability, almost like electrons can be in multiple places at the same time. That made me think, hmm, could electrons exist more as smears instead of hard-edged particles? A smear can be in more than one place at a time. The only difference is a smear doesn't have a hard edge like a spherical particle would. It would sort of "blend" matter and space. I'm also wondering if perhaps smears would demonstrate wavelike properties that hard-edged particles can't. Is there any knowledge out there that states that matter particles either must have a hard edge or can't have a hard edge?
| This is one of the key results of quantum field theory: particles are not single points, they are disturbances in quantum fields that are spread out over space. Typically the disturbance is not spread out very much, otherwise it looks more like what we know as a wave than a particle. The technical term for what you're calling a "smear" is a wavepacket.
| {
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How did Rutherford's gold foil disprove the plum pudding model? What stops one of the two following scenarios from happening, consistent with the plum pudding model?
*
*The $\alpha$ particle, attracted by the electrons on the outer shell of the pudding, orbits nearly parabolically around the atom, causing the near-180 degree deflection angle seen.
*The $\alpha$ particle hits a plum pudding atom directly, and because the atom consists largely of positive charge, it is deflected by nearly 180 degrees.
Followup: did they know anything about the spacing of these plum-pudding atoms? Did they expect them to be lined up in a grid-like way; difficult to penetrate?
I am a physics undergraduate, I've only taken an intro QM class, it would be nice if that were kept in mind while answering.
| Based on thomson's model, all of the alpha particles should go through or reflect back but in Rutherford's experiment, it was more of in between, some reflecting back and some going through, disproving the theory.
| {
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Cooling a satellite Satellites are isolated systems, the only way for it to transfer body heat to outer space is thermal radiation. There are solar panels, so there is continuous energy flow to inner system. No airflow to transfer the accumulated heat outer space easily. What kind of cooling system are being used in satellites?
| The satellite itself can do with radiative cooling but some instruments on board, e.g., IR sensors, require temperatures as low as than 4 K for which Helium dewars are used. Bolometers require even lower temperatures (in the mK range).
A good summary is available here.
| {
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Can the physical properties of the EM field be described directly from the 4-gauge potential? I'm trying to make an argument that classically, the EM field is considered a more 'real' physical quantity than the potentials, and am tempted to say that the fact that the field carries energy & momentum is evidence of this. Now I am wondering this: Is it possible to determine the energy or momentum of the electromagnetic field 'object' from the magnetic & vector potential in a way that isn't just directly equivalent to calculating the field first?
| You can find the Hamiltonian of the potentials if you know the Lagrangian, or you can calculate the energy in reference to the work done on charges in an electric field, which the electric potential sort of comes about secondarily, because it is related to the work. The calculation is done here. As for the equivalence between the two, are you familiar with the Aharonov Bohm effect?
| {
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Time slowed by gravity If time moves more slowly on Earth (due to our proximity to a gravitational body) than for someone orbiting Earth in a spaceship, yet the opposite occurs in the frequently cited "twin paradox" of the earth-bound twin vs. the twin in the fast-moving spacecraft, then my question would be this--is the slowing of "time" the same as the slowing of molecular/atomic motion?
| You are correct, time dilation will occur for someone staying on the earth surface relatice to somebody in space. And yes, everything that can be measured by a clock, excepth the speed of light, will be perceived as slower, including molecular motions.
| {
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Introducing angular momentum for the first time to a class What is the best way to introduce the notion of angular momentum to a class without making it appear an unnecessary and artificial construction?
| Just two cents: I assume you already introduced Newton's laws, you can say that is something like "When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force"
yu can explain that from the other two Newtons laws it can be shown that the first law naturally applies also to objects that move on circular motion, not only in a stright line. Then explain that this amount of rotation implies that if the objects contract due to internal forces (no external influences), the amount of circular motion is still constant, but the angular speed needs to augment because now it has to cover a smaller circumference with the same amount of rotation. Show the example of the ballerina that speeds up as she contracts her arms.
| {
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A particular coordinate transformation of a metric tensor So, this was a problem set question for my GR class due yesterday, and I can't for the life of me solve it, it seems I am missing something very trivial. Either the given answer is wrong, or I am.
The given line element is,
$$ds^2 = \left(1 - \frac{r^2}{\alpha^2}\right)dt^2-\left(1 - \frac{r^2}{\alpha^2}\right)^{-1}dr^2-r^2d\theta^2-r^2\sin^2(\theta)d\phi^2$$
where $\alpha > 0$.
Then I am supposed to do a coordinate transformation using,
$$r = \rho\exp(2\tau/\alpha) \\ t = \tau - \frac{\alpha}{2}\ln(-\alpha^2+\rho^2\exp(2\tau/\alpha))$$
And I am supposed to get the following metric,
$$ ds^2 = d\tau^2 - \exp(2\tau/\alpha)\left(d\rho^2+\rho^2d\theta^2+\rho^2\sin^2(\theta)d\phi^2\right) $$
But I can't seem to arrive at that answer. To start off with baby steps, I tried the following:
$$g_{m'n'} = \frac{\partial x^m}{\partial x^{m'}} \frac{\partial x^n}{\partial x^{n'}} g_{mn} $$
$$g_{\tau\tau} = \left(\frac{\partial x^m}{\partial \tau}\right)^2 g_{mm} = 1 $$
$$g_{\tau\tau} = \left(\frac{\partial t}{\partial \tau}\right)^2 g_{tt} + \left(\frac{\partial r}{\partial \tau}\right)^2 g_{rr} = 1$$
This seems to be the correct coordinate transformation law for a tensor, but doesn't seem to give the right result - that is $1$.
I am sure I'll understand what I did wrong if I can't simply solve for the first component of the transformed metric. But I can't see where I went wrong.
| There appears to be a typo in the coordinate transformation for $r$: it should be
$$
r = \rho\,\exp(\tau/\alpha).
$$
With this in mind, your reasoning is correct; if you work out the partial derivatives, you'll find that $g_{\tau\tau} = 1$, and you can work out $g_{\rho\rho}$ in a similar way.
P.S. also note that $t = \tau - \frac{\alpha}{2}\ln(r^2-\alpha^2)$.
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What is the difference between electromagnet and solenoid? What is the difference between electromagnet and solenoid? Both these terms seem as the same thing to me. The only difference that I can find seems to be that an electromagnet contains a soft iron core. I'm sure there must be some other difference between the two and I hope someone can clear this matter up for me.
| A solenoid is an electromagnet but an electromagnet needn't necessarily be a solenoid.
A wire carrying electric current is an electromagnet. In fact, the very first electromagnet was a horseshoe shaped. So a suitable definition of electromagnet would be, anything which produces a magnetic field around itself when a current passes through it is called an electromagnet.
But most of the times, the electromagnets that we prepare consist of large number of coils with a core inside the coil (usually a solenoid). The only function of adding a core and increasing the number of turns is to increase the strength of the magnetic field.
Solenoid can be defined as a coil wound into a tightly packed into a helix. Both electromagnets as well as solenoids can have magnetic cores. In fact, most simple inductors are basically a solenoid with a magnetic core to enhance its inductance.
To summarize, an electromagnet is anything which has the ability to produce a magnetic field using electricity whereas a solenoid is a tightly wound coil, you can save that solenoid is a type of electromagnet.
| {
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How are band gap energy, dielectric constant (permittivity) and resistance related to each other? The following three properties are related to current flow:
*
*Band gap energy
*Dielectric constant
*Resistance
I would expect them all to have the same trend (i.e. higher band gap energy would cause higher dielectric constant and higher resistance), but this is not the case. For example:
*
*SiO2 is a very good insulator (band gap of 9eV), but its dielectric constant is very low ($\epsilon_r=3.9$), compared to many materials. Why so?
*In this ref (page 8) a consistent inverse correlation between $\epsilon$ and $\rho$ for Si is presented. Why is it inverse? I would expect them to rise together.
| Dielectric constant is based on relative capacitance to vacuum. And it is usually calculated by using static electric plates. You put two parallel plates in vacuum and measure the capacitance. Then you do the same, but put the medium you want to measure instead. Divide the new capacitance of the new medium with the capacitance of vacuum:
$ϵ = C_x/C_v$
Since capacitance is charge divided by voltage:
$C=q/V$
The charge will be the same on both setup. Though the voltage will not. It depends on the medium between the two plates. If the molecules in the medium are polarized it is easier to mediate static electric field across the medium. That means more energy is stored in each charge. If not it is less easy to mediate static electric field and less energy stored per charge. In this case the latter. So a medium that is easier to mediate electric field has a lower dielectric constant and vice versa.
| {
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What limits the doping concentration in a semiconductor? Si and Ge can be blended in any ratio, $\mathrm{Si}_x\mathrm{Ge}_{1-x},\ 0\le x\le 1$. So do
InxGa1-x.
So what exactly causes doping impurities inside Si/Ge/etc. to saturate at $\sim 10^{-19}\ \mathrm{cm^{-3}}$?
| In the chemistry of liquids and solids, some combinations are "miscible", meaning they mix in any ratio (like water + alcohol or silicon + germanium). Other combinations are not (like water + oil or phosphorus + silicon), in which case there is a certain "solubility", and if you try to put in more than the "solubility limit" it prefers to segregate into separate phases (e.g., little nano-crystal inclusions can form).
Fun fact: You can actually put in more dopants than the solubility limit! ...But only if you use tricks like ion implantation + pulsed laser melting, where the solid freezes so fast that the molecules get stuck in place and cannot segregate even though segregating is the more stable configuration.
It's not terribly surprising that silicon & germanium are miscible, because they are both happiest in the exact same crystal structure (the diamond lattice), which in turn is because they have same number of valence electrons, i.e. the same column of the periodic table. For the same reason, they don't dope each other.
Likewise, it's not terribly surprising that silicon and phosphorus are not miscible because they do not prefer the same crystal structure. Phosphorus is happier when bonded to three neighbors, whereas it has four neighbors when it is shoehorned into silicon's diamond lattice. So if enough phosphorus is in the same place, they will rearrange into a different kind of lattice they like better.
| {
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Why is Graphene So Strong? There has been a lot of news about Graphene since its discovery in 2004. And as we are all told it is a revolutionary material which is very strong, conductive and transparent.
But what is it about the structure of Graphene which makes it so strong?
| Graphene's strength mainly comes from the strong covalent bonds of the carbon atoms.
Graphite is made of layers of graphene but it is weaker because the layers making up graphite are bonded to each other through London forces hence why the layers can slide past each other and the material is soft. These weak inter-layer London forces provide a weak point in graphite's structure which doesn't exist in graphene.
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What is the origin of CMB fluctuations? I have read somewhere that CMB (cosmic microwave background radiation) fluctuations in temperature are linked to mass distribution fluctuations in the early universe (at ~350000 years after Big Bang, which is of course when the cosmic radiation was emitted), and that is used to explain the formation of large structures (galaxies, clusters of galaxies..). Why is that so?
|
The CMB (cosmic microwave background) is a snapshot of the oldest light in our Universe, imprinted on the sky when the Universe was just 380,000 years old. It shows tiny temperature fluctuations that correspond to regions of slightly different densities, representing the seeds of all future structure: the stars and galaxies of today.
The anisotropies are very small,
The cosmic microwave background radiation is an emission of uniform, black body thermal energy coming from all parts of the sky. The radiation is isotropic to roughly one part in 100,000: the root mean square variations are only 18 µK, after subtracting out a dipole anisotropy from the Doppler shift of the background radiation .
One needs a model and the dominant one is the Big Bang model
This light stopped interacting with electrons and nucleons at 380.000 years after the BB,
before gravitational attraction formed stars and galaxies. The great uniformity of temperatures, at the level of 10^-5, cannot be explained thermodynamically because there was no uniform communication over the then universe ( due to special relativity) of the particles and energy forms before the photon separation, to homogenize the soup. This uniformity forced the inflation period in the very beginning of the Big Bang model. The rapid inflation homogenized the primordial soup so that we end up with the observed CMB spectrum.
The CMB has been used to model a homogeneous early inflation period to explain the CMB observation. Once the CMB homogeneity was modeled by the inflationary period, the homogeneity in the density of cluster of galaxies and galaxies also follows.
and that is used to explain the formation of large structures (galaxies, clusters of galaxies..).
The CMB does not explain the formation of large structures, those are explained by statistical mechanics and gravitational forces as the universe cooled from the BB . It explains the homogeneity , which is still under study.
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Induced EMF of a rectangular loop should be zero? Considering the shape of a rectangular loop in a changing magnetic field:
The induced $\epsilon$ would be zero? Since a rectangular loop is a combination of wires in series to create such a shape. Each wire in this loop induces $\epsilon$ opposes the other, and they should each cancel out?
Here is the diagram adjusted with polarities:
EDIT:
Examples of induced $\epsilon$ canceling out:
A -
B -
Where there are two separate conductors that are wired in series together, each in the same magnetic field, that experience the same flux change over the same time period.
| You can only calculate the EMF induced from a close loop, instead of "a conductor" as you illustrate in your example A and B.
We know $\mathcal E = -\frac{d\Phi_B}{dt}$, where $\Phi_B=BA$.
EMF induced in a "conductor" makes no sense because we cannot find the surface $A$ to calculate the magnetic flux $\Phi_B$ through the conductor.
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Finding the energy lost due to non-conservative forces I stomped across this question and would very much appreciate any form of clarification.
A otter 75kg slides down a hill starting from rest. Hyp = 8.8, height = 6.5, final speed of otter = 9.2 m/s. And it wants me to find how much energy was lost due to non-conservative forces on the hill?
I guess my question is what the term non-conservative means in this context. I am not asking anyone to do this for me, I merely ask for some advice or even which formulas I could use to solve this. Is friction a non-conservative force? If so, do I just have to find how much work was done by friction?
| As a first estimate you can evaluate energy deficit in conversion from initial potential energy to a final kinetic energy:
$$ \Delta E_{~lost} = mgh_0 - \frac {m{v^{~2}_f}}{2} $$
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Is Parity really violated? (Even though neutrinos are massive) The weak force couples only to left-chiral fields, which is expressed mathematically by a chiral projection operator $P_L = \frac{1-\gamma_5}{2}$ in the corresponding coupling terms in the Lagrangian.
This curious fact of nature is commonly called parity violation and I'm wondering why? Does this name make sense from a modern point of view?
My question is based on the observation:
A Dirac spinor (in the chiral representation) of pure chirality transforms under parity transformations:
$$ \Psi_L = P_L \Psi = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} \rightarrow \Psi_L^P = \begin{pmatrix} 0\\ \chi_L \end{pmatrix} \neq \Psi_R$$
Chirality is a Lorentz invariant quantity and a left-chiral particle is not transformed into a right-chiral particle by parity transformations.(The transformed object lives in a different representation of the Lorentz group, where the lower Weyl spinor denotes the left-chiral part.)
In understand where the name comes from historically (see the last paragraph) but wouldn't from a modern point of view chirality violation make much more sense?
Some background:
Fermions are described by Dirac spinors, transforming according to the $(\frac{1}{2},0) \oplus (0,\frac{1}{2})$ representation of the (double cover of the) Lorentz group. Weyl spinors $\chi_L$ transforming according to the $(\frac{1}{2},0) $ representation are called left-chiral and those transforming according to the $(0,\frac{1}{2})$ representation are called right-chiral $\xi_R$. A Dirac spinor is (in the chiral representation)
$$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}$$
The effect of a parity transformation is
$$ (\frac{1}{2},0) \underbrace{\leftrightarrow}_P (0,\frac{1}{2}),$$
which means the two irreps of the Lorentz group are exchanged. (This can be seen for example by acting with a parity transformation on the generators of the Lorentz group). That means a parity transformed Dirac spinor, transforms according to the $(0,\frac{1}{2}) \oplus (\frac{1}{2},0) $ representation, which means we have
$$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^P = \begin{pmatrix} \xi_R \\ \chi_L \end{pmatrix} $$
Now we can examine the effect of a parity transformation on a state with pure chirality:
$$ \Psi_L = P_L \Psi = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} \rightarrow \Psi_L^P = \begin{pmatrix} 0\\ \chi_L \end{pmatrix}$$
This means we still have a left-chiral spinor, only written differently, after a parity transformation and not a right-chiral. Chirality is a Lorentz invariant quantity. Nevertheless, the fact that only left-chiral particles interact weakly is commonly called parity violation and I'm wondering if this is still a sensible name or only of historic significance?
Short remark on history
I know that historically neutrinos were assumed to be massless, and for massless particles helicity and chirality are the same. A parity transformation transforms a left-handed particle into a right-handed particle. In the famous Wu experiment, only left-handed neutrinos could be observed, which is were the name parity violation comes from. But does this name make sense today that we know that neutrinos have mass, and therefore chirality $\neq$ helicity.
| Okay, I think I have an idea why the terminology is used, but I think this argument makes little sense:
The Lagrangian term describing weak interactions is of the form
$$ \bar \Psi \gamma_\mu P_L \Psi W^\mu $$
Under parity transformations $ \Psi \rightarrow \gamma_0 \Psi$ and $ \bar \Psi \rightarrow \bar \Psi \gamma_0 $, therefore
$$ \bar \Psi \gamma_\mu P_L \Psi W^\mu \rightarrow \bar \Psi \gamma_0 \gamma_\mu P_L \gamma_0 \Psi W^\mu $$
We can transform the parity transformed Lagrangian, by using the explicit form of $P_L = \frac{1-\gamma_5}{2}$ and $\{\gamma_5, \gamma_\mu \}=0$ :
$$ \bar \Psi \gamma_0 \gamma_\mu P_L \gamma_0 \Psi W^\mu = \bar \Psi \gamma_0 \gamma_\mu \gamma_0 P_R \Psi W^\mu = \bar \Psi \gamma_\mu P_R \Psi W^\mu $$
which shows that the weak interaction term in the Lagrangian is not invariant under parity
transformations. I think this argument is wrong!
The above discussion misses that under parity transformations $\gamma_5 \rightarrow - \gamma_5$. If we take this in account the corresponding term in the Lagrangian is invariant und weak interactions are invariant under parity transformations.
We can see this, because for an ordinary Dirac spinor we have the projection operator:
$$P_L \Psi = \begin{pmatrix}1 & 0\\ 0 &0 \end{pmatrix} \begin{pmatrix} \chi_L \\ \xi_R\end{pmatrix} = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} = \Psi_L $$
and for the parity transformed Dirac spinor $\Psi^P$ the left-chiral projection operator is
$$P_L^P \Psi^P = \begin{pmatrix}0 & 0\\ 0 &1\end{pmatrix} \begin{pmatrix} \xi_R \\ \chi_L\end{pmatrix} = \begin{pmatrix} 0\\ \chi_L \end{pmatrix} = \Psi_L^P $$
The parity transformed Dirac spinors live in a different representation and therefore we need different projection operators. The discussion above shows that under parity transformation
$$ P_L \rightarrow P_L^P = \frac{1+\gamma_5}{2} = P_R$$
and therefore the parity transformed in the Lagrangian reads:
$$ \bar \Psi \gamma_0 \gamma_\mu P_R \gamma_0 \Psi W^\mu = \bar \Psi \gamma_0 \gamma_\mu \gamma_0 P_L \Psi W^\mu = \bar \Psi \gamma_\mu P_L \Psi W^\mu $$
which shows the invariance.
| {
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Is time travel impossible because it implies total energy in the universe is non-constant over time? I have always argued with my friends regarding Time Travel that it is impossible. My argument has been that it will destroy the theory that all the energy in the universe is constant since when one travels to a different time, the universe at that time requires extra energy for accommodating the extra person. Similarly the total energy of the universe of that person's current time will be lesser.
I would like to know whether I'm thinking correctly? Has anybody ever experimented or proved anything in similar veins?
| I am not a physicist but all you experts seem to have not understood some basics. Your molecules are not unique - they have always existed as something else. When you go back in time your molecules are already existing as a tree or a bird or grass or Napoleon bleeding Bonaparte, etc. So if you appear in the past what are you made of? Every molecule in the universe has already been assigned in that time period therefore you must be made of brand new molecules that didnt exist before. It's like having a 100 piece set of lego: You can use the 100 pieces to make a motorbike or to make a rocket ship but if you send the rocket ship back in time to when the lego was a motorbike, all of a sudden you have 200 pieces of lego. Or am I missing something?
| {
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What is the cause for the validity of Newton's third law? What, specifically, causes newton's third law? For instance, if I push on a wall, why is it that I experience a force in the opposite direction?
I seem to vaguely understand that is has something to do with electronic repulsion or molecular compression (maybe that's completely wrong, I don't know). As a related question, what would happen if two objects that were /infinitely/ immovable (at the molecular level, it cannot be broken or compressed) collided with each-other?
| In John's answer he states:
But we know that momentum is conserved, ...
So we can derive Newton's third law from the conservation of momentum.
But you may ask where does momentum conservation come from?
Noether's Theorem
If we apply Noether's Theorem with the condition that the Lagrangian is invariant over space then we get that there is a quantity (momentum) which is conserved.
If you construct your Lagrangian such that it varies from place to place you may get a situation which does not obey the Newton's third law. However this is never necessary and rarely desired.
| {
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Do the fields exist without electric charges? I read in an old book on electrodynamics by Pauli that
theoretically there does not exist any need of charges to be there. Fields can even exist without the charges but still independent fields have not been observed still. EMWs are also produced from any accelerated charges.
I want to know that have physicists found fields without electric charges today? I also want to know that if there actually does not exist any field without directly or indirectly originated from charges then there is something missing in the Maxwell equations or it is not like that?
I think that until we can experimentally prove the existence of the fields without charges we cannot tell the Maxwell equations a complete theory of the electrodynamics because it allows the fields without charges but if there exist some nature of the electromagnetic entity that it requires charges for fields then it is not explained in these equations.
| I don't know cases in which the e.m. field can be produced without charges, however I can tell you how to produce an e.m. field without electric charges. Take a neutral particle, meet it with its antiparticle, have them clash, and you'll get gamma rays. The latter are electromagnetic.
Also, gamma emission from excited nuclei, has not much to do with the fact that the nuclei contain protons. The gamma radioactivity originates in eliminating the excess of energy by excited nuclei (where the nuclear field dominates).
If we can say or not that gamma emission from nuclei is without charges, I am not sure. But, in the case of annihilation the charges for generating e.m. field are charges of another field.
(A good question would be here, are a particle and antiparticle opposite charges of some field? For the gravitational field, they are just masses.)
Good luck !
| {
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What will happen to a permanent magnet if we keep the same magnetic poles of two magnets close together for a long time? What will happen to permanent magnet's magnetic field or magnetic ability if we keep same magnetic poles of two permanent magnet for long time?
Will any magnetic loss happen over the long period of exposure or does the magnetic strength remain the same?
Sorry if my logic is wrong. Please explain this.
| If we keep two magnets with same poles together, then they'll become weaker overtime. I once deliberately did it to find out what would happen.
This could be explained with Magnetic Domain Theory. In magnets, the magnetic domains are aligned in the same direction, giving them a strong magnetic field. If two magnets are kept with like poles facing each other,
... the effect will be similar to applying an external magnetic field to the magnet in direction opposite to the magnetic field of the magnet. This will result in the magnetic domains trying to align with the external magnetic field, and in this case, as the magnetic fields of both magnets are equal, some of the domains will disorient, thus weakening the magnetic field of each magnet,
| {
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Simple example showing why measurement & interaction are different Does someone know of a clear (pedagogical) example where one can really see(with the math) where interaction and measurement are not synonymous in quantum mechanics?
*
*I know that every measurement involves a certain interaction with the outside world (e.g. momentum gain from a photon), which results in the system collapsing into one of its eigenstates, meaning a pure state.
*On the other hand, it is also known that not all interactions result in collapsing into eigenstates of the system, so they are in principle very different from what we call "measurement".
It would be definitely nice to also see a bit of the math behind, maybe for simplicity just restricting it to operator algebra and showing how measurement and interaction are defined, shedding a clear light on their difference. I must admit, from a purely physical point of view, I don't know their difference either.
| Quantum optics demonstrates the existence of interaction-free
measurements: the detection of objects without light—or
anything else—ever hitting them.
Paul Kwiat, Harald Weinfurter and Anton Zeilinger SciAm November 1996
http://www.arturekert.org/sandvox/quantum-seeing-in-the-dark.pdf
| {
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Solar spectrum units Why is intensity $I$ on a graph of the solar spectrum always showed in units of $[\mathrm{W/m^2/nm}]$ instead of simply $[\mathrm{W/m^2}]$? (The y-axis on the graph.)
It is apparently shown as intensity per wavelength, but why add this extra specification? For me it just complicates matters (it is not clear to see at which wavelength the intensity is greatest e.g.), so what is the point?
| Intensity has units of watts per area:
$$
\left[I\right]=\rm\frac{W}{ m^2}
$$
where the area is the surface area of the emitting source (in this case, the sun). This tells you the total amount of radiation present (over all wavelengths).
The extra factor of 1/nm in your plot gives the spectral irradiance:
$$
\left[\mathcal E\right]=\rm \frac{W}{m^2\,nm}
$$
This tells you the intensity at each particular wavelength.
| {
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Length contraction in cyclic space Consider a flat universe with at least one finite cyclic spatial dimension: travel x meters in one direction, and you will end up back where you started.
For an object that is of small size relative to the scale of the cyclic dimension, relativistic length contraction ought to work out just fine; the cyclic nature of the space doesn't matter, and the object appears contracted in its direction of motion.
For sufficiently large objects, however, there appears to be a paradox. Consider a solid rod of length x, oriented along the cyclic dimension, so that it wraps around the universe and reconnects with itself. Topologically, it's a circle, but everywhere straight and flat. If the rod is accelerated along its length, it should appear to contract; this, however, would make it not long enough to span the cyclic space; thus, one should expect a discontinuity where two ends of the rod will break apart. There is, however, no unique location at which this discontinuity could occur.
So, what's going on? Is a flat cyclic spacetime simply not possible? Or am I missing some deeper understanding of special relativity?
| Why do you assume the length needed to "span the cyclic space" would be the same in different frames? With cyclic universes it helps to think of them as equivalent to an infinite universe where matter just repeats cyclically, see the "tiling diagram" with the bee and the spider on this page. From this perspective, it's clear the width of a given tile shrinks as well--cyclic universes necessarily involve a globally preferred reference frame where the width is maximum, see this paper for example.
| {
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QFT and violation of Heisenberg uncertainty principle In some QFT books is said that a free electron can emit a virtual photon as long as it reabsorbs the photon and returns to its original state within a time:
$$\Delta t<\dfrac{\hbar}{2\Delta E}$$
That inequality DOES VIOLATE the Heisenberg Uncertainty Principle. Why is that POSSIBLE? If it were said in a time
$$\Delta t\geq \dfrac{\hbar}{2\Delta E}$$
I would not be so puzzled.
| The uncertainty principle is still true in its usual form, but it refers to your knowledge of the state. Suppose your state is just one electron, you can confirm this if you observe the system for a time $\Delta t$ and you don't see additional particles. However, due to the uncertainty principle you can only measure particles that increase the energy of the system above the bound $\Delta E \geq \frac{\hbar}{2\Delta t}$. Therefore, virtual particles below this energy threshold can exist, you are just not able to observe them.
| {
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How are these types of time dilation related? How are these two phenomena related (if at all):
1. Gravitation slowing down time
2. High speed slowing down time
| In relativity there's no objective frame-independent way to compare the rate two clocks at different locations are ticking--different coordinate systems can give different answers (ultimately this is due to the relativity of simultaneity). There is also no frame-independent notion of speed, so you can't say in any objective sense that clocks moving at high speed slow down, though you can say that relative to any given inertial frame, clocks with greater speeds in that frame tick slower in that frame. The most objective way to talk about differences in time elapsed is to compare clocks at the same location, move them apart, and then bring them back together to compare at a common location. In this case, a coordinate-independent consequence of gravitational time dilation could be seen if one clock moved closer to a source of gravity like a planet, spent some time at a shorter distance, and then moved back out to reunite with a clock that remained farther away. And a coordinate-independent consequence of velocity-based time dilation could be seen if one clock moved inertially between the meetings, while the other accelerated to turn around once they had moved apart for some time (see the twin paradox).
Ultimately, the difference in time elapsed in both scenarios can be derived from the rule that a time-like geodesic path between events in spacetime (the closest equivalent to a 'straight line' path between the events) is the one that locally maximizes the proper time. But if you aren't familiar with terms like "geodesic" and the notion of evaluating "proper time" along a path using the "metric" of a given spacetime, this would require a fair amount of explanation. You could try a conceptual introduction like General Relativity from A to B to learn about the basic ideas involved here.
| {
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Half-integer eigenvalues of orbital angular momentum Why do we exclude half-integer values of the orbital angular momentum?
It's clear for me that an angular momentum operator can only have integer values or half-integer values. However, it's not clear why the orbital angular momentum only has integer eigenvalues.
Of course, when we do the experiments we confirm that a scalar wavefunction and integer spherical harmonics are enough to describe everything. Some books, however, try to explain the exclusion of half integer values theoretically. Griffiths evokes the "single valuedness" argument, but he writes that the argument is not so good in a footnote. Shankar says that the $L_z$ operator only is Hermitian when the magnetic quantum number is an integer, but his argument isn't so compelling to me. Gasiorowicz argues that the ladder operators don't work properly with half-integer values. There are some low impact papers (most of them are old) that discuss these subjects, although they are a little bit confusing.
So, basically, my question is: Does anyone have a decisive argument on why do we exclude the half-integer values from the orbital operator spectrum?
| First of all, notice that from the general theory of angular momentum, the eigevalues $m$ of $J_z$ are integer if and only if $j$ is integer because $$m = -j, -j+1,\ldots, j-1, j\:.$$
At this juncture notice that, passing from Cartesian coordinates to spherical ones, you find $$L_z= -i \partial_\phi\:.$$ From direct inspection, using this expression, you see that the eigenfunctions of $L_z$ are in $\mathbb{Z}$. Use in particular the fact that the eigenfunctions must be periodic in $[0, 2\pi]$:
$$-i\partial_\phi f(\phi) = m f(\phi)$$
implies, for some constant $C$,
$$f(\phi) = Ce^{im \phi}$$
Since $f(0)= f(2\pi)$, the only possibility is $m = 0, \pm 1, \pm 2, \ldots$ and thus $j$ is integer as well.
It is worth stressing that it is not possible to answer this question relying only on physical arguments. Physically speaking, there is no way to see the phase $-1$ associated to semi-integer $j$ for rotations of $2\pi$, since $|\psi\rangle$ and $-|\psi\rangle$ represent the same quantum state.
This fact, in addition to the superposition principle leads to the superselection rule of the angular momentum.
| {
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