Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Why is the mean density the same for all nuclei? Tell me if this is a correct theory? So the radius $R$ of the nucleus is directly proportional to $A^{1/3}$ (the nucleon number).
As $$V = \frac 43 \pi r^3,$$ this makes $V$ directly proportional to $R^2$. Also, as the nucleon number increases, the mass also increases and as the masses of protons and neutrons are similar you could say that the mass of the nucleon is directly proportional to the nucleon number.
If you put all of this together, you get the mass of the nucleon being directly proportional to the volume where the constant is the density.
Thus, that is why the density is constant for all nuclei?
| Indeed, this is a non-trivial fact about the nuclei, as opposed to atoms. The atoms are held together by the Coulomb forces, and their density increases (i.e. they become more compact) as the charge of the nucleus and the number of electrons increase.
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Eddy current confusion Every source tells that eddy currents are produced by change in magnetic flux in conductor, but according Gauss' law for magnetism net flux over a closed surface is 0, then how magnetic flux will change to produce eddy current?
| Assume you have a coil winded along z axis and a large metal plate put in xy plane. Then applying AC voltage across the terminals of coil eddy current induced on the metal but in this case charges are moving and using Gauss' Law considering the change in the magnetic apparently flux changes. With another setup, for instance a metal disk is again in xy plane rotates under uniform magnetic field toward $\hat{k}$ direction and magnetic field covers small portion of the disk. In this case to visualize the eddy current, consider atoms as hard spheres and when the disk rotates each atom with their charges rotates as well but under the magnetic field atom's position barely changes conversely to the regime which goes under the magnetic field changes for small time difference. Basically, for the induced current on disk depends on the scanned area change by magnetic field. So, applying Gauss' Law may be hard but it works (even for moving charges).This link may help you to understand better with it's pictorial explanations.
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What is the Unruh effect for an accelerated observer in heat bath? The Unruh effect states an accelerated observer in flat Minkowski spacetime sees excited states in a heat bath with the Unruh temperature.
Then, when the initial rest observer is in a heat bath with the uniform temperature $T$ in the Minkowski spacetime, what is the temperature the observer will see after some acceleration? Is it $T+T_{\mathrm{Unruh}}$?
| Because the velocity of the observer has no effect on the temperature of the heat bath (reservoir) or, the other way round the temperature of a get bath is not dependent on the velocity of the bath, the answer is yes.
Of course at relativistic speeds, the mass of the bath will increase. But one can always find a reference frame in which it's velocity is zero. If the frame is accelerating together with the observer, the bath will appear to be in a gravitational field. This too will have no influence on the temperature.
Note that The Minkowskski spacetime is used mostly in SR since it has no intrinsic curvature. Puting different M-spaces together (for a mass with a continuously varying velocity) creates a curved spacetime.
This is essentially the same as asking if the temperature of a box filled a gas changes when it moves. The answer to this question is no.
Only for enormous accelerations, the Unruh effect will cause a change in temperature for the observer measuring the temperature.
an accelerating thermometer (like one being waved around) in empty space, subtracting any other contribution to its temperature, will record a non-zero temperature, just from its acceleration
Taken from this Wikipedia article. In this article, it is also stated:
It is currently not clear whether the Unruh effect has actually been observed, since the claimed observations are disputed
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Ceiling fans with just one large winglet? I am from India and in India ceiling fans have generally three wings.
Today while laying on my bed a question came in my mind.
First of all, I know that if we use just a single wing at a time of same dimensions as each wings of a three winged fans have then it will reduce the amount of air being pushed down.
But what if we just weld the edges of the three wings to make a single wing ? Will we get the same amount of air as we get generally from a three winged fan ? Which factors determine the amount of air being pushed down ?
I think by this technique the amount of air being pushed downward due to the curvature of the three wings (as a whole) will not be affected. And we can get exactly the same amount of air.
| A fan blade is just a wing. At the tip, air spills round creating a vortex in its wake. This wastes energy and reduces the efficiency of the wing.
A long, thin wing has a smaller tip, and hence smaller losses, than a short, fat wing of the same area. So the long, thin wing is more efficient. In other words, you need a smaller area to do the same job.
A ceiling fan would be a little more efficient if it had one long, thin blade. But the blade would be very long. Many smaller blades are more practical.
But if blades are close to each other they interfere, reducing efficiency again.
For a ceiling fan, three blades is the best compromise between small size and high efficiency.
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Magnetic, geographic and geomagnetic poles
Can someone explain me in simple terms where and what are the magnetic, geographic and geomagnetic poles?
Some sites say that magnetic north pole is in the south and thus it attracts the south pole of the magnetic needle while some say the north pole of the needle points south. In each place, I seem to get a different angle between the magnetic and geographical axis (10°, 11°, 17° etc.). And I don't get clearly what the geomagnetic pole is.
I'm utterly confused, please help me out.
This question slightly addresses my doubt but the answers contradict each other.
Thanks in advance.
| The geographic poles are the North Pole and the South Pole of the Earth. These are not the actual magnetic poles, they are the axial poles. The magnetic poles are the actual poles of the Earth's magnetic field, but are not antipodal as they do not pass through the Earth's center. The geomagnetic poles are antipodal as they pass through the center of the Earth to represent a dipole bar magnet at Earths center. The magnetic north pole attracts the north pole of a bar magnet, or compass needle, so the magnetic north pole is actually the south pole of Earth's magnetic field. You may get contradictory locations for the magnetic and geomagnetic poles as they move slightly over time.
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Universe and Anti-Universe pair Two years ago Latham Boyle, Kieran Finn and Neil Turok, in their work CPT-Symmetric Universe, proposed that the pre and post bang eras constitute a universe/anti-universe pair.
For Anti-universe in general we mean a universe rich in antimatter, that is matter composed of anti-particles, which are particles whose time of the world line is in contrast with the time in our spacetime, we can therefore say that in the hypothetical anti-universe flows an anti-time; Can we equally speak of anti-space and therefore anti-spacetime?
Since matter and anti-matter annihilate each other, can we hypothesize that spacetime and anti-spacetime annihilate each other as matter and anti-matter do?
With anti-space is it possible to consider a negative dimensional space, as hypothesized for example by Parisi and Sourlas in their 1979 work Random Magnetic Fields, Supersymmetry, and Negative Dimensions, or Pedrag Cvitanovic in his two works (1981, 1982) Negative Dimensions and the Emergence of E7 Symmetry in Supergravity and Spinors in Negative Dimensions?
I know it is a very speculative question to which obviously it is not possible to give a certain answer, but I find it interesting the possibility of talking about it and having some opinion.
| This cosmology is one in which there are two independent expanding regions, one in the $+t$ direction from the origin and one in the $-t$ direction. The thermodynamic arrow of time points away from the origin, and there's an exact symmetry (CPT) that exchanges the two regions, so they're indistinguishable and we could be in either one. There's no anti-space or anti-spacetime. It's just a spacetime that is "twice as large" as the usual one.
There's a technical sense in which antiparticles are time reversals of their regular-particle counterparts, but it's not related to the arrow of time. We know experimentally that antiparticles don't have a reversed arrow of time. The bi-directional arrow of time in models like this one is related to the special low-entropy state at $t=0$ and not to predominance of matter or antimatter.
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Why doesn't static friction point down the incline on a ball rolling down an incline?
In the figure above in which the sphere rolls without slipping down an incline, why is static friction directed upward? Static friction is meant to oppose impending motion so it should oppose the object's tendency to slide, and gravity would cause it to slide down in the absence of friction. Doesn't this mean that friction should point down since a downward force causes the sphere to move up?
| The static force pointing upward doesn't mean the ball is accelerated upward. It's the force of gravity that does the job. It pulls on the ball, and if the ball doesn't slip, the static friction causes a torque, together with the force of gravity, which makes the ball roll downward.
| {
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Does a linearly accelerated charge radiate? Does a linearly accelerated charge radiate? And if yes, is the electromagnetic waves emitted detectable from both its non-inertial frame and from other observing inertial frame? And If yes, doesn't this contradict with Einstein Equivalence principle?
|
Does [a] linearly accelerated charge radiate?
"It depends". Accelerating relative to whom?
Is the electromagnetic waves emitted detectable from both its
non-inertial frame and from other observing inertial frame
No.
Doesn't this contradict with Einstein Equivalence principle
This is indeed an apparent paradox in GR.
The solution to the paradox is to accept and acknowledge that radiation has no absolute meaning. Which is the reason of the "it depends" to the first question. The emission and detection of radiation depends both on the motion of the source (radiation field) and on that of the observer.
An inertial observer, stationary with respect to the accelerating charge, will indeed detect the radiation.
But a comoving observer, to whom the charge is at rest, will not detect any radiation. Simple reason being that yes, in their frame the charge is not accelerating. The mathematical reason is that, when you put this in Rindler coordinates, the charge and its field are enclosed in a horizon.
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Force on a relativistic laser source Consider a laser source emitting a laser beam in the horizontal direction. If we are given the power of the laser source, we can calculate the force on the laser beam.
Now consider that the laser beam is moving horizontally with a 0.5c. Would the force on the laser source change by the emitted laser beam?
Since the frequency of the emitted laser beam is increased (momentum of photons is increased) and the time interval between two emissions is decreased, the force on the laser source should increase
Is the reasoning correct?
| To someone riding alongside with the laser source (or in the same system of reference, in relativistic terms) the laser is behaving like it always did, with no increase/decrease in power output.
But the scenario of an observer experiencing a light source moving at relativistic speeds have a name - Doppler beaming, also known as relativistic beaming. Quoting Wikipedia, it's
[...] the process by which relativistic effects modify the apparent
luminosity of emitting matter that is moving at speeds close to the
speed of light.
So Doppler-related frequency shift would cause an observer that sees the laser source approach at .5c to experience a higher energy output due to blue-shifting, while observers seeing the laser move away would experience the exact opposite due to red-shifting:
Relativistic Doppler effect. A source of light waves moving to the right, relative to observers, with velocity 0.7c. The frequency is higher for observers on the right, and lower for observers on the left. Source
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Why does the electric field have non-zero curl for magnetic monopoles? In Griffiths’ Introduction to Electrodynamics, he asks what changes would need to be made to Maxwell's equations to accommodate the existence of magnetic monopoles. Now, it is clear to me that the Gauss’s law and Ampere’s law must be left untouched. I also understand why the divergence of $\vec{B}$ should be $\alpha_{0}\rho_{m}$, where $\alpha_{0}$ is some constant to be determined experimentally and $\rho_{m}$ is the density of monopoles.
What I don’t understand is why the curl of $\vec{E}$ must be $\beta_{0}\vec{J}_{m}$, where $\beta_{0}$ is some constant.
The idea that moving electric monopoles produce a magnetic field is an experimental fact. Why must we assume that the same holds for magnetic monopoles as well? Is the symmetry of Maxwell's equations enough of a motivation for us to be sure that the electric field will now have a non-zero curl in the statics regime?
I get that such symmetries often motivate the discovery of such properties, but does it guarantee that the electric field curls?
| With the standard caveat that all theory is predicated on experimental validation, I would strongly expect magnetic charge to obey a continuity equation. If you make the modifications you suggest without changing Faraday's law to include magnetic current, then you would find that $\frac{\partial \rho_m}{\partial t}=0$, implying that the total magnetic charge in any given region could never change.
Note that while Ampere's law is an experimental observation, the absence of the electrical current term similarly implies via Gauss' law for electric charge that the electric charge density of the universe is static, which would be intuitively upsetting were it not so easily demonstrated to be false.
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How does light, which is an electromagnetic wave, carry information? We see an object when light from a source strikes the object and then reaches our eyes. How does light, which is an electromagnetic wave, gets encoded with the information about the object? Do the individual photons get encoded with this information or is it the wave nature of light that gets modified to carry information about the object?
Also, if light hits an object and then another on the way to our eyes, does it only carry information from the very last interaction it had? How is the information due to all previous interactions erased (if it is indeed)?
|
How does light, which is an electromagnetic wave, gets encoded with
the information about the object?
There are several stages to this. Initially there has to be a light source emitting photons. This can be the object itself, but is more likely to be a separate light source such as the sun or a light bulb, usually a 'white' or 'wide spectrum of wavelengths' light source.
Photons from the light source hit the viewed object whereupon some of the wavelengths are absorbed and some re-emitted (reflected), depending on the colour of the object. These re-emitted (reflected) photons are then imaged by the eye lens and reach a sensing cell (rods or cones) on your retina. Your brain knows that each rod or cone can only get stimulated by photons entering your eye lens at a unique angle. In this way the brain builds up a map of the scene.
What's more, the 3 colour receptors in the retina can also detect which wavelengths got absorbed and compares it to the overall level from all your colour receptors. In ths way the brain is able to determine the colour of the object which sent the photons in at each angle. - i.e. the colour of each point in the scene.
Also, if light hits an object and then another on the way to our eyes, does it only carry information from the very last interaction it had?
Each prior interaction can vary the relative wavelength absorption (colour), intensity, and direction of the reflected light. Other factors such as polarisation can also be affected, but the eye can't detect that directly.
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If gravity is curvature of space why are more massive objects "heavier?" How does curved space explain why a denser object of the same shape and volume feels heavier?
| If we are only dealing with weights in a localized place at the surface of the earth, it is possible to use the principle of equivalence: the physical properties are the same as being in a spaceship in the outer space with an acceleration $g$.
Heavier objects here on earth would be equally heavier there. It is easy to see that the acceleration is the same for all objects, because the ship is really accelerated. Of course, as $F = ma$, more mass means more weight.
The metric at the surface of earth is such that the covariant acceleration is $g$ for a body at rest, as explained here.
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Meaning of Eigenvalues for position operator To each observable in quantum mechanics there is an operator corresponding to it. I don't understand what's the meaning of the eigenvalues of the $\hat{x}$ operator. Since $\hat{x}$ is hermitian, eigenvalues correspond to real numbers, what's their physical meaning? If they describe a particle localized at a particular point, isn't it in opposition with the statistical nature of quantum mechanics?
Second: Since position and momentum operators do not commute, the eigenvectors of the Hamiltonian are usually different from the eigenvectors of both position and momentum operators. But I see in books applying the $\hat{x}$ to the $\psi_n$a representing the autostates of the Hamiltonian operator. How do I find the autostates for $\hat{x}$ if they are not the same obtained by solving Schrödinger equation?
| Eigenvalues are the values that are measured in the experiment, i.e. eigenvalues of $\hat{x}$ are the values of the position obtained when measuring it. Every measurement will produce a different result, i.e. a different eigenvalue, unless the system was prepared in an eigenstate of the measured quantity - this is the statistical nature of quantum mechanics. Thus, after $N$ measurements we have a sample average
$$
\bar{x} = \frac{1}{N}\sum_{i=1}^N x_i,
$$
whose value approaches the one estimated in quantum mechanics, $\langle \hat{x}\rangle$.
(All this looks trivial, but the problem of modern physics is that QM is often taught to students who has never taken a course in statistics.)
The state of the system does not have to be eigenstate of all the operators, which is pretty much the point of the uncertainty principle. Thus, an eigenstate of the Hamiltonian operator has well-defined energy, but uncertain position and often uncertain momentum. For example, the eigenstate of a harmonic oscillator gives a Gaussian distribution of position, i.e. the emasurements $x_i$ measured above will be distributed, as if they come from a Gaussian distribution, with variance $\sigma_x^2 = \langle \hat{x}^2\rangle - \langle \hat{x}\rangle^2$.
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Doubt on net acceleration during non-uniform circular motion During non-uniform circular motion, the direction of net acceleration is not in the direction of the centripetal acceleration, then why does a particle still move in a circular path, please explain.
| Let's look at a position vector in general polar coordinates
$$\vec r(t)=r(t)\pmatrix{\cos\theta(t)\\\sin\theta(t)}$$
Now define
\begin{align}
\hat r&=\pmatrix{\cos\theta\\\sin\theta}\\
\hat \theta&=\pmatrix{-\sin\theta\\\cos\theta}
\end{align}
to make our lives easier. I dropped the time dependence but remember it's still there. You can check that
\begin{align}\dot{\hat r}&=\hat\theta \dot\theta\\\dot{\hat \theta}&=-\hat r \dot\theta\end{align}
where a dot indicates a time derivative. A general acceleration then becomes
\begin{align}\ddot{\vec r}(t)=&\hat r(\ddot r-r\dot\theta^2)+\\&\hat\theta(r\ddot\theta+2\dot r\dot \theta).\end{align}
Again this is a nice exercise to proof. For circular motion the radius is constant so $\dot r=0$. What's left of the general acceleration is
\begin{align}\ddot{\vec r}(t)=&\hat r(-r\dot\theta^2)+\\&\hat\theta(r\ddot\theta).\end{align}
The $-r\dot\theta^2$ term is the usual centripetal acceleration which points towards the center. You might recognise it if you plug in $\dot\theta=\frac v r$.
The $r\ddot\theta$ term is new. It points along the path of the particle and together with the $-r\dot\theta^2$ term it makes sures the radius stays constant. Note that if $\dot\theta$=constant this term drops out and we have regular old uniform circular motion.
EDIT I'll provide a slightly easier to read explanation.
In circular motion the velocity of a particle is always at 90 degrees with its radius. Like in this picture
When the speed says the same we have uniform circular motion. In that case the acceleration point directly at the center. We can decompose the acceleration into two components: one that points towards the center called 'centripetal acceleration' or $a_c$ and one that points along the velocity of the particle called 'tangential acceleration' or $a_t$.
In uniform circular motion we have $a_t=0$ since the acceleration is only towards the center. Also the centipetal acceleration is given by $a_c=\frac{v^2}r$.
If we accelerate in the tangential direction the speed of the particle increases. In that case the centripetal acceleration must increase to compensate, because $a_c=\frac{v^2}r$ and $r$ is constant. It's easier to see if you write it as $r=\frac{v^2}{a_c}$. If we make the speed twice as large then $a_c$ must get 4 times as big.
So long story short, it is possible to accelerate in the tangential direction but to do that you must increase the centripetal acceleration in a precise way to keep the same radius. Similarly we must decrease $a_c$ if we decelerate in the tangential direction.
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If we repeatedly divide a solid in half, at what point does it stop being a solid? Suppose I have some material in solid-state (say), I cut it into two parts. Take the first cut it into two parts, take the first cut it into two parts, and then repeat this again and again. There will be a point when the substance loses its solid property. I'm interested in this point.
I realize we don't have to go until we break it into two molecules but the state will come a little sooner. This thought experiment is a little bit crazy (and at a point impossible) but please consider this and correct me if there is a flaw in my thinking.
| Suppose you start in the thermodynamical equilibrium of the solid at a given temperature and suppose "cutting" means also moving apart from each other by the mean free path in the gas-phase. For doing this you have to apply work. If you are finished with cutting (say you did this in the order of 3*Avogadro number of times) you end up with a non-equilibrium gaseous state, to do so you had to invest an amount of work equal to the heat of sublimation, but the temperature of the surrounding heat bath will still be a temperature where the substance is solid in the TD equilibrium, and that will relax pretty fast back to the solid state under release of energy (heat).
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Stokes law and speed at equilibrium It is stated that the equation leading to Stokes law describes an equilibrium state, such as the upwards force is equal to the downward force:
$$ 6\pi r\eta v = \frac43\pi r^3 (d_1 - d_2)g $$
Now $v$ is velocity, and velocity at equilibrium is $0$, and this is the breaking point of my understanding. What is the nuance here?
| (a) Your equation does not lead to Stokes's law but $incorporates$ Stokes's law. The law states that a sphere of radius $r$ moving at speed $v$ through a fluid of viscosity $\eta$ experiences a resistive force given by
$$F_\text{res}=6 \pi \eta r v$$
and is in the opposite direction to the body's velocity. It holds provided that there is streamline flow of the liquid relative to the sphere.
(b) Your equation applies to when a body, falling through a fluid, has reached a constant velocity (provided that flow remains streamline). The condition for constant velocity is that upward and downward forces on the sphere are balanced.
The upward forces are the Stokes's law force and the Archimedean upthrust given by
$$F_\text{Arch}=\tfrac43 \pi r^3 d_2 g$$
in which $d_2$ is the fluid density. The downward force is the pull of gravity,
$$F_\text{grav}=\tfrac43 \pi r^3 d_1 g$$
in which $d_1$ is the density of the material of the sphere.
As we have said, when a steady velocity if reached
$$F_\text{res}\ +\ F_\text{Arch}\ =\ F_\text{grav}$$
Substituting from formulae above and re-arranging gives the equation you've quoted.
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Fourier Transform of $1/k^4$ I am dealing with a higher derivative theory problem and I have to perform the following integral,
\begin{equation}
\int \dfrac{d^3k}{(2\pi)^3}\dfrac{e^{i{\bf k}\cdot {\bf r}}}{k^4}
\end{equation}
This is because i have to solve something like $(\nabla^2)^2f=K\delta^3({\bf r}).$
I know that if the exponent in the denominator was a 2, the result is the Coulomb potential $1/(4\pi r)$ (see https://en.wikipedia.org/wiki/Common_integrals_in_quantum_field_theory), but I didn't found anything about a 4 in the denominator.
Does anybody know the result of this integral?
| We can proceed exactly as in the case of the Coulomb potential i.e. Fourier transform of $1/k^{2}$. So, performing the same steps as in Coulomb case, we have
$ \begin{split}
\int \frac{d^{3}k}{(2\pi)^{3}} \frac{e^{i\vec{k}.\vec{r}}}{k^{4}} &= \frac{1}{(2\pi)^{2}}\int_{0}^{\infty} \frac{dk}{k^{2}} \left[\frac{e^{ikr} - e^{-ikr}}{ikr} \right] \\
&= \frac{1}{4\pi^{2}ir}\int_{-\infty}^{\infty}dk \frac{e^{ikr}}{k^{3}}
\end{split}
$
To perform the final integral, we can simplify as
$ \begin{split} \int_{-\infty}^{\infty}dk \frac{e^{ikr}}{k^{3}} = (\pi i)Res(\frac{e^{ikr}}{k^{3}}) = -\frac{\pi ir^{2}}{2} \end{split} $
Here, since the pole is on the contour we take only half its contribution when using the residue theorem. Using this we have finally,
$\begin{split}\int \frac{d^{3}k}{(2\pi)^{3}} \frac{e^{i\vec{k}.\vec{r}}}{k^{4}} = - \frac{r}{8\pi} \end{split}$
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Measurement of force (statics) body 1 and 2 are pulling the balance by 20 N
what does the balance read?
same question but without 2.
for 1st question I thought the balance will read zero force because they are two opposite forces of the same magnitude but it's wrong. the correct answer is 20N I don't know why
| For every weighing scale the force you exert is always counterbalanced by an equal and opposite force - that's the third Newton's law. If the scale lies on the ground, your weight would be opposed by the normal force coming from the ground, which is fixed; the same would happen with a Newton meter (or a spring balance), hanging down hinged on a wall. But if you suspend the spring meter and pull it toward yourself, it will consequently accelerate with your hands. You need an equal strength from the opposite side to fix the balance and let it measure the force, which is in this case 20 N.
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Why do we use Eigenvalues to represent Observed Values in Quantum Mechanics? One of the postulates of quantum mechanics is that for every observable $A$, there corresponds a linear Hermitian operator $\hat A$, and when we measure the observable $A$, we get an eigenvalue of $\hat A$ as the result.
To me, this result seemed to come out of nowhere. While I could understand representing an observable $A$ by a linear operator $\hat A$, I can't understand why must the results of measuring $A$ have to be an eigenvalue of $\hat A$. Is it possible to better motivate this postulate?
Edit: Since my question might be a little vague, let me try to rephrase it - how can one motivate this postulate to a student first being introduced to Quantum Mechanics? Are there experimental results, for example, which can be used as motivation?
| You might want to have a look at the ideas of Quantum Darwinism. I am not sure how popular this thoughts are, so decide for yourself.
As far as I understand there an attempt is made to explain why certain states are measured, based on how "stable" they are compared to other states when interacting with the measurement device and the environment.
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Perpetual motion: Conditions for rolling a wheel sliding on a bar This is a basic question about energy conservation and classical mechanics:
Question: Under what situations can this motion be perpetual?
*
*Without gravity and without frictions.
*Without gravity and with frictions.
*With gravity and without frictions.
*With gravity and with frictions.
*Others setup (please state the setup)
*Impossible to be perpetual
Gravity (say) is along the vertical $y$ direction, with a constant gravitational force and a linear gravitational potential $V(y)=mgy$.
| Neglecting GR effects, the answer is clearly meant to be (1) for all speeds and (3) provided the inner gear is moving fast enough, because gravity is a conservative force.
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Confusion about Lorentz Coordinate Transformation A normal Lorentz coordinate problem might say:
At $t=t'=0$, two coordinate systems $S$ and $S'$ have their origins coincide with the $S'$ system moving with speed $v$ in the $+x$ direction relative to $S$. If event 1 happens at $x=a$, $t=0$ in the $S$ system then when/where does this event happen in the $S'$ system. I know that $x'=\gamma(x-vt)$ and I have no confusion with that equation, but in problems like these, I am confused specifically about the time. Since $t=t'=0$ and $t=0$, I would imagine that the time in the prime system would also be 0. But when one uses the Lorentz transformation $t'=\gamma(t-vx/c^2)$, $t'$ comes out as a non-zero negative number. How do we reconcile this?
| Consider Einstein's train thought experiment that demonstrates non-simultaneity. Let's say the train moves left to right. The lightning strikes at the front and back of the train occur simultaneously for the platform observer (S frame) but for the train observer (S' frame) the strike at the front of the train occurred first. If S and S' are set up so that the two frames coincide with t = t' = 0 at the back of the train when the strike there occurs, we have t = 0 in the S frame for both strikes but in the S' frame the strike at the front would have occurred at t' < 0. In your example, the two two events in the S frame are (x,t) = (0,0) and (a,0). These are simultaneous in the S frame so we know they are not simultaneous in the S' frame and the Lorentz transformation give us the t' values in S'. The (0,0) event also occurs at (0,0) in S', but the (a,0) event does not occur at t' = 0.
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Which heat capacity is used in the heat equation? The heat equation is often written as $\frac{\partial T}{\partial t} = \frac{\kappa}{c} \nabla^2T$ where $\kappa$ is the thermal conductivity and $c$ is a heat capacity per volume.
I often see $c$ written as $c_P$ implying that it is the heat capacity (per unit volume) for a system held at constant pressure, but I was wondering if this was necessary of not?
I understand that in most 'everyday' experimental examples, pressure will be the variable that is held constant, and that for liquids and solids there isn't much difference between $c_P$ and $c_V$ anyway. However, in theory, can the heat capacity in this equation be with whatever variable you want to hold constant (so could be either $c_P$ or $c_V$ depending on your situation?
I also found this similar question but I couldn't find a definitnive answer to my question in their answers: In deriving the heat transfer equation, why do we use heat capacity at constant pressure?.
| Both options ($C_p$ or $C_v$) are approximations.
In the general situation, a non-rigid solid material will expand non-uniformly and therefore will have a time-dependent stress distribution, which stores internal strain energy.
The simplest assumption is that the object is not mechanically constrained and that the non-uniform internal strain energy is small. Therefore the internal pressure in the solid remains approximately constant and $C_p$ is the relevant thermal capacity.
For a fluid, the internal pressure is uniform (ignoring the weight of the fluid and any other inertia forces) and the correct choice is more obviously dependent on whether the volume is constrained or not.
If the object is constrained to have constant volume, $C_v$ would be a better approximation, since it takes account of the internal strain energy (assuming the internal stress field is uniform, of course).
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Many-world interpretation of double slit experiment Different interpretations of young's double slit experiment is available, I read Copenhagen and Feynman's path integral interpretation of double slit experiment; former uses the idea of wave function and later uses the idea of infinite paths and sum over the weight factors of each path obeys causality.
I Googled lot of time to see how double slit experiment is explained using Many-world interpretation of quantum mechanics, but unfortunately I don't find anything.
It would be great helpful if someone give a nice explanation that, how famous double slit experiment is explained using Many world interpretation of quantum mechanics?
| Mathematically, the path integral is simply a way to derive the wave function.
The three interpretations you mention can broadly be summarised:
*
*Copenhagen The wave function is probabilistic and the selection of any given outcome is otherwise inherently random. Any underlying reality is inaccessible to science. Shut up and calculate.
*Path integral The probability function is the sum of every possible path throughout spacetime. Quite why one particular path wins out in any given measurement is not directly addressed; "hidden variables" are sometimes proposed but lack any scientific basis.
*Many worlds The paths are real. For each possible path (or unique outcome), the real world divides and each reality follows one of those outcomes. The extent to which different paths with the same observed outcome might first split into parallel worlds, and subsequently re-merge back into one reality, has varied in response to criticisms.
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Time for an object's afterimage to disappear at the event horizon My understanding is that as someone/something falls into a black hole, they would get dimmer and dimmer until disappearing entirely as they cross the event horizon. Most non-radiant objects would probably become invisible to outside observers well before hitting the event horizon. But if we had something very bright (like a hot young star) hitting a black hole's event horizon, about how long would we see its "afterimage floating near the horizon" before it disappeared? (Assume the star's trajectory into the black hole was directly along our line-of-sight distance to the black hole itself.)
| The time will be proportional to the light crossing time of the black hole, which is less than a millisecond for a stellar-mass black hole, or around a day for the M87 black hole.
I don't have a source for the following derivation and it's possible I made a mistake.
An object radially infalling into a Schwarzschild black hole satisfies $d^2r/ds^2 = -GM/r$. Integrating this once gives $dr/ds = -\sqrt{r_s/r-r_s/r_0}$. For simplicity I'll take $r_0=\infty$ (the object falls in from infinity). Arbitrarily choosing $r=r_s$ at $s=0$, the solution is $r = r_s (1 - \frac32 s/r_s)^{2/3}$.
I'll use Eddington-Finkelstein outgoing coordinates $$ds^2 = (1-r_s/r)\,dv^2 + 2\,dv\,dr$$ because the redshift as seen by someone at rest at infinity is just $1{+}z = dv/ds$. Plugging in the formula for $dr/ds$ gives us $$1 = \left(1-\frac{r_s}{r}\right)\left(\frac{dv}{ds}\right)^2 - 2\,\frac{dv}{ds}\,\sqrt{\frac{r_s}{r}}$$ which has the solution $dv/ds = \left(1-\sqrt{r_s/r}\right)^{-1}$. Plugging in the formula for $r$ gives us $$\frac{dv}{ds} = 1{+}z = \left(1-\left(1-\frac{3\,s}{2\,r_s}\right)^{-1/3}\right)^{-1}$$
I'd rather have $z(v)$ than $z(s)$ so I multiply by $ds/dz$ to get $$\frac{dv}{dz} = 2\,r_s\,(1+z)^3/z^4$$ $$v(z)=2\,r_s\left(\ln z - 3/z - 3/2z^2 - 1/3z^3\right) + \text{const}$$
I doubt that's invertible. At any rate, you can see that at late times ($z\gtrsim 10$) the redshift increases exponentially with a time constant of $2\,r_s$ (33 hours for the M87 BH). It takes about $4\,r_s$ to get from $z=e$ to $z=e^2$, about $9\,r_s$ to get from $z=1$ to $z=e$, and about $44\,r_s$ to get from $z=1/e$ to $z=1$.
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1-dimensional Heat Equation I have to solve the following differential equation:
$$ \partial _t u(x,t) = D \partial ^2_x u(x,t) $$
with the initial condition $$ u(x,0)=\exp \left( -100^2 \left( x-\frac{1}{2} \right) ^2 \right) .$$
The $x$ and the $t$ Interval is [0,1].
The Boundary Conditions are $$u(0,t) = u(1,t) = 0.$$
I tried to use Fourier Transformation but I dont know how to deal with the initial condition.
| As others have already pointed out, there exist many methods for solving this equation (separation of variables, integral transforms, Green's function approach, etc.) One thing that is worth noting is that here we are dealing with a problem with initial conditions. Thus, if we pursue the solution using Fourier transform, then it makes sense to use this transform for both $u(x,t)$ and $u(x,0)$ in respect to $x$, but to use the Laplace transform in respect to $t$ - this way the initial condition is incorporated seamlessly.
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Optics viewing angle and distance Almost all optics like microscope, telescope.. require observer putting their eyes very close to the optics to see.
Can the same optics designed so observer can see like viewing TV screen, i.e. from a larger distance like 30cm from optics? Or is it impossible?
| There are a few laws of optics that come into play here. Probably the most significant one is simply this: if you step back 30 cm from the exit aperture, the aperture now subtends a very small angular field of view. Since all the light coming from the telescope must pass through that aperture, it can't take up a large part of the space your eye can see! (There are additional optics laws -- look up terms like "exit pupil" and "angular magnification" -- which explain why moving even a cm away makes the image look tiny inside the eyepiece)
Now, by comparison, if you look at the optics of a projector, that lens is designed to produce an image at some distance. You, the viewer, can look at the entire image on the screen because it takes up a large part of your field of view. But if you were standing at the image screen, looking at the projector, all you would be able to see is that small piece of the image which hits your eyeball.
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Why amplitudes are rational functions? In Bootstrap and Amplitudes: A Hike in the Landscape of Quantum Field Theory there are few statements about analytical structure of amplitudes.
I want to understand statement:
Tree amplitudes must be rational functions of Mandelstam
variables.
What is the reason for such a property of tree level amplitudes? What is the simplest explanation of such a fact?
Comments about other analytical properties are also very appreciated!
| @Anonjohn has given a good answer. I would like to add a bit more. I think, that the author of statement tacitly assumes dealing with massless theory with a linear dispersion relation (kinetic term of form $\phi \Box \phi $) : Why does nature favour the Laplacian?. So each propagator is of the form:
$$
\frac{1}{(\sum_{i \in I} k_i)^2}
$$
The analytical properties are crucial in determining the exact expression for tree amplitudes in massless theories. For instance, the proof of famous Parke-Taylor formula by BCFW https://arxiv.org/pdf/hep-th/0501052.pdf - is based on them.
For a good introduction I recommend these notes, written by the same author as the article you cite - https://arxiv.org/pdf/1308.1697.pdf.
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How to convince myself that capacitance cannot be a function of voltage? My textbook states that:
...The capacitance $C$ depends only on the geometrical configuration (shape, size, separation)of the system of two conductors. [As we shall see, later, it also depends on the nature of the insulator (dielectric) separating the two conductors.]...
Now I wanted to know if these were the only two factors on which the capacitance of a conductor depends. So I tried thinking about it from the basic definition, which is:
Capacitance is the ratio of the change in electric charge of a system to the corresponding change in its electric potential. (Source: Wikipedia)
i.e.,
$$C= \frac {dQ}{dV}$$
where, $V$ is the potential of the conductor with respect to zero potential at infinity.
Now, I can not think of why $C = f(V)$ isn't a possible scenario, assuming that shape size, etc remains constant.
So:
*
*Why cannot $C$ be a function of $V$?
| $C$ can be a function of $V$, for example the capacitance of a varicap diode is controlled by the voltage across it.
However the voltage does not directly set the diode capacitance. What it does is to control the separation of the charge layers. It is this varying separation which results in the capacitance change.
Thus, the capacitance is a function of voltage only because the separation is a function of voltage and the capacitance is a function of that separation.
This kind of secondary effect is the only way that voltage can affect capacitance; it has to alter one of those basic physical parameters.
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Why doesn't planet Earth expand if I accelerate upwards when standing on its surface? According to General Relativity I am being accelerated upwards by planet earth while writing this question. But a curious person on the the other side of the planet relative to me would have the same experience. That means we are accelerated in opposite directions, although earths diameter do not seem to increase. How can this be?
| It comes down to a definition of acceleration. Acceleration is most universally appreciable as a force application contradicting an object's natural position or trajectory. Notice that this does not require that the object move -- only that it is being affected by a force, as in 'experiencing pressure'.
So by this definition an object at apparent rest on a table top is being forced by the solid table surface, and feels the pressure of this force throughout its form, and so on.
It also helps to appreciate gravity as an electromagnetic phenomenon, as the definition of acceleration also applies to (ferro-)magnetic forces. When you see two strong magnets pulling or pushing each other, it appears that they are exerting a force, as though expending energy...
But to the magnets their unhindered relative motion represents a state of rest given their natural atomic states. Energy expense is only experienced by the person holding the magnets apart/together against the natural tendency, and by the magnets when they are prevented from their natural relative motion (including if/when they impact).
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Why is Schrodinger equation taught while it does not describe an electron? Strictly speaking, it is "wrong" because it does not describe spin-1/2 particle like an electrons. Why in every QM textbook is it taught, not as a historical equation, but as a current equation?
| I think you are confusing the treatment of relativistic (spin-$\frac{1}{2}$ particles) electrons as compared to the non-relativistic case. The Schrodinger equation can perfectly describe the properties of the non-relativistic electron. The Dirac equation describes the interactions of relativistic electrons (and other spin-$\frac{1}{2}$ particles).
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Rotating current carrying loop Consider a circular loop of wire fixed on the rim of a wheel. This wire carries a current 'i' in it. When the wheel is at rest, which basically means that the current carrying loop is at rest, the magnitude of magnetic field at the center is, say B1. If I set the wheel in motion with a constant angular velocity with the center of the wheel at rest and without changing the current in the loop, which implies that the current carrying loop rotates about its center point, will the magnitude of magnetic field change at the center point?
I think that it will change because number of charges passing through a unit cross section which is at relative rest with respect to the center point changes and hence current effectively changes and hence magnetic field changes.
Is this correct?
| I'm thinking that with the wheel rotating, the motion of the positive charges will constitute an electric current that offsets the change in the motion of the free electrons.
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When a sphere is in pure rotation, will all particles in its surface have same linear speed? Imagine this sphere to rotate about its diameter, from the centre to the point of surface if we take all of them have equal distance that is 'r(radius of the sphere)'.So same linear speed right? I looked up many sites but they all say "The linear speed v = ɷr. That means the particles at different r will have different linear speed".I am not able to digest it. Can anyone please make me understand in a better way?
| Linear speed of a point on a rotating object will depend on the angular speed and distance between the point and the axis of rotation. All points on a sphere are equidistant from the center of the sphere, but are not equidistant from an axis of rotation through the sphere. Consider the earth spinning about its axis - the poles have no linear movement whatsoever (they simply spin in place), while points on the equator are moving at 460 m/s in order to complete their 40,000km rotational circuit once every day.
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What will happen if I multiply a ket vector by a complex number? I was reading Zettili’s Quantum Mechanics book. There I have seen
when a ket (or bra) multiplied by complex number, we also get a ket (or bra)
But how do we infer this by mathematics?
| Two things happen, depending on the magnitude and phase of the complex factor. Both can be seen by examining what happens when you insert the factor $f$ in $\mid\Psi\rangle = f\mid\psi\rangle$.
Your primary observables are always bra-ket pairs, so calculate $\langle\Psi\mid\Psi\rangle=\langle\psi\mid f^*f\mid\psi\rangle = |f|^2\langle\psi\mid\psi\rangle$. ($f^*$ is applied to the bra, since it's the complex conjugate of the ket.) That is, multiplying a state by a complex value multiplies the number of particles by the square of the $f$'s magnitude. The phase of $f$ can only be observed via interference with the original $\mid\psi\rangle$, since phase is generally unobservable.
Since operators like mass and momentum are linear, increasing the number increases mass & momentum proportionately. If there are self-interactions in the energy or other operators, those can change in non-linear fashion.
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Will any solution to the wave equation be a wave in reality? In the mathematical sense, a wave is
any function that moves. In that sense we can consider that any function that complies with the wave equation (let's consider in one dimension to simplify things) will be a wave. But is this kind of thinking physically valid? That is, any function that satisfies the differential equation can give us information about some phenomenon in reality? or the definition of what a wave is has to go beyond complying with the wave equation. I appreciate your time
|
In the mathematical sense, a wave is any function that moves.
Not so, unless one defines the variables of the function, the above has no meaning. Wave equations are used in order to model observations of nature.
What is a wave? From sound and water waves we come to an association with sine and cosine variational behavior. Wave equations are differential equations whose elementary solutions are sinusoidal .
In classical mechanics and electrodynamics, waves are seen as the sinusoidal solutions, and involve the energy carried by the wave in space as a function of time.
When dimensions become very small, compatible with h, the Planck constant the individual "particles" electrons etc., can be described sometimes like classical billiard balls, and at the same time they exhibit a randomness, which when accumulated displays interference and other wave characteristics, which give rise to the statement "particle/wave duality".
See this answer of mine for more details.
So no, not all solutions of wave equations correspond to physical reality. One chooses the variables to be used in the wave equations so that they correspond with observables.
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Bernoulli Pipe Flow Equation Physical Meaning I have just been introduced to the Bernoulli Equation for fluid flow. However, I am unable to understand why the pressure and velocity are inversely proportional.
Because, as the fluid goes through a smaller cross-sectional area, the flow velocity increases, but due to the Bernoulli equation this increase in velocity means that there is a decrease in pressure. However, if say the fluid is not an ideal fluid, when it goes through a smaller cross section, it compresses, and hence is more "focused" therefore, it should surely mean that the pressure is increased. Furthermore, using the fundamental pressure equation: P = F/A, a smaller cross-section will mean that there is more pressure.
I have seen all of the mathematical proof (the derivation of the Bernoulli equation), observed it in practice (using a U-tube manometer) and read some analogies to GPE to KE, but I still don't underdstand, physically how this can happen.
If someone has a good explanation, I would love to hear it, thanks.
| If the pressure in the small diameter, where the flow has greater velocity, was equal or greater, we could take a derivation from that region and inject the fluid back in the region of big diameter.
In this case, the average velocity in the region of big diameter would increase, leading to increased velocity in the region of small diameter and so on. Energy would be created from nothing.
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Is $U^\dagger(R)\hat{H}U(R)=\hat{H}$ always true? Consider a Rotation transformation on momentum state,
$$U^\dagger(R)\hat{\mathbf{p}}U(R)=R\hat{\mathbf{p}}$$
Now the question is whether,
$$U^\dagger(R)\hat{H}U(R)=\hat{H}\,?$$
Here, $\hat{H}$ is the Hamiltonian of a free particle. Is it always true? Is there any counter examples?
My attempt:
\begin{align}
U^\dagger(R)\hat{H}U(R)&=\frac{1}{2m}U^\dagger(R)\hat{\mathbf{p}}^2U(R)\\
&=\frac{1}{2m}U^\dagger(R)\hat{\mathbf{p}}U(R)U^\dagger(R)\hat{\mathbf{p}}U(R)\\
&=\frac{1}{2m}(R\hat{\mathbf{p}})(R\hat{\mathbf{p}})
\end{align}
Is this always true that $$\frac{1}{2m}(R\hat{\mathbf{p}})(R\hat{\mathbf{p}})=\frac{1}{2m}\hat{\mathbf{p}}^2\, ?$$
If it is why? If not when it is not?
Note: This is an exercise from Coleman's course 253a (https://arxiv.org/abs/1110.5013). See equation (1.8) there. It would be better if the answer is provided using his notations.
| If the operator $U$ is indeed the unitary operator, then
$U U^{\dagger}= U^{\dagger} U = 1$
then expression you have (which appears to be applied incorrectly) above will reduce back to its original form
$\frac{1}{2m} U^{\dagger} p^2 U = U^{\dagger} H U = H$
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Magnetic field around a current carrying wire and Special Theory of Relativity I understand how the magnetic field and electric field (around a current-carrying conductor) swap their roles depending on the frame of reference, due to the effect of length contraction (in the special theory of relativity); however, this question has been troubling me.
Consider two charges separated by a vertical distance and both moving horizontally with equal velocities (with respect to the ground frame). In the ground frame, we can observe both a magnetic effect (each moving charge produces a magnetic field, in which the other charge is moving) and an electric effect of each upon the other. But in the charges' frame, there is only the electrostatic force (of the same value as that of observed in the ground frame because the charge and the vertical length are invariant). How is this possible? What is the actual value of force between the charges?
| In a frame with two equal charges (a source charge and a test charge) separated by $y$:
The electric field from the source at the test charges is:
$$ \vec E = k_e \frac q {y^2} \hat y$$
and the force on the test charge is:
$$ \vec F = q\vec E = k_e \frac {q^2} {y^2} \hat y$$
If we boost this by $\vec v = -v\hat x$ so that the two charges appear to move in the $+x$ direction, then the electric field at the test becomes:
$$ \vec E' = \gamma(\vec E + \vec v \times \vec B)-(\gamma-1)(\vec E\cdot\hat v)\hat v = \gamma \vec E$$
There is also a magnetic field:
$$ \vec B' = \gamma(\vec B - \frac{\vec v \times \vec E}{c^2})-(\gamma-1)(\vec B\cdot\hat v)\hat v = \gamma\frac v {c^2} ||E|| \hat z$$
The Lorentz force law
$$ \vec F' = q(\vec E' + \vec v' \times \vec B) $$
gives:
$$ \vec F'=q(\gamma \vec E - \frac{v^2}{c^2}\vec E) = \vec F/\gamma $$
So the 3-force is not a Lorentz invariant. Note that when the charges appear to be moving, the electric field attraction is stronger, but it is mitigated by an opposing magnetic force.
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Why can we set $c$ and $\hbar$ to 1 when it changes the result? So in my QFT course, my professor said that you can set $c$ and $\hbar$ to 1. And he gave us an example:
$$E = mc^{2}$$
And then set $c = 1$:
$$E = m$$
This seems completely ludicrous to me to do. Doesn't it change the result? Why can this be done and why isn't it wrong?
I mean, $E = mc^{2}$ gives you one answer and $E = m$ gives you another, completely different answer!
| If you ask two doctors how much you weigh and one doctor says “you weigh $100\text{ kg}$” and the other doctor says “you weigh $220\text{ lbs}$”, would you claim that they have given you completely different answers? No. They gave you the same answer using different units.
Setting $c$ or $\hbar$ to 1 is simply choosing to use units where those quantities are 1. For example units of years for time and units of light years for distance in the case of setting $c=1$
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Does the second law of thermodynamics imply that renewable energy also leads to global warming/climate change? So I have this (possibly dumb) question about the implications of the second law of thermodynamics to the use of renewable energy for the world, so please bear with me. Here goes: Apart from the finiteness of fossil fuels (FFs), which is obviously an issue, the main problem for sustainability in burning FFs for energy is that it leads to the release of greenhouse gases. These trap a lot more of the incoming solar energy than otherwise would have been the case, leading to global warming. OK. So we need to move to solar, wind, nuclear, etc.
But the second law implies that we can only extract some finite amount of energy for useful work, and the rest invariably goes to heat (right?). Solar panels, batteries, wind farms, etc. all presumably leak all unconverted energy into heat. And a lot of this energy is coming from 'outside' the biosphere. So doesn't that mean that even with renewable sources, we will inevitably leak heat into the biosphere leading to at least some global warming?
So isn't it really only a difference of scale in terms of the warming that is caused by greenhouse gases from FFs vs. from renewables? Won't we always heat up the surface, no matter how efficient we get?
Also, increasing efficiency could (and I think tends to) lead scaling up our use of energy, so that the total waste heat generated might still increase. Is it possible to keep the biosphere at the same approximate temperature even if billions more people start using the same amount of energy per capita as say a modern wealthy European does?
So finally, are there 'sinks' for all this excess heat that we could tap into? Space is at ~4 K, right? So can we use space as an infinite heat sink?
| There is no increase in heating from solar or wind power, because they're using energy that's already in the system. Yes, it all eventually gets converted to heat--but it would anyway. Sunlight not falling on solar panels would fall on something else; wind hitting wind turbines would otherwise run into buildings, trees, or mountains.
| {
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Charge conjugation of fields This page on Wikipedia says, "In the language of quantum field theory, charge conjugation transforms as -
*
*$\psi \Rightarrow -i\big(\bar{\psi} \gamma ^0 \gamma ^2 \big)^T $
*$\bar{\psi} \Rightarrow -i\big(\gamma ^0 \gamma ^2 \psi \big)^T $
Could someone explain the reason behind choosing this specific mathematical transformation for charge conjugation (since the only thing it's supposed to do is invert the sign of all charges to convert a particle into its anti-particle)? Any relevant sources are also welcome.
| The charge conjugation symmetry is the proper time reversal symmetry.
It becomes an exact symmetry of relativistic quantum systems if we combine with space-time time reversal and parity, due to unitarity.
About the techinical details you are raising I think that this transformation looks complicated just in you notation, afterwards you are hiding the indices.
I think that in four dimensions there is a suitable choice of indices ($SL(2,\mathbb{C})$ indices) where the charge conjugation reduces to:
$$\psi^{\alpha}\leftrightarrow \overline\psi^{\dot\alpha}\,\quad \alpha=1,2\;\quad \dot\alpha=3,4\,.$$
| {
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If you keep the center of gravity of two objects on each other would you never be able to separate them? To find the attractions between planets and stuff like that, you use the center of gravity/mass to apply to Newton's equation. So even if those planets collided into each other, you could separate them if you give enough force, because $r$ (distance between the center of gravity/mass of each planet) in the gravitation equation is not $0$ therefore $r^2$ is not $0$. But the problem comes when you put the centers of gravity/mass of two objects on each other. Then $r$ is $0$, $r^2$ is $0$ and when you divide by $r^2$ (in the gravitation equation), you’re dividing by $0$ which means the gravity is infinite; i.e you'll never be able to separate them. Now you might say that there will never be such an instance where the two centers of gravity/mass will never be on each other, but consider this-
Two hoops, one 1/2 in radius of the other, placed on a table such that the circumference of those 2 hoops are parallel (like a train track that goes in perfect circles). The center of mass of the bigger hoop will be at the very center of the area (circle) enclosed by the bigger hoop. The same goes for the second, smaller hoop. The center of mass of each hoop will lie on the same point. So does that mean no matter how much you tried, you'll never be able to separate them? This question has been puzzling me for ages so help would be great.
| The idea that the force between two spherical bodies goes as $1/r^2$ is only valid outside of the bodies.
Once you're inside the bodies, things are different. If the bodies are of uniform density, the "shell theorem" applies, and the force goes to zero as $r$ goes to zero. (It may not be obvious, but if you work out the math, there is no net gravitational force from any of the mass outside of your $r$, so as $r$ gets smaller so does the mass, and the mass gets smaller faster than $1/r^2$.)
The argument is similar for the hoops, but the math will be more complicated.
But, in the end, being very close to the center of mass without actually being very close to an actual mass will never result in very large forces.
| {
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Prove that the elements of the dual frame of an IC POVM cannot be positive I've read a claim about POVMs in my lecture notes, which I fail to prove. Hence, I would be grateful if some of you have some hints for me /can help me.
Let $\{N_i\}_{i=1}^{d^2}$ be an informationally complete POVM on $Herm(V)$, where $V$ is a Hilbert space. We define its dual frame $\{D_j\}_{j=1}^{d^2}$ by requiring $Tr[N_i D_j] = \delta_{ij}$. The claim is that at least one of the $D_j$'s is not positive.
Well, we know that $\forall i \in\{1,...,d^2\}: ~ N_i \geq 0$ and $\sum_{i=1}^{d^2} N_i = I$. Furthermore, as we said that we deal with an IC POVM, we have $\text{Span}\{N_i\}_{i=1}^{d^2} = \mathcal{L}(V)$ and we have the dual-frame condition $Tr[N_i D_j] = \delta_{ij}$. In addition, as the dimension of $\text{Herm}(V)$ equals $d^2$, we know that $\{N_i\}_{i=1}^{d^2}$ is a basis.
I tried to find an argument, why we find at least one $k \in\{1,...,d^2\}$ such that $D_k$ is not positive semi-definite, i.e. has at least one negative eigenvalue. But this turned out to be not very constructive...
Alternatively, I tried to construct a contradition by assuming that both $\{N_i\}_{i=1}^{d^2}$ and $\{D_i\}_{i=1}^{d^2}$ are positive, but (as you might have guessed, as I am asking this question here) I failed to prove it that way.
Is anyone having an idea how I can prove this statement?
Thank you in advance four your help!
| Suppose $D_k \geq 0$ for all $k$. Notice that all the $D$ operators have unit trace.
For two PSD matrices it holds that, $Tr(AB) \leq Tr(A)Tr(B)$.
This implies that, $Tr(N_k D_k) \leq Tr(N_k)$ for all $k$
Now we know that $Tr(N_k D_k) = 1$. So if all the dual operators were PSD, we would have the $Tr(N_k) \geq 1$ for all $k$.
This clearly contradicts the POVM property $\sum_i N_i = I$. We can see this by taking trace on both sides. The RHS has trace $d$, while the LHS would have a trace of at least $d^2$
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How do wavelengths add-up? If I have 3 LED lights of following colors and wavengths:
Red 650 nm
Blue 450 nm
Green 550 nm
All three are placed side by side and turned ON.. then what will be the wavelength of the combined light that my eye will see? Most probably the combined color will be 'whitish' because the mixing of RGB colors give white color.
| Wavelengths are one to one only with the spectral colors, and wavelengths do not add up to a new one, they remain distinct .
It is the perception of color which is a biological function, due to the receptors of the eye, which creates new hews and colors when more than one wavelength falls on the retina of the eye.
The perceived color will depend on the combination of wavelengths according to the plot above.
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How much energy is in the Universe as photons? The mass of Universe in kilograms is about 10 power 53.
But how much energy exists in the Universe in form of photons?
And if they would all be converted into mass, how much mass that would make?
Dark matter and dark energy are out of the question.
| Mass of the "observable" universe is about 10 to the power of 53. However, the answer of your question would be irrelevant, since what we have observed and measured is practically from the past. For example, the measured mass of a star in 1 billion light years from earth has added up to the calculation to get the "10 to the power of 53" number, but right now that star probably doesn't exist; and part of it's mass has converted to photons through nuclear fusion.
| {
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Is the Green function of electromagnetism a scalar or a tensor? When I check classical electromagnetism books Maxwell equations
\begin{equation}
\Box A^\nu (x)=\frac{4\pi}{c}j^\nu (x)
\end{equation}
can be solved using a scalar Green function $G(x,x')$
\begin{equation}
A^\nu (x)=\int G(x,x')j^\nu (x')d^4x'
\end{equation}
where the Green function satisfies
\begin{equation}
\Box G(x,x')=\frac{4\pi}{c}\delta^4(x-x')
\end{equation}
Examples of this are Jackson, eq. 6.48 on sec. 6.5. Also, on "The classical theory of fields" by Landau, on eqs. 62.9 and 62.10 he uses the scalar green function as well.
This immediately feels strange, since the 4-potential $A^\nu(x)$ could, in theory, have different boundary conditions for each component and a scalar Green function simply doesn't have enough degrees of freedom to accommodate that. Evenmore, in the context of quantum field theory, the photon propagator (which is essentially the Green function) is a tensor $\Pi_{\mu\nu}$ so I'm confused about the nature of the Green function in classical electromagnetism: Is the scalar Green function $G(x,x')$ the most general Green function or in a general case we need a tensorial Green function $G_{\mu\nu}(x,x')$?
Note: This question is explicitly about classical electromagnetism, I'm using the quantum field theory propagator as an example to show my confusion but the question applies to the classical theory.
| Index $\nu$ has nothing to do with the equation itself. The equation does not know whether $A$ and $j$ are scalars, 4-vectors on spinors. The equation itself is scalar type, so its Green function can be scalar only.
Another argument: let say $A^\nu$ is a solution for $j^\nu$. Then $C^\mu_\nu A^\nu$ must be solution for $C^\mu_\nu j^\nu$. If $G^\mu_\nu$ is the general green function then we have (I skip integrals to shorten the notations and imply same indices summation)
$$
a)\quad A^\mu = G^\mu_\nu j^\nu
\qquad
b)\quad C^\mu_\lambda A^\lambda = G^\mu_\lambda C^\lambda_\nu j^\nu
$$
what immediately gives us for any matrix $C$
$$
C^\mu_\lambda G^\lambda_\nu = G^\mu_\lambda C^\lambda_\nu
$$
i.e. $G^\mu_\nu \propto \delta^\mu_\nu$.
| {
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Why can’t gravitons distinguish gravity and inertial acceleration? If gravitons mediate the gravitational force, couldn’t the detection of gravitons by an observer be used to distinguish whether they are experiencing gravitational acceleration vs. inertial acceleration, contradictory to general relativity? If this is not the case, and detection of gravitons can not be used to distinguish gravity from other acceleration, shouldn’t acceleration affect the way objects interact with the gravitational field?
Obviously, this can not be correct, so what am I missing?
| In the language of quantum field theory, the gravitational force is self-coupling, which means that gravitons carry gravitational charge and can feel the gravitational field.
In the language of general relativity, sufficiently small gravitational waves are perturbations of spacetime travelling in a curved background spacetime, and will follow null geodesics in that spacetime, which means that they will curve in that spacetime in exactly the same way light does.
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Why don't we use the concept of axis of mass in place of center of mass? Being a high school student, I read the concept of center of mass and it was written in my book that
When a spinning ball is projected with some velocity , then all the points on the ball have complicated paths except the center of that ball which follows the well known parabolic trajectory. And hence we define that point as center of mass.
However, I think that all the points on the any axis about which the ball is spinning follow the parabolic trajectory and are not influenced by spin .
Edit : Most of the answers argued that the rotation axis may change because of torque but the main point to note here is that we can't differentiate between two axis in case of a sphere since it is symmetric from all sides and also that a sphere can't rotate about more than one axis at a time . So saying that it will rotate about different axis is I think meaningless.
So is it okay to define axis of mass in place of center of mass for sphere or other symmetric bodies or am I wrong somewhere ? If not, give a proper reason.
| If the ball has a single axis of rotation, then all points on that axis will not rotate (that's pretty much the definition of "axis of rotation"). However, it's possible for a ball to have more than one axis of rotation simultaneously, so while their claim is misleading in that it implies that all non-center points always have complicated paths, it is true in the sense that only the center of mass is guaranteed to have a simple path.
As for your question of whether we can define an "axis of mass", that is clearly impossible in the case of a sphere; since the spherical is ... well, spherically symmetric, there's no way to distinguish a particular axis. There is one axis distinguished by the rotation, but that axis is particular to that motion, and not an inherent property of the sphere. Even if there is a particular axis that it is rotating around, there are an infinite number of axes that it could be rotating around.
For objects in general, it is possible to distinguish three principal axes.
| {
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Effect of coloured filters on white light I am concerned about the accuracy of some information in a science textbook which I would like to clarify please. When white light is shone through a blue filter, only blue light will pass through. When the emergent blue light is passed through a red filter, no light gets through, because there is no blue light left. This makes perfect sense.
However, what would you expect to see, and why, if white light is shone through:
*
*a yellow filter, followed by a blue filter?
*a blue filter followed by a yellow filter?
*a yellow filter followed by a red filter?
*a red filter followed by a green filter?
*a green filter followed by a yellow filter?
| My guess is that the author of the textbook has the common misconception that there are only three colors—red, green, and blue—and yellow is red+green. Therefore, you're expected to answer that the red+yellow filters will let through red light, and the green+yellow filters will let through green light, and the others will let through nothing.
That isn't actually true. Light comes in a continuous range of wavelengths; filters can in principle block any of those wavelengths and let any others through; and there are many different combinations of wavelengths that we'll perceive as red, yellow, green, or blue.
On this page I found a chart of the transmission spectra of some color filters:
These aren't canonical transmission spectra of filters of those colors; they're just examples of particular filters made by one company. There's no yellow filter on the chart, but you can probably imagine that it's similar to the orange filter, but shifted farther to the left.
You can see that even the red and blue filters together will transmit some light, in the 600-700nm range, which will probably look red or a bit orangish. It will be dim, though. The green and blue filters together will transmit a larger amount, which may appear blue-green in color. And so on. Again, these are just examples; you can't be sure how any two filters will appear in combination unless you know their transmission spectra and the spectrum of the light you're filtering.
| {
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How is a free theory defined? In field theory, I've seen a free theory described as
*
*A field with the specific Lagrangian density ${\cal L}=|\partial\phi|^2+m^2\phi^2$
*A field whose equation of motion yields a linear set of solutions
*A field with non-interacting i.e. free normal modes
The first seems too specific, the second seems too general, and the third seems ill-defined. I was hoping that these three could be extended to solve any of those problems or if there is some way to unify, say, the first and the second then maybe that final description would strike right.
| I believe the problem with all the definitions is that they are not well defined. I can always make a highly non linear field redefinition and make a free quadratic Lagrangian appear interacting. That's why the best way to define a free field theory is to say that it's S matrix must be unity.
| {
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Reduced density matrix: Derive or postulate? Let us consider a quantum mechanical system of interest S that interacts with the environment E. Then, the reduced density matrix
$$
\hat \rho_\mathrm{S} = \mathrm{Tr}_\mathrm{E} \{ \hat \rho \}
$$
is the partial trace over the environment, where $\hat \rho$ denotes the density matrix of the complete system (S + E).
Is this expression postulated or can it be derived (from basic postulates, such as the set given by Nielsen and Chuang for example)?
| Usually this expression is neither postulated nor derived, but rather a serves as a definition of the reduced density matrix.
For more context let me add:
Reduced density matrix typically appears in certain contexts, notably, when you consider a system coupled to a bath, and want to trace out the degrees of freedom of the bath. One then typically shows that such a reduced density matrix possesses all the properties of a true density matrix, necessary for the complete description of the system of interest.
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Why does a capacitor act as a frequency filter? What is it about a capacitor which allows it to filter frequencies?
I understand the construction of a high-pass RC filter, and the mathematics behind it, but I'm struggling to find an explanation of the physics behind the phenomenon.
In my mind I can picture the broad spectrum signal hitting the capacitor, but I feel like the "output" behaviour would be mush, not a controlled and predictable behaviour. I'm not a physicist, but I'd like to understand this problem better.
What is the physical behaviour which allows a capacitor to act as a high or low pass filter?
| A capacitor along with a resistor can act like a filter because its impedance is frequency dependent and by division of voltage between resistor and capacitor it works.
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Why is there a density instead of mass in the Navier-Stokes Equation, if it's analogue to Newton's Second Law? I read in Ian Stewart's 17 Equations that Changed the World book that Navier-Stokes equation (I know it's not exactly a scientific book, but still, I'd like clarification on what is wrong if it's the case):
$\rho \left( \dfrac{\partial v}{\partial t} + v \cdot \nabla v\right) = -\nabla p + \nabla \cdot T + f$
is an analogue of Newton's Second Law, for fluids.
Newton's second Law states force is equal to mass times acceleration, and I can see that inside the parenthesis in NS equation there are derivatives of velocity, so it is acceleration. The RHS are the forces on the fluid. But so why is $\rho$ (density) multiplying acceleration instead of $m$? What is the property of fluids that allows for this?
| The Navier-Stokes equation describes the motion of some infinitesimal volume of the fluid. That is we divide the fluid up into tiny volumes $dV$ and the equation tells us how these tiny volumes move. The overall motion of the fluid comes from combining the motions of all these tiny volumes.
So the equation is really:
$$ \rho \left( \dfrac{\partial v}{\partial t} + v \cdot \nabla v\right) dV = -\nabla p dV + \nabla \cdot T dV + f dV $$
and of course $\rho dV$ is the mass of the volume element $dV$, so the left hand side is mass times acceleration just like Newton's second law.
However we don't want the equation to depend on the exact value of $dV$, and we will be taking the limit $dV \to 0$ anyway, so we divide through by $dV$ to get the form of the equation that Stewart gives.
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How does one determine which signs to take for the Gradient Wind Equations? Under geostrophic balance, one can write
$$\frac{V^2}{R}+fV-fV_g=0$$ where $V:$wind speed, $V_g:$ geostrophic wind speed, $f:$ Coriolis parameter, and $R:$ radius of curvature.
Solving for $V$, we can get a relationship between $V$ and $V_g$ as
$$V=-\frac{fR}{2}\pm\frac{\sqrt{f^2R^2+4fRV_g}}{2}$$
How does one then determine when to use $\pm$ for cyclonic and anti-cyclonic flows?
I was able to find a solution from these slides 40-41 and I am also aware that the $R$ can be positive and negative, and plays a part in being physically meaningful as seen here. However, I fail to understand why and how the $\pm$ signs come into play. To be clear, I am referring to the $\pm$ between the two terms on the RHS.
| It's a general problem that quadratic equations have two roots when only one is expected. One will then be unphysical and we have to decide which is the one we want.
In this case we can proceed as follows: When $|R|$ becomes very large (either positive or negative) then the centripetal acceleration becomes negligible and $V\to V_g$. The sign has to be chosen to make this happen. So (assuming that $f$ is positive) for $R$ positive we must take the postive square root and for $R$ negative we must take the negative root. To see that this is so it helps to write your quadrtic solution as
$$
V=-\frac{fR}{2} \pm |fR| \frac {\sqrt{1+4V_g/fR}}{2}\\\approx -\frac{fR}{2} \pm |fR|(1+V_g/fR+\ldots)
$$
The last expansion is valid when $|R|$ is large.
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Does mass really exist? In elementary physics, I have learned mass as the amount of matter (naively) and energy as the ability to do work. Now we know that they interchangeable by famous Einstein's equation:
$$E=mc^2$$
It seems from here that they are interchangeable, but then I watched this video, which says, in a nutshell, that the mass actually never existed. We are actually measuring energy all along.
Now there are some subtleties I'm facing through this concept. Why do I need the concept of mass now?
Edit: In video, he says that at most deeper level particles -- like protons -- get their energy from quark potential energy. So that means at this level you don't have mass but energy. So can't we change everything's concept, so that everything is energy?
| Mass is the proportional coefficient between force and acceleration. In everyday life it's necessary,but in models without particle-like things you can throw it away.
In some particle models it is interchangeable with energy, but the buildup of the model needs its concept.
| {
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Properties of the dot product Suppose we have three vectors $\textbf{A}$, $\textbf{B}$ and $\textbf{C}$. If $\textbf{A}\cdot\textbf{C}=\textbf{B}\cdot\textbf{C}$, does that mean that $\textbf{A}$ must be equal to $\textbf{B}$? If so, can this property be proven?
Though the question is mainly mathematical, it has occurred to me a number of times when studying physics and I'll like a good explanation.
Now, the fundamental theorem for gradients states that
$$ V (\textbf{b}) - V (\textbf{a}) = \int_\textbf{a}^\textbf{b}(\nabla V)\cdot d\textbf{l}, $$
so
$$ \int_\textbf{a}^\textbf{b}(\nabla V)\cdot d\textbf{l} = -\int_\textbf{a}^\textbf{b}\textbf{E}\cdot d\textbf{l}. $$
Since, finally, this is true for any points $\textbf{a}$ and $\textbf{b}$, the integrands must be equal:
$$ \bbox[10px,border:1px solid black]{\textbf{E} = -\nabla V.}\tag{2.23} $$
As an example of such a case, I have added an excerpt from Griffiths' Introduction to Electrodynamics. In the calculations, it was assumed that ${\textbf{E}}$ is equal to $-\nabla V$ based on the fact that $\textbf{E}\cdot d\textbf{l}=-(\nabla V)\cdot d\textbf{l}$ .
| If for given $\vec A$ and $\vec B$ the equality $\vec A\cdot\vec C = \vec B\cdot\vec C$ holds for all vectors $\vec C$, or at least for a set of generators (say, a basis), then we can conclude that the two vectors are equal, otherwise we can't.
I will try to make it plausible: If we take the standard basis $\{\vec e_x, \vec e_y, \vec e_z\}$ for vector $\vec C$, then we get
$$\vec A\cdot\vec e_x = \vec B\cdot\vec e_x$$
$$\vec A\cdot\vec e_y = \vec B\cdot\vec e_y$$
$$\vec A\cdot\vec e_z = \vec B\cdot\vec e_z$$
But $\vec A \cdot \vec e_i=A_i$, the i-th component of the vector. So we have just shown that $A_x = B_x$, $A_y=B_y$ and $A_z=B_z$ and thus $\vec A = \vec B$.
On the other hand, let us assume that the equality holds for the first two basis vectors but doesn't for the last one, $\vec e_z$, then we know that the first two components of $\vec A$ and $\vec B$ coincide but the z components don't and thus the two vectors aren't equal.
The approach works for any basis, not necessarily the standard basis. Then you will get components with respect to the given basis -- if they all coincide, then the vectors also coincide.
It's not a rigorous proof but maybe helps to make the statement intuitive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can someone explain the measurement problem with little bit of mathematics? Can someone mathematize the statement of the quantum measurement problem? I am only interested in the statement of the problem (and not its solutions). Thanks.
Still confused. Stated in this way (as in the current answers), the measurement problem seems funny to me. The measurement of an operator $A$ on the state $|\psi\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$, either gives the eigenvalue $a_0$ associated with the state $|0\rangle$ or $a_1$ associated with $|1\rangle$. Isn't this natural? How does it make sense to get a superposition after a single measurement? What on earth does that mean? What would the result of a single measurement be if superposition were retained?
| The actual act of measurement and the subsequent collapse of the wave function is not a dynamical process and hence has no mathematical equations to quantify or describe this process.
This is why there are various interpretations of the measurement/collapse process.
| {
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"source": "stackexchange",
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Why does power increase as a constant force accelerates a body? If a constant force is being applied to a body, without any other external forces, F = ma says that that body will accelerate at a constant rate. This acceleration will continuously increase the body's velocity.
According to P = Fv, since the force is constant and the velocity is continuously increasing, the power required by the force will continuously increase.
I understand all the maths, but am trying to get a better intuitive understanding of this. I cannot seem to come to terms with the fact that a constant force will need to supply an increasing power. What is this power being represented by, if the force is constant? What typical inner-workings of such a force would require its power to increase, even though its ultimate "output" is the same? What actually constitutes "power" and "force" at the "force-side" of things?
| There is no internal mechanism, imagine a ball falling in a constant gravitational field. It basically means that the same force acts for a longer distance (not a longer time) if the speed is larger. Intuitively, for a conservative force like gravity this means that you get potential energy faster
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can there be waves in different fields? I am a physics fan, but I am not a physicist. I don't even know that the question I have asked make sense or not.
There can be waves in the gravitational field. So I would like to know if there can be waves in other types of fields, such as electromagnetic field, Higgs field etc.?
|
Can there waves in different fields?
Yes. You have already heard about waves in the gravitational field. They were first detected about five years ago!
Waves in the electromagnetic field include light waves, radio waves, microwaves, etc. These electromagnetic waves simply differ in their frequency and wavelength. Otherwise, they are essentially similar because they are all waves in the same EM field and all correspond to the particle called the photon.
Electrons act like waves in an electron field. Quarks act like waves in a quark field. The Higgs particle acts like a wave in the Higgs field.
All quantum particles can act like waves in their respective fields. You can read about wave-particle “duality” here. Physicists discovered that quantum objects can behave like waves or particles, depending on what you do with them!
The Standard Model of particle physics has seventeen different kinds of fields, waves in which correspond to the seventeen known kinds of elementary particles: 6 quarks, 3 charged leptons, 3 neutrinos, 1 photon, 1 gluon, 1 W boson, 1 Z boson, and 1 Higgs. (If you also count different “colors” and antiparticles, you get more than 17, but this is a common way of counting the particles and their fields.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/591159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Orbit with crash Let's assume I shoot an object from a high tower horizontal to the earth's surface. As far as I understand, depending on the velocity I will get different types of orbits. With decreasing velocity I will go from
*
*hyperbolic orbit where the focal point is the earth's center
*parabolic orbit where the focal point is the earth's center
*elliptic orbit where the earth's center is that focal point that is nearest to the tower
*circular orbit where the earth's center is the center of the circle
*crash orbit
My question is about the last orbit. If earth was transparent for the thrown object,...
Would orbit 5 be an ellipse with the center of earth being at the focal point furthest from the tower?
|
transparent for the thrown object
Worth mentioning an alternative to point-mass orbits here. Assuming the Earth was literally transparent to the object, but all the mass was still there, distributed over the full volume.
Assuming uniform density, we now have a central force field where gravity is proportional to $r^1$ due to the shell theorem, unlike the point-mass case where gravity is proportional to $r^{-2}$.
Such force fields also have stable periodic orbits (actually the only other exponent than $r^{-2}$ that has them), but with some notable differences.
*
*All orbits are elliptical with no exception, as there can be no escape trajectories.
*Elliptical orbits have the centre of the Earth at their geometric centre, not at a focal point, meaning you end up at the surface again at the opposite side of the planet.
In practise, however, the density of the Earth is not uniform, so we don't have a neat $r^1$ force field for tunnel orbits.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/591295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Canonical Transformation in Quantum Phase Space I am looking for a unitary representation $\hat T$ of the following canonical transformation
\begin{align}
q_1&\rightarrow q_2 &p_1&\rightarrow p_2\\
q_2 &\rightarrow -q_1&p_2&\rightarrow -p_1
\end{align}
which is a 90°-rotation in the $(q_1,q_2)$-subspace of a 4-dim phase-space. It is therefore a point-transformation, since it does not mix positions and momenta. $\hat T$ acts as
$$
\hat T \hat q_1 \hat T^\dagger =\hat q_2 \quad \hat T \hat p_1 \hat T^\dagger =\hat p_2\\
\hat T \hat q_2 \hat T^\dagger =-\hat q_1 \quad \hat T \hat p_2 \hat T^\dagger =-\hat p_2
$$
One guess of mine is
$$
\hat T = e^{-i( p_1(q_1-q_2)-p_2(q_2+q_1))}
$$
but I do not know a way of proofing it, apart from expanding the exponentials and then computing everything brute-force, e.g. in
$$
\bigg( \sum_n^\infty \frac{i^j( p_1(q_1-q_2)-p_2(q_2+q_1))^n}{n!}\bigg) \hat q_1 \bigg( \sum_m^\infty \frac{i^m( p_1(q_1-q_2)-p_2(q_2+q_1))^m}{m!}\bigg) = \hat q_2.
$$
One would have to commute $\hat q_1$ to the left, which seems ridiculously laborious to me. Is there an easy way to find $\hat T$ for such a point-transformation? And if one must resort to guessing, is there an easy way to proof that what one has found acts in the right way?
I am deeply grateful for any help!
| Skip the silly hats--everything is an operator.
Observe the obvious invariants
$$
I=q_1^2+ q_2^2, ~~~ J= p_1^2+p_2^2.
$$
Observe the hermitian operator
$$
r=q_1p_2-q_2p_1
$$
commutes with both of them, so it's worth considering its effect on your four variables,
$$
[r, q_1]=iq_2 \\ [r, q_2]=-iq_1 \\ [r, p_1]=ip_2 \\ [r, p_2]=-ip_1.
$$
But this is the precise rotations you are after a π/2 rotation for, so
$$
T= e^{-i\pi r/2}
$$
will do the trick, by the Hadamard identity,
$$
T q_1 T^\dagger = q_1 + (-i\pi/2) [r,q_1] + \frac{1}{2!} (-i\pi/2)^2 [r,[r,q_1]]+... \\
= q_1 \cos\pi/2 +q_2 \sin \pi/2= q_2,\\
T q_2 T^\dagger =- q_1, \\ T p_1 T^\dagger = p_2, \\ T p_2 T^\dagger =- p_1.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/591531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is the proof that the product of a linear operator with it's adjoint always given in terms of the inner product? The proof that $LL^\dagger$ is Hermitian, where $L$ is a linear (but not necessarily Hermitian) operator, is usually given in terms of the inner product, i.e:
\begin{align}
\langle \phi |LL^\dagger |\psi \rangle &= \langle \psi |(LL^\dagger)^\dagger |\phi \rangle^* \\
&= \langle \psi |(L^\dagger)^\dagger (L)^\dagger |\phi \rangle^* \\
&= \langle \psi |L L^\dagger |\phi \rangle^*.
\end{align}
Why is it not sufficient to write, simply:
\begin{align}
(LL^\dagger)^\dagger = (L^\dagger)^\dagger (L)^\dagger = LL^\dagger~?
\end{align}
| When you write
$$(L^†L)^† = (L^†)(L^†)^† = L^† L$$
you are implicitly using two facts, that $(AB)^† = B^† A^†$ and that $(A^†)^† = A$.
If you already accept those facts then the second version is fine. But if you want to prove those two facts you will need to consider the inner product.
You must consider the inner product because the adjoint is only defined in relationship to a particular inner product. The adjoint of an operator $A$ is defined as the operator $A^†$ (if it exists) such that
$$\langle A\chi,\psi\rangle = \langle \chi, A^†\psi\rangle$$
for all $\chi,\psi$. The inner product is integral to the definition, and any proof of universal properties of the adjoint must use this definition.
One way of seeing that the adjoint really does depend on the choice of inner product is to note that same operator may be self-adjoint with respect to one inner product but not with respect to a different one!
For example, consider two inner products on $\mathbb{R}^2$. The first is $\langle y , x\rangle_1 = y^T x$ and the second is $\langle y ,x\rangle_2 = y^T \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix} x$. Then consider the operator represented by the matrix $A =\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$ You can check that $\langle y , Ax\rangle_1 = \langle Ay , x\rangle_1$ but $\langle y , Ax\rangle_2 \ne \langle Ay , x\rangle_2$
| {
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"source": "stackexchange",
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Importance of center of mass for ceiling fan I am learning about center of mass these days. While doing so, I encountered the equation
$$\sum {F_{external}} = ma_{com}$$
Then, it was written that with the help of this equation, it can be clearly seen that if the blades of ceiling fan are not kept at 120° the center of mass won't be at the center of the fan and the fan would start moving in a conical pendulum.
I am not getting why it will move in a conical pendulum?
I tried browsing the net but couldn't find answer. So I request you to please give me a hint, so that I can understand why the fan will move in a conical pendum.
| The answer to your question can be found here. It is a process called precession which is a change in the orientation of the rotational axis of a rotating body ... other words, if the axis of rotation of a body is itself rotating about a second axis, that body is said to be precessing about the second axis.
Because the blades are not arranged symmetrically, this results in an unbalanced net torque while the fan is rotating.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Black rings in 4D gravity In Introduction to Black Hole Physics by Valeri P. Frolov, Andrei Zelnikov there is a discussion of gravity solutions. They present some examples of solutions with non-spherical horizon topology:
But such solutions are known in 5D.
About the possibility of such solutions in 4D authors say:
In 4D it is believed that there are no stationary solutions for vacuum black holes with horizons consisting of several disconnected components.
What about black ring solutions in 4D?
Is it impossible due to "no hair theorem"? If not, why?
Or maybe exist some other theorems which prohibit such solutions?
| Yes, in 4D the topology censorship theorem seems to rule them out:
Every causal curve extending from past null infinity to future null
infinity can be continuously deformed to a curve near infinity.
Roughly speaking, this says that an observer, whose trip begins and ends near infinity, and who thus remains outside all black holes, is unable to probe any nontrivial topological structures.
That is, if you somehow made a ring you cannot detect that it is a ring by shining light through it; the photons will either miss it or be absorbed, and you cannot shoot one through the ring.
A real shame.
| {
"language": "en",
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"source": "stackexchange",
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Why topologically non-trivial materials are robust againist any external perturbations or defects? Topologically non-trivial materials are insensitive to perturbations or defects. How can I prove it mathematically?
I thought of making the first-order perturbation term zero.
$$\left< \psi \right|H'\left| \psi \right>=0$$ Where $H'$ is the perturbation applied.
But I am unaware of the starting assumptions or conditions to be applied. Can anyone help me with any hint or answer to prove why topological materials are insensitive to perturbations?
| For the sake of this explanation, let's concentrate on systems that have a spectral gap (not the most general scenario but it shall do).
Let $P$ be the Fermi projection of some topological material $H$ such that its Fermi energy is placed inside of a spectral gap of $H$. We have the Riesz formula $$ P = -\frac{1}{2\pi\mathrm{i}}\oint(H-zI)^{-1}\mathrm{d}z $$ where the contour of the integral encloses the spectrum below the gap.
If we perturb $H\mapsto H'$ such that the two Hamiltonians share a common gap, we have the formula (using the same contour) \begin{align} P' &= -\frac{1}{2\pi\mathrm{i}}\oint(H'-zI)^{-1}\mathrm{d}z \\ &=P -\frac{1}{2\pi\mathrm{i}}\oint \left((H'-zI)^{-1}-(H-zI)^{-1}\right) \mathrm{d}z \\
&=P -\frac{1}{2\pi\mathrm{i}}\oint (H'-zI)^{-1}(H-H')(H-zI)^{-1}\mathrm{d}z
\end{align} and so, assuming that $H,H'$ are semibounded (from below) we get the estimate $$ \|P-P'\| \leq C\|H-H'\|\,. $$ where the constant $C$ depends on the size of both gaps and lower bound of the spectrum (of $H$ and of $H'$).
Since we can express the Chern number as a function of $P$ which is moreover continuous w.r.t. this operator norm, this shows the desired stability.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Destructive interference Take the most simple academic example for interference. Since it is not any real experiment, one can have shocking contradictions.
For example: 2 monochromatic plane waves with (parallel) amplitudes propagating in the same direction. The Poynting vectors of the 2 waves without superposition are always constant. Once superposed, the resulting Poynting vector is constant, but is dependent on the phase difference.
Thus, how can we explain the energy balance? If there is an energy redistribution it may be easy perhaps, but when the three values are constant in space... What is the right explanation? –
| The energy balance is indeed an interesting problem. For a monochromatic plane wave the source is an infinite sheet of sinusoidal current.
It is not trivial but is straightforward to calculate the Poynting vector for this arrangement. When you do so you find that energy propagates away from the current sheet with equal power density on both sides of the sheet. When you further calculate $\vec E \cdot \vec J$ at the current sheet itself you find that the work done by the current is equal to the radiated power. So the conservation of energy holds.
Now, Maxwell’s equations are linear and translation invariant, so you can simply shift the current sheet some distance to get two current sheets. The total field from the sum of the two current sheets is simply the sum of the fields from each sheet.
However, although the fields add linearly, the energy is not linear. So you could take a current sheet which by itself produces waves with some given power density $P_1$ and a second sheet which by itself produces a power density $P_2$ and when you add them together you get waves with a power density $P\ne P_1+P_2$.
The key is to recognize that the two sources affect each other. If you calculate the work done by the first sheet you will find that $\vec E \cdot \vec J \ne P_1$. In other words, the presence of the second source changed the work needed by the first source to produce the same current.
Such sources are called coupled, and this coupling can be damaging to RF power amplifiers driving coupled antennas. The power density of the two waves is different from the sum of the original waves, but it matches the power produced by the coupled sources.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/592672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Matsubara sum with log term How do I compute the Matsubara sum
$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right)?$$
If I have sums like
$\sum_n \frac{1}{i\omega_n -m}$, I can sum it up by calculating the sum of residues of the function $\frac{1}{z-m}g(z)$ at the poles where $g(z)=\begin{cases}
\frac{\beta}{\exp (\beta z)+1} \text{ for Fermions}\\
\frac{\beta}{\exp (\beta z)-1} \text{ for Bosons}
\end{cases}$
But how do I do I compute in this case where there is a $\log$ term and there are no poles.
| S=$\sum_n\ln(-i\omega_n+\epsilon)=\sum_n\ln(i\omega_n-\epsilon)+C$
(usually this is an action and the constant is irrelevant. This transform is unneccessary just for convenient)
$=\mathrm{Res}\left\{\ln(z-\epsilon)g(z)\right\}$
when $g(z)$ is what you've mentioned. Then the problem is to evaluate this integral. We could select the branch cut as $(-\infty,\epsilon)$ and the contour $-\infty+i\delta\to\epsilon\to(small\;circle\;around\;\epsilon)\to\epsilon\to-\infty-i\delta\to(circle\;in\;infinte\;distant)$ and
$$S=\frac{1}{2\pi i}\int_C\ln(z-\epsilon)g(z)=\frac{1}{2\pi i}\int_{-\infty}^{\infty}(\ln(z^+-\epsilon)-\ln(z^--\epsilon))g(z)$$
using $g(z)=\xi\partial_z\ln(1-\xi e^{-\beta z})$ and integrate
by part
$$S=-\frac{\xi}{2\pi i}\int_{-\infty}^{\infty}\ln(1-\xi e^{-\beta\epsilon})\left(\frac{1}{z+i\delta-\epsilon}-\frac{1}{z-i\delta-\epsilon}\right)$$
and the relation $\lim_{\delta\to0}\frac{1}{x+i\delta}=\frac{1}{x}-i\pi\delta(x)$
we could get the result.
Reference: Altland,Simons Condense Matter Field Theory
| {
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Problem with the proof that for every timelike vector there exists an inertial coordinate system in which its spatial coordinates are zero I am reading lecture notes on special relativity and I have a problem with the proof of the following proposition.
Proposition. If $X$ is timelike, then there exists an inertial coordinate system in which $X^1 = X^2 = X^3 = 0$.
The proof states that as $X$ is timelike, it has components of the form $(a, p\,\mathbf{e})$, where $\mathbf{e}$ is a unit spatial vector and $\lvert a \rvert > \lvert p \rvert$. Then one considers the following four four-vectors:
\begin{align*}
\frac{1}{\sqrt{a^2 - p^2}}(a, p\,\mathbf{e}) & & \frac{1}{\sqrt{a^2 - p^2}}(p, a\,\mathbf{e}) & & (0, \mathbf{q}) & & (0, \mathbf{r})\,,
\end{align*}
where $\mathbf{q}$ and $\mathbf{r}$ are chosen so that $(\mathbf{e}, \mathbf{q}, \mathbf{r})$ form an orthonormal triad in Euclidean space. Then the proof concludes that these four-vectors define an explicit Lorentz transformation and stops there.
For me this explicit Lorentz transformation is represented by the following matrix.
\begin{bmatrix}
\frac{1}{\sqrt{a^2 - p^2}} a & \frac{p}{\sqrt{a^2 - p^2}} & 0 & 0 \\
\frac{p}{\sqrt{a^2 - p^2}} e^1 & \frac{a}{\sqrt{a^2 - p^2}} e^1 & q^1 & r^1 \\
\frac{p}{\sqrt{a^2 - p^2}} e^2 & \frac{a}{\sqrt{a^2 - p^2}} e^2 & q^2 & r^2 \\
\frac{p}{\sqrt{a^2 - p^2}} e^3 & \frac{a}{\sqrt{a^2 - p^2}} e^3 & q^3 & r^3 \\
\end{bmatrix}
However, multiplying the column vector $(X^0, X^1, X^2, X^3)$ by the matrix above does not seem to yield a column vector whose spatial components are zero.
What did I miss?
|
The 3+1-Lorentz transformation is
\begin{align}
\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t
\tag{01a}\label{01a}\\
c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right)
\tag{01b}\label{01b}\\
\gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{01c}\label{01c}
\end{align}
and in differential form
\begin{align}
\mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt
\tag{02a}\label{02a}\\
c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right)
\tag{02b}\label{02b}
\end{align}
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$
For time-like vectors
Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace
\begin{equation}
\mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}
\tag{03}\label{03}
\end{equation}
under the assumption
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c
\tag{04}\label{04}
\end{equation}
For space-like vectors
Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace
\begin{equation}
\mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}
\tag{05}\label{05}
\end{equation}
under the assumption
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c
\tag{06}\label{06}
\end{equation}
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$
ADDENDUM
*
*From \eqref{02a} and \eqref{03} if
\begin{equation}
\mathbf{X}\boldsymbol{=}
\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\mathbf{x}\\
\\
x_4
\end{bmatrix}\,\quad \texttt{with }
\left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{>}0
\tag{07}\label{07}
\end{equation}
is a time-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity
\begin{equation}
\boxed{\:\:\mathbf{u}\boldsymbol{=}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:}
\tag{08}\label{08}
\end{equation}
its space component is zero
\begin{equation}
\mathbf{X}'\boldsymbol{=}
\begin{bmatrix}
x'_1\\
x'_2\\
x'_3\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\mathbf{x}'\\
\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\boldsymbol{0}\\
\\
x'_4
\end{bmatrix}
\,\quad \texttt{with }
x'^{2}_4\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3
\tag{09}\label{09}
\end{equation}
*From \eqref{02b} and \eqref{05} if
\begin{equation}
\mathbf{X}\boldsymbol{=}
\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\mathbf{x}\\
\\
x_4
\end{bmatrix}\,\quad \texttt{with }
\left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{<}0
\tag{10}\label{10}
\end{equation}
is a space-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity
\begin{equation}
\boxed{\:\:\mathbf{u}\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:}
\tag{11}\label{11}
\end{equation}
its time component is zero
\begin{equation}
\mathbf{X}'\boldsymbol{=}
\begin{bmatrix}
x'_1\\
x'_2\\
x'_3\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\vphantom{x'_1}\\
\mathbf{x}'\\
\vphantom{x'_3}\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\vphantom{x'_1}\\
\mathbf{x}'\\
\vphantom{x'_3}\\
0\vphantom{x'_4}
\end{bmatrix}
\,\quad \texttt{with }
\Vert\mathbf{x}'\Vert^2\boldsymbol{=}x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\boldsymbol{-}x^2_4
\tag{12}\label{12}
\end{equation}
| {
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Why thin metal foil does not break like a metal stick? Consider a metal stick, say iron or aluminum. From the experience, even if it's resilient, bend it forward and backward a couple of times, it would be broken.
However, consider a thin iron foil or thin aluminum foil. From the experience, we know that it could be bend forward and backward for almost as many time as time was permitted.
How to explain this in solid states? Why was it that the thin foil seemed to be much more deformable than stick?(Does it has anything to do with the fact that in the normal direction, the metallic bound was weak?) Why thin foil doesn't break?
| "Thin" is a relative term, but let's assume we're talking a foil that's 0.01 to 0.02 mm thick (i.e., kitchen aluminum foil). Let's also assume that our foil is a soft alloy -- i.e. nearly pure aluminum or iron.
If I take a piece of kitchen foil and I bend it with my hands, the bend radius is likely going to be no less than 5mm. So the ratio between bend radius and thickness is something like 500:1.
If I take a 1cm square bar of 1100 aluminum and I bend it on a 5m radius, it'll survive a lot of bending. Ditto a 1cm square bar of 1020 steel.
This is because the amount of stretch the material must undergo is small, and it's because the material is soft.
I know from experience that you see the same effect with wires. To break thin material, you must bend it using an instrument (like pliers that have a nice sharp edge) that makes the radius on the order of the material thickness. Do that, and it'll break. Keep the radius large with respect to the material thickness, and it won't.
| {
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Regarding the velocity of waves in even dimensions A few years ago I asked on Reddit about the behavior of wave propagation in even and odd dimensions. I received this answer:
"The answer lies in the solutions to the wave equations. Essentially, in odd dimensions a wave will propagate at a single characteristic velocity $v$, while in even dimensions it propagates with all velocities $<v$."
Another user added: "If you interpret the mathematics strictly, the speeds are all strictly less than $v$."
This article, however, says in the second paragraph: "Of course, the leading edge of a wave always propagates at the characteristic speed $c$."
For that reason, I was wondering, is that information on Reddit correct? Does the wave, in even dimensions, propagate with all speeds less than $v$, or does it propagate with all speeds equal or less than $v$?
Edit:
The original comment (which is linked above) refers to the wave equation in this manner (direct quote):
“(I think the wave equation can approximately be written as v2 d2 /dx2 - d2 /dt2 = 0 in terms of v, at least up to some dimensionless constant)”
| I assume a wave equation $(\Box +m^2) f = 0$. This is simply the wave version of Einstein's famous relation $E^2 = m^2c^4 + p^2c^2$. Thus $v \in [0,c)$ for $m\neq0$ and $v=c$ for $m=0$ for any positive number of space dimensions.
| {
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How to solve this problem only using the kinematics of rotational motion? This is a problem from my introductory physics textbook:
A wheel of moment of inertia $I$ and radius $r$ is free to rotate about its centre. A string is wrapped over its rim and a block of mass m is attached to the free end of the string. The system is released from rest. Find the speed of the block as it descends through a height $h$.
The answer in my book utilizes energy considerations, reasoning that "the gravitational potential energy lost by the block must be equal to the kinetic energy gained by the block and the wheel." Thus,
$$mgh=\frac{mv^2}{2}+\frac{Iv^2}{2r^2}$$
$$\rightarrow v=\sqrt{\frac{2mgh}{m+I/r^2}}$$
My question is, is it possible to solve this question using only the equations of kinematics of rotational motion, viz. $$\omega = \omega_0+\alpha t$$ $$\Delta \theta=\omega_0t+1/2\alpha t^2$$ $$\omega^2=\omega_0^2+2\alpha \Delta \theta$$
As an analogy, consider the case when the wheel was massless. Then, the equation from energy considerations would have been: $$mgh=\frac{mv^2}{2}$$
Solving the above, we get, $$v=\sqrt{2gh}$$
When we use the equation $v^2=u^2+2gh$, setting $u=0$ for the system starting from rest, we again get, $v=\sqrt{2gh}$.
| The case with a massless wheel is a bit boring, because it is just the case without a wheel, i.e., a free falling body.
In classical mechanics, there is always multiple routes that lead to the same answer, so you could start from the kinematics equations if you can argue that the angular acceleration $\alpha$ is constant. However, if you don't start from the energy point of view, you have to work from a force point of view. So, what you have to do, is draw the forces on your sketch (you have a sketch right?). Then, what you need is the tension in the string. The tension force $T$ exerted by the string on the mass, is the same as the tension force exerted by the string on the wheel (why?).
Since this is a homework question, I'll leave the rest to you (hint: determine $\alpha$ based on what you know about $T$). The force and energy approach will lead to the same answer. However, in some cases, one approach is much easier to analyze than the other.
| {
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Determine the direction of electric and magnetic field in for plane EM wave A problem states that
Measurement of the electric field (E) and the magnetic field (B) in a plane-polarized electromagnetic wave in vacuum led to the following:
$$
\begin{array}{ll}
\frac{\partial E}{\partial x}=\frac{\partial E}{\partial y}=0 & \frac{\partial E}{\partial z}=-\frac{\partial B}{\partial t} \\
\frac{\partial B}{\partial x}=\frac{\partial B}{\partial y}=0 & \frac{\partial B}{\partial z}=+\frac{\partial E}{\partial t}
\end{array}
$$ ,then what can one conclude about the directions of the electric and magnetic fields?
The two equations on the left side imply that the electric and magnetic fields are constant in the xy-plane. So our wave is a plane wave in the XY plane and therefore the direction of propagation is perpendicular to the xy plane.
But this does not specify the precise direction of the electric and magnetic fields.Neither am I able to use the other time derivative equations to conclude anything.
Could anyone please help me or hint me .
Also this is not a homework problem as I am self studying the subject and I will be grateful for any help.
Thank you.
| The direction of the fields (perpendicular to the direction of propagation) is determined by the source of the wave. If you consider a positive charge oscillating back and forth, the motion of the charge introduces a transverse component into the (preexisting) electric field (in the direction of motion). The motion also produces a magnetic field (wrapped around the direction of motion, as predicted by a right-hand-rule). Both of these disturbances are at a maximum when the velocity is a maximum, and they move away from the charge with a velocity predicted by Maxwell's equations. It would appear that the cross product BxE shows the direction of propagation.
| {
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Two Cylinders on Ramp Suppose I have two cylinders: a light one and a heavy one. Now, I let the cylinders roll down a ramp without slipping. My question is, which one will get to the bottom of the ramp first, and why?
| There's an intuitive proof that two cylinders differing in only density will roll with the same speed: just superimpose two cylinders with the same mass. The cylinders will roll at the same rate. But together they make a cylinder of twice the mass, that is also rolling with the same rate. As a practical matter, giving a meaning to "superimpose" can be a bit complicated, but that doesn't affect the proof from an intuitive standpoint.
| {
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Why is gravity considered a negative vector in a pendulum question? In many places and tutorials, gravity is often considered as a negative vector.
I am confused as to why is that? I though I was missing something from trigonometry but it was just negative in first place. It is pointing downwards, but if we negate that vector, then it should move up.
I need some explanation about this sign.
Picture:
| Other answers are good, but I think there is a misunderstanding of the OP that needs to be addressed.
There is often an ambiguity on the meaning of $g$, that can be used for two related things:
*
*The scalar $g$, that it's about $9.81 \text{m}/\text{s}^2$
*The vector $\vec{g}$, which points downward. $\vec{g}=(0, 0, -g)$
Please notice that in the convention used in the picture in the question, $g$ is a positive number but the vertical component of $\vec{g}$ is negative, therefore $-g$.
| {
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Kinetic energy and curvature In quantum mechanics, the kinetic energy of a particle described by the wave function $\psi$, is related to the curvature of the $\psi$. This is easily seen, but I have confused my self with the negative sign. That is: $\hat{T} = -\frac{\hbar^2}{2m}\nabla^2$, is the kinetic energy operator. So what I gather is, that the greater the curvature of $\psi$, the lower the kinetic energy, due to the minus sign. I know this can not be right.
| What you mean by 'curvature' is often negative. Take the sine wave $\psi(x)=A\sin kx$. Larger $k$ means larger curvature. By taking the 1D kinetic energy operator we get
\begin{align}
-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)&=-\frac{\hbar^2}{2m}\left(-k^2 A\sin kx \right)\\
&=\frac{\hbar^2k^2}{2m}\psi(x)
\end{align}
So your intuition is still right: greater curvature means greater kinetic energy.
Edit: so to expand a bit more on the definition on curvature. There are multiple ways to define curvature but a natural one is to parametrise a curve $\mathbf{r}(s)$ in terms of its path length and consider the second derivative $\mathbf {r}''(s)$. See also https://en.wikipedia.org/wiki/Curvature. For a function the signed curvature becomes
$$\kappa_{\text{signed}}=\frac{f''(x)}{(1+f'(x)^2)^{3/2}}$$
We don't want to consider this generalised curvature that works for any curve; we only want to look at the second derivative. But we can still adopt this sign convention. This gives
$$\kappa_{\text{signed}}=f''(x)$$
This is positive when the function is concave upward (happy smiley) and negative when the function is concave downward (sad smiley). In the Schrödinger equation we have, like you mentioned, the following equation for the second derivative
$$\psi''(x)=-\frac{\hbar^2}{2m}(E-V)\psi(x)$$
Because $\psi$ turns up on both sides the constant $E-V$ just tells you about whether the function curves towards the x-axis or away from it. Convince yourself of the following picture.
In the $E>V$ we have the classically allowed region. Here solutions kind of look like sine waves. The region $E<V$ is the classically forbidden region. Here solutions look like exponentials but since the states must be normalizable in practice this means they must decay to zero.
| {
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Why is the tension on both sides of an Atwood machine identical?
The field forces $F_{g1}$ and $F_{g2}$ push down on Block 1 and Block 2, respectivley, where
$$F_{g1}=m_1g$$$$F_{g2}=m_2g$$
Since the pully system reverses the direction of each force, wouldn't the following be true?
$$T_1 = F_{g2} = m_2g$$$$T_2 = F_{g1} = m_1g$$
And since $m_1 \neq m_2$, wouldn't $T_1 \neq T_2$?
My textbook states that tension is the same throughout the whole string, but I can't wrap my head around why this is so. If $m_1 \neq m_2$, how could the same force $T$ accelerate both of them an equal amount? Wouldn't Block 2 require a greater force?
|
Since the pully system reverses the direction of each force, wouldn't the following be true?
T1=Fg2=m2g
T2=Fg1=m1g
Technically you are correct. But a few changes have to be made to your statement.
The blocks are not at rest and both travel with acceleration $a$. We can assume that $M_1$ goes up and $M_2$ goes down. Since $M_1$ goes up, it's weight increase ($W_{M_1}=M_1(g+a)$). While $M_2$ goes down, its weight decreases ($W_{M_2}=M_2(g-a)$)
$\therefore T_1=W_{M_2}=M_2(g-a)$ and $T_2=W_{M_1}=M_1(g+a)$(modified version of your quoted equations).
Now, $T_1$ supports $W_{M_{1}}$,
$\therefore T_1=W_{M_{1}}=M_1(g+a)=T_2$
| {
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Can coldness be converted to heat energy? We know that the heat can be converted into heat energy with the help of thermoelectric generators, but why can't we generate energy from coldness?
Like the temperature of the universe in 1 K, can this be used in the near future to be used as an energy resource for probes or satellites?
Here is the link to the article that made me think about this. Somewhere in the middle it is written that scientists can harness the cold energy using some active input method.
I think this article is poorly written.
| Strictly speaking, heat is not converted into energy - instead heat is energy. A thermoelectric generator is sometimes loosely described as turning heat into energy, but what actually happens is that a temperature difference between a heat source and a cold sink (usually the surrounding environment) causes heat/energy to flow between them and some part of this heat/energy is used to do work e.g. to generate a current which drives an electric motor etc.
If everything was at the same temperature, no matter whether this was $1$ K or $1$ million K, you could not use the heat/energy to do work because there would be no temperature difference and so no flow of heat/energy.
| {
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Spin part of the angular momentum from the Lagrangian For fermions of spin $1/2$ the angular momentum has following form:
$$
\mathcal{J}_z = \int d^{3}x \ \psi^{\dagger} (x) \left[i(- x \partial_y + y \partial_x) + i\sigma^{xy} \right] \psi(x)
$$
Here the first term is orbital part and the latter one is the spin part of angular momentum.
However, in the general prescription for derivation of the angular momentum:
$$
\mathcal{J}^{ij} = \int d^{3} x (x^i T^{0 j} - x^j T^{0 i}).
$$
I cannot see, where the spin part can actually arise.
For example for the photon field, where:
$$
T^{\mu \nu} = F^{\mu}_{\alpha} F^{\mu \alpha} - \frac{1}{4} g^{\mu \nu} F_{\alpha \beta} F^{\alpha \beta} .
$$
This procedure seems to provide only the orbital part of the angular momentum, and no the spin. Or it is implicitly included in this expression?
For the spin $1/2$ field the term $\sigma^{xy}$ emerges, when one considers corrected energy tensor https://en.wikipedia.org/wiki/Belinfante%E2%80%93Rosenfeld_stress%E2%80%93energy_tensor. (formula 6 in https://arxiv.org/abs/1508.06349).
However, for the spin-1 theory this expression incorporates everything and is the most symmetric one.
I would strongly appreciate any help and comments!
| The symmetric EM tensor does contain the spin. There is a discussion by Michael Berry about this, and also some in Wikipedia here
| {
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Integrating cross products I know that while integrating dot product to two vector quantities along a line integral, the limits of the integration implicitly takes care of the direction in which we integrate from here and here.
But would this be true in case of cross products? Would the limits of the line integral of a cross product implicitly take care of the direction?
| In general, limits on an integral over a subset of $\Bbb R^n$ implicitly take care of integration direction. (The case $n=1$ is familiar; if $a<b$ the directions for $\int_a^bfdx,\,\int_b^afdx$ are obvious.) It doesn't matter whether what's integrated is a univariate dot product, a $3$-dimensional cross product or in general a $k$-dimensional integrand, viz. $(\int_SVd^nx)_i=\int_SV_id^nx$ for $1\le i\le k$.
| {
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Why does ponytail-style hair oscillate horizontally, but not vertically when jogging? Many people with long hair tie their hair to ponytail-style:
Closely observing the movement of their hair when they are running, I have noticed that the ponytail oscillates only horizontally, that is, in "left-right direction". Never I have seen movement in vertical "up-down" direction or the third direction (away-and-back from the jogger's back). Why is the horizontal direction the only oscillation?
| As @nuclear-hoagie said, the ponytail is basically a pendulum, so vibrations up and down are not really possible (or are much more complicated) as the hair would have to elongate itself in order to store the kinetic energy from the hair falling down so it cannot release it later. However the sideways and back and forth vibrations can be maintained almost for free as they do not involve stretching the hair. I guess that you don't see the back and forth vibration for a reason somewhat similar to the up and down case, when the hair stops agains the back it loses all its kinetic energy and cannot continue the vibration.
You can look for parametric resonance in vertically vibrated pendulums. Basically when the point from which the pendulum is hanging is forced to oscillate vertically at double the natural frequency of the pendulum, an bifurcation occurs.
The configuration of equilibrium (the hair hanging down vertically) is no longer stable. It is still a point of equilibrium however any slight missaligment makes the pendulum deviate from its resting point and oscillate at the natural frequency.
If you do it with a solid pendulum you can even reach a limit where the normally unstable equilibrium (the pendulum pointing upwards) becomes stable.
There are demonstrations of this dynamic stabilization on youtube, and basically is the same process but in the extreme case:
https://www.youtube.com/watch?v=5oGYCxkgnHQ
| {
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Can spacetime be curved even in absence of any source? Einstein's equation in absense of any source (i.e., $T_{ab}=0$) $$R_{ab}-\frac{1}{2}g_{ab}R=0$$ has the solution $$R_{ab}=0.$$
But I think $R_{ab}=0$ does not imply that all components of the Riemann-Christoffel curvature tensor $R^c_{dab}$ be zero (or does it?). From this can I conclude that spacetime can be curved even in absence of any source?
| What you're asking about is referred to as a vacuum solution to the field equations. This does not mean that there is no mass anywhere, rather that we are considering a region of our curved spacetime in which there is no mass.
The Schwarzschild solution for instance is a "vacuum solution" because we are considering the region outside of the central mass in which there is no matter, but in which the curvature is non-zero.
You are correct that the vanishing of the components of the Ricci tensor does not imply the vanishing of the components of the full Riemann tensor. $R_{\mu\nu}=0$ is a vacuum solution, ${R^\alpha}_{\beta\mu\nu}=0$ is flat spacetime.
| {
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We know a human cannot run on water, but could a much stronger and faster animal? We know a human cannot run on water, but could a much stronger and faster animal?
This viral video shows a moose running across a body of water:
https://www.youtube.com/watch?v=K5-0d00hV1c
Some say the video is fake. Others think the water must be shallow. We probably won't resolve these issues here.
My question is: could this actually happen (in deep water)?
I imagine that in theory, a sufficiently strong animal could kick the water really hard and if it moved across the surface really fast, it could stay above the water.
But could a real animal with a moose's strength and weight do it?
| Humans, always overlooking the world of the small. It is as vast as the big world, hence strength is relative. So the answer is: insects. As long as they are not winged, etc. which of course can be sure death-trapping appendages, many insects are too light to break the surface tension of water. Yet if you want to narrow it down to strength, one of the mightiest creatures on Earth are: ants.
| {
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3 pulley atwood machine with 2 blocks and 1 free-moving pulley I'm supposed to use the Lagrangian approach in the solution. I've just recently started classical mechanics course and it's still somewhat confusing to me.
The problem I am working on is as follows: "2 masses $m_1$ and $m_2$ are connected with a line across 3 pulleys of which the middle one has mass $M$ and can move freely in the up-down direction. The line is frictionless and massless. Find the acceleration of $m_1$ and $m_2$."
What I've figured on my own so far:
*
*I've marked velocities of the blocks and the middle pulley and I've figured out that $$ \dot x_3 = \frac{\dot x_1 + \dot x_2}{2} $$
The reason is that if the section of the string from which $m_1$ is suspended shortens by $x_1$ , and that from which $m_2$ is suspended by $x_2$ , then the central loop lengthens by $x_1+x_2$ .
Since this loop consists of two vertical sections (and a semicircular one around the pulley whose length stays constant), this means that the pulley descends a distance of $\frac{x_1+x_2}{2}$ .
*I wrote down the kinetic and the potential energy: $$T=\frac{1}{2} m_1 \dot x_1^2 + \frac{1}{2} m_2 \dot x_2^2 + \frac{1}{2} M (\frac{\dot x_1 + \dot x_2}{2})^2$$ $$U=g(m_1x_1-M\frac{\dot x_1 + \dot x_2}{2} + m_2x_2)$$
*At this point I could write down the Lagrangian as $$L=T-U$$
This is where I'm somewhat stuck, as I don't think I can write down the Lagrange's equation yet as I've got both $x_1$ and $x_2$ in there, and I would expect to only have one $x$ and it's derivative.
As I've said, I'm new to lagrangian mechanics and I've got zero intuition on how to approach such problems so I'd be more than welcome for any clue.
| There is one Lagrangian for both $x$'s, but there is a separate Euler-Lagrange equation for each of $x_1$, $x_2$.
| {
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Physical meaning of $\frac{π}{8}$ in Poiseuille's equation Recently I read about the Poiseuille's equation which relates the flow rate of a viscous fluid to coefficient of viscosity ($\nu$), pressure per unit length($\frac{P}{l}$) and radius of the tube ($r$) in which the fluid is flowing. The equation is
$$\frac{V}{t}=\frac{πPr^4}{8\nu l},$$
where $V$ denotes volume of the fluid.
One thing which I don't understand in this equation is that why do we have the constant term as $\frac{π}{8}$ . Is there some physical significance of this specific number here or is it just a mathematical convention?
| Poiseuille's law is the result of doing a a force balance on the fluid, applying Newton's law of viscosity between the fluid velocity gradient radially and the shear stress, solving for the fluid axial velocity distribution, and integrating the velocity distribution to get the volumetric flow rate. The $\pi/8$ comes in from integrating the velocity over the circular area.
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Radiation Within a Faraday Cage - Can It Escape? so we all know that the Faraday Cage prevents most EM radiation from entering the Faraday Cage. But what about, if we place radiation within the Faraday Cage - can it escape outside? I sort of recall working in class on a lab and finding that yes, it does escape - because the charges go to the outer surface - but I am having trouble finding a source to verify this, and I am using this concept for a project. Many thanks!
| In a case of cage being a closed shell made of perfect conductor, no radiation should be present outside, because charges on the inner surface will move in such a way that all radiation is reflected from the inner walls back to the insides. They work like a mirror for all frequencies.
This is why perfectly reflecting cavity is used in analyses of equilibrium radiation, or why microwave oven has to be made of metallic walls - both to prevent radiation inside to leak outside.
In practice, some radiation always gets out because there are no known materials that are perfect conductors for all frequencies.
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Intrinsic relations between two solutions Recently I've encountered a problem stated as follows:
A smooth bowl has the shape of a paraboloid. The equation of the cross-section with the $(x, z)$-plane is $x^2= 2R_Az$, where $R_A$ is the curvature radius at point $A$. One releases at a point with height $z=h$ a point mass with an initial velocity equal to $0$. What is in this case the magnitude of the force exerted by the particle on the bowl when it passes point $A$? Neglect air friction.
I've come up with two solutions to this problem, one needs some naive knowledge of curvatures and one doesn't:
Solution #1:
By Law of conservation of energy,
we have $$mgh=\frac{mv_A^2}{2}$$
Hence $$v_A^2=2gh$$
Since the mass experiences a centripedal force pointing towards the centre of curvature at point $A$,
we have ( N is the contact force ):
$$N-mg=\frac{mv_A^2}{R_A}$$
Therefore
$$N=mg(1+\frac{2h}{R_A})$$
Solution #2:
The displacement vector of the particle is $$\vec{s}(t)=x(t)\vec{i}+\frac{1}{2R_A}x^2(t)\vec{k}$$
(where ${x(t)}$ is the horizontal displacement)
Therefore the acceleration is $$\vec{a}(t)=\frac{d^2\vec{s}}{dt^2}=\ddot{x}(t)\vec{i}+\frac{1}{R_A}\Big(\dot{x}^2(t)+x(t)\ddot{x}(t)\Big)\vec{k}$$
At point $A$, $\ddot{x}(t)=0$, Therefore $$\vec{a}=\frac{1}{R_A}\dot{x}^2(t)=\frac{v_A^2}{R_A}$$
Hence both of my solutions give the same value of acceleration and hence the same results.
My questions are: Is there a rigorous explanation to the validity of my first solution? Is it just a coincidence that the acceleration at $A$ is identical to the centripetal force related to the curvature circle? Are there any "intrinsic links" between the two solutions? Thanks in advance.
| In your first solution, you have assumed that the centripetal acceleration is $v^2$/R (based on what you know about circular motion). In the second, you have shown that, in this situation, the vertical acceleration is $v^2$/R at point A.
| {
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Near point and focal length of the eye What is the relationship between near point and focal length of the eye?normally in a convex lens when the object is kept at 2f we get the same size image as shown . A healthy human eye can see objects without any trobles. Then is the near point of the eye, 2f of the eye lens? Im really confused about focal length of eye lens and near pont.
| As I recall, a relaxed lens of a healthy eye puts the image of nearby (but not close) objects on the retina of the eye. For distant objects, the muscles around the lens must make minor adjustments to the shape of the lens (unless you are far-sighted). For nearby objects, the muscles must work harder to shorten the focal length of the lens. When the object is brought in so close that it can no longer be kept in focus, it is inside of the near point.
| {
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What is the velocity $u^\mu$ in the stress-energy-tensor of a perfect fluid? I am currently learning about fluid dynamics in special relativity. We defined the stress-energy-tensor of a perfect fluid to be
\begin{equation}
T^{\mu \nu} = (\rho + P) u^\mu u^\nu + P g^{\mu \nu}.
\end{equation}
We said that $u^\mu$ is the 4-velocity of the "MCRF" of the fluid, that is the frame in which the bulk velocity is zero, i.e.
\begin{equation}
\sum_a u_a^\text{MCRF} = 0.
\end{equation}
Where the summation goes over all particles.
It is however possible to derive the Euler-equation (in the non-relativistic limit) from the conservation of energy-momentum $T^{\mu\nu}_{\quad,\nu}= 0$,
\begin{equation}
\rho (\partial_t \mathbf{v} + (\mathbf{v} \cdot \nabla) \mathbf{v})) = - \nabla P.
\end{equation}
I have trouble imagining that the bulk velocity should depend on space-time coordinates. I guess you could in principle define the bulk velocity to be zero at spaces other than the fluid and $u^\mu$ at the point (or volume) of the fluid. But then the Euler equation seems to be strange. The Euler equation usually treats the velocity as a vector field of the fluid and gives insight into the internal dynamics of the fluid. How can that happen with the bulk velocity which contains no information about the internal flows?
Could someone explain to me how I can interpret $u$ or $\mathbf{v}$ as a vector field in the sense of how the Euler equation treats them?
| The MCRF is not a single frame: there is a different frame at each spacetime point. The summation for the bulk velocity doesn't run over all particles: it runs over a small region surrounding the chosen point, much smaller than the typical distances we will be considering in our study of the fluid, but much larger than the distance between molecules or the mean free path or whatever. It's a local bulk velocity.
There's nothing relativistic about this: it's exactly the same definition of fluid velocity as the one we use in classical fluid dynamics.
| {
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Functional derivative in Faddeev Popov method (Lorenz Gauge) When applying Faddeev and Popov method (am using Peskin and Schroeder as reference), we use the identity:
$$1=\int \mathcal{D}\alpha(x)\delta(G(A^\alpha)) \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) \tag{9.53}$$
to write
$$ \int \mathcal{D}Ae^{iS[A]}=\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)\int\mathcal{D}\alpha(x)\int\mathcal{D}Ae^{iS[A]}\delta (G(A^\alpha))\tag{9.54}$$
When we use the Lorenz gauge, we obtain:
$$G(A)=\partial^\mu A_\mu+\frac{1}{e}\partial^2\alpha(x)$$
My questions is: how do we obtain the following:
$$ \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) = \det\left(\frac{1}{e}\partial^2\right) $$
I am also confused about how can we obtain an operator by taking a functional derivative, if someone could give an intuitive explanation of this it would be highly appreciated!
| $$\begin{align}{\rm Det} \left(\frac{\delta G}{\delta \alpha}\right)
~=~&\int {\cal D}c{\cal D}\bar{c}\exp\left(\int \!d^4x \int \!d^4y ~\bar{c}(x)\frac{\delta G(x)}{\delta \alpha(y)}c(y) \right) \cr
~=~&\int {\cal D}c{\cal D}\bar{c}\exp\left(\int \!d^4x \int \!d^4y ~\bar{c}(x) \frac{1}{e}\partial_x^2\delta(x-y) c(y) \right)\cr
~=~& {\rm Det} \left(\frac{1}{e}\partial^2\right).\end{align}$$
| {
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Shouldn't some stars behave as black holes? Some of the "smaller" black holes have a mass of 4-15 suns. But still, they are black holes.
Thus their gravity is so big, even light cannot escape.
Shouldn't this happen to some stars, that are even more massive? (mass of around 100 suns)
If their mass is so much bigger, shouldn't their gravity be also bigger? (So they would behave like a black hole). Or does gravity depend on the density of the object as well?
| If the visible matter became enough dense to be concentrated inside its Schwarzschild radius, it becomes a BH.
Until their inner pressure withstand the gravitation they stay being stars.
| {
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Applications of representation theory of finite group in physics? Well, I have just finished my study on basic representation theory of finite group from a pure math course. After tortured a lot by abstract constructions, I would like to know the real application of this theory, however, it seems to me not many topics in physics are related to representations of finite group. Anyone could provide some examples on the application?
| The moonshine phenomenon is a deep subject involving conformal field theory and string theory that is based on the observation of some relationships between the representation theory of finite groups of "big size"(but finite) and modular forms (obtained as partition functions of some CFTs).
References:
Miranda Cheng - Progress on moonshine
TASI lectures on moonshine
Miranda Cheng - Umbral Moonshine and String Theory
| {
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Writing a simple sum with einstein notation Is there any shorthand using the Einstein notation to write a simple sum?
$X = X_1 + X_2 + X_3 ....$
Can I write: $X = X_i$ ?
I don't think so because the definition states that only indices that appear twice is summed over. What is the correct way to write a simple sum like the example above?
| The Einstein summation notation includes repeated indices (i.e. summation notation). For vectors (and tensors), $X$ would usually mean the full object, $X = X^{i} e_{i} = X^{0}e_{0} + X^{1}e_{1}...$ etc. The $X^i=X^i (x)$ are the components of the vector and $e_i$ is some basis. The $i$ deonotes the components, and this is summed in the above equation. But we can talk just about the components $X^i$, which means $(X^0, X^1, ... X^n)$.
So yes, you can write $X^i$, but it doesn't mean what you've written down. This is just four-vector notation, but not specifically Einstein summation notation. The sum you wrote is more like the Euclidean norm, which as the other comment says, it's easiest notated as $\sum_{i}^{n} X^i $.
| {
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Mass of the fundamental particles I have always wondered that how do scientists measure the mass of a fundamental particle.
Obviously they can't weigh it in a conventional machine we use to weigh other things in our daily life.
And do they use the formula:
$$E^2=m^2c^4+p^2c^2$$
| The mass of particles can be measure in a variety of ways depending on what's the particle is( and How precise measurement you want). For instance, an electron's mass can be measure as follows
The electron rest mass can be calculated from the Rydberg constant $R_∞$ and the fine-structure constant $α$ obtained through spectroscopic measurements. Using the definition of the Rydberg constant:
$$R_\infty=\frac{m_ec\alpha^2}{2h}$$
so $$m_e=\frac{2hR_\infty}{c\alpha^2}$$
There are other methods, I'm just referencing them. Here
How to measure electron mass? Another.
One most famous way for the measurement is what is called: Shell renormalization scheme
In quantum field theory, and especially in quantum electrodynamics, the interacting theory leads to infinite quantities that have to be absorbed in a renormalization procedure, in order to be able to predict measurable quantities. The renormalization scheme can depend on the type of particles that are being considered. For particles that can travel asymptotically large distances, or for low energy processes, the on-shell scheme, also known as the physical scheme, is appropriate. If these conditions are not fulfilled, one can turn to other schemes, like the Minimal subtraction scheme.
Lastly yes! If the particle is moving, You must use
$$E^2=(pc)^2+(m_0c^2)^2$$
| {
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Why a fan make us feel colder instead of warmer in a cold room? Stay in a cold closed room, turn on a fan directed at you, it makes you feel colder instead of warmer. why?
Here is the definition of heat:
Matter exists in different physical forms – solids, liquids, and
gases. All matter is made of tiny particles called atoms, molecules, and
ions. These tiny particles are always in motion – either bumping into
each other or vibrating back and forth. It is the motion of particles
that creates a form of energy called heat (or thermal) energy that is
present in all matter...
so before turning on the fan, the air molecules are relatively still and not moving much. After turning on the fan, they are moving a lot faster so the motion of the particles is more and it should create heat, but why can't we feel the heat on our body and skin when standing in front of the fan? (at least in my experience)
| Others have given wonderful answer. But I think one more point is necessary here.
Why does a wet cloth become dry when the wind is blowing ?
Actually if we take a closed beaker half filled with water and vacuum in the upper half then the upper half is not vacuum after sometimes. This is because some of the higher energy molecules in the water manage to break the influence of bonds and fills the vacuum area.
After some more time, the amount of water remaining in the lower half becomes fixed. So we say that the liquid water is with equilibrium with the vapour state.
Now coming back to your question, when the fan was switch off , the air near to your skin was in almost equilibrium with your skin but when the fan was switched on, it created an unbalance situation leading to decrease in pressure out and thus more liquid comes out from your skin for stability and by evaporating they take away energy and you feel cool.
Hope it helps .
| {
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Relativistic energy of harmonic oscillator What is the relativistic energy of an harmonic oscillator:
$$\frac{m_0 c^2}{\sqrt{(1-\frac{v^2}{c^2})}}+\frac{1}{2}kx^2$$
Or
$$\frac{{m_0 c^2}+\frac{1}{2}kx^2 }{\sqrt{(1-\frac{v^2}{c^2})}}$$
I think the first one is true but I need an exact logic or derivation.
| First of all the speed $v$ is the property of the particle. The points on spring move at different speeds. So you can't write like the 2nd one. Even the 1st one is wrong. Also practically speaking the Hooks law works only for small distances and velocities, and after that it fails. If you assume there exists a mass less spring(technically I should say it has negligible kinetic energy otherwise it should always move at speed $c$) which follows Hooke's law at all distances and speeds, then since relativistic contraction does not induce any stresses, we should divide the spring into infinitesimal parts and find their infinitesimal lengths (measured in the lab frame) at the instant.
We can interpret the dx in Hook's law for an infinitesimal spring($k_i$ is the spring constant of that infinitesimal part)
$$dF=k_idx $$
as the difference between proper lengths of an infinitesimal part at an instant(obtained by multiplying the infinitesimal lengths (measured in the lab frame) by $\gamma$ of that infinitesimal spring at that instant). So it will become somewhat complex. But still the 1st equation is a better approximation than the 2nd one.
| {
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How can a bicycle keep its momentum while changing direction? I was riding my bicycle downhill. When the street leveled I realized I had forgotten something so I started turning around without pedaling, pretty much using the momentum I had acquired. Soon I was moving in the complete opposite direction, still without having to pedal.
I couldn't help but thinking how that was possible.
I understand the momentum has a direction arrow and somehow I changed that arrow to the opposite direction. Of course I was going much slower by the time I completely turned around, but still moving.
*
*Is the friction of the wheel against the floor somehow like tiny collisions that "bounce" that momentum force gradually changing that direction? (To the expense of losing energy quicker, thus, speed)?
*I used the bike example only because that's when the doubt popped up, but the same I can ask about a ball rolling down and going through a loop or a skier going down a hill and jumping up in a 90 degrees (relative to the downhill part) ramp
What's the physics and maths behind the momentum change to the opposite or perpendicular direction of a moving object when no extra force is applied?
| Your momentum is not conserved because there are external forces acting on the you + bicycle system. Though there are several external forces, the main one is friction.
The friction from the ground will will provide a torque and force on your bicycle, hence your angular momentum will not be conserved.
| {
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How can we find the total weight of a pulley apparatus? I recently have been trying to solve this problem, which asks me to find T3 assuming that g = 9.8:
My idea was to find the acceleration of the two blocks in order to calculate the apparent weights of the 40 and 26 kilogram masses, and add the apparent weights to 54*9.8.
We can use Newton's Second Law to find a:
$F = ma$, so we first derive equations for T1 and T2:
$T2 - 40(9.8) = 40(-a)$, so $T2 = 40(9.8 - a)$
$T1 - 26(9.8) = 26(a)$, so $T1 = 26(9.8 + a)$
Now, Torque = I * (alpha), so
$r*T2 - r*T1 = (0.5mr^2)(\frac{a}{r})$
$T2 - T1 = 0.5ma$
$40(9.8 - a) - 26(9.8 + a) = 0.5*54*a$
$14*9.8 = 93*a$
Thus, $a$ is approximately $1.475$ meters per second squared.
Now, we know that the acceleration of Block 2 is -a, and the acceleration of Block 1 is a. Thus, Block 2's apparent weight is $m * a$ = 59 and Block 1's apparent weight is $m * (-a)$ = -38.35.
Thus, the total weight of the system would be equal to
$54*9.8 + 59 - 38.35$ = 549.85.
However, this answer is wrong, and I don't know why. Can somebody please point me in the right direction?
| You haven't taken into account the actual weight of the masses.
$$W_{sys}=-(W_1+W_2+W_P)+(M_2-M_1)a=-(M_1+M_2+M_p)g+(M_2-M_1)a$$
Putting the values, You have calculated :
$$W_{sys}\approx 1155$$
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Wave function boundary condition in scattering problem The the boundary condition for a wave function in a scattering problem is
$$\psi_{\boldsymbol{k}_{1}}(\boldsymbol{r}) \underset{r \rightarrow \infty}{\rightarrow} A\left(\exp \left(\mathrm{i} k_{\mathrm{i}} \cdot \boldsymbol{r}\right)+f(\theta, \phi) \frac{\exp (\mathrm{i} k r)}{r}\right)$$
Why people impose this boundary condition in the wave function? what is the physical intuition to impose this boundary condition?
Actually what I am not understanding more, is imposition that the scattered wave should be spherical. Why are we certain that the outgoing wave is spherical?
| Any correctly posed mathematical problem involving differential equations requires boundary conditions (initial conditions are also a kind of boundary conditions). Otherwise it simply cannot be solved, although the issue is often glossed over in not very mathematically rigorous physics textbooks.
When it comes to the Schrödinger equation, one can distinguish two important types of problems: the eigenvalue problems and the scattering problems. The examples of the former are a particle in a square well (with obvious boundary conditions) or the harmonic oscillator (with the boundary conditions at infinity - otherwise we would have more solutions than just usual ones with the Hermit polynomials). Note that these are usually supplemented by the normalization condition.
Scattering problems draw their inspiration from scattering problems in classical physics - for example, a problem of an asteroid passing near the Earth and being deflected by it. Note that even in this classical physical problem, one cannot strictly distinguish the states before, during and after the collision, since the gravitational potential has infinite range. In quantum mechanics this situation is compounded by the fact that the wave solution should exist everywhere in space. The ansatz presented in the question is a solution for a particle with a definite momentum (hence the incident plane wave) being scattered by a centrally symmetric potential (hence the centrally symmetric outgoing solution). In QFT-oriented texts such scattering experssions are motivated by analyzing the time evolution from the distant past, $t=-\infty$, to the distant future, $t=+\infty$, but introductory texts often do not go into such depths... and sadly do not such problems as scattering by a rectangular barrier or tunnelinga s scattering problems at all.
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General relativity modifies Newton's inverse square law of gravity. Why do many people do experiments to test the inverse square law? General relativity may induce the so-called post-Newtonian correction to the inverse square law of gravity. For details, please refer to chapter 9 of Weinberg's Gravitation and Cosmology.
However, there are many research papers about testing the inverse square law on small scale (millimeter range). For example, the following two papers.
*
*S.-Q. Yang et al., "Test of the Gravitational Inverse Square Law at Millimeter Ranges", Phys. Rev. Lett. 108, 081101 (2012).
*E.G. Adelberger, B.R. Heckel and A.E. Nelson, "Tests of the Gravitational Inverse-Square Law", Ann. Rev. Nucl. Part. Sci. 53: 77-121, 2003, arXiv:hep-ph/0307284.
What are they testing? Where is the correction due to general relativity?
| Newton acknowledged the $1/r^2$ behaviour his laws expected of point masses and outside spherically symmetric densities would experience $1/r^3$ corrections (which primarily matter at short distances) due to factors such as the central mass being spheroid, and he showed this still leads to elliptical orbits; it just makes them process. Prior to general relativity, we knew of several complications, including gravity quadrupoles. All these do is change the overall $1/r^3$ coefficient, and general relativity changes it again, but the only planet in our Solar System for which this mattered observationally in 1915 is Mercury.
Modern tests are instead concerned with very long-range corrections to $1/r^2$. The outermost stars in galaxies haven't been flung out by their own centripetal acceleration, but the known amount of gravity from baryonic matter can't explain that. While dark matter is the preferred explanation, another is that gravity is surprisingly strong on that length scale.
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Why do low-energy waves produce heat, but high-energy waves do not? Radio waves, microwaves and infrared are known to produce heat and even cause burns, while visible light and ultraviolet are not. This seems counterintuitive to me, as the latter contain the highest amount of energy.
Why is this? Does it have to do with the quantity of waves rather than the energy?
| We need to clear something up first, which is the difference between photon energy and energy delivered by an absorbed wave. Photon energy is what you’re referring to in your question, but photon energy has nothing to do with total energy delivered. For example, 1 Watt is pretty bright for a laser, but it’s pathetic for your kitchen microwave.
So why would you use a microwave to heat things rather than a UV laser of the same average power? Low-photon-energy, long-wavelength waves tend to get absorbed a deeper depth into the material. Also, long wavelengths are difficult to pinpoint because of diffraction. So with long wavelengths, you have a large and deep absorption volume, which leads to general warming.
In contrast, short wavelengths can be better focused, and they tend to be absorbed near the surface. So the energy delivered is localized, tending to ablation and burning of the material. It’s still heat, but just delivered differently.
And, of course, everything I’ve written is a gross generalization.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/598277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why do we bend little forward for rising from Chair? How does torque arises when not bending forward and how does it gets cancelled while bending little forward? Please give me a video explanation or figurative explaination if possible.
| Try sitting straight and push the arms of your chair to get up. You will notice that you don't always need to bend forward to rise from the chair instead a reaction force is exerted on you by the chair which helps you to get up.
Similarly, in case you are trying to get up without using your hands you bend your body to generate a force which will try to push the floor backwards which in turn produces a reaction force which helps you to get up.
All you need is a reaction force which will give you a lift.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/598365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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