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Object with friction in circular motion caused by a string A physics problem in my textbook reads:
A 0.40kg ball, attached to the end of a horizontal cord, is rotated in
a circle of radius 1.3m on a frictionless horizontal surface. If the
cord will break when the tension in it exceeds 60N, what is the
maximum speed the ball can have? How would your answer be affected if
there were friction?
Obviously the first question is easy to calculate. But the second one gave me some trouble. The book answer states that friction would not affect the problem, however I believe it would. In order for a limp cord to accelerate the ball in uniform circular motion, the force moving the cord would have to go in a circle of its own if there were friction. Below I drew a picture of my idea of the problem. (The text in the middle says center of rotation).
You can see that the net force has to be greater than it would otherwise have been because of friction. Assuming that the book answer is wrong though, I have another question about the diagram I drew. Would the force of friction act tangent to the circle as I have indicated below? Or would it behave differently?
| You understand the case with no friction. Let's look at the case with friction.
There are two forces acting on the ball: tension and friciton.
The friction comes from the interaction between the string and the talbe, and the direction of friction is always opposing the direction of relative motion of the two objects in contact. So the friction force points along the tangent to the circle. The strength of the friction force is proportional to the weight of the ball which is constant in this problem.
The tension force arises from the interaction of the ball with the string. It is important to note that the tension force must always point along the string, so that here the force points to the center of the circle. The magnitude of the tension force must be the centripetal force for the circular motion: $mv^2/r$.
Notice two things. One, that the tension force can't possible slow down the ball since it is directed perpendicular to the ball's direction of motion. Two, that is ok since the table is the one that is responsible for slowing down the ball through friction.
Now we have our answer. Since the force from the string is the same whether or not there is friction, the maximum speeds for the two cases will not be different.
| {
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Deriving the Lorentz force from velocity dependent potential We can achieve a simplified version of the Lorentz force by
$$F=q\bigg[-\nabla(\phi-\mathbf{A}\cdot\mathbf{v})-\frac{d\mathbf{A}}{dt}\bigg],$$
where $\mathbf{A}$ is the magnetic vector potential and the scalar $\phi$ the electrostatic potential.
How is this derivable from a velocity-dependent potential
$$U=q\phi-q\mathbf{A}\cdot\mathbf{v}?$$
I fail to see how the total derivative of $\mathbf{A}$ can be disposed of and the signs partially reversed. I'm obviously missing something.
| Velocity-dependent potential is not strictly a potential. Lagrange equations say that
$$\frac{d}{dt}\frac{\partial L}{\partial \bf{v}} = \frac{\partial L}{\partial \bf{r}}$$
You have $L = L_0 - U$ where $L_0$ corresponds to free motion (e.g. $L_0 = mv^2/2$ or $L_0 = -mc^2\sqrt{1-(v/c)^2}$).
If $U$ does not contain $\bf{v}$ you have ${\partial L}/{\partial \bf{v}} = \bf{p}$ and so $\dot{\bf{p}} = -\nabla U$.
In this case, however $U$ contains $\bf{v}$ so on the you have
$$\frac{d}{dt}\left({\bf{p}} + q\bf{A}\right) = -\nabla U$$
| {
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About vector form of friction I read a text on mechanics and in the chapter on friction, there written that the kinetic friction is in the form
$$f_k = \mu_k F_N$$
where $f_k$ is the kinetic friction, $\mu_k$ is the kinetic friction coefficient, $F_N=mg$ is the normal force due to the weight of the object. In another chapter, there mentioned the vector form of the force. But it is confusing that if $f_k = \mu_k F_N$, should the vector form written as $\vec{f}_k = \mu_k\vec{F}_N$? It doesn't look right to me since $\vec{F}_N$ is downward but $\vec{f}_k$ is along horizontal. So what's the right way to write the vector for, of kinetic friction? How does people know that $f_k = \mu_k F_N$? From experiment?
I am guessing that if the object is moving at the direction $\hat{\mathcal{d}}$, so the vector form of kinetic friction should be
$$
\vec{f}_k = -\mu_k(\vec{F}_n\cdot\hat{\mathcal{d}})\hat{\mathcal{d}}
$$
Is that correct?
| You're very close. The vector giving the friction force has magnitude $\mu_k F_N$ and is opposite the direction of travel, so it can be written as the product of $\mu_k F_N$ with a unit vector pointing in the direction opposite the direction of travel. Since the velocity $\vec v$ is in the direction of travel, the unit vector $-\vec v/v$, where $v$ is the object's speed, points opposite to the direction of travel, so the force of friction can be written as
\begin{align}
\vec f_k = -\mu_k F_N \frac{\vec v}{v}
\end{align}
The issue with your last expression is that $\vec F_N\cdot \hat d = 0$ since the normal force is perpendicular to the surface, and the direction of travel is parallel to the surface.
| {
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Why should nature of light(or any quantum object) depend on observation? We know that, in the double slit experiment, observation changes the behavior of a quantum object, that it behaves like a particle when observed and a wave when not observed. But why should its nature depend on observation? What if we didn't exist and hence no observation...? The nature of the quantum objects must remain same, right? Why is it based on observation? Am I trying to understand the wave-particle duality in a wrong way?
| It seems that you stumbled upon the measurement problem, where if you don't observe or measure a quantum object yet, it moves in all directions.
Its nature depends on observation as if you measure it, it will disturb or interfere with it as you measure it as it changes it's direction to a specific one, the environment it is in will also make it interact with the environment and washes away the behaviour of quantum objects.
So basically, quantum objects that are measured are interfered so that it then changes it's direction to a specific way other than all directions.
| {
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Surface gravity of Kerr black hole I'm going through Kerr metric, and following the 'Relativist's toolkit' derivation of the surface gravity, I've come to a part that I don't understand.
Firstly, the metric is given by
$$\mathrm{d}s^2=\left(\frac{\Sigma}{\rho^2}\sin^2\theta\omega^2-\frac{\rho^2\Delta}{\Sigma}\right)\mathrm{d}t^2-2\frac{\Sigma}{\rho^2}\sin^2\theta\omega \mathrm{d}\phi \mathrm{d}t+\frac{\Sigma}{\rho^2}\sin^2\theta \mathrm{d}\phi^2+\frac{\rho^2}{\Delta}\mathrm{d}r^2+\rho^2 \mathrm{d}\theta^2$$
With
$$\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2,$$
$$\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}$$
The Killing vector that is null at the event horizon is
$$\chi^\mu=\partial_t+\Omega_H\partial_\phi$$
where $\Omega_H$ is angular velocity at the horizon.
Now I got the same norm of the Killing vector
$$\chi^\mu\chi_\mu=g_{\mu\nu}\chi^\mu\chi^\nu=\frac{\Sigma}{\rho^2}\sin^2\theta(\Omega_H-\omega)^2-\frac{\rho^2\Delta}{\Sigma}$$
And now I should use this equation
$$\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu$$
And I need to look at the horizon. Now, on the horizon $\omega=\Omega_H$ so my first term in the norm is zero, but, on the horizon $\Delta=0$ too, so how are they deriving that side, and how did they get
$$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta$$
if the $\Delta=0$ on the horizon? Since $\rho$ and $\Sigma$ both depend on $r$, and even if I evaluate them at $r_+=M+\sqrt{M^2-a^2}$ they don't cancel each other.
How do they get to the end result of $\kappa$?
| $$\nabla_\nu(-\chi^\mu\chi_\mu) =\partial_\nu(-\chi^\mu\chi_\mu)\\
=\frac{\rho^2 }{\Sigma}\partial_\nu \Delta + \Delta \partial_\nu(\frac{\rho^2 }{\Sigma})-(\Omega_H -\omega)^2 \partial_\nu(\frac{\Sigma\sin^2{\theta}}{\rho^2})$$
Now use the horizon condition you will get
$$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2 }{\Sigma}\partial_\nu \Delta $$
Since $\chi^\mu$ is null on horizon and a null vector is normal to itself so $\chi^\mu$ must be proportional to the normal of the horizon. A constant r surface has a normal $\partial_\mu{r}$. So
$$\chi_{\mu}=C\partial_\mu{r}$$
Now our job is to find C. It can be found easily
$$g^{\mu\nu}\chi_{\mu}\chi_{nu}=C^2g^{\mu\nu}\partial_\mu{r}\partial_\nu{r}\\
=C^2g^{rr}$$
So
$$C^2=\frac{\chi^{\mu}\chi_{\mu}}{g^{rr}}$$
So after the algebra is done take the horizon limit and you will find C. The rest are just few lines algebra.
| {
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By saying a physical state has some 'symmetry', what do we really mean? Here our arguments are restricted to the realm of the Projective Symmetry Group(PSG) proposed by Prof. Wen,
Quantum Orders and Symmetric Spin Liquids. Xiao-Gang Wen. Phys. Rev. B 65 no. 16, 165113 (2002). arXiv:cond-mat/0107071.
and the following notations are the same as those in my previous question, Two puzzles on the Projective Symmetry Group(PSG)?.
When we say the projected physical spin state $P\Psi$ has some 'symmetry', e.g., translation symmetry, there will be two understandings:
(1) After a translation of the mean-field Hamiltonian $H(\psi_i)$, say $DH(\psi_i)D^{-1}$, the physical spin state is unchanged, say $P\Psi'\propto P\Psi$, where $\Psi'$ is the ground state of the translated Hamiltonian $DH(\psi_i)D^{-1}$.
(2) $D(P\Psi)\propto P\Psi$.
I would like to know: are the above understandings equivalent to each other? Thanks in advance.
| I just found that I asked a naive question and I can answer it by myself now.
(1) and (2) are equivalent to each other. Because if $\Psi$ is a ground state of $H(\psi_i)$, then $\Psi'=D\Psi$ is the ground state of $DH(\psi_i)D^{-1}$, and $[P,D]=0$, therefore $D(P\Psi)=P\Psi'$.
Remark:
More generally, when we talk about any kind of symmetry of the physical state, the identity $[P,A]=0$ is the reason for the equivalence between (1) and (2) statements. Where the unitary(or antiunitary, e.g. time-reversal) operator $A$ represents the corresponding symmetry.
| {
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Symmetry breaking in Bose-Hubbard model According to Landau's symmetry breaking theory, there is a symmetry breaking when phase transition occurs.
*
*What is the symmetry breaking of superfluid-Mott insulator transition in Bose-Hubbard model?
*Why metallic state to Mott insulator state transition in Fermi-Hubbard model is not a phase transition, but a crossover.
| The Mott transition in the Bose-Hubbard model is a quantum phase transition. From the point of view of field theory, that does not change much compare to standard (finite-temperature) phase transitions. The main difference is that you now have to take into account the quantum fluctuations which correspond to the "imaginary time" direction in addition to the d dimensions of space. It also means that there are at least two control parameters (i.e. parameters that have to be fine-tuned to have the transition), a non-thermal control parameter (such as the hopping amplitude or the density) and the temperature (which must be zero by definition).
Other than that, you can use Landau theory to understand the transition (which is second-order) at zero-temperature. The disordered phase is the Mott insulator, and the ordered one is the superfluid, where the non-zero order parameter is the condensate density (I will only talk about the 3D case, which is the simplest, as I won't have to deal with BKT phases). The broken symmetry is the usual one for Bose-Einstein condensate : the U(1) symmetry. One can then show that there is two universality classes, depending on the way the transition is made (at constant density or with a change of density at the transition).
Now, at finite temperature, things are different. First, the Mott insulator does not exist anymore, as a finite temperature can excite particles and one gets a finite compressibility (or conductivity). That might correspond to the cross-over you're talking about in the fermionic case. On the other hand, the superfluid exists at least up to a critical temperature.
| {
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Quantum entanglement as practical method of superluminal communication As I understand it (from a lay physics perspective), quantum entanglement has been experimentally demonstrated - it is a reality. As I understand it, you can measure something like the spin of an electron and know that its entangled pair will, in that same instant, no matter where in the universe it is, have the opposite spin.
This would not seem to have any utility as the foundation of a superluminal communications device. Is this true, or has it been established that is there some aspect of quantum entanglement that can ultimately lead to the development of such a device.
In other words: is superluminal communication via quantum entanglement an open scientific question, has it been settled as an impossibility, or is it currently more of an engineering problem than a scientific one?
| To rephrase Danu answer, you can not use the correlation of these entangled particles before you have exchanged some information with another (subluminal) device.
The main problem comes from the fact that the outcome of a measurement is random, so there is no way to agree beforehand on how to interpret a measurement done by one of the party involved in this superluminal communication.
| {
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Is there an easy way to get water at roughly 70°C in our kitchen? Some green tea requires to pour water at 70°C. I have no thermal sensor or kettle with adjustable temperature with me.
Do you know a way to get water at roughly 70°C like “boil water and wait for x minutes” or “mix x part of boiling water with 1-x part of fresh water” ?
| Assumptions:
*
*we have a volume of water $(v_1)$ at $100^{\circ} C$ (boiling water)
*we have a volume of water $(v_2)$ at $20^{\circ} C$ (tap water)
*we want a volume of water $(v_1+v_2)$ at $70^{\circ} C$
*the pot that will finally hold the $70^{\circ} C$ water will consume a negligible amount of energy from the water (generally a bad assumption. Might have to account for an addition $5^{\circ} C$ of temperature loss)
The energy $(Q)$ required to heat a body of water is given by:$$ Q = m \cdot c \cdot \Delta T$$
The mass of water is the same as volume. When mixed, volume $1$ of water will need to increase by $50^{\circ} C$ and volume $2$ of water will decrease by $30^{\circ} C$. Using our assumption of no energy loss we can setup the equation $$Q_1 = Q_2$$
$$v_1 \cdot c (\text{Water}) \cdot 50 = v_2 \cdot c (\text{Water}) \cdot 30$$
Our ratios of tap water to boiling water will be: $$\frac{v_1}{v_2} = \frac{3}{5}~.$$ e.g., $1~$ L of boiling water needs $600~$ml of tap water and will result in $1.6 ~$L of $70^{\circ} C$ water.
| {
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Definition and motivation for Twistors What are Twistors? Why are they important?
This particular statement in Wikipedia is intriguing
According to Andrew Hodges, twistor space is useful for conceptualizing the way photons travel through space, using four complex numbers.
| Your question can be answered at many levels. I'll keep it simple; I'm not very well versed with the grand picture.
Twistors provide an efficient (and possibly natural) means to encode the kinematics of massless particles and the resulting conformal symmetry.
For a nice and clear introduction, check Witten's lecture notes from PiTP 2004:
http://www.sns.ias.edu/pitp2/2004/schedule.html
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What are the assumptions of the Navier-Stokes equations? I wanted to model a real life problem using the Navier-Stokes equations and was wondering what the assumptions made by the same are so that I could better relate my entities with a 'fluid' and make or set assumptions on them likewise. For example one of the assumptions of a Newtonian fluid is that the viscosity does not depend on the shear rate. Similarly what are the assumptions that are made on a fluid or how does the Navier-Stokes equations define a fluid for which the equation is applicable?
| Most importantly, the Navier-Stokes equations are based on a continuum assumption. This means that you should be able to view the fluid as having properties like density and velocity at infinitely small points. If you look at e.g. liquid flows in nanochannels or gas flows in microchannels you could be in a regime where this assumption breaks down. As far as I know there is no hard limit for the continuum assumption, but the Knudsen number is a useful indicator.
Additionally there is, as @ShuchangZhang mentioned, an assumption on the nature of the stress in the fluid. Although I am not sure whether you could really call this an assumption or whether it should be considered a theory (like the NS equations itself).
The strongest assumptions are typically not in the Navier-Stokes equations themselves, but rather in the boundary conditions that should be applied in order to solve them. To give an example, whether the no-slip boundary condition (fluid velocity at the wall equals wall velocity) or the navier slip boundary condition (fluid velocity equals a scaled velocity gradient at the wall) has been a much debated subject, in particular for hydrophobic surfaces (see e.g. Phys. Rev. Lett. 94, 056102 (2005) and references therein and thereto)
| {
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Frames of reference: Inertial and accelerated - and jerked, snapped, crackled and popped? There are inertial frames of reference and the accelerated frames of reference, but are there any frames of references w.r.t. higher order derivatives of velocity? [1] [2]
For example, jerked frames of reference, snapped frames of reference, crackled frames of reference and popped frames of reference and so on?
| A frame of reference does not need to be inertial though, for a non-inertial frame of reference, there is, at any instant, a momentarily co-moving (inertial) reference frame or MCRF
Now suppose that a particle does accelerate. In that case, we can have
an inertial frame at any event in the particle’s life by defining the
momentarily comoving reference frame or MCRF for short. This is a
reference frame that, at a given event, has the same velocity as the
particle. If the particle is accelerating, then the MCRF will change
from one event to the next, but at each point it is always an inertial
frame.
Note that there is no requirement that the acceleration is uniform.
| {
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Is such an orbit possible that allows a satellite on Earth and another on Mars to always be in direct line-of-sight? If not always, how much "most of the time" could it get?
Or would a multi-satellites setup be more suited?
| The best way to minimise loss of sight would be to have the satellites orbiting at a very high altitude, e.g. beyond the distance of the moon. It would then be very rare for the satellites to be eclipsed by the planets, or moon. Such eclipses could also be easily avoided with a small orbit correction or by synchronizing the orbit in such a way that potential eclipses never happen
The only case where you would be caught out would be when Mars passed behind the Sun as seen from Earth but such events are very rare. If the orbits were higher than the radius of the Sun (700 Mm) it would be possible to avoid even the Sun occultations with careful planning and some orbit adjustment.
If the satellites have to be nearer their planets you could keep them orbiting around the axis joining the planets so they are always in sight, but more fuel would be required to keep them in those orbits as the planets move.
| {
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Is inflation deterministic? In some theories inflation is supposed to be able to turn quantum fluctuations into macroscopic inhomogeneities.
I don't understand how an isolated system such as the universe can undergo such a random transformation : if at the beginning the universe is in a state $A$, quantum mechanics says that $A$ will evolve to $B=UA$ with $U$ being a unitary operator, and general relativity is also a deterministic theory.
So does inflation suppose that the universe is not isolated or does it use some modified theories which include randomness?
| Ok I ask the question to some specialists and it just seems that indeed the evolution of the whole universe is not unitary in inflation theory.
| {
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Time it takes two oppositely charged particles to collide I think this is supposed to be a simple problem but I am having a hang up converting it to a one-body problem. It's one-dimensional. +q and -q a distance d apart, held stationary then let go at t=0. The potential is V(x)=kq^2/x. If I turn it into a one body problem, then m-->m/2, but how do i interpret the new x? Both particles are moving toward each other, so they travel a distance d/2 before colliding. I am guessing the relevant equation will be $t = {\sqrt{\frac{m}{2}}} \int \frac{dr}{\sqrt{E - V(x)}}$
What concepts am I lacking? I think this is supposed to be really easy, but it's not for me.
edit, so x is now the relative distance between the two particles so it should be like one particle traveling the whole distance d ? I get a negative value, but is that acceptable? Something like
$t=\frac{\sqrt{m}}{2} \int_d^0 \sqrt{\frac{d}{kq^2}} \frac{dx}{\sqrt{1-d/x}}$ And that isn't giving me a very good answer when I calculate it.
| The negative sign doesn't have to do with electrostatics, it's a problem with a global sign which comes from the square root.
In a 1D problem, you can write
$$t_f-t_i=\int_{x_i}^{x_f}\frac{\text dx}v,\ \text{where}\ v^2=\frac2m(E-V(x)).$$
In the present case, though, with $x$ decreasing as $t$ increases, you need to take the negative sign for $v=\cfrac{\text dx}{\text dt}$. Thus if the particles are a distance $d$ at time zero and collide at time $T$, you should write
$$T=-\int_d^0\frac{\text dx}{\sqrt{\tfrac2m(E-V(x))}}>0.$$
Easy!
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How does a pressure suit work? I recently learnt that a suit called pressure suit is worn by fighter plane pilots to prevent red-outs and black-outs. And it seems to be work by -
"..applying pressure to selective portions of the body."
How do these suits work; i.e. by what means, selective portions of the body are pressurised?
Do astronauts wear these while takeoffs, and also F1 drivers?
| The suits are designed to provide protection from the temperatures +100F to -90F.
When pressurized to 3.5 pounds per square inch (roughly equivalent to the atmospheric pressure at 35,000 feet), the suit can help to avert symptoms of decompression sickness (the "bends").
Above about 62,000 feet, the liquid in pressure suit tissues could turn to gas and expand dangerously, a condition called ebullism, but the suit will maintain pressure around his body to prevent such expansion.
The exterior of the suit is made of a material that is both fire retardant and an insulator against extreme cold.
A "controller" is the "brain" of the suit. Only the size of a hockey puck, it's an extremely reliable mechanism for maintaining pressure automatically at various altitudes.
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Quantum tunneling effect in a potential of the kind $V(x)=A\frac{x^2}{1+x^4}$ Given a potential: $$V(x)=A\frac{x^2}{1+x^4}$$ with $A\gt 1$ and a quantum particle inside the well around the point $x=0$. I'm stuck on the calculation of the transmission and reflection coefficients for this particle vs. its energy.
| Here's a not-so-clever answer.
The plot of the function is shown below.
The red line denotes the energy of the particle being tunneled which expressed in terms of A. The black line denotes the max value of the potential which is A/2.
The task is to evaluate the transmission coefficient of the particle through one of the bumps of the potential.
According to the WKB approximation the tunneling transmission coefficient across a given barrier is given by.
To evaluate the integral, taylor expand the square root in equation 1 around the point x = 1. And one would arrive at (for 0 < c < 0.5).
Now, the limits of the integral are determined by the points at which the line U(x) = cA (Energy of the particle) intercepts the bumps of the curve. The integral of the square root in equation 1 must be evaluated between these points because the square root will give rise to imaginary numbers at all other points. To obtain the values of x at which the line U(x) = cA intercepts the bumps, one must solve the 4th power polynomial equation.
The four roots are given by
Two of these roots/intercepts are on LHS bump and the other two are on RHS bump. Since we are only interested in the intercepts on one of the bumps we select only the positive roots which correspond to the intercept of U(x) on the RHS bump.
The the above values in equation 5 become the limits of the integral in eqtn (1).
Now to complete the problem one must integrate all the terms in equation 2 with respect to x and plug in the limits of the integral given in equation 5, which is a routine (and yet tedious) task. The result can be substituted in equation 1 to obtain the transmission coefficient.
I believe the process becomes easier if c is known. The general equation for all values of c (c < 0.5) becomes rather large and messy.
References:
1. A. Messiah (1991), "Quantenmechanik 1", Degruyter, 1991.
*G. Squires, (1995). "Problems in quantum mechanics", Cambridge University Press, Cambridge, UK.
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Question on field strength tensor in YM just a quick question on $F_{\mu\nu}^a$. I'm correct to think $F_{\mu}^{\mu,a}$ vanishes, aren't I? (Just want to make sure...) My reasoning is as follows:
The derivative terms cancel anyways - that's obvious - so the only "critical" term of $F_{\mu}^{\mu,a}$ is $f^{abc}A_{\mu}^b A^{\mu,c}$ but this vanishes because the combination of A's is symmetric but the $f$ totally antisymmetric. Am I right?
| Yes, $$\sum_{\mu} F_{\mu}{}^{\mu}~:=~\sum_{\mu,\nu}F_{\mu\nu} g^{\nu\mu}~=~0$$ vanishes because it is a trace of a product of a symmetric and an antisymmetric tensor. It is irrelevant for the argument that $F_{\mu\nu}$ is Lie algebra valued.
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Why does vacuum have a nonzero characteristic impedance towards electromagnetic radiation? On Wikipedia, the impedance of free space $Z_0$ is defined as square root of the ratio of the permeability of free space $\mu_0$ to the permittivity of free space $\epsilon_0$, i.e.
$$Z_0 = \sqrt{\mu_0 / \epsilon_0} \, .$$
The value is approximately 377 Ohms.
Now impedance is described as an impeding effect to flow of something, it makes more sense for electric current travelling in a wire where the characteristic impedance of the line (as the line consists of capacitance and inductance per unit length) prevents the flow of AC/DC flow.
Why in the world does free space have a characteristic impedance? That makes no sense to me. Wires makes a lot of sense, but free space having 377 ohm of impedance is too much and not clear why such a value exists.
| Since a finite fluctuating electric field creates a finite fluctuating magnetic field (and vice versa), their ratio must also be a finite value. In other worlds, that being a finite quantity is only a consequence of the existence of light in vacuum.
Now that value itself is merely a consequence of the definition of units you choose, so it being $376.730... \Omega$ isn't something very interesting by itself.
However, the fact that it doesn't depend on frequency is very important (and visible to the naked eye) : that is because both $\epsilon_0$ and $\mu_0$ are fundamental constants. In other words, vacuum is a non-dispersive medium ; that is why we see all colors from an object arriving at the same time.
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Dehn twists and topological order I am trying to understand notion of a "Dehn twist" and how it relates to topological order.
In particular refering to http://arxiv.org/abs/1208.4834 it is stated that
Xiao Gang Wen's paper on "Topological Order in Rigid States" (http://dao.mit.edu/~wen/pub/topo.pdf) is supposed to provide an introduction to "non-abelian adiabatic Berry phases associated with Dehn twists for a U(1) Chern Simons Theory". Skimming through the respective paper however I could not find the notion of a "Dehn twist" appearing at all?
Maybe it appears under a different name or it is not given a name at all?
I would be very happy for any support.
Best.
| Any oriented closed surface is a torus with g holes (for an actual torus g=1, for a sphere g=0, etc.), where g is called the genus. Associated to these surfaces is the mapping class group, which is the group of equivalence classes of homeomorphisms (topological isomorphisms) of the surface to itself, where two such mappings are considered equivalent when they can be continuously deformed into each other. Dehn first proved that for an orientable genus g surface, this group is generated by what are now called Dehn twists. A Dehn twist is easy to understand: take a surface, cut open a tube, twist it around a full turn, and glue all points back to their original positions. This defines a map from the surface to itself mapping a point on the surface to the corresponding point on the twisted surface.
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How to establish relation between flow rate and height of the water column of the tank? Suppose a water tank has 1" diameter drain at the bottom and is filled with water up to one meter height above the drain. What time it will take the tank to drain out completely. Now say, the tank is filled up to two meter height above the drain then what time it will take the tank to drain out? Will it be double or less than it? Can we establish a relation between flow rate for the given height of water column?
| The issue is, what is the flow velocity?
It is not simple, and it depends on pressure, which is proportional to height.
For low pressure, viscosity will dominate, and velocity will be proportional to pressure.
For high pressure, velocity will be proportional to square root of pressure.
In any case, the geometry of the opening matters.
The general subject is Orifice Flow.
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What is the most efficient way to travel between tidally locked binary planets? Suppose that there are two planets of roughly the same volume and mass are orbiting each other. What would be the most efficient way to travel from one to the other? In other words, what kind of orbits, and connecting rocket-engine burns, would require the least amount of $\Delta v$ to do this transfer?
It seems unlikely, but could a rocket positioned at the center of the near side of one of the planets just simply travel upwards? Or is it pretty much the same maneuver as getting to the Moon? If so, would having the center of rotation closer to the center of the two bodies have an effect on fuel consumption?
| I worked out the outline of thee solution to this problem using Newtonian mechanics and vectors. This is not an easy problem because it is a three body problem. For simplicity I have made a few assumptions: 1. Mass of the spaceship is negligible and the orbits are perfect circles.
Working out the final solution requires some tedious calculation. Hopefully, the solution gives you an idea about how to solve it rather than the exact answer.
I believe that this question can be solved using Lagrangian and Lagrange multiplayer in an easier way.
The outline of the solution is here.(https://i.stack.imgur.com/lMGKJ.jpg)
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Nature of frictional force I was thinking about a situation where a person in standing on the ground with some friction. The frictional force is directly proportional to the normal reaction acting on him by the ground. Assume that he leaned forward i.e his center of mass is not vertically up the point of contact of him with the ground.Then, does the frictional force change?
My thoughts:Although the person has leaned forward the force $mg$ acting on him will be vertically downward. Since normal reaction is always normal to the surface of contact its magnitude will remain same and thus frictional force will remain same.But, I have seen athletes starting the running race with a leaning forward position which would be mostly for increasing the friction between their shoes and the track. So, I'm in a dilemma. Please help.
| When the person is standing perfectly upright, the frictional force on him is actually zero and the normal reaction is $mg$. The frictional force is the horizontal component of the contact force and normal reaction is the vertical component. Therefore, unless there is a horizontal force, there will not be any friction whatever may be your composure.
In the second case when your COM moves forward, the normal reaction now tries to prevent toppling over by shifting its position and the line of action so that it can balance torques about the centre of mass. This is possible only in certain positions. Try to see the dynamics of toppling over to understand how normal reaction tries to balance the torque of weight acting at centre of mass, and under what conditions can it be balanced.
In case of athletes, balancing and optimizing their composure for best configuration of COM is very important, though I don't think that bending forward has anything to do with reduction or increasing the friction.(but rather to optimize Aerodynamic and classical forces)
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Is it possible for the entropy in an isolated system to decrease? As far as I can tell, the concept of entropy is a purely statistical one. In my engineering thermodynamics course we were told that the second law of Thermodynamics states that "the entropy of an isolated system never decreases". However, this doesn't make much sense to me.
By counter-example: Consider a gas-filled isolated system where the gas has maximum entropy (it is at equilibrium). Since the molecular motion is considered to be random, at some point in the future there will be a pressure gradient formed by pure chance. At this point in time, entropy has decreased.
According to Wikipedia, the second law purely states that systems tend toward thermodynamic equilibrium which makes sense. I then ask a) is the second law as we were taught it wrong (in general), and b) what is the use of entropy (as a mathematical value) if it's effectively an arbitrary definition (i.e. what implications can we draw from knowing the change in entropy of a system)?
Thanks in advance for your help.
| By the Poincaré recurrence theorem, an isolated system is guaranteed to return arbitrarily close to its initial state after a sufficiently long time.
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Do orbitals overlap? Yes, as the title states: Do orbitals overlap ?
I mean, if I take a look at this figure...
I see the distribution in different orbitals. So if for example I take the S orbitals, they are all just a sphere. So wont the 2S orbital overlap with the 1S overlap, making the electrons in each orbital "meet" at some point?
Or have I misunderstood something?
| The orbitals shown in the figure are the different eigenstates of the electronic wave functions derived from the solution of Schrodinger equation. One of the postulates of QM states that these eigenstates are independent in a free atom.
Orbitals do overlap when two atoms are close together, e.g. in a molcule and the degree of overlap corresponds to the type of bonding (ionic, covalent etc.).
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How fast does light travel through a fibre optic cable? The principle behind a fibre optic cable is that light is reflected along the cable until it reaches the other side, like in this diagram:
Although I know that the light is slowed down somewhat because it's not going through air, I've always wondered about another factor: what about the fact that the light path is zig-zagged rather than straight? Doesn't that significantly increase the distance that the light has to travel? If so, by how much does it slow down the time the light takes to travel through the cable?
| We can make a 1st-order approximation by assuming the following:
*
*$L=3$ m is the length of the fibre optic cable
*$d=3\cdot10^{-6}$ m is the diameter of the cable
*the cable is perfectly straight
*$\theta=0.785$ rad (~45$^\circ$) is the angle of reflection inside the cable
*photons are classical balls
*reflection is perfectly elastic
*photons still travel at $c$
Simple geometry shows that the particle travels $h=\frac{d}{\sin\theta}=4.24\times10^{-6}$ m over a linear distance of $x=3\cdot10^{-6}$ m. Do this a million times, you find that the photon traveled 4.24 meters instead of 3 meters!
Given speed of light in vacuum, it would take 14.1 nanoseconds for the photon to travel the reflected path, whereas it would take 10.0 nanoseconds to travel 3 meters linearly. Both the distance & duration are about 40% increases!
Since $L=3$ m and $t=14.1\cdot10^{-9}$ s, then the "linear" photon speed in the fibre optic cable is $v_{\gamma,fo}=2.13\cdot10^8$ m/s, a reduction of about 30%.
EDIT
As per the request in the comment, using $2c/3$, the reflecting photon would take 21.2 nanoseconds to travel the cable while the linear photon travels the distance in 15 nanoseconds. This would then lead to $v_{\gamma,fo}=1.42\cdot10^8$ m/s (instead of $\sim2\cdot10^8$ m/s) which is still a reduction of about 30%.
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How to find the wavefunction that solves an infinite square well with a delta function well in the middle? Solutions for the wavefunction in an infinite square well with a delta function barrier in the middle are easily found online (see here for an example). I am wondering what the wavefunction is for an infinite square well with a delta function well in the middle. The setup is the bottom of the infinite square well is defined to be zero energy. I realize that there will be two situations, one where the particle's energy is less than zero and will therefore be bound to the delta function well and one where the particle's energy is greater than zero and is bound to the infinite square well. What are the wavefunctions for these two situations?
| Consider an infinite square with free region $[0, L]$. Place a delta function potential at $L/2$ with strength $\alpha$. Then, Schrödinger's equation is
$$E\psi = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x ^{2}} + \alpha \delta(x- \frac{L}{2})\psi$$
First, note that the delta function is zero everywhere except at the center, which means that the energy eigenstates everywhere else are given by $\psi = A \sin (kx + \phi)$. Furthermore, we know that $\psi(0) = \psi(L) = 0$. Thus, we know that on the left hand side of the delta, we have $\phi = 0$, and on the right hand side, we have $\phi = - kL$. Thus, for $0<x<L/2$, we have $\psi = A \sin(kx)$, while for $L/2 < x < L$, we have $\psi = B \sin (k(x-L))$. The wave function must be continuous at $L/2$, so this guarantees that $A=-B$.
We can derive a restriction on $k$ by integrating Schrödinger's equation over an arbitrarily small region around $x = L/2$. Since the wave function is continuous, the left hand side drops off, and we're left with:
$$0 = -\frac{\hbar^{2}}{2m}(\psi^{\prime}_{+} - \psi{\prime}_{-}) + \alpha\psi(L/2)$$
Or, more concretely,
$$\psi^{\prime}_{+} = \frac{2m\alpha}{\hbar^{2}}\psi(L/2) + \psi^{\prime}_{-} $$
Taking the derivative and substituting, we find:
$$-Ak\cos(kL/2) = \frac{2m\alpha}{\hbar^{2}}A\sin(kL/2)+Ak\cos(kL/2)$$
Finally, this gives us the transcendental equation
$$\tan(kL/2) = - \frac{\hbar^{2}k}{m\alpha}$$
Noting that we still have, as in the finite square well case, $E = \frac{\hbar^{2}k^{2}}{2m}$, all solutions have $E > 0$, irrespective of the sign of $\alpha$, although simple grasping shows that there are infinite numbers of solutions to this equation, independent of the sign of $\alpha$ (though the sign of and value of $\alpha$ will shift where those solutions are quite dramatically).
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Why are we not affected by the radiation of the radioactive decay going inside the Earth? I was reading the question Why has Earth's core not become solid?, and one of the answers says that
The core is heated by radioactive decays of Uranium-238, Uranium-235, Thorium-232, and Potassium-40
Why are we not affected by the radioactive emission of the condition below? Is this due to the fact that there is a very thick layer of mantle and crust between us and the core? Or I am wrong and we suffer from it's radiation in everyday life up to some extent?
| Alphas and betas are stopped by a small amount of material. Gammas are more penetrating, but a gamma ray with a typical energy would be stopped by something like 10 cm of rock.
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2N Fermions $\stackrel{?}{=}$ N Bosons We know that we do have composite particles (for example Atoms) made of fermions or bosons or mixture of them with fermionic or bosonic statistics. So why can't a gas of $2N$ fermions become a gas of $N$ bosons and condense to the lowest state at low temperature (just like what happens in superfluidity and superconductivity)?
| It can. This is exactly what happens when Helium-3 becomes superfluid. It's also what happens in superconductivity, which you mention in your question, when electrons combine into Cooper pairs.
Well, it's not exactly what you ask since neither liquid Helium-3 nor electrons are a gas. It's very unlikely a gas of fermions could pair up to form a gas of bosons because the energy required to disrupt the pair is typically very low, and requires temperatures at which the gas would liquify. The nearest to a gas would be a fermionic condensate, which was first made in 2004.
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Compressible flow - subsonic to supersonic and the 2nd law of thermodynamics I'm reading chapter 16 in Fluid Mechanics by Kundu.
It is stated (figures 16.16 and 16.17) that in a constant area duct flow with heating or friction, to go from subsonic conditions to supersonic violates the 2nd law of thermodynamics. How can that be possible? How does a fluid attain supersonic speeds in the first place? Somehow, one must be able to get ambient quiescent air to supersonic velocities (it's done all the time). I know that you cannot generate a normal shock by going from subsonic to supersonic, only the other way around, but why can't you speed a fluid from subsonic conditions to supersonic in the duct? In that case, there just would not be any shock generated? Would that violate the 2nd law?
I'm missing something here...
|
Somehow, one must be able to get ambient quiescent air to supersonic velocities (it's done all the time).
Of course, and that same chapter already told you how it could be done: in convergent–divergent nozzle.
The statement
... the upper left
branch of the solution $M_2 > 1$ when $M_1 < 1$ is inaccessible because it violates the
second law of thermodynamics
applies only to constant area ducts with friction and heating.
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Why is moment dependent on the distance from the point of rotation to the force? The formula for moment is:
$$M = Fd$$
Where F is the force applied on the object and d is the perpendicular distance from the point of rotation to the line of action of the force.
Why? Intuitively, it makes sense that moment is dependent on force since the force "increases the intensity". But why distance? Why does the distance from the line of action of the force to the point of intensity affect the moment?
I am NOT looking for a derivation of the above formula from the cross product formula, I am looking for intuition. I understand how when I am turning a wrench, if the wrench is shorter its harder to turn it but I don't understand WHY.
Thanks.
| The best definition of torque (or moment) is the work per unit angle of rotation (in Joules per radian) that can be done by a force which is acting in a manner that tends to cause a rotation. This implies that you want the component of the force which is acting along an arc, times the arc length, divided by the angle (in radians). But the arc length s = rθ, and s/θ = r. So you end up multiplying a force component by a radius.
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Quantum Thrusters and Warp Drives Any reservations about the potential of this (given in the link below/title above)?
Does it seem like a helpless attempt or something which might have the potential of developing into something real?
http://news.discovery.com/space/quantum-thruster-warp-drive-physics-130823.htm
| The idea is very much possible by using exotic matter through quantum waves amongst other things of course. Besides the energy was scaled down in caparison to Voyager 1 size and achieves this by using a ring around the ship. Sure this may not be feasible now ,but so many other things were considered impossible before they were created. The equation that made the warp drive even possible was created by the physicist Alcubierre and later made in the realm of possibility by Dr. Harold White. One could speculate that the negative energy required could be made by using the Casmir effect or a antimatter reactor using plasma or lasers. The argument that it can be feasible at any point is speculative and cannot be dismissed as an entirely impossible idea.
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Why can't we perfectly focus light-abberations aside I don't understand why there is necessarily a diffraction limitation on optical systems. Where does this limitation in focusing light come from?
| Well I suppose by perfectly focus, you mean to a mathematical point. And if we could do that, Heisenberg's principle implies that the momentum uncertainty would be infinite. So by the time we looked where the point was supposed to be, it would have moved to someplace else, in fact it could be anywhere at all, and we would never find it.
The problem is that the wavelength is not zero.
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How accurate is Newtonian Gravity? I know that really fast moving things need Relativity rather than Newtonian physics.
I also know the quirk of the Mercury´s orbit.
But just how much more accurate is General Relativity than Newton´s Law of Gravitation for predicting say the orbit of Earth or Neptune?
Can the "slingshot" effect where we use another planet´s gravity to accelerate a space probe be done with Newton or does that require General Relativity?
Is the speed of Jupiter (18 km/s I think) fast enough to make a difference in the accuracy of GR v Newton´s Law of Gravity?
| There are several different questions embedded in here. The answer to all of them is "Newton's theory is accurate to great precision, and beyond measurement accuracy for most of your examples".
The key point is that Newtonian physics fails when, roughly, the quantity $v/c > .1$ or $\frac{GM}{c^{2}r} > .1$. You can calculate both of these quantities for the cases of Earth and Jupiter, and you will find that your answer is quite small.
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Why does Newton's third law exist even in non-inertial reference frames? While reviewing Newton's laws of motion I came across the statement which says Newton's laws exist only in inertial reference frames except the third one. Why is it like that?
| $\newcommand{fp}[0]{\vec{F}_\textrm{phys}}$ $\newcommand{fn}[0]{\vec{F}_\textrm{non-inertial}}$ $\newcommand{fab}[0]{\vec{F}_{AB}}$ $\newcommand{fba}[0]{\vec{F}_{BA}}$In a non-inertial frame, every object feels the physical force $\fp$, that it felt in the inertial frame, plus a force $\fn$. The non intertial force felt by an object may depend on its mass, position, time, and possibly other things. An objects acceleration is then given by $m \vec{a} = \fn + \fp$. Thus newton's second law, $m \vec{a} = \fp$, breaks down, and you need a correction for the non-inertial forces.
Let's look at newton's third law. It says $\fab= -\fba$. We know this holds true in the inertial frame. If we transform these forces to a non-inertial frame, the transformed coordinates will be different, but because of the way coordinate transformations work, it will still be true that $\fab = -\fba$ in the transformed coordinate system. Thus newton's third law still holds.
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How much energy does it take to simply run forward? I'm interested in tracking as much data about my runs as I can in an effort to get faster, and while I can easily estimate energy expenditure during an uphill run due to the change in elevation, I can't estimate energy expenditure due to the requirements of just keeping in motion. I could estimate wind resistance, creating a simple cylindrical model and estimating drag, but obviously it would take energy to move forward on a flat surface in a vacuum as well.
Where would I start for this? What would be my variables? I know for a bicycle it's pretty simple to account for rolling friction, but I'm not sure what I'd need to account for. The question: What makes running so much less energy-efficient than bicycling? seems to downplay the importance of vertical motion in energy dissipation.
I guess to further boil down the question I'd ask: where does the energy go when you run?
| Most of the energy of running is used to move the legs of the runner. This isn't very efficient because the legs are heavy and are being quickly accelerated and deaccelerated from the running speed. The leg muscles are very powerful but runners hit a maximum speed when most of the power of their muscles is used up moving their legs back and forth. Bicycling is much more efficient because the legs don't have to be accelerated nearly as much, and the gearing makes better use of the power of the muscles.
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Calculating Rotational Inertia Using Parallel Axis Theorem I am working on the following physics problem and have run into some trouble
The figure above shows particles $1$ and $2$, each of mass $m$, attached to the ends of a rigid massless rod of length $L_1 + L_2$, with $L_1 = 20cm$ and $L_2 = 80cm$. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) particle $1$ and (b) particle $2 \space ?$
My Approach:
So I first considered the net torque when the system is at rest so that I could get to the angular acceleration using the equation $\tau_{net} = I \alpha$. Given that particle $2$ would induce clockwise motion, it is given a negative sign while particle $1$ is given a positive sign because it induces counter-clockwise motion so $\tau_{net} = F_{t2}r_2 – F_{t1}r_1 = I \alpha$. (Where $F_{ti}$ represents the tangent force acting on particle $i$)
Solving for alpha we have that $\alpha = \frac{F_{t2}r_2 – F_{t1}r_1}{I} = \frac{mgL_2 - mgL_1}{I}$. The next step is then to find the rotational inertia. Now after consulting with my solutions manual I see that this can be found by simply treating the fulcrum as the axis of rotation, but I didn't see this approach when solving the problem. Instead I used the parallel axis theorem $I = I_{com} + Mh^2$. Now even though this approach is a waste I'm trying to figure out why I didn't arrive at the same answer anyway, so I've included my work computing the rotational inertia in this way.
Computing Rotational Inertia Using Parallel Axis Theorem
First I computed the center of mass of the rod as follows: $x_{com} = \frac{m_w * 0 + m_w* 0.80 m}{m_w + m_w}= 0.4m$ (Note: I use $m_w$ to denote mass while I use $m$ to denote distance). Next I computed $I_{com}$ as follows: $I_{com} = \Sigma \space m_{wi} \cdot \space r_i^2 = 2\space m_w \cdot (0.4 m)^2$.
And by the parallel axis theorem $I = I_{com} + Mh^2 = 2\space m_w \cdot (0.4 m)^2 + 2\space m_w (0.2m)^2$. After inputting this value into the original equation for angular acceleration I arrive at an invalid value. What have I done wrong? Also, how do we know definitively when to use the parallel axis theorem?
Any help understanding my problem would be appreciated greatly
Note: When either rotational inertia value is inputted into the angular acceleration formula the masses will cancel so using the correct approach and canceling the mass $\frac{I}{m_w} = (0.2m)^2 + (0.8m)^2 = 0.68 m^2$ while using my original approach I have $\frac{I}{m_w} = 2((0.4 m)^2 + (0.2m)^2) = 0.4m^2$
| Your problem is in the calculation of the COM. You first need to define the origin from where, you will get the COM distance by using the formula. Let the left end(particle 1) be the origin. Now, defining all distances from this origin:-
$$x_{com}=\frac{1}{M_{Total}}\Sigma{m_i r_i}=m_w(L_1+L_2)/2m_w=0.5m$$ WHICH MEANS THE com is exactly at the centre of the rod, $0.3M$ right of the fulcrum.
(unlike the $o.4m$ you got).
Now, for $I_{com}=2m_w(0.5)^2$ which gives $$I_{fulcrum}=I_{com}+M_{total}(0.3)^2=2m_w(o.5^2+0.3^2)$$
which should give you the same answer as in the solutions manual.
| {
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Oscillation of a Bose Einstein condensate in a harmonical trap We were asked to try to make a theoretical description of the following phenomenon:
Imagine a 2D Bose Einstein condensate in equilibrium in an harmonical trap with frequency $\omega$. Suddenly the trap is shifted over a distance a along the x-axis. The condensate is no longer in the center of the trap and will start oscillating in the trap.
First I thought about using a 2D trial wavefunction in the Gross-Pitaevski equation or the hydrodynamical equations for condensates, but then we were told that we should actually look at how the energy of the condensate depends on certain parameters (position, width,...) and do something with the fact that, for small deviations of such a parameter, a second order expansion can be made, which will introduce a restoring force.
This makes sense for classical motion, but in this case it confused me, cause I don't know if the energy that is meant here is the original potential energy of the harmonic trap or the Gross-Pitaevskii energy that is calculated with the GP energy functional. This last one, that was calculated in an earlier exercise for a variational Gaussian wave function, turned out to be $E = \hbar \omega \sqrt{1+Na_s}$ (with $a_s$ the scattering length for the interaction energy) and so it doesn't even depend on the position.
Does anyone has got any idea how I should start or approach this theoretical description?
| From the way the question is worded, I would assume you can treat this system as a BEC wavepacket in a potential barrier, where the potential barrier is given by your harmonical trap. Since the trap is harmonic, this is in analogy to the well-known vibrational wavepackets. I would also recommend reading this paper for some more insight:
http://arxiv.org/pdf/quant-ph/9708009v1.pdf
| {
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Infinite Wells and Delta Functions In considering a delta potential barrier in an infinite well, I can just enforce continuity at the potential barrier-it doesn't have to go to zero. Why then does it need to go to zero at the walls of the infinite well? These two cases seem to be very similar to be, I even feel like the well wall is equivalent to a summation of delta functions... Where is my logic faulty?
| You are confining the particle into a region of $|x|< a$ with an infinite potential that extends infinitely far for $|x|\geq a$:
Image source
Since we are confining the particle to a particular region (by applying a potential outside this region), you will never find the particle outside this region, so the wave function, $\psi$, must be zero starting at $|x|=a$.
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Expansion of the Universe: Conversion of gravitational potential energy to kinetic energy? Suppose there is an object floating in space which over time begins to fall toward the source of a gravitational field. As it falls, its motion happens to be such that it gets locked in orbit around the source with a greater velocity than it had before it 'began to fall'. So it's gravitational potential has been converted to kinetic energy. According to relativity, this increased speed should increase the gravitational potential of the object (and therefore the object + the original source of the field), correct?
Does this mean that as a result of this, the expansion of the Universe should slow down slightly (because there is now a slightly greater gravitational potential in the Universe)? And if so, would that imply that the conversion from gravitational potential to kinetic energy is in a sense a conversion between the bulk kinetic energy of the expanding Universe and the local kinetic energy of a test mass?
| Gravitational potential of universe can't be defined now because the universe has been expending due to this we can't find the exist value of gravitational potential.
As the diameter of universe has being expend,gravitational potential become decrease & the gravity due to sun on all the planet would be decrease, because sun is the centre and all eight planet revolving around it. Sun has the great centrifugal force.
According to Einstein diameter of universe given by the equation:
$$D_u=\frac{c^2}{g_u}$$
where $D_u$ is diameter of universe, $c$ is the velocity of light and $g_u$ the gravitational potential of universe.
| {
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Anderson localization in 1d, 2d and 3d Why in 1d and 2d systems, all states will be localized for infinitesimal disorder, but in 3d only states with energy lower below mobility edge will be localized?
| *
*1D : all states are localized
*2D : all states are localized ; the length scale of localization
grows exponentially with E and marginal dimension for the Anderson
transition
*3D : mobility edge ; finite localization for $l_B <~ \lambda/(2\pi)$
Some good references
*
*LSP et al., Phys. Rev. Lett. 98, 210401 (2007)
*J. Billy et al., Nature 453, 891 (2008)
*M. Piraud et al., Phys. Rev. A 83, 031603(R) (2011)
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When can a surface charge density exist? In my syllabus about electromagnetism, they state:
"This surface charge density will not always be present, e.g. when considering two non-conducting dielectrics such surface charge density remains absent. However, at a perfect conductor, a surface charge density will be present. If one of the media (or both) carry a conduction current, a surface charge can also be present (explain why!)"
I don't really see though, how this can be explained. They seem to imply that there is a relation between the conductivity and the possibility of a surface charge, but I can't figure how they are related.
They also state somewhere earlier that there can't be a surface current between two lossy dielectrics. This seems to be analogous to my previous question. I thought that this was due to the fact that dielectrics don't conduct current, but I wonder if someone knows a better explanation here too (that perhaps explains why they explicitly mention that the dielectrics are lossy).
| A simple line is as follows:
*
*Conductor is an equipotential volume. If there were potential difference between any two points, free charges would flow to compensate for this difference, hence produce currents. If there is a current, it produces heat. However, due to energy conservation the heat cannot be produced forever. Hence over time all the currents have to stop. And this is possible only when there is no potential difference between any two points in the conductor.
*Hence $\vec E=-\nabla \phi=\vec 0$ inside the conductor.
*Hence $\nabla\cdot\vec E=4\pi \rho=0$ everywhere throughout the volume.
One cannot write $\nabla \cdot \vec E$ on the surface, because $\vec E$ is not continuous there. However, surface charge can make $\vec E$ be non-zero outside the boundary and zero inside the volume.
The reasoning applies to conductors and conducting materials, but does not apply to dielectrics. However, this does not mean, that dielectrics cannot have surface charges.
| {
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Charged spheres - help with method to work these out? Can anyone demonstrate how to get the answers to these questions? I'm just interested in the method I need to use in order to obtain the correct answer no matter what the values are.
Three small spheres are placed at fixed points along the x-axis, whose positive direction points towards the right.
Sphere A is at x = 36.0 cm, with a charge of –8.00 μC.
Sphere B is at x = 41.0 cm, with a charge of 9.00 μC.
Sphere C is at x = 46.0 cm, with a charge of –3.00 μC.
a) Calculate the magnitude of the electrostatic force on sphere B.
b) Sphere B is now removed:
What would be the magnitude of the electric field at the point where sphere B was located?
c) Sphere B is still missing. Give the x-coordinate of the point on the x-axis where the field due to spheres A and C is zero
| Since the charges are distributed on spheres they can be considered as point charges.
Both the charges A and C will attract charge B so the force is given by subtracting these two forces vectorially
The electrostatic field due to A and C at point of B comes out to be 18 × 10^6 towards A.
Correspondingly the force experienced by B would be 162N towards A.
The field due to both charges would be zero at 6.2 cm from A towards B which would be 42.2 cm.
Now the interesting part is that even if we suppose that the charges are held together by some unknown uninterferring forces we do not bother to think about the induction i myself was too much bothered by this till sometime back when i read an article that the induction that could actually change your answer and that too drastically comes at extremely close distances. Normally it is just negligible, i would encourage you to read this articld here as it would clarify even more that why I could solve your question so easily without the need of some supercomputer to account for the induction.
http://www.nature.com/news/like-attracts-like-1.10698
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Does a heavy body move with the slightest force on a frictionless surface? If I apply horizontal force on a body resting on the ground, my force will be opposed by the frictional force and the body will accelerate at the point where my force exceeds the force of friction = $\mu\, \mathrm{N}$ ($\mathrm{N}$ being the normal and $\mu$ being the coefficient of friction). In this case, the threshold value will be $\mu mg$ where $m$ is the mass of the resting body since $\mathrm{N} = mg$.
Is the following statement then true: Regardless of the mass/weight of the body, if the body is placed on a frictionless surface, the body will move with the slightest force?
| When a force is applied to a body initially at rest for a finite amount of time, the velocity of that body is given by conservation of momentum:
$$m\Delta v = F\Delta t$$
When initial velocity is zero, $\Delta v = v$ - the final velocity is equal to the change in velocity. So we can write
$$v = \frac{F\Delta t}{m}$$
From this it can be seen that regardless of the size of $F$, $m$ and $\Delta t$, as long as all of them are finite, there will be a final velocity $v$.
Whether you can actually measure this velocity is a separate topic...
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Intuitively, how can the work done on an object be equal to zero? To my understanding the work done on an object is defined mathematically as:
$$W = \vec{F}\cdot\vec{S}=|\vec{F}||\vec{S}|cos\theta$$
This, I understand. My problem is that I don't understand that if the angle $\theta$ is 90 degrees how can the work done by $\vec{F}$ on the object is zero. For example; say you have a particle and the direction of the displacement is directly to the right, and you also have a force vector acting on the particle that is straight up(like the normal force on a box that is standing on a flat surface). How is it possible that the force vector is not doing any work? Must the particle not take a different route because of the force vector acting upward on the particle, like if you add the vectors together?
There has to be something wrong with my reasoning, but what is it?
| Because it's not any work, but the work done by a force that produces a displacement.
In the scenario you describe, somehow that force is not doing any work on the particle. This could be because the particle is restricted by another force to not go perpendicular and then the sum of forces in the perpendicular direction is zero.
In the second scenario, with the box and the normal force, it's the same. That force doesn't do any work since in the direction of that force there is zero movement. Which is analogous to say that the cosine of the angle between the displacement and such force is 90°.
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Does our existence cost us energy? When something needs to inform its presence, such as the electromagnetic presence of charged particles , or the gravitational presence of particles due to their mass. Is this made by sending information of its existence propagated in space via some kind of electromagnetic waves or hypothesised gravitational waves.
So as a part of telling others of their presence to others, do the particles constantly lose their energy ? I understand that they could also gain energy in same way, but if a particle was completely isolated would it lose its existence and turn into pure energy over time ?
I understand that the mass and energy would together remain constant, but my question is that whether something would happen to a particle or would it remain a particle ?
|
Anything when it needs to inform its presense such as electromagnetic presense of charged particles and gravitational presense of particles due to their mass does so by sending information of its existencs propogated in space via means of electromagnetic waves or hypothesised gravitational waves.
Not true. Maxwell's equations have wave solutions and static solutions. They are two different things. Similarly, the Einstein field equations have wave solutions and static solutions, and they're different things.
So as a part of telling others of their presense to others, do the particles constantly lose their enegy ?
No. Static fields don't transport energy.
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Structure factor of crystals (X-ray crystallography) How can one prove that the degree of each node in a distance graph must be at least four in order to obtain a unique solution to an exact distance geometry problem with sparse distance data?
The example in the Wikipedia article has three degrees for each node, which is less than four, but I am guessing the information derived in this case is not the same as what I am trying to prove; i.e. it is not a unique solution (http://en.wikipedia.org/wiki/Distance_geometry).
I am pretty confused, and any help would be greatly appreciated!
| The smallest possible case where you get ambiguity is that of 5 nodes. In the complete graph on these nodes every node has degree four. If we remove a single edge we get ambiguity (in general): it contains a complete graph on 4 nodes (the edges of a tetrahedron) and the fifth node is connected to three of the others, but not the fourth. If we reflect the node in the plane of those three nodes, the distances in the graph don't change, but those in the arrangement do.
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Angular Momentum of Two Non-interacting Particles I'm reading a book (An Introduction to Mechanics by Kleppner) where they calculate the angular momentum $l$ of a system of two non-interacting particles, but I don't understant what are they doing.
Consider two non-interacting particles with $m_1$ and $m_2$ moving toward each other with
constant velocities $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$. Their paths are offset by distance $b$, as shown in the sketch.
In general the energy of a system of two particles, relative to the center of mass, can be written
$$E=\frac{1}{2}\mu v^{2}+U(r), \qquad (1)$$
being $\mu$ the reduced mass. Or, using $v^2=\dot r^2+r^2 \dot \theta ^2$ and $l=\mu r^{2}\dot{\theta}$,
$$E=\frac{1}{2}\mu\dot{r}^{2}+\frac{l^{2}}{2\mu r^{2}}+U(r), \qquad(2)$$
So the book calculates the angular momentum $l$ of the system using (1) and (2) and the fact that for a system of two non-interacting particles $U(r)=0$.
The book says:
The relative velocity is
$$\mathbf{v}_{0}=\dot{\mathbf{r}}:=\dot{\mathbf{r}}_{1}-\dot{\mathbf{r}}_{2}=\mathbf{v}_{2}-\mathbf{v}_{1} $$
with $\mathbf{v}_{0}$ is constant since $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ are constant (which I think it's right).
The energy of the system relative to the center of mass is (see equation 1)
$$E=\frac{1}{2}\mu v_{0}^{2}$$
or (from equation 2)
$$E=\frac{1}{2}\mu\dot{r}^{2}+\frac{l^{2}}{2\mu r^{2}}$$
Now here comes the argument I really don't understand:
Argument a): When $m_1$ and $m_2$ pass each other, $r=b$ and $\dot r = 0$ (But the book said earlier that $\mathbf{v}_{0}=\dot{\mathbf{r}}$, a constant vector in time!). Hence
$$\frac{l^{2}}{2\mu b^{2}}=\frac{1}{2}\mu v_{0}^{2} \qquad (3)$$
I'm confused because I could apply the same condition in an arbitrary point, say $r=2b$, then $\dot r=0$, and (3) wouldn't be valid.
So, my question is: Is the argument a) wrong?
If not, please help me to clarify the ideas behind this.
| Elaborating on Trimok's comment, you are mixing up $\dot{\mathbf r}$ and $\dot r$. $\dot{\mathbf r}$ is the rate of change of the separation vector, while $\dot r$ is the rate of change of the separation distance.
We can quickly see they are different. By definition, we know that $ \mathbf r = r \,\hat{\mathbf r} $. Take the time derivative of this:
$$ \dot{\mathbf r} = \dot r\, \hat{\mathbf r} + r \, \dot{\hat{\mathbf r}} $$
In this case, the direction between the two is changing, i.e. $\dot{\hat{\mathbf r}}\neq 0$, so both terms on the right-hand side of this equation contribute to the value of $\dot{\mathbf r}$. In particular, $\dot{\mathbf r}$ can remain constant while $\dot r$ changes, so long as the above equation is obeyed.
| {
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Why does the specific thrust of an ideal turbojet drop with increasing compressor pressure ratio? As the pressure ratio increases in an ideal turbojet (fixed flight Mach number), the specific thrust, $\frac{F}{\dot{m_{air}}}$ rises, reaches a peak for small pressure ratios and then starts to decrease with increasing pressure ratios in the compressor.
It seems counter-intuitive to assume that the specific thrust should decrease with increasing compressor pressure ratio since we're generating a greater pressure gradient, therefore should be able to generate more thrust.
| Increased compressor pressure ratio means increased temperature in inlet of the turbine.
Turbine has limits $T_{max}$ on inlet temperature (turbine could melt). To keep temperature in operational limits the fuel/air mass ratio $f$ is lowered at high pressure ratios.
Lowered fuel/air mass ratio $f$ means that less chemical energy is released per mass unit of the air.
| {
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Resolution of identity in interacting QFT $\mathbf{Background:}$ Consider a free scalar field $\phi$ ($\mathcal{L}_0 = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi + \frac{1}{2}m^2 \phi^2$).
In the Hamiltonian viewpoint, this system has a Hilbert space $\mathcal{H}_0$ (the Fock space). We can write down a resolution of the identity in $\mathcal{H}_0$ (in the Schrodinger picture):
$\begin{align}
I = |0 \rangle \langle 0 | + \int dp\ |p \rangle \langle p | + \frac{1}{2!} \int dp_1 dp_2\ |p_1 p_2 \rangle \langle p_1 p_2 | +\ \cdots
\end{align}$
$\mathbf{Question:}$ If we add an interaction term to the Lagrangian (such as $\lambda \phi^4$), is $I$ still the identity operator in the new Hilbert space $\mathcal{H}$?
(To be clear, I mean exactly the same $I$, with the free vacuum and free particle states.)
$\mathbf{Motivation:}$ The analogous thing holds in QM. A non-relativistic particle moving in a potential $V$ has a Hilbert space $L^2({\mathbb{R}})$, so by Fourier transform, $\int dp\ |p \rangle \langle p |$ is a resolution of the identity. (Where $\langle x | p \rangle = \frac{1}{\sqrt{2\pi}} e^{i p x}$ are free particle states.) The Hilbert space is the same regardless of $V$.
| Yes, that's still a resolution of the identity.
But knowing this isn't likely to make your life easier. The observable algebra for the free theory and the observable algebra for the interacting theory are generated by unbounded operators. These algebras do not have the same domains of definition. The 'free-particle' resolution of the identity you wrote down is convenient when you're dealing with the domain where the free field operators have a well-defined action, but it's generally singular when you're trying to study vectors which live in the domain of the interacting field operators.
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Explain reflection laws at the atomic level The "equal angles" law of refection on a flat mirror is a macroscopic phenomenon. To put it in anthropomorphic terms, how do individual photons know the orientation of the mirror so as to bounce off in the correct direction?
| According to quantum electrodynamics (QED), light can be thought of as going along all paths. However, the only paths that do not experience destructive interference are those in the neighbourhood of paths with stationary (e.g., minimal) action (time), which, in your case, is the "equal angles" path.
I strongly recommend reading Feynman's QED: The Strange Theory of Light and Matter. In the link you'll also find a link to video.
So, with QED in hand, anthropomorphically, photons don't need to know where to go, because they go everywhere. :)
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Does antimatter curve spacetime in the opposite direction as matter? According to the Dirac equation, antimatter is the negative energy solution to the following relation:
$$E^2 = p^2 c^2 + m^2 c^4.$$
And according to general relativity, the Einstein tensor (which roughly represents the curvature of spacetime) is linearly dependent on (and I assume would then have the same mathematical sign as) the stress-energy tensor:
$$G_{\mu \nu} = \frac{8 \pi G}{c^4}T_{\mu \nu}.$$
For antimatter, the sign of the stress-energy tensor would change, as the sign of the energy changes. Would this change the sign of the Einstein tensor, causing spacetime to be curved in the opposite direction as it would be curved if normal matter with positive energy were in its place? Or does adding in the cosmological constant change things here?
| See also: What is anti-matter?
Currently there is no reason to believe/require antimatter has negative mass. It should therefore behave exactly the same in a gravitational field.
The matter-antimatter distinction is pretty arbitrary. We found protons/neutrons/electrons first, so particles of the same families that exhibit similar behavior are "matter", and those with certain properties (charge, baryon number, or something else, depending on the family) as opposite would be antimatter. We could call positrons as matter and electrons as antimatter and nothing would change except for our definition of lepton number (and the labels of the muon/tau).
When Dirac calls it a negative energy solution, he's looking at the case where we have a sea of ground state matter, and we excite one. The "hole" left behind by the excited particle behaves like the particle itself, but can recombine with an excited particle with no net energy change so one can view it as having a negative energy.
In this case, the hole does have negative mass because it is in a "sea" of positive-massed particles, and removing these leads to a hole with negative mass. And it behaves similarly from the POV as gravity.
In the general case, an antiparticle has the same energy as a particle.
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How does that Boltzmann distribution interact with entropy? In an ideal gas, the Boltzmann distribution predicts a distribution of particle energies $E_i$ proportional to $ge^{-E_i/k_bT}$.
But, doesn't entropy dictate that the system will always progress towards a state of maximum disorder? In other words the system evolves towards a macro-state which contains the maximum possible number of indistinguishable micro-states. This happens when all particles have the same energy, which seems to contradict the Boltzmann distribution.
I'm pretty sure I've misinterpreted entropy here, but I'd be please if someone could explain how!
| In any system in equilibrium, the entropy of such system is the maximum given a set of constrains. If you think of a microcanonical ensemble, the total energy is fixed while in an canonical ensemble of particles the temperature is the one being held constant.
This distribution probability you mention, is for a canonical situation. Given that the temperature is being held fixed, the many different microstates available for such macrostate are given by that exponential function which depends of the energy of the particles and the temperature.
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Why would an object appear a different size when in water? A friend of mine has a homework question and we're having some trouble figuring out what physical mechanisms come into play for this.
An underwater swimmer sees a spherical air bubble that appears to have a diameter $d=$ 1.5 cm. What is its actual diameter?
We are having trouble, specifically, thinking of physical mechanisms that would change the apparent sizes of objects when seen underwater. Is it caused by refraction at the curved surface of the bubble?
| Objects do appear larger (or equivalently nearer) underwater when wearing a mask or goggles. See the image below for confirmation of this fact. Why is this?
The interface between the water and your mask obeys Snell's law which can be written, in the small angle approximation, as
$$
n_1\theta_1=n_2\theta_2.
$$
Since air has an index of refraction of essentially 1 and water has an index of refraction of 1.33 the angle from which the rays of light reach your eyes is larger than the angle they would in air. This makes the angular size larger to your eyes which makes the object look larger relative to how they would look in air. This effect is shown qualitatively in the ray diagram below. The index of refraction of the glass interface does not play a role as long as 1) the thickness is much smaller than the distance to the object and 2) the two surfaces of the glass are parallel to each other.
You can get an approximate answer as to how much larger things would look by assuming that the distance between your mask and the object is much larger than the distance between the mask and your eyes. In this case the angle which the ray hits the mask from is roughly the same as it would be in air, and the angle it hits your eye with is simply $n_2/n_1=1.33$ times that. So, the approximate magnification is 1.33 in water. For objects which are closer up you would need to relax the small angle approximation as well as take the distance between the mask and your eyes into account.
| {
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How does energy transfer between B and E in an EM standing wave? I'm trying to understand how an electric field induces a magnetic field and vice versa, its associated energy, as well as relating it to my understanding of waves on a string.
Using a standing wave as an example, I came up with the equations
$\vec{E}=A\sin(\omega t)\sin(kx)\hat{y}$
$\vec{B}=\frac{Ak}{w}\cos(\omega t)\cos(kx)\hat{z}$
I checked them against Maxwell's equations, and they're self-consistent. At time 0, this reduces to:
$\vec{B}=\frac{Ak}{w}\cos(kx)\hat{z}$
Since the electric field is 0, based on the Poynting vector, there's no energy transfer at this time. At this time, at a node where $\vec{B}=0$, there's neither electric field nor magnetic field. If there's no energy transfer, and no energy stored in either field, then how can an electric field exist at this point at some time later? How is the energy stored, or transferred from elsewhere?
| The variation of the fields over a quarter of a period looks something like this.
In the left diagram it is the magnetic field which is storing the energy of the system whilst in the right hand diagram it is the electric field.
The exchange between the electric field and magnetic field follows Maxwell's equations - Faraday and Ampere.
As with all standing waves you can think of then as the superposition of two travelling waves carrying energy and in this example at a node the net transport of energy is zero for all time whereas at other positions it is the average value of the Poynting vector over half a period which is zero.
| {
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Does the moon affect the Earth's climate? So, this morning I was talking to a friend about astronomical observations, and he told me that lately there has only been good weather when there was a full moon in the sky, which was a shame.
I jokingly said: 'maybe there's a correlation!', but then I started thinking: wait, if the moon can affect the oceans, why shouldn't it also make an impact on the atmosphere, which is just another fluid. So... are there atmospheric tides? Does the moon affect the weather or the climate in a significant way?
| There are tides in rocks and those tides affect volcanoes and volcanoes can affect climate.
Scientists have not yet found any correlations between land tides and earthquakes but they have found a relationship between the tides and volcanic eruptions because of the movement of magma or molten rock inside volcanoes (USGS). 1
| {
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A simple conjecture on the Chern number of a 2-level Hamiltonian $H(\mathbf{k})$? For example, let's consider a quadratic fermionic Hamiltonian on a 2D lattice with translation symmetry, and assume that the Fourier transformed Hamiltonian is described by a $2\times2$ Hermitian matrix $H(\mathbf{k})=a(\mathbf{k})\sigma_x+b(\mathbf{k})\sigma_y+c(\mathbf{k})\sigma_z $ and has a finite energy gap, then the Chern number $N$ can be determined.
If $H(-\mathbf{k})=H(\mathbf{k})$ holds for all $\mathbf{k}\in BZ$, then the Chern number $N$ is always an even number, am I right? This seems to be true from the geometrical interpretation of Chern number as a winding number covering a unit sphere, but I have not yet found a rigorous mathematical proof.
Remark: The necessary condition finite energy gap ($\Leftrightarrow$ The map $(a(\mathbf{k}),b(\mathbf{k}),c(\mathbf{k}))/\sqrt{a(\mathbf{k})^2+b(\mathbf{k})^2+c(\mathbf{k})^2}$ from BZ(2D torus) to the unit sphere is well defined) is to ensure that the Chern number/winding number is well defined.
| I just found a relative rigorous argument supporting my conjecture:
The Chern number $N=\frac{1}{2\pi}\int _{BZ}b(\mathbf{k})$, where $b(\mathbf{k})$ is the Berry curvature. Since $H(-\mathbf{k})=H(\mathbf{k})$, it's easy to show that $b(-\mathbf{k})=b(\mathbf{k})$, accordingly, we can divide the $BZ$ into two halfs called $BZ_1$ and $BZ_2$, therefore, $N=\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})+\frac{1}{2\pi}\int _{BZ_2}b(\mathbf{k})=2\times \frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$. Now there are two ways to show that $\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$ is an integer,
(1)Due to the relation $b(-\mathbf{k})=b(\mathbf{k})$ and periodic structure of the $BZ$, the half Brillouin zone $BZ_1$ is topologically equivalent to a sphere, and the 'flux' through a sphere(closed surface) $\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$ must be quantized as an integer;
(2)Since $H(-\mathbf{k})=H(\mathbf{k})$, the eigenfunction $\psi(\mathbf{k})$ is also even(i.e. $\psi(-\mathbf{k})=\psi(\mathbf{k})$), then the Berry connection $\mathbf{a(k)} \propto \left \langle \psi(\mathbf{k})\mid \bigtriangledown_{\mathbf{k}} \psi(\mathbf{k})\right \rangle$ is odd, thus, it's easy to show that $\int _{BZ_1}b(\mathbf{k})=\oint _ {\partial BZ_1}\mathbf{a(k)}\cdot d\mathbf{k}=0$(where $\partial BZ_1$ is the boundary of $BZ_1$), however, to be consistent, the number '0'(Berry phase) here should be understood as $2\pi\times integer$.
Remark: The key point in our argument (1) is that the points $\mathbf{k}$ and $-\mathbf{k}$ are equivalent, and hence the half $BZ$ is topologically equivalent to a sphere which is a closed surface.
| {
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What constitutes displacement current? In the chapter electromagnetic waves I was introduced with the concept of displacement current inside a capacitor. Since the region inside the capacitor is a dielectric there is no charge carriers in it. Then, what constitutes displacement current if it flows over there?
I know that it is a type of electric current formed by changing electric field.Conduction current involves flowing electrons and there is no electrons capable of moving about in a dielectric. But, then what is actually happening when we say displacement current?
| Displacement current is not formed by a changing magnetic field. Rather, it is due to a changing electric field (between the two plates of a capacitor, maybe).
| {
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Thin lens formula Can someone help me or guide me how the thin lens formula:
$$\frac{1}{s_1}+\frac{1}{s_0}=\frac{1}{f}$$
can be proven?
I was trying to prove it on my own using similar triangles, only to fail.
| A lot depends on how much you want to learn, just to teach elementary (geometrical) optics. There are many good books, but often they cover too much that you aren't interested in.
By far the best textbook on actual geometrical optics, relating to lens design, including aberrations etc, is one that as a mathematician, you would enjoy for yourself.
Applied Optics & Optical Design, By A. E. Conrady written in 1926 , taught all of the world's good optical designers of WW-II, on all sides of the conflict. It is fully mathematical, but high school level. It's a two volume Dover Press paper back, that you can buy at Amazon, or B&N for peanuts. If you get a free account for the Internet Archive, you can also borrow it hourly.
High school, algebra, geometry, and trigonometry required. NO calculus needed, and it will show you lens formulae that are much more useful, than the one you give above.
I should add, that all of the computations are done with logarithm tables, which was the best in those days, but now you simply use a calculator.
I managed to convince a good number of employers for over 50 years, that I knew a little bit of optics; solely because of Conrady.
His daughter, Hilda Conrady Kingslake, was the wife of chief Kodak optics guru, Rudolph Kingslake, and she was for years, historian for the Optical Society of America. Sadly, both are now doing optics in the clouds.
| {
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Can air bubbles sink at extreme depths? I was thinking earlier about air bubbles in water. if you had a bubble of air (say in a balloon) then as you take it down in water the bubble shrinks because of the pressure and because it is compressible. This means its density increases.
Water on the other hand being incompressible remains the same density at whatever depth you are at (I don't know if it is truly incompressible but I am assuming if it is at all compressible this is insignificant).
Does this mean that there is a depth at which an air bubble would be denser than the surrounding water and thus sink instead of float?
Would the gas at this density be doing things like turning into a liquid? Is the depth such that this is not even vaguely realistic? I feel like the situation is probably a lot more complicated than it is in my head which says that you would get air bubbles sinking at sufficient depth?
| We need a phase diagram for the gas in the balloon. I found one for CO2
The pressure at the bottom of the ocean can be estimated as 1 atmosphere every ten meters depth. For 4000 meter that is 400 atmospheres.
Temperatures at the bottom of the ocean are above icing, a few C, so from the diagram a balloon with CO2 released by a bathysphere at a suitable depth , would start turning to liquid as the pressure equalizes with the water pressure from about 100 meters (above 10 atmospheres), for temperatures usual to the sea.
Since CO2 is heavier than H2O I would expect it to sink to the bottom.
Edit after comments:
Better charts readable with accuracy can be found in this IPCC report: and the comment below is right that higher atmospheres are necessary for CO2 liquid at around zero temperatures, over 200. In addition for sinking the density versus curvature curve in the same link has to be consulted where we see that the two phase region extends at these temperatures up to 900 atmospheres, so sinking will not take place before then.
| {
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Classical Wave Equation - Approximations I don't understand the derivation of the wave equation given below -
$$T \sin (\theta _1) - T \sin (\theta ) = T\tan (\theta _1 )-T\tan (\theta ) = T \left. \left(\frac{\partial f}{\partial z} \right|_{z + \Delta z} - \left. \frac{\partial f}{\partial z}\right| _z \right) = T \frac{\partial ^2 f}{\partial z^2} \Delta z$$
I understand that the small angle approximation was used, but I'm at a loss for figuring out we turned $\tan$ into a derivative, and then after made it become a second derivative.
The derivative of $\tan \theta$ is of course $\sec \theta$ which is equal to $\frac{1}{cos \theta}$, which was taken with respect to $\theta$, maybe there's a way to use the chain rule to find $\partial _z f$?
| $tan(\theta) = \frac{opposite}{adjacent} = \frac {rise}{run} = gradient$
| {
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Chinook Helicopter Torque The Chinook Helicopter has 2 rotors to counteract the torque generated by spinning the blade.
Theoretically, could you use a smaller "back" rotor that is farther away from the main rotor to achieve the same result, ie no twisting?
| The interchangeability of rotors for tourque might be nearly impossible because of the ma in f=ma.The rotor speeds would have to continuously vary ideally.Now if it were just a question of lift it would be in aronautical engineering. But then again as they say ,nothing is really impossible,you just have to find a way of doing it.
| {
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Proper name for a thermodynamic process with constant internal energy $U$ Back in the day I learned that a few special thermodynamical processes have special names.
For example, if one keeps $P$ constant, the process is called isobaric, if one keeps $T, V$ or $S$ constant, one gets, correspondingly, isothermic, isochoric or isentropic processes. Similarly, if one keeps $\dfrac{\mathrm{d} \ln P}{\mathrm{d} \ln \rho}$ constant during the process, it is called polytropic, and if $\delta Q = 0$ at any time, the process is called adiabatic.
Now, the question: what is the process called, if one keeps internal energy $U$ constant?
| The process with constant internal energy is called free expansion.
| {
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One way insulation? I know from basic physics lessons that a box painted black will absorb heat better than a box covered in tin foil. However a box covered in tin foil will lose heat slower than a black box.
So what is the best way to conserve the temperature of a box? (aiming for 0 degrees Celsius inside the box when it's -60 outside).
I mean would painting the outside of the box black, and having tin foil on the inside work? So the box can absorb heat better (black paint) and the tin foil making it harder for heat to escape?
| make 1 box, with the inside and outside as reflective as possible. make a second box the same way except bigger. using magnets on all sides to 'levitate' box 1 inside of box 2. suck all the air out of box 2.
| {
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Initialize a traveling wave in a 1D gas? I am trying to initialize a traveling wave for a 1d simulation as one can see from the attached figure.
Such that it will be traveling to the right.
However, I cannot initialize the right velocity profile, and this makes the initial pressure distribution tends to be more uniform to reach the same pressure of the surrounding fluid !
Can any one provide some support?
| We are trying to get the simple wave solution, so one can assume the dependence of the functions defining the solution (namely $u$, $p$ and $\rho$) only on a single combination of variables $x$ and $t$. In case of weak sound wave this combination would be $x - c t$, but nonlinear effects would makes this more complicated. Nevertheless, we still can choose one of the functions, for example $\rho$, as an independent variable on which the other two would depend and write
$$
\rho = \rho(x,t),\quad p= p(\rho), \quad u = u(\rho) .
$$
We can than substitute these into continuity equation and Euler equation:
$$
\dot{\rho}+\rho' u + \rho \frac{d u}{d\rho} \rho ' = 0, \tag{1}
$$
$$
\frac{du}{d\rho}\dot{\rho}+ u \frac{du}{d\rho} \rho' + \frac{c^2(\rho)}{\rho}\rho ' =0,\tag{2}
$$
where $\rho'= \dfrac{\partial \rho}{\partial x}$, $\dot{\rho}= \dfrac{\partial \rho}{\partial t}$. The local speed of sound is defined by $c^2(\rho)= \dfrac{d p}{d \rho}$ and could be found using adiabatic equation for an ideal gas.
For the initial conditions on the velocity $u$ we could solve (1) and (2) for $\frac{du}{d\rho}$ (also eliminating $\dot{\rho}$):
$$
\dfrac {du}{d\rho} = \pm \frac{c(\rho)}{\rho},
$$
where two sign choices correspond to simple waves traveling to the right (+) and left (-). Integrating we obtain:
$$
u = \pm \int \frac{c}{\rho} d\rho = \pm \int \frac{dp}{\rho c}.
$$
The final explicit result could be obtained by using the adiabatic process equation: $p \rho^{-\gamma}= p_0 \rho_0^{-\gamma}$.
| {
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Thermionic Emission: Kinetic Energy Distribution of Emitted Electrons I am having a conceptual problem understanding the kinetic energy of thermionically emitted electrons.
I know that in order to escape the surface the electrons must have energies of at least the Fermi Level plus the work function ($\varepsilon \ge \varepsilon_F + \phi$). However, is all that kinetic energy converted to potential energy once it leaves the material so that once it leaves the surface an electron has kinetic energy $T = \varepsilon - \varepsilon_F + \phi$?
| The electron maintains the total energy $\epsilon$ that it had while inside the material. Inside the material it has some kinetic energy and some potential energy, but we don't really care exactly what those are since we know their sum. The point is, once it enters the vacuum, its total energy is still made up of kinetic and potential energy. Since the vacuum potential energy is $U = \epsilon_F + \phi$ just outside the surface, then indeed the kinetic energy is $T = \epsilon - U = \epsilon - \epsilon_F - \phi$. Of course if there are any electric fields in the vacuum, then the kinetic energy of the electron will change as it moves away from the surface.
A simple way to understand it is that the minimum kinetic energy is zero, since the electron can't go into the vacuum if that would give it less than zero kinetic energy. The actual distribution of energies is a complex topic from what I understand, and material dependent.
| {
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Molecular rotation - Energy levels for an asymmetric molecule For a molecule with spherical symmetry, the energy level of rotation for quantum number $J$ is:
$$E(J)=\frac{J(J+1)\hbar^2}{8\pi^{2}I}$$
"$I$" is the Moment of inertia for the molecule ($I_{x}=I_{y}=I_{z}$ for a molecule with spherical symmetry)
For an asymmetric molecule however, $I_x$, $I_y$, $I_z$ are all different. What would then be the expression for energy levels in terms of quantum numbers? Thanks.
| Download the file at https://www.dropbox.com/s/5l3lvqbal1ch2gm/Asymmetric%20Top.nb and run in Mathematica.
Executing the command en[3,0], for example, yields the energies and vector representations of the $J=3,m=0$ wavefunctions in the basis $$\{|3,-3,0\rangle,|3,-2,0\rangle,|3,-1,0\rangle,|3,0,0\rangle,|3,1,0\rangle,|3,2,0\rangle,|3,3,0\rangle\}$$ where the notation $|J,k,m\rangle$ is used. If you do not provide numerical values for $A_e,B_e,C_e$ then the code will attempt to yield exact analytic forms when it computes the eigensystem, which becomes impossible for $J>4$ or so and the program will stall. If you provide numerical values, it executes nearly instantly regardless of $J$. If you know your rotational constants, this can be used to find all the energy-levels quickly, or at least, the energy levels in the absence of rovibronic coupling.
| {
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Why are high voltage lines “high voltage?” If I have two spheres of the same size and one sphere has a small amount of charge compared to the other that has a lot more charge, then clearly the sphere with the larger charge has a larger voltage (relative to the ground). My question is do high voltage power lines have a lot more charge that is placed on them? Is that what gives them the high voltage? I think I have a grasp of the step up stations that use transformers to kick up the voltage from power plants.
This question seems almost silly to me but I have been struggling with this for a long time. I’ve done several searches online and I am not able to find answers. If there is a link that someone can provide, I much appreciate it.
| The voltage in electric cables has almost nothing to do with the amount of electric charge on or in some portion of the cable.
The voltage is a measure of electric potential.
Charge carriers are present in all conductors, even those that have no voltage across them. Applying a high voltage does not alter the number of charge carriers in any section of cable.
The presence of voltage (i.e. an electric field) is what causes the randomly moving charge carriers to, on average, slowly drift in one direction - but the voltage is not produced by the presence of charge carriers in the conductor.
There are smaller effects but these are distractions from the fundamentals.
Most power lines on land are AC, but most undersea power lines are DC, so AC effects can be disregarded when considering the fundamentals of what is meant by voltage or high voltage and it's relation to charge.
| {
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Angular Displacement If something is rotating about a point and it covers a complete circle, should we take its angular displacement as 360 degree or 0?
Please give link to some established material on this subject which defines your answer, whether it is that it should be taken as 0 or 360 degrees.
Question that led to this problem : The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. 1. What is its angular acceleration, assuming the acceleration to be uniform, 2. How many revolutions does the wheel make during this time.
In the solution to above question the author(s) for solving part 2 have used equation of motion and mentioned that they have obtained angular displacement for it, it basically implied that the equations of motion used for rotation provide angular displacement which does not become zero on returning to original point which is not followed when we consider it analogous to displacement of linear motion.
Addendum : Even here ( Angular displacement and the displacement vector ) the selected answer says that on completing the circle, angular displacement is 360 degrees, is there some established text to support this ?
Similarly here ( http://www.ask.com/question/calculating-angular-displacement ) something else is said in the answer, is then angular displacement ambigous and hence has no correct definition, if there is please guide me to some established text.
| An angle is a ratio. The ratio of arc length to radius. So if you want to consider zero arc length or full circle is really up to you. It depends on the situation. If you are using the angle for orientation it does not matter, but if you are using it for angular displacement it depends on the initial conditions.
| {
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Could there have been two "Big Bangs"? A couple of years ago, I remember seeing a documentary on the big bang theory. The theory presented was that to explain the cosmic microwave background radiation, there needed to have been two big bangs. Is this theory legitimate? I've tried searching for details without success.
My question is essentially, is it possible for there to have been two swift metric expansions of space that were the "Big Bang"?
| It is not really said that there were two big bangs. However there is a theory that multiple big bangs occurred during the beginning of the universe. Each of these gave rise to multiple bubble universes. So whether two big bangs happened, well it could be legitimate however it is not widely accepted.
| {
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Infinitesimal Lorentz transformation is antisymmetric The Minkowski metric transforms under Lorentz transformations as
\begin{align*}\eta_{\rho\sigma} = \eta_{\mu\nu}\Lambda^\mu_{\ \ \ \rho} \Lambda^\nu_{\ \ \ \sigma} \end{align*}
I want to show that under a infinitesimal transformation $\Lambda^\mu_{\ \ \ \nu}=\delta^\mu_{\ \ \ \nu} + \omega^\mu_{{\ \ \ \nu}}$, that $\omega_{\mu\nu} = -\omega_{\nu\mu}$.
I tried expanding myself:
\begin{align*}
\eta_{\rho\sigma} &= \eta_{\mu\nu}\left(\delta^\mu_{\ \ \ \rho} + \omega^\mu_{{\ \ \ \rho}}\right)\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\
&= (\delta_{\nu\rho}+\omega_{\nu\rho})\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\
&= \delta_{\rho\sigma}+\omega^\rho_{\ \ \ \sigma}+\omega_{\sigma\rho}+\omega_{\nu\rho} \omega^\nu_{{\ \ \ \sigma}}
\end{align*}
Been a long time since I've dealt with tensors so I don't know how to proceed.
| Note that if you lower an index of the Kronecker delta, it becomes the metric:
$\eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho}$
And in your last step you got a wrong index. It should be $\omega_{\rho\sigma}$, not $\omega^{\rho}_{\sigma}$.
Then, the metric terms cancel and you neglect cuadratic terms.
That should be enough to solve it.
| {
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What is negative Energy/Exotic Energy? So I have been researching around a little as I am highly interested in Astrophysics and I came across an energy I have never heard of before; negative energy also commonly known as exotic energy. Now I started to research this however I found the concept rather hard to grasp due to a simple lack on information around on the Internet.
Could somebody kindly explain (if possible using real life analogies) what exactly negative energy is or at least the whole concept/theory behind it.
| Negative energy occurs when the energy level for a given space is below that which is considered zero energy. A zero energy space is not really zero but is always full of some virtual particles popping in and out of existence.
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Can sound be propagated without initial mechanical interference? I have researched up a little on sound, and it seems that sound is a mechanical wave that propagates through the air as energy, and that is how we hear it through our ears.
Depending on the medium's density, the type of wave, and the amplitude, among other factors, we determine how loud it will sound, etc.
So my question is how does a speaker do this with just a diaphragm, two magnets, etc.?
|
In the future, please search for these answers first by e.g. looking
up "how speakers work". A quick search will bring up what you want.
An electromagnet in the speaker cone is usually fixed next to a permanent magnet. Electricity is channeled through it to create a magnetic field which, by changing polarity, can flip the poles of the magnetic field. As the polarity of electricity changes, the electromagnet's field fluctuates. This causes it to vibrate because of the interference from the permanent magnet, which has a constant magnetic field. The result is a mechanical wave which oscillates through the air.
Just to be clear, you are not hearing the kinetic energy which moves the molecules in a wave, but you are interpreting the vibrations as it reaches and subsequently begins to affect your ear drum.
| {
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Mean kinetic energy according to the WKB approximation
Show that in the WKB approximation, the mean kinetic energy $T_{n1}$ in a bound state $\psi_n$ in a potential $V(x)$ is given by
$\langle T_n \rangle = \frac{1}{2}\left(n+\frac{1}{2}\right) \frac{dE_n}{dn}$
This is a homework problem and I'm having trouble understanding what it's asking.
*
*What does the notation $T_{n1}$ mean?
*What is the meaning of the expression $\frac{dE_n}{dn}$? Bound states should have discrete spectra.
Could someone provide a clearer statement of this problem?
| The notation $T_{n1}$ may come from the fact that the mean is also referred to as the "first moment", this is what the number $1$ as a lower label might stand for. For reference, see the Wikipedia article on moments in mathematics.
Regarding the derivative of the Energy: you are supposed to formally take the derivative of some expression with respect to some variable, even if it is a discrete one.
| {
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Will the heat flow of Joule heat be different, if the Joule heat is dissipated in a material that has a temperature gradient beforehand? Let us assume one dimensional heat transfer, for example a finite length wire starting at point $0$ and ending at point $\ell$. If the current passes the wire, the Joule heat $I^{2}R$ will be generated and dissipated into the wire and its thermal surroundings. Had the wire had a constant temperature $T$, the half of the power $I^{2}R / 2$ will be passing the left end, the other half will be passing the right end.
Will the situation change if the non-zero temperature gradient $\nabla T $ is present before the Joule heating starts? I cannot grasp, which principle has "higher priority" in this case - be it either principle of dissipation of heat which should be considered "a random walk" or the second thermodynamic principle which states that on average more heat will flow from colder to hotter parts.
Motivation for this question are heat transfer equations in thermoelectricity. Thank you in advance for any answer of insightful comments!
| As mentioned by Programmer, saying that if the wire temperature is constant then half of the heat will flow in either direction is incorrect. It really depends on the boundary conditions on either end of the wire (since it is of finite length). Assuming that the wire is at a spatially uniform temperature (not constant in time) and has same boundary conditions at both ends, the Joule heating will simply raise the temperature of the wire - uniformly. Heat cannot flow unless there is a axial temperature gradient in the wire. The temperature of the wire can then simply be modeled as:
\begin{equation} mc\frac{dT}{dt} = I^2 R \end{equation}
This assumes that the wire is not losing heat to its surrounding, otherwise a convection term needs to be added to the above equation. Next, say that the Joule heating starts with the wire already having a lengthwise temperature gradient. The flow of heat and development of temperature profile will be governed by:
\begin{equation} \frac{m}{l}c\frac{dT}{dt} = -kA\frac{dT}{dx} + I^2\frac{R}{l} \end{equation}
Again, the direction in which heat will flow really depends on your two boundary conditions.
| {
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Reference request for exactly solved models in statistical mechanics Can someone recommend a textbook or review article that covers exactly solved models in statistical mechanics, such as the six- or eight-vertex models? If there is literature at the undergraduate level, that would be ideal. I'm only familiar with Baxter's classic text on the subject, but this is tough reading for an undergraduate student.
| How familiar are you with exact solutions to the 2D Ising model? If you don't know that forwards and backwards, I would probably start with some textbooks that cover that material in depth. This would teach you some of the basics of transfer-matrix techniques and some other tricks about estimating eigenvalues, the thermodynamic limit, and so on. The book that I like is Plischke and Bergersen's Equilibrium Statistical Physics (3rd Edition). Plischke and Bergersen devote a whole chapter to solving the 2D Ising model, and work through it in a lot of detail. One prerequisite is some Quantum Mechanics - the solution that they present uses Pauli matrices and some creation and annihilation operators. They also do an Ising model on a fractal lattice (and on a chain, of course), both examples that can be solved exactly. This textbook won't teach you six- or eight-vector models directly, but if you work through it, Baxter will be a lot easier.
There are scanned copies floating around on the net - just google it if you want a preview.
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Photons from stars--how do they fill in such large angular distances? It would seem that far-away stars are at such a distance that I should be able to take a step to the side and not have the star's photons hit my eye. How do stars release so many photons to fill in such great angular distances?
| A very non-technical answer, but in trying to get your head around this, have you thought about the speed of light?
The angle distended by the star on your eyeball (or by your eyeball on the star) is very small. So it seems like a very tiny region of space must be 'full of photons' for the star to be constantly visible, and since the point where you are standing is not special, all similar regions must be equally 'full'. But the region in question is actually a very narrow beam whose length is the speed of light multiplied by the time that images persist in our vision. If the latter is 50ms, the length of the column is 15,000 km - the diameter of the earth. In this there would need to be a few dozen photons for the star to be marginally visible iirc.
Not a rigorous explanation I know, but it might help reconcile your intuition with the science.
| {
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Gradient of a two-component field I have a two-component field:
$$\phi(\vec{x}) = \left( \begin{array}{c} \phi_1(\vec{x}) \\ \phi_2(\vec{x}) \end{array} \right)$$
with $\phi^T = (\phi_1, \phi_2)$. And I am trying to evaluate:
$$(\vec{\nabla}\phi)^T(\vec{\nabla}\phi)$$
I'm pretty sure the result of that expression is supposed to be a scalar, but I can't figure out why. $\phi$ is a 2x1 matrix, which means $\vec{\nabla}\phi$ must be the result of a matrix multiplication between a Nx2 and 2x1 matrix. You would think N would be 3, since $\vec{\nabla} = \frac{\partial}{\partial x}\vec{x} + \frac{\partial}{\partial z}\vec{z} + \frac{\partial}{\partial z}\vec{z}$, but then why are there 2 columns? Not sure what to do here so any assistance would be quite helpful.
(I'll mention that the full thing I'm trying to evaluate is $\mathcal{L} = \frac{1}{2}\left(\dot{\phi}^T\dot{\phi}-m(\vec{\nabla}\phi)^T(\vec{\nabla}\phi)\right)$. $\mathcal{L}$ is a scalar, right?)
| When you have a matrix $\Phi = \begin{pmatrix} \phi_1\\ \phi_2\end{pmatrix}$, with one column and two rows, and its transpose matrix $\Phi^T = \begin{pmatrix} \phi_1 & \phi_2\end{pmatrix}$, with one row and two columns, the product of the two matrix $\Phi^T \Phi$ is a matrix $P$ with one column and one row:
$P =\Phi^T \Phi = \begin{pmatrix} \phi_1 & \phi_2\end{pmatrix}\begin{pmatrix} \phi_1\\ \phi_2\end{pmatrix} = (\phi_1^2+\phi_2^2)$
Because this matrix $P$ has one row and one column, it may considered as a scalar.
For your particular problem, you have :
$(\vec{\nabla}\phi)^T.(\vec{\nabla}\phi) = \sum\limits_{i=1}^N ({\partial_i}\phi)^T({\partial_i}\phi) = \sum\limits_{i=1}^N ((\partial_i \phi_1)^2 + (\partial_i \phi_2)^2)$
The first equality comes from the definition of the inner product and the gradient, and the second equality comes from the definition of the transpose operation $T$, and the manipulation of these matrices, as seen at the beginning of the answer.
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Archimedes' principle for two liquid layers
Problem: I have a cylindrical vessel of height $H$ and radius $R$. There are two liquid layers in the vessel. The first has density $D_1$ and height $h_1$, the second has density $D_2$ and height $h_2$. The second liquid is floating on the first liquid (thus $D_2 < D_1$) and they are both entirely within the vessel ($h_1+h_2 \le H$).
Now I have a cube of density $D$ and edge length $n$. The cube fits into the vessel ($n \le \sqrt2R$) but it might not be completely inside it ($n>H$ is possible). I place it into the vessel such that one side is horizontal (the cube is not lopsided). Some liquid might overflow as a result. How do I use Archimedes' principle to calculate the topmost liquid level after placing the cube? I can ignore damping and other physical effects.
EDIT: Note that the only values I know are $H, R, D_1, h_1, D_2, h_2, n$ and their abovementioned constraints.
My workings: I tried using the formula directly, where $\text{weight of cube}=\text{weight of displaced liquid}$. I first get $Dn^3=D_2n^2s_2$ where $s_2$ is the height submerged in the second liquid. If $s_2 \le h_2$ then the cube is only submerged in the second liquid. So the answer is $\min(H, h_1+h_2+{n^2s_2\over\pi R^2})$, accounting for overflows. This case is simple.
But if $s_2>h_2$ then the cube can be submerged in the first liquid too. I thought of subtracting the weight of the second liquid displaced to obtain the height submerged in first liquid, but realize that I don't know that. The weight displaced depends on the final liquid height, which can be affected by overflows, whether the cube fits totally within the two liquids, etc. In fact, it can be the case that after the liquid level rises, the cube now displaces more of the second liquid, causing it to not be submerged in the first (is that even possible?) It seems very messy and I have no idea how to start. :(
Does anyone have a nice solution to this problem?
| This will help for sure. Floating between two liquids
You only have to calculate the height. And the information you get for this is enough.
I used this formula to write a program in a contest in codeforces. The problem is same as yours. No change."Cocktail"
| {
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Electromagnetism: Conductors Even though the thermal velocity of electron in a conductor is comparatively high, the thermal velocity is not responsible for current flow? Why is this the case?
| Let's change that to "thermal velocity by itself is not responsible for current flow".
Thermal differences can produce current flow. And just like with electricity, a potential or change in thermal energy is required to convert the form of the energy to something else.
The reason heat cannot simply generate electric current is that in a material used to conduct electricity the construction is not designed to take advantage of temperature differentials. This is because no one can know exactly what kind of temperature differentials an object may be under prior to construction. If so than one would design the object with such heat conduction advantages.
Also...
Conductive objects that are cooled, tend to conduct even better. This fact demonstrates that heat is not the source of conductivity.
This is explained with the electron acting as a wave.
| {
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The velocity of a cloud? I noticed an unusually fast moving cloud this morning.
My questions:
*
*What is the average velocity of a cloud on Earth?
*What is the greatest ever recorded cloud velocity?
*What factors affect the velocity of a cloud? (e.g. do they experience inertia?)
| It looks like there can be a difference of up to 5 m/s between cloud velocity and wind velocity (http://journals.ametsoc.org/doi/abs/10.1175/1520-0450%281976%29015%3C0010%3AWEFCMP%3E2.0.CO%3B2 ).
| {
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Why is the decay of a neutral rho meson into two neutral pions forbidden? Why is the decay of a neutral rho meson into two neutral pions forbidden? (Other modes of decay are possible though.)
Is it something with conservation of isospin symmetry or something else? Please explain in a bit more detail.
| As far as I understand, due to conservation of angular momentum, the resulting system of neutral pions would need to have angular momentum 1, therefore, the identical neutral pions would be in an anti-symmetric state, which does not seem possible as they are bosons. Note that a neutral rho meson can decay into two neutral pions and a $\gamma$, although this decay is suppressed.
| {
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Physics of the inverted bottle dispenser
When you invert a water-bottle in a container, the water rises and then stops at a particular level --- as soon as it touches the hole of the inverted bottle. This will happen no matter how long your water-bottle is. I understand this happens, because once the water level touches the hole, air from outside cant go inside and therefore there is nothing to displace the water that falls out of the container.
Now according to the laws of pressure ---- the pressure at the water level must be same everywhere --- whether it's inside the water bottle or outside. And that must be equal to the atmospheric pressure. Therefore the pressure of the water column + air column inside the inverted bottle must be equal to the atmospheric pressure.
What I dont understand is, no matter how long a bottle you take, the water level will always stop at the hole. So that means that no matter how long a bottle you take, the pressure of the water column + air column inside the water bottle will be equal to the atmospheric pressure. How could this be possible?
Also I'd like to let you know that, if you pierce the upper part of the bottle with a small pin, then the water level rises and overflows out of the container. I'm assuming air from outside rushes in and pushes the water out.
| Other answers are right, but let me put it without math:
Water can't come out of the bottle if air can't go in.
(Except: if the water in the bottle is so tall that water can come out even if air can't go in.)
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Relationship between power and max. speed I'm talking about the maximum speed if let's say I have a car with the power $P = 1000 \text{W}$ and a force of friction of $5 \mbox{N}$ acting in the opposite direction. After some googling I found that the maximum speed is given by $P=Fv$, where $P$ is the power, $F$ is the force, and $v$ is the velocity.
I understand that $W= Fs$ and that $P = W/t$ and $s/t$ is $v$, so yes I understand where the equation comes from, however wouldn't this be the average speed and not the maximum speed? And the force of friction is not the force that's doing the work, so why is it used in the equation?
I hope I've made my question clear enough, thank you in advance!
| *
*The expression $P=Fv$ expresses a relation between the instantaneous power, the force and the velocity. You don't have to average for it to be true.
*In your case, the velocity in constant. This implies that the net force is zero. Hence, the force propelling the car is equal and opposite to the friction force. We can then use the magnitude of the friction force.
| {
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If a spaceship was pulled toward a sun, would it spin? I was watching a movie. A spaceship was forced into "warp speed". The co-ordinates could not be set. The spaceships trajectory was that of a nearby sun. Forcing the spaceship to power down was the solution. Now out of "warp speed" and with no computer aid (steering etc) the spaceship was seen to be spinning toward the sun trapped in its gravitation field.
My question is, would the spaceship (typically aerodynamically shaped) spin toward the surface? My opinion is no. The spaceship would just fall flat due to the surface area provided at the bottom of the fuselage
| The only way for a falling object to be made to rotate and translate is if there was a separate force causing this rotation.
In an atmosphere this is a net force on one side of the craft whose surface area (and therefore drag) is the highest, causing this part of the craft to rotate away from the direction the entire craft is translating. Essentially different parts of the craft have different terminal velocities in different orientations relative to the falling direction, and will rotate in free fall until an equilibrium orientation is achieved.
In the absence of an atmosphere a similar mechanism is tidal force--an apparent unequal gravitational force on different parts of the same object due to a large massive body. The craft has to be extremely huge or long in order to experience tidal forces, as one end of the craft is closer to the planet than the other end. The closer end falls faster, causing the craft to rotate into the vertical direction. If the craft is a perfect sphere it will just cause the closer side to tidally lock in position while it falls--no rotation at all.
So in space, free spinning in free fall is not due to the free fall.
| {
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Physical intuition for independence of components of velocity in derivation of Maxwell–Boltzmann distribution Maxwell derived the shape of the probability distribution of velocity of gas particles by starting with just two assumptions.
These are:
*
*The probability distribution is rotation invariant.
*The components (of velocity of a gas particle) in the direction of the coordinate axes are statistically independent.
And the rest is lovely deduction, but I found that as a layman I don't have any physical intuition as to why the second assumption is plausible. Is there an intuitive explanation behind the second assumption? If not, is there a way to derive the second assumption from a set of more plausible-looking assumptions?
| I'm just going to quote Wikipedia here:
For the case of two colliding bodies in two dimensions, the overall velocity of each body must be split into two perpendicular velocities: one tangent to the common normal surfaces of the colliding bodies at the point of contact, the other along the line of collision. Since the collision only imparts force along the line of collision, the velocities that are tangent to the point of collision do not change. The velocities along the line of collision can then be used in the same equations as a one-dimensional collision.
Key point: During a collision, the components of velocity perpendicular to the axis of contact (which is random) do not, for example, decrease when the other increases. Each collision selects an axis at random and perturbs the velocity in that direction.
| {
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What is the physics of a spinning coin? When we spin a coin on a table, we observe 2 things:
*
*It slows down and stops after sometime.
*It does not stay at just one point on the table but its point of contact with table changes with time.
I was trying to explain quantitatively this but I am stuck at how to take frictional torques into account. Any help will be appreciated.
| I think that if you spin "perfectly" (i.e., such that the rotational axis is normal to the surface and goes through he centre of the coin), is only a rotation movement with friction. This motion is unstable though, so, the axis tilt a little bit and this cause a rotation in the axis itself, the precession. The point of contact will be moving with the precession, maybe you can calculate its position by geometrical arguments, although it should be a circular/spiral/cycloid movement (if you see in the coin a movement towards a given direction, this is solely because of the way you made if spin or the coin or because the table has a tilt or imperfections).
I don't know your level of knowledge, but for a complete description you need knowledge of Hamiltonian dynamics, rigid body and Euler angles, so basically a course of classical (a.k.a. analytical) mechanics. A very common, related, problem is the problem of the spinning top, the difference here is that the contact point is material, so there you have to see if you have to see if the contact point slips or not (if not, it creates a rotation in the axis normal to the coin).
Personally I think that it is a complicated but somehow treatable problem (with a lot of patience).
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Are coherent states of light 'classical' or 'quantum'? Coherent states of light, defined as
$$|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle
$$
for a given complex number $\alpha$ and where $|n\rangle$ is a Fock state with $n$ photons, are usually referred to as the most classical states of light. On the other hand, many quantum protocols with no classical analog such as quantum key distribution and quantum computing can be implemented with coherent states.
In what sense or in what regime should we think of coherent states as being 'classical' or 'quantum'?
| It is all about what meaning you put into the words "quantum" and "classical". Fock space and elements of this space are notions that belong to quantum theory of radiation and have no direct relation to states of radiation in classical electromagnetic theory, so the coherent state may be called "quantum" with good reason.
However, coherent states have properties very similar to those of harmonically-oscillating standing waves of electromagnetic field as used in classical theory of microwave cavities, so they are often called "classical-like" quantum states.
| {
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Visible Lines for Hydrogen Say H atoms are excited to 4th level, n=4 that is, how many lines do we see? How to decide the number of the lines?
| An excited state electron may transition to any lower level. From n=4, the electon could go to n=3, n=2 or n=1. Of these 3 transitions, only n=4 to n=2 (wavelength 486nm) is visible light. n=4 to n=1 is ultraviolet and n=4 to n=3 is infrared.
The wavelengths of the transitions are given by the Rydberg formula.
http://farside.ph.utexas.edu/teaching/qmech/lectures/node82.html
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Question on energy mass conversion I have a question regarding the energy-mass conversion. Well, when a particle starts moving with a speed comparable to that of light, its (relativistic) mass increases that means some matter is created and that too of the same particle...energy being converted to mass is ok but how does energy perceive what atoms it has to form? Say I take a stone to a high speed, then constituents of stone is formed. And if I perform same thing with another substance, its constituents are formed..How? Energy can be converted to mass but a mass of what? Does that mean we can create matter of any desirable substance?
| If you accelerate a stone to relativistic speeds, no new atoms are created in the stone. There is constant amount of atoms. If new atoms were created, that would mean than these atoms have to disappear when you decelerate the stone. And now think about someone accelerating together with the stone. From the perspective of accelerating observer, the stone does not get heavier. So from their perspective no new atoms would be needed.
The quantity that is increasing during acceleration is the total energy. Relativistic mass (that's the "mass" which is increasing) is in fact nothing else but the kinetic energy. The amount of atoms is unchanged. The kinetic energy can be calculated into kilograms with the relation E=mc^2, but such result is not really useful. What matters is the energy. That's why the concept of relativistic mass (which increases with speed) is not used. Only the rest mass (whcih does not change with speed) is important.
The increase of energy manifests itself as higher resistance to further acceleration, which somehow vaguely relates that to mass (inertial mass namely), but using the concept of relativistic mass is not useful. The total energy is what matters.
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Why are roofs blown away by wind? Whenever there are high winds, such as in storms, thin metal roofs on sheds as well as concave roofs on huts are sometimes blown away.
One explanation provided to me said that the higher velocity of the air outside causes the air pressure above the roof to decrease and when it has decreased to a certain extent such that the air pressure above the roof is lesser than the air pressure beneath the roof and due to some kind of osmosis, the air particles move from the area of higher pressure (beneath the roof) to the area of low pressure. In this process, the roof is blown away.
Another explanation, specifically about the thin metal roofs, said that it was blown away due to the lift caused by the air and this is the same kind of lift you get when you blow on paper.
Both these explanations puzzle me.
What really bothers me is the basis of the first one, how can an increase in velocity cause pressure to drop? I can't seem to correlate that with the Force per unit area definition of pressure.
Please, oh great physicists of the internet, help me and every other ordinary person to understand how and why roofs get blown away.
| High speed winds are accompanied by reduced air pressure
So high pressure from inside the house pushes roof to low pressure and gets blown away.
| {
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How can one see that the Hydrogen atom has $SO(4)$ symmetry?
*
*For solving hydrogen atom energy level by $SO(4)$ symmetry, where does the symmetry come from?
*How can one see it directly from the Hamiltonian?
| It's because there is another vector quantity $A_i$ conserved in addition to the angular momentum $L_i$. Furthermore, the commutation relations of $A_i$'s and $L_i$'s are those of $SO(4)$. See for instance this reference : http://hep.uchicago.edu/~rosner/p342/projs/weinberg.pdf
| {
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Why do solar cells have a window layer on top of the absorber layer and not below it? In solar cells there is a p-n junction. P-type semiconductor (for example CdTe) is often absorber layer because of its carrier lifetime and mobilities. In case of CdS/CdTe,* CdS is n-type window layer and everywhere it is said that it should be very thin and has large band gap – not to absorb any light and let it go through to the p-type absorber (that is why it is called a window layer).
But why should it be on top of the absorber layer and not below it?
If n-type layer is below, the light can hit the p-type absorber directly.
I have some ideas that it is related to the distance between the place of absorption and p-n junction, but I am not sure.
Image by Alfred Hicks/NREL (source).
*A similar design is used in CIGS, CZTS and other thin film solar cell designs; this question applies to all of them - solar cells with a p-type absorber and an n-type window layer
| When the light strikes the P-substrate, it excites an electron. This electron either is absorbed back into the P-substrate, or it can move into the N-substrate and gets absorbed there. Once the electron has moved into the N-substrate, due to the PN junction, the easier path to balance the charges is to push electrons through an external circuit.
It makes more sense to generate excited electrons near the N-substrate, than to generated the excited electrons on the side away from the N-substrate. More excited electrons can transfer to the N-substrate if it is nearby.
| {
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Lorentz boost matrix in terms of four-velocity As I understand it, the value of a 4-vector $x$ in another reference frame ($x'$) with the same orientation can be derived using the Lorentz boost matrix $\bf{\lambda}$ by $x'=\lambda x$. More explicitly,
$$\begin{bmatrix}
x'_0\\
x'_1\\
x'_2\\
x'_3\\
\end{bmatrix}
=
\begin{bmatrix}
\lambda_{00}&\lambda_{01}&\lambda_{02}&\lambda_{03}\\
\lambda_{10}&\lambda_{11}&\lambda_{12}&\lambda_{13}\\
\lambda_{20}&\lambda_{21}&\lambda_{22}&\lambda_{23}\\
\lambda_{30}&\lambda_{31}&\lambda_{32}&\lambda_{33}\\
\end{bmatrix}
\begin{bmatrix}
x_0\\
x_1\\
x_2\\
x_3\\
\end{bmatrix}
$$
I have seen examples of these components written in terms of $\beta$ and $\gamma$, which are defined as
$$\beta=\frac{v}{c}$$
$$\gamma=\frac{1}{\sqrt{1-\beta\cdot\beta}}$$
where $v$ is the 3-velocity and $c$ is the speed of light. My question is this: How can the components of $\lambda$ be written in terms of the 4-velocity $U$ alone?
I know that $U_0=\gamma c$ and $U_i=\gamma v_i=\gamma c\beta_i$ for $i\in\{1,2,3\}$, but I'm having trouble deriving the components for $\lambda$ using the matrices based on $\beta$ and $\gamma$. An example of one of these matrices can be found at Wikipedia. How can I rewrite this matrix in terms of $U$ alone?
| Take units $c=1$.
You have $U_0^2-\vec U^2=1$, that is $\gamma^2(1-\beta^2)=1$. With some basic transformations, you will get : $\frac{\gamma - 1}{\beta^2}= \frac{\gamma^2}{\gamma + 1}$
Now, from your Wikipedia matrix, you have obvious term, $ U_0 =\gamma , U_i =\gamma \beta_i$
You have $(\gamma -1) \frac{\beta_i\beta_j}{\beta^2} = \frac{\gamma^2}{\gamma + 1}\beta_i\beta_j = \frac{U_i U_j}{U_0 + 1}$
Finally, $1 + (\gamma -1) \frac{\beta_i^2}{\beta^2} = 1 + \frac{U_i^2}{U_0 + 1}$
| {
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As the universe ages, will we see more stars or less? After a very long time will we see more stars (due to the fact that more light is get to us) or less stars (as the universe expends and light have to pass larger distance)?
In general, can stellar objects go outside of the scope of the observable universe or is it only growing with time? Calculations are always better than just discussion.
Does it have any connection to dark matter?
| I red some times ago about a scenario in which the number of visible objects is becoming smaller and smaller. This is basically due to the Hubble's law: the further two objects are, the faster they move away from each other and when the speed exceed the speed of light, no news can come from them any more. If you take into account that the expansion of the universe is accelerating (the Hubble constant is increasing) this becomes even worse.
I don't like to think that a lot of things in the universe are constantly dropping outside my light cone, however this is what our current observations suggest and is probably still better that to have the universe to compress and finally squeeze on itself.
| {
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If friction disregard area, why the direction you drag a long object matters? I am talking here about dry friction between solid objects, for example a ruler and a table, not anything lubricated or fluid.
I noticed that with a ruler and a table for example, if you drag the ruler like it was a knife, it is much easier to do it than if still holding the ruler in the same position you drag it sideways (like if you are scrapping something).
Also I noticed that when I am washing dishes, if I leave two metal objects (ie: to flat metal areas) in contact, they don't move much, but the same objects, if I try to find deformations that make the area of contact between them smaller, then they can be easily pushed around from rest, or spun.
This also applies to tyre sizes (ie: for dramatic effect, dragster cars with HUGE rear tyres and tiny front tyres).
Also I did some experiments with a paper, ie: holding it down with a finger make it much easer to slide than if I make sure more of its area is in contact, but doing that also is a downforce on the paper, so I guess I can sum that on the normal force.
My best guess is that it all has to do with the normal force, but I am not sure at all... Can someone quench my curiosity here?
| Deformities in the surface cause more friction. As it's rightly said friction is a necessary evil, friction between the surface helps in forward movement as well. Like when we walk on a rough surface we get more forward push.A slippery and smooth surface is really difficult to walk on. A very nice article with Free Body Diagram is here, which you can check pls: How friction helps in walking
| {
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Would a black hole created on the surface burrow through the crust? If scientists created a microscopic black hole with an initial mass of one ton on the surface of the earth, would the gravitational attraction to the center be enough for it to "burrow" until it eats its way through the crust? It seems like there would be a bad outcome. How dense would rock have to be to form a barrier?
| Black holes this small will have very high Hawking temperature:
$$
T_H = \frac{\hbar c^3}{8 \pi G M k_B} \approx 10^{20}\,\text{K},
$$
So, before this black hole can fall down even the diameter of an atom it will evaporate through Hawking radiation. As a result, the 1 tonne of black hole mass would be converted into the energy of very high energy particles of all kinds.
Some of these particles, such as neutrinos, gravitons and weakly interacting particles (if they exist), will fly away without interacting with anything on Earth, others would interact, resulting in the formation of the giant fireball analogous to hydrogen bomb explosion. The yield will be hundreds times larger than the Tsar Bomba explosion, so potentially it can wipe out the considerable part of any continent, but most likely will not end life on Earth.
| {
"language": "en",
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A question about T duality Normally T-duality is introduced perturbatively by computing world-sheet spectrum of fundamental strings, and one of the conclusions is that it switches between momentum mode and winding mode of fundamental strings. My question is that does this switching extend beyond fundamental strings, namely, is it true that T-duality simply turns any momentum modes (for example, those carried by D-strings) into winding fundamental strings? If yes, is there a convenient way to see why is this the case?
| Yes, whenever the momentum is conserved and T-duality holds, T-duality must map a conserved quantity such as this momentum to another conserved quantity, i.e. the string winding number in this case, and this fact is independent of the carrier of the momentum or the winding charge. In the general nonperturbative case, you shouldn't think about the charges as some "convoluted results of some particular behavior of some particular object" but as about conserved quantities generating symmetries that exist regardless of any spectrum of the objects.
This still leaves the question whether T-duality is valid nonperturbatively. In all the known simple enough supersymmetric string vacua where it is provable in perturbation theory, it also holds nonperturbatively. We can't rigorously prove this statement in the universal situation because we're lacking the universal nonperturbative definition of string theory. However, we may prove it for many vacua by various tools – argue it is true for vacua with the maximal supersymmetry which produce SUGRA with exceptional noncompact (in M-theory: discrete U-duality) symmetries (T-duality is a subgroup); we may prove it for IIA and $E_8\times E_8$ heterotic string theory using BFSS Matrix theory, and so on.
| {
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"url": "https://physics.stackexchange.com/questions/90265",
"timestamp": "2023-03-29T00:00:00",
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What is the reasoning behind the Hill Sphere? According to Wikipedia, Hill Sphere is: the volume of space around an object where the gravity of that object dominates over the gravity of a more massive but distant object around which the first object orbits.
True as this may be, it just mathematically supports a phenomenon that has been observed but it does not give reason or logic as to why does this happen in the first place. I mean why should the gravity of a less massive object dominate the gravity of a more massive one?
I wasn't aware of the Hill Sphere until recently when I was trying to visualize the orbits of different celestial bodies. The Hill Sphere comes closest to explaining why the moon orbits the Earth, more than it orbits the Sun and why the Earth orbits the sun, more than it orbits the center of our galaxy. By this logic all celestial bodies within the Gravitational pull of the center of our galaxy should directly be orbiting the center.
My argument is that if the Hill sphere of the Sun is as large as the solar system itself, any object within this sphere should be orbiting the sun. Why was the moon caught into the earth's gravitational pull in the first place when it had a much stronger pull from the sun?
The answer to this would also eventually clarify why the earth orbits around the sun and not the center of the milky way.
| "I mean why should the gravity of a less massive object dominate the gravity of a more massive one?"
Maybe because it's closer?
"The strength of the gravitational force between two objects depends on two factors, mass and distance."
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Experimental evidence for non-abelian anyons? Since non-abelian anyons have become quite fashionable from the point of view of theory. I would like to know, whether there has actually been experimental confirmation of such objects.
If you could cite original literature, that would be great!
| As far as I know we do not yet have definitive verification of non-abelian statistics which would indicate the existence of non-abelian Anyons. The latest results I know about are the ones mentioned by akhmeteli and An, et. al. "Braiding of Abeliana and Non-Abelian Anyons in the Fractional Quantum Hall Effect." Arxiv:1112.3400. However, I do not believe either are accepted as definitive proof.
You are correct in that there are other candidate systems proposed, such as $p+ip$ superfluids, superconductors, and other systems. Some good references and also a review seem to be given in these review articles. As I understand it, the big push to observe evidence of Majorana fermions. That we do have some pretty good evidence for. Pairs of Majorana fermions are supposed to realize non-abelian statistics, but this has not yet been implemented and observed (to my knowledge).
As a final note, we do have evidence (see here and here) of abelian anyonic statistics, however I am under the impression that there may be some controversy about this.
| {
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