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How do gravitons and photons interact? First of all, I am a noob in physics (I‘m a computer scientist) and started reading Hawking‘s „A brief history of time“. In Chapter 6 he says that “electromagnetic force [...] interacts with electrically charged particles like electrons and quarks, but not with uncharged particles such as gravitons.”
My question now: how come that extremely massiv object are able to bend light (e.g. we are able to see distant stars that are behind the sun)? I mean, how can gravitation (actually gravitons) affect photons if gravitons are not charged?
I know that there are some questions here that go in the same direction but as I‘m a noob in physics, I don‘t quite get the answers.
I‘d appreciate if someone had a laymen‘s explanation for this that not necessarily covers all different aspects (I might pose some follow-up questions) but explains the essence.
Thanks to y‘all!
| Gravitational lensing (light being "bent" by stars for instance) relates to general relativity, the graviton is a theorised particle in quantum field theory. There is currently no complete and accepted theory of quantum gravity which connects the two.
In general relativity the curvature of spacetime alters the path of particles and causes what we call gravity. This means even massless particles like photons have their paths changed since they still travel through spacetime.
| {
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Where does the energy of a spinning diver go? Ok, before we get to my question, I will describe the problem. You have a diver who jumps off a bridge with a slight angular speed about his center of mass (figure 1). While in the air, he curls into a ball. Only gravity acts on the person while he is jumping, and since gravity acts on the center of mass of the person, gravity provides no external torque and hence the angular momentum of the person is conserved. He then uncurls and dives into the water with the exact same angular velocity he started with. During this whole process, imagine that no heat is generated (a.k.a human body is 100% efficient, no air resistance, etc.).
So this is where I get confused. Intuitively, I think that if you end up the same angular velocity, then its like the diver never curled up in the first place and instead had the same angular velocity the whole time. So, the change in kinetic energy should be (mass of person) x (height travelled) x (gravitational field) due to the work done by gravity. But, it takes energy to curl up and un-curl. The internal chemical energy must be converted to something, but it seems as if it's not converted to kinetic energy (which I assume is the only thing that it could be converted to (might be a wrong assumption)). So, where does the internal chemical energy go?
| You already have your answer:
But, it takes energy to curl up and un-curl.
Your muscles do work to move your body. That is where the "chemical energy" you are asking about goes.
| {
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Can a very small piece of material be superconducting? The existing theory of superconducting seems to be based on statistical mechanics.
Can an ultrasmall piece of material, like a quantum dot with very few atoms (like a small molecule), be superconducting?
For example, can a cubic of 3 * 3 * 3 = 27 copper atoms be superconducting?
What is the minimum n for a cubic of $n*n*n$ copper atoms to be able to be superconducting?
Can a few unit cells of a complex high temperature superconducting material be superconducting?
If so, then maybe some calculation from first principles can be done on such a piece of material as a molecule to understand the exact mechanisms of high temperature superconducting.
If not, can some first principle calculation on such a small piece of material be done to find some pattern that lead to a possible theory of high temperature superconducting?
| First you need to define what "being superconducting" means for a finite-size (small) system.
From a theoretical point of view, superconductivity (as many other broken-symmetry phases) is defined in terms of long-range order, that is some correlation function $<\rho(r)\rho(0)>$ stays finite as $r$ goes to infinity. Then, for a finite-size system, one needs a criterion to decide how small it can be, in order for the concept of long-range order to make any sense.
From an experimental point of view, you probably need a system large enough to be able to perform a measurement of resistivity or of magnetic flux expulsion.
So the answer is yes, it can be superconducting, but it should not be too small. How small it can be, probably depends from case to case. A few unit cells, however, is for sure a too small system size for the concept of superconductivity (or any other phase of matter) to make any sense.
| {
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Why is the Force of Gravitational Attraction between two “Extended” bodies proportional to the product of their masses? Newton’s Law of gravitation states that force of attraction between two point masses is proportional to the product of the masses and inversely proportional to the square of the distance between them. I know that the force of attraction between two spheres turns out to be of the same mathematical form as a consequence of Newton’s law. But I am not able to prove how the force between any two rigid masses is only proportional to the product of their masses (as my teacher says) and the rest depends upon the spatial distribution of the mass. So $F$ is ONLY proportional to $Mmf(r)$ where $f(r)$ maybe be some function based on the specifics of the situation.
| If to be summarized in short - you need to apply and solve second Newton law equation for two-body problem :
$$ \vec F_G = \mu \, \vec r^{\,\prime \prime} $$
Where $\mu$ is two-body system reduced mass :
$$
\mu ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}}
$$
Btw, it's interesting to note that reduced mass has reciprocal additive property :
$$ {\frac {1}{\mu }}={\frac {1}{m_{1}}}+{\frac {1}{m_{2}}} $$
Reduced mass helps to analize two-body problem as it were just 1 single body. And $\vec r$ is displacement between bodies.
That's why gravity force is proportional to mass product of both bodies. (I.e. product increases faster than sum of masses). Another way which is helpful to physical intuition is to check the moment of inertia of binary system :
Which is :
$$
I={\frac {m_{1}m_{2}}{m_{1}\!+\!m_{2}}}x^{2}=\mu x^{2}
$$
| {
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Actual meaning of refraction of light The definition of refraction which I found on wikipedia is
In physics, refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium.
But in the below case, there is no change in direction of light. So, is this also refraction?
| Actually refraction is the phenomenon in which speed of light changes when it passes from one medium to another. The change in direction can be explained by Fermat's principle which you can check on Wikipedia.
| {
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Does bending your arm in space require any energy? Since your are weightless in space, your arm has no weight, right? Does this mean that bending it in space requires no energy? Why or why not?
| The short answer is yes, bending your arm in the weightlessness of space still requires energy. You are correct that bending your arm does not require us to overcome the weight of your arm, but we do have to overcome its inertia. Inertia refers to the sluggishness that massive objects have (even in the weightlessness of outer space) given how much matter they have.
A bit of explanation to clarify things: we could just as well ask does it take any energy to push a 100,000 lbs. asteroid a distance of five feet while in the weightlessness of space?
I'm sure you'd agree that the asteroid doesn't weigh anything at all in outer space. But that doesn't mean you can just float up and thump it with your finger and expect it to take off at near light-speed. You still have to push on it (and expend energy in doing so) to overcome its inertia.
The simple summary of all of this is that when we try to move objects in outer space, we don't have to fight against any gravitational weight, but we still have to overcome the basic sluggishness that massive objects have (i.e. inertia).
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Doppler effect of light when it's windy I think I understand the classical doppler effect in sound, where the equation is non-symmetric whether the source of the observer is moving because the speed of medium where sound wave propagates is different according to each of the observers.
I think I also understand why doppler effect is symmetric with light since the speed of "the medium" where light propagates is the same for both observers, meaning we need special relativity to explain the doppler effect of EM waves in a vacuum.
But I struggling to make an eqution to describe the doppler effect of light in an actual realistic moving medium.
What is the frequency shift of light between the source and the observer if wind is blowning at 1/3 of $c_0$, flowing towards the observer.
I have to somehow take in to effect the slowdown of light, the lenght contraction of space as well as the fact that for two observers, the light is now travelling at different speeds. The source is here glowing his laser beem in a lenght-contracted medium. It gets even stranger if you change the wind to water and assume the water is moving faster than the speed of light in water.
On a nano-level, the slowdown of light is caused by the delay in absorption and emmitance speeds of photons in. If the wind is blowing, it is moving those tiny photon-emmiting molecules in space thus causing a classical doppler shift as well.
|
I have to somehow take in to effect the slowdown of light, the lenght contraction of space as well as the fact that for 2 observers, the light is now travelling at different speeds.
It is considerably easier than that. You simply use the relativistic velocity addition formula adding the speed of the medium in the frame of interest and the speed of light in the medium. So in your case the speed of the air is 0.3 c and the speed of light in air is 0.9997 c. Using the relativistic velocity addition formula we get:
$$\frac{u+v}{1+vu/c^2}=0.9998 c$$
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Should the thermos flask better be half full or half empty? Every evening I am preparing hot water for my two year old son wakes up in the night to get his milk. We use a rather bad isolation can for this. It is a typical metal cylinder shaped can holding half a liter. If I put cooking hot water into it, I know that about 5 hours later it will have room temperature already, but it does the job as my son typically wakes up 2 or three hours after I go to bed, and so he gets his milk temperated.
As I need only about 200ml then to mix up his milk, I was asking myself if it is better to only fill in that amount of hot water or to fill up the whole can.
I guess losing temperature has much to do with the amount of water but also with its surface touching the (colder) room air outside.
With no idea anymore of what my old physics teacher told me twenty five years ago I hope you could share some wisdom for my little story here. Thanks in advance ;)
| The idealised formula (lumped thermal analysis) for a cooling object, according to Newton's Cooling Law is:
$$T(t)=T_{\infty}-(T_{\infty}-T_0)\exp\Big(-\frac{t}{\tau}\Big)$$
where $\tau$ is the characteristic time:
$$\frac{1}{\tau}=\frac{U A}{m c_p}$$
with:
*
*$T(t)$ is the temperature of the object in time $t$
*$T_0$ is the object's starting temperature and $T_{\infty}$ the surroundings' temperature
*$A$ is the surface area of the object exposed to the surroundings
*$m$ is the mass of the object, with specific heat capacity $c_p$
*finally, $U$ is the overall heat transfer coefficient. Better quality thermos flasks will generally have lower values of $U$
From this we can conclude that for larger $m$ the rate of cooling will be slower. Note however that greater mass usually also implies greater $A$, thereby somewhat offsetting the mass-effect.
Following in the footsteps of @probably_someone (comments) I'll explore $\frac{A}{m}$ for the following shape:
With $m=\rho V$ and $\rho$ (density) a constant we can evaluate $\frac{A}{V}$ instead:
$$V=\frac{\pi D^2}{4}H+\frac12 \frac 43 \pi \Big(\frac{D}{2}\Big)^3=\frac{\pi D^2 H}{4}+\frac{\pi D^3}{12}=\frac{\pi D^2(3H+D)}{12}$$
$$A=\frac12 4\pi \Big(\frac{D}{2}\Big)^2+2 \pi \Big(\frac{D}{2}\Big)H=\frac12 \pi D^2+\pi D H=\frac{\pi D(2H+D)}{2}$$
This gives us a result for $\frac{A}{V}$ of:
$$\frac{A}{V}=\frac{6(2H+D)}{D(3H+D)}$$
Here $D$ is a constant and only $H$ increases with $m$. Because the coefficient $3$ in the denominator, instead of $2$ in the numerator, as $H$ increases $\frac{A}{V}$ does indeed decrease which points to lower cooling rate, as expected.
Another consideration. Assume we half-fill the flask with boiling water. Due to steam formation it is reasonable to expect the entire flask to reach approximately the same temperature. In that case, again as an approximation, the total surface area $A$ could be used.
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Is the average force calculated from $F(x)$ the same as that calculated from $F(t)$? Say a force is doing work on an object in one dimension. I could calculate the average force over the distance with
$$\frac{1}{\Delta{x}}\int_{x_1}^{x_2} F(x) \text dx$$
If I also formulated force as a function of $t$, I could calculate the average force over the total time period with
$$ \frac{1}{\Delta{t}} \left| \int_{t_1}^{t_2} F(t) \text dt \right|$$
Obviously, the integrals themselves are not equal. One represents work, and the other represents impulse; $ \Delta{p} \neq \Delta{K} $.
However, if $x_1$ corresponds to $t_1$ and $x_2$ corresponds to $t_2$, could I set the above expressions equal to one another? Does averaging over the interval make them equal? Put another way, I've never seen texts differentiate between the $F_{ave}$ in $W = F_{ave} \cdot \Delta{x}$ and the $F_{ave}$ in $\Delta{p} = F_{ave} \Delta{t}$. Is there a difference? If not, I'm thinking they can relate work to $\Delta{t}$.
| To show constructively that these quantities are not equal in general, let's consider an object undergoing simple harmonic motion over a quarter-cycle (i.e., from the equilibrium position to maximum displacement.) In this case, we have
$$
F(x) = - k x, \qquad x(t) = A \sin \omega t, \qquad t \in [0, \pi/(2\omega)],
$$
with $\omega = k/m$ (though this won't matter in the end.)
The average over displacement is
$$
\langle F \rangle_x = \frac{1}{A} \int_0^A (-kx) \, dx = - \frac{1}{A} \left[ \frac{1}{2} kx^2\right]_0^A = - \frac{1}{2} k A,
$$
while the average over time is
$$
\langle F \rangle_t = \frac{2 \omega}{\pi} \int_0^{\pi/2\omega} (-k A \sin \omega t) \, dt = - \frac{2 k A \omega}{\pi} \left[ - \frac{\cos \omega t}{\omega} \right]_0^{\pi/2\omega} = -\frac{2}{\pi} k A.
$$
Since $\pi \neq 4$,[citation needed] these results are different.
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Expression and explanation of quantum mechanical harmonic oscillator I am currently studying the textbook Infrared and Raman Spectroscopy, 2nd edition, by Peter Larkin. In a section entitled Quantum Mechanical Harmonic Oscillator, the author says the following:
Fig. 2.6 shows the vibrational levels in a PE [potential energy] diagram for the quantum mechanical harmonic oscillator. In the case of the harmonic potential these states are equidistance and have energy levels $E$ given by
$$E_i = (v_i + 1/2)h \nu \ \ \ \ v_i = 0, 1, 2, \dots$$
Here $\nu$ is the classical vibrational frequency of the oscillator and $v$ is a quantum number that can only have integer values. This can only change by $\Delta v = \pm 1$ in a harmonic oscillator model, and thus a transition will be forbidden unless the initial states differ by one quantum of excitation. The so-called zero point energy occurs when $v = 0$ where $E = \dfrac{1}{2} h \nu$ and this vibrational energy cannot be removed from the molecule.
I have two questions relating to this:
*
*Throughout this text, it seems that the author is not a fan of commas, so I'm unsure if "the so-called zero point energy occurs when $v = 0$ where $E = \dfrac{1}{2} h \nu$[,] and this vibrational energy cannot be removed from the molecule", or whether the author meant it (and it is correct) as written. In other words, is the fact that this vibrational energy cannot be removed from the molecule a requirement for the zero-point energy to occur, which is what is meant as written, in the absence of the comma, or is $E = \dfrac{1}{2} h \nu$ the only requirement for the zero-point energy to occur, and the fact that "this vibrational energy cannot be removed from the molecule" is an afterthought, and not a requirement for the zero-point energy to occur, which would be the case with the comma?
*Why does the expression for $E_i$ contain the $\dfrac{1}{2}$? It is not immediately clear why this expression would need to include this, so I'm curious as to the physical/mathematical reason for this.
I would greatly appreciate it if people would please take the time to clarify these two points.
| If we have a Harmonic oscillator with Hamiltonian
$$\hat{H}=\frac{\hat{p}^2}{2m} + \frac{m\omega^2x^2}{2},$$
then, by solving the Schrödinger equation, we obtain its energy states as
$$E_n = \hbar\omega(n+\frac{1}{2}), n = 0, 1, 2,...$$
We could remove $\frac{1}{2}$ by shifting the energy origin (and sometimes it is done), but note that $E=0$ is the minimal possible energy of the corresponding classical oscillator. So $\frac{1}{2}$ is kept to underscore this distinction, which is consistent, because the averages of $\langle x^2\rangle$, $\langle p^2\rangle$ remain different from zero.
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A uniformly polarized sphere Say there is a polarized sphere with polarization density $\vec{P} = \alpha \hat{r}$. How can I tell if the electric field outside of the sphere will also be radial? I see in many places that it is taken as obvious, but why is it?
*Edit: rephrase
| Pay attention to the fact that the sphere you are considering is not uniformely polarized: only the magnitude of the polarization density is uniform, while its direction changes at each point. In particular the direction of $\vec{P}$ is radial. Therefore the system has a symmetry with respect to the center of the sphere and it is possible to say that the electric field outside the sphere has to have the same symmetry. This is why it is radial.
Note that if $\vec{P}$ is taken "really" uniform on the sphere (i.e. it has same magnitude and direction at each point on the sphere) the external electric field outside is not radial: it is a dipole field.
| {
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Equation of motion for a particle under a potential in special relativity The equation of motion of a particle in Newtonian mechanics in 3D under an arbitrary potential $U$, is written as $$m\frac{\mathrm{d}^2 \mathbf{r}}{\mathrm{d} t^2}=-\nabla U.$$ Now, my question is, how can this be generalised to Special relativity?
I know that the naive answer, $$m\frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=-\partial^{\mu} \Psi$$, where $\Psi$ is some relativistic generalisation of potential energy, cannot work, since every four force $K^{\nu}$ has to satisfy $K^{\nu} \dot{x}_{\nu}=0$, where dot indicates derivative with respect to proper time, so for this shows that the above naive generalisation cannot work, unless $\Psi$ is a constant, which makes it physically useless.
How can one solve this caveat, in order to obtain a physically useful generalisation that works in special relativity?
| In this context, the question that I posted yesterday seems to be relevant. That is,
Is there anything like relativistic potential energy? If not, why? We know relativistic force along the direction of velocity, involving γ3. We also have the standard expression relating force and potential energy in Newtonian mechanics, F=−dV/dx where V is a function of x. I don't see any reason why this can not be applied for relativistic case too.
Hence I ask, "can we get potential energy in the relativistic case using F=−dV/dx? If not, why?"
My intuition tells me that there is no reason to find fault with this expression and it should apply for relativistic case also. If it's application leads to problems, this only shows that all is not well with relativity and it contains a flaw that needs to be eliminated. Please note that we should not think that the relativity theory is fully correct just because it predicts many things successfully.
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What does CERN do with its electrons? So to get a proton beam for the LHC, CERN prob has to make a plasma and siphon off the moving protons with a magnet. Are the electrons stored somewhere? How? I don’t mean to sound stupid but when they turn off the LHC, all those protons are going to be looking for their electrons. And that’s going to make a really big spark.
| Just to add an electrical engineering answer to these good physics answers, insulators are never perfect. In school we talk about perfect insulators, but in practice everything has some level of conductivity. Air itself has a resistivity of somewhere on the order of $10^{16}\Omega-m$, for example.
Usually when doing a process that generates ions is done, one side is "grounded," which means we let the charge flow into the ground. The earth can take an enoremously large number of these electrons before the electrostatic forces between them start to add up. And that gives time for these electrons to re-pair with protons.
Failing that, combining the electrical charges into the earth means our effect is combined with the electrostatic effects at the planetwide level. For example, the aurora borealis is a product of a massive stream of charged particles coming from the sun at a magnitude most of us cannot even fathom!
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What does it mean to say the CMBR has a “temperature”? We say the cosmic microwave background radiation “has a temperature” of 2.7K.
Does this mean it has a temperature in the way we say the air around a warm lightbulb has a temperature of 120° or in the sense that the mix of electromagnetic waves given off by the lightbulb are like those emitted by a blackbody at 3000K?
Is the temperature of the electromagnetic field at some point in space the temperature “of the space” in which the field exists, or the temperature of a black body that would emit the spectrum of the EM field in the space if the blackbody were there at that moment.
I hear people speak as if the first, but I think they mean the second. Yes?
| It means the mix of electromagnetic waves "given off" by the CMB (really: "that constitute the CMB") are like those emitted by a blackbody at 2.7K.
Temperature is a property of an ensemble of particles (photons in this case, could be air molecules, or even virtual particles) but not of space itself.
Yes, people speak as if the first but mean the second.
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Angular momentum of the earth We know the tidal waves are decreasing the spin rate of the earth which causes the days to longer, so as the angular momentum of the earth decreases it means it rotational kinetic energy also decreases since energy is always conserved the translational kinetic energy of earth must increase now right? Then that would cause number of days in a year to decrease as we right?
|
since energy is always conserved the translational kinetic energy of earth must increase now right?
Energy is conserved, but kinetic energy is not necessarily conserved. In this case, the tidal drag converts some of the kinetic energy of the earth-moon system into thermal energy.
It's very similar to simple friction in other contexts. If you slide a book across a table, it comes to a stop. Momentum (of the earth-book system) is conserved, but kinetic energy is not.
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What happens to an inductor if the stored energy does not find a path to discharge? Suppose an inductor is connected to a source and then the source is disconnected. The inductor will have energy stored in the form of magnetic field. But there is no way/path to ground to discharge this energy? What will happen to the stored energy, current and voltage of the inductor in this case?
| The current will flow back and forth between the end points of the wire after the potential difference is removed. It is like a wave traveling back and forth on a string. This is because all of the electrons are behaving as one wave (wave function) in the superconducting regime. If there is no heat loss this will cont. forever. In reality, electron standing wave on the string will damp out eventually. It will behave like an antenna and it will radiate its initial energy as an EM wave.
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Why is (anti)-holomorphicity considered in physics? As a mathematician, holomorphicity is an extremely good property that provides rigidity, finite dimensionality, algebraicity. etc to whatever theory that's considered. I'm curious about why (anti-)holomorphicity is considered in physics.
As an example, apparently holomorphic representations of $SL(2,\mathbb{C})$ apply to physics, while weaker kinds of representations are of interest in math as well.
As another example, 2D conformal field theory models fields as meromorphic functions. Though conformality almost equals holomorphicity in 2D, I'm still curious why picking "conformal" in the beginning?
Q. Do people consider them because they have well-developed mathematical background and thus allow us to say something interesting, or there are deeper reasons behind it?
| Scale invariance is common in physical systems at phase transitions. If the characteristic length of a system is small in one phase (disordered) and infinite in another phase (ordered), then typically the system is scale invariant at the transition between the two phases. A non-trivial physical observation is that in many cases scale invariance (together with invariance under rotations and translations) implies conformal invariance. And in two dimensions, conformal transformations are described by holomorphic functions.
Therefore, in two-dimensional conformal field theory, the appearance of holomorphic functions is the mathematical manifestation of physical symmetries: ultimately, scale invariance.
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Derivation of Lorentz transformation I am reading Lorentz transformation from Robert Resnick book. It is given that
$x'= a_{11}x + a_{12}y + a_{13}z +a_{14}t$
$y'= a_{21}x + a_{22}y + a_{23}z +a_{24}t$
$z'= a_{31}x + a_{32}y + a_{33}z +a_{34}t$
$t'= a_{41}x + a_{42}y + a_{43}z +a_{44}t$
While solving coefficients of $y'$, as the two frames are coinciding at t=0 and one of the frame is moving with $v$ velocity in $+x$ direction. So we can say that $y'\,axis$ is always perpendicular to x and z axis. So $y'$ has nothing to do with x and y so $a_{21}$ and $a_{23}$ are zero.
But, why $a_{42}$ and $a_{43}$ are zero?
How to use the postulates to derive these coefficients which I have asked because other coefficients are given in other books also. But these coefficients are not explained explicitly. Please give me some approach.
| You can think of Lorentz transformation similar to a rotation in $x$ and $t$ plane ( if the velocity is along x axis) , so basically its just intermixing time and only $x$ component of space. So $a_{42} $ and $a_{43} $ will be zero as there is no velocity along $y$ and $z$ axis.
| {
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Why do we need a lens in the delayed-choice quantum-eraser experiment of Kim? This is the delayed-choice quantum-eraser experiment of Kim diagram described on Wikipedia https://en.wikipedia.org/wiki/Delayed-choice_quantum_eraser
Why do I need a lens in front of the D0? Why we should focus photons? The classic Double-slit experiment does not require a lens.
What pattern will we see on D0 if we remove the lens from the experimental setup? Why don't we see interference fringes?
| From the paper Taming the Delayed Choice Quantum Eraser by Johannes Fankhauser:
The lens in front of detector D0 is inserted to achieve the far-field limit at the detector and at the same time keep the distance small between slits and detector.
So the reason is that using the lens allows them to keep D0 closer to the slits.
Different patterns by different focus:
*
*From the same paper:
*From an experimentally correct simulation
And even closer peaks from Kim et al's paper's results:
| {
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Question about an Exercise with Time-Dependent Hamiltonian I've been recently been assigned this exercise:
Consider two spin 1/2 particles which are coupled through a time dependent interaction:
$$ H(t) = a(t) s_1 \cdot s_2 $$
where $a(t)$ is a function which is constant in the interval $[0,T]$ and zero elsewhere.
The system is in the state $|+,->$ for $t \to -\infty$.
The exercise then asks various questions about the probability of finding the state in another state for $t\to +\infty$.
It appears as if this exercise is extremely easy as it can be solved exactly for any state in the $|S,M_S>$ base of eigenstates of $S^2, S_z$. More precisely, in this base the Time-dependent Schrodinger Equation becomes a system of four decoupled first-order linear differential equations in the coefficients of $|\psi(t)>$ in this base, since $H(t)$ is diagonal in this base:
$$ H(t) = a(t) \bigg[ \frac{S^2}{2} - \frac{S_1^2}{2} - \frac{S_2^2}{2} \bigg] = \frac{a(t)\hbar^2}{2} [S^2 - 3/2] = \frac{a(t)\hbar^2}{2}
\begin{pmatrix}
1/2 &&&& \\
& 1/2 &&&\\
&& 1/2 &&\\
&&& -3/2
\end{pmatrix} $$
where the states are ordered as such: $|1,1>,|1,-1>,|1,0>,|0,0>$. It's pretty easy from here since $a(t)$ is either constant or zero, which means the coefficients evolve with an imaginary exponential in the interval $[0,T]$ and stay constant elsewhere. More precisely:
$$ |\psi(t)>\;\; =
\begin{pmatrix}
b_1(t) \\ b_2(t) \\ b_3(t) \\ b_4(t)
\end{pmatrix} \Rightarrow i\hbar\frac{d}{dt}|\psi(t)> = H(t)|\psi(t)>\; \Rightarrow
\begin{cases}
i\hbar \frac{db_j}{dt}(t) = \frac{a(t)\hbar^2}{4}b_j(t) & j=1,2,3\\
i\hbar \frac{db_j}{dt}(t) = -3\frac{a(t)\hbar^2}{4}b_4(t)
\end{cases}$$
and finally:
$$ b_j(t) =
\begin{cases}
b_j(0) & t<0\\
b_j(0)e^{-i a_0\hbar t/4} & 0\leq t\leq T\\
b_j(0)e^{-i a_0\hbar T/4} & t>T\\
\end{cases} \quad j=1,2,3 \qquad
b_4(t) =
\begin{cases}
b_4(0) & t<0\\
b_4(0)e^{i 3a_0\hbar t/4} & 0\leq t\leq T\\
b_4(0)e^{i 3a_0\hbar T/4} & t>T\\
\end{cases}
$$
and in the case $|\psi(-\infty)> = |\psi(0)> = |+,->$ the initial conditions give $b_1(0) = b_2(0) = 0$ and $b_3(0) = b_4(0) = 1/\sqrt{2}$, which means it will oscillate between different linear combinations of $|1,0>,|0,0>$ or equivalently $|+,->,|-,+>$.
Then the probability of finding the system in a given state is a simple scalar product between 4-dimensional vectors.
Is this the case or am I forgetting something? It's been the easiest homework assigned so far and I find it very strange. Also, why would the exercise specify the state of the system for $t\to\pm\infty$ since the system only evolves between $t = 0$ and $t = T$, should it not be for $t<0$ and $t>T$ instead? What I mean is it seems like unnecessary detail, since the state changes only in a finite amount of time.
EDIT: Inserted the calculations necessary to find the time evolution of the system in any given initial state.
| You are right that at $t=0$ the system is still in state $|+,-\rangle$, since $a(t)=0$ for $t<0$ and this state is an eigenstate of the Hamiltonian. However, for $t>0$ it is not an eigenstate anymore, but a superposition of the two eigenstates $|1,0\rangle$ and $|0,0\rangle$. Each of these evolves with different time exponent, so the result at $t=T$ may be rather different from what you had at $t=0$. Note that the problem is solvable, even if $a(t)$ is an arbitrary function in $[0,T]$.
| {
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Why is surface tension measured in units of milliNewtons per meter? Rather than square meter(s)? Why is liquid surface tension written in units of mN/m, or milliNewtons per meter?
The related concept of surface energy for solids uses units of milliJoules per square meter.
| You're correct that the two concepts are indeed representing the same quantity. If the surface tension of a surface is $T$ then it would need $T l dx$ amount of work to move its boundary of length $l$ by a distance $dx$. Since $l dx$ would be the increase in area $dA$, the amount of work needed to increase the area of the surface by a unit area would be simply $T$ (and this is what surface energy is, by definition, the amount of work needed to increase the surface area by a unit amount). So, the two quantities are the same.
However, there is nothing wrong in the units you mention. Since $\text{Joule=Newton}\cdot\text{ meter}$, the two units you mention are identical.
| {
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How does a water jet hitting a wall move parallel to the wall if momentum is conserved? Classical mechanics says that if I throw a ball with velocity perpendicular to the wall and it collides elastically with the wall with a velocity $v_0$, then it bounces back with the same velocity $v_0$.
However, if I shoot a beam of water perpendicular to the wall, in most cases it will not deflect back perpendicular to the wall instead it gains velocity perpendicular to the initial velocity and continues to move on the surface. Isn't this a violation of conservation of momentum since for any small molecule inside the beam of water we had no momentum in the perpendicular direction to get started with?
| If a tennis ball hits a wall, it bounces normally, but with several balls in a continuous jet, the outcome is different. Many bouncing balls will hit incoming ones, and the result will be a scattering pattern. The advantage of the direction parallel to the wall is that it is a free path from new coming balls.
I think that the process with water jet is similar.
By the way, in a elastic collision against a wall, the moment is not conserved if we forget to include the Earth recoil in the system. The point of elastic collisions is that the ball kinetic energy is conserved.
| {
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Examples of path integral where path of extremal action does not contribute the most? I have learnt that by doing a saddle point approximation in the path integral formulation of quantum mechanics, the classical action (extremal action where $\delta S=0$) is the one that contributes the most, hence seeing how classical physics arises from quantum physics.
The question is: is there any example in quantum physics (especially QFT) where the most contribution does not come from the path of extremal action? I know that the proof using the saddle point approximation seems very general, but I was thinking that there might be some peculiar/interesting terms in the action (such as topological) where the path with $\delta S=0$ do not contribute the most?
| The answer by @Accidental FourierTransform has already mentioned the instantons, but let me give a more pedestrian example and a more pedestrian view on the question.
When we use a path integral to describe the motion of a single quantum particle, the saddle point approximation corresponds to what is otherwise known as quasiclassical approximation, and the justification for it is literally taking $\hbar\rightarrow 0$. (As a trivia: Landau & Livshitz in their QM book literally derive quasiclassical approximation from what they call action using eikonal approximation without ever referring to path integrals.)
This provides us with a simple hint of where the saddle point approximation may not work: where quasiclassical approximation doesn't. A well-known example is the tunneling between two degenerate potential minima, where the correct splitting of the energy levels can be recovered via the alraedy mentioned instantons (which can be also viewed as a saddle-point approximation after transformation to the Eucledean space.)
| {
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Is this a typo in Peskin's QFT? In ''An intro to QFT (2018)'' chapter 3, Peskin does the following:
Let me introduce some notation first, let $v^s_k=\begin{pmatrix}\;\;\,\sqrt{k\cdot\sigma}\,\xi^{-s}\\-\sqrt{k\cdot\bar{\sigma}}\,\xi^{-s}\end{pmatrix}$ be a bispinor for the negative-energy solution to the Dirac eq. with momentum $k$ and spin state $\xi^{-s}$.
From previous chapters we know $\,\xi^{-s}\equiv-i\sigma_2(\xi^s)^*$ and $\,\sigma_2^*=-\sigma_2\,$, then, $\,(\xi^{-s})^*=-i\sigma_2\,\xi^s$. We also know $\,(\sqrt{k\cdot\sigma^*}\sigma_2=\sigma_2\sqrt{k\cdot\bar{\sigma}}\,)\,$ and $\,(\sqrt{k\cdot\bar{\sigma}^*}\sigma_2=\sigma_2\sqrt{k\cdot\sigma}\,)$ so we can compute $(v^s_k)^*$ as
$(v^s_k)^*=\begin{pmatrix}-i\sqrt{k\cdot\sigma^*}\sigma_2\,\xi^s\\\;\;i\sqrt{k\cdot\bar{\sigma}^*}\sigma_2\,\xi^s\end{pmatrix}=\begin{pmatrix}-i\sigma_2\sqrt{k\cdot\bar{\sigma}}\,\xi^s\\\;\;i\sigma_2\sqrt{k\cdot\sigma}\,\xi^s\end{pmatrix}=\begin{pmatrix}0 & -i\sigma_2 \\ i\sigma_2 & 0\end{pmatrix} \begin{pmatrix}\;\;\,\sqrt{k\cdot\sigma}\,\xi^s\\-\sqrt{k\cdot\bar{\sigma}}\,\xi^s\end{pmatrix}=-i\gamma^2 u^s_k$.
Here, $u^s_k$ is the bispinor of the positive-energy solution with momentum $k$ and spin state $\xi^s$.
We see that $\:\boxed{\,(v^s_k)^*=-i\gamma^2 u^s_k\,}\;$ but my question comes now as Peskin proceeds saying that the following expressions follow immediately from it:
$u^s_k=-i\gamma^2 (v^s_k)^*\;$ and $\;v^s_k=-i\gamma^2 (u^s_k)^*$.
How is that possible? They don't even hold for $\,u^s_k\neq0$. Did I missed something?
| Schwartz uses negative signature, so $(-i\gamma^2)^2=1$.
| {
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Green Function in Open Quantum Systems Imagine an open quantum system interacting with an environment that admits a density matrix (Markovian) description in terms of Lindbladians ($c$ and $c^\dagger$). Is there a meaningful way to define a single particle Green function for this system up to the time the steady state is achieved, but without knowing any more details about the environment-system interaction?
Finally, how do steady state Green functions eventually become meaningful quantities to look at within a reduced density matrix description? For example, see here.
Background confusion: Single particle Green functions could be thought of as propagation amplitudes of particles/holes from one time to another. In a closed quantum system, one would sum all amplitudes leading from the initial to the final state, which is indeed the right thing to do. However, in an open quantum system with just a reduced density matrix (Lindbladian) description and one that exchanges matter, it feels like there is no way to sensibly decide how to coherently sum up amplitudes for different processes (eg. a process in which particle vanishes, then comes back from the environment etc.), esp. when starting from some initial random density matrix. But then to have a meaningful steady state Green function, some things must be magically falling into place in the steady state $-$ what am I missing? Any pointers would be greatly appreciated.
| Certain open system processes can be described within the Greens function formalism by adding imaginary terms to the Hamiltonian, namely all processes where something only exits the open system and nothin comes back from the environment. So, basically decay processes. Other non-unitary processes such as dephasing cannot be directly accounted for by a Green function.
I suppose that in your case in the steady-state no such non-unitary evolution remains and thus the dynamics can be described by Green functions.
| {
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Why do scientists need to measure extremely small intervals of time? Why do scientists need to measure extremely small intervals of time? Why is it necessary?
| It all depends on the phenomena one is studying. If you need to make an analysis about the general pick up of a car engine, i.e. how much time it take to go from 0 to 100 Km/h, one does not need to bother about femto-seconds rather seconds will suffice. However, for the time scales involved in atomic transitions or sub-atomic interactions have very short time scales. Hence, there one needs to have faster measuring device to have relevant data set to that phenomena.
One can think of it like Shutter speed of the camera. If you want to capture a moving object then you need a fast shutter speed. The faster the object is, the commensurate shutter speed needed to capture the object in motion. Otherwise, with slow shutter speed and faster body all you get is a hazy picture. This is the closest analogy i could think of.
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Is my friend right about omitting $c^2$ in world famous tiny equation? I know $E = mc^2$ says that inertial mass of a system is equal to the total energy content of a system in its rest frame. My friend told me the $c^2$ can be omitted from this equation because that's just an `artifact' when measuring inertia and energy in different units.
Is he right?
| It is very common to "omit" constants, especially in theoretical physics.
Actually, there is a system based only on physical constants ($c,\hbar,\epsilon_0,G ...$). Physical units in this system are called natural units and we can "normalize" those constants by choosing a system in which their value is 1, they are called purely natural units (or just natural units again).
You can read more here: Natural system of units in General Relativity and this other question maybe usefull too.
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$\gamma_5 \gamma^\sigma$ expressed with Levi-Civita tensor We have that $\gamma_5 = -\frac{i}{4!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho \gamma_\sigma$. Using this, what approach would be suggested in showing that $\gamma_5 \gamma^\sigma = \frac{1}{3!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho$? You would think that the solution is that
\begin{equation}
\gamma_5 \gamma^\sigma = -\frac{i}{4!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho \gamma_\sigma \gamma^{\sigma} = -\frac{i}{4!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho \times 4 = \frac{1}{3!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho,
\end{equation}
which makes no sense...
| Indeed, you are right that conflating a saturated dummy index with a free index makes not sense.
However, you know that
$$
\gamma_5 \gamma^\kappa= \tfrac{1}{2}[ \gamma_5 ,\gamma^\kappa ],
$$
so what happens if you substitute the quadrilinear expression in the commutator? You know the leading, quintilinear, term vanishes. So only trilinears may survive. You can easily compute them. What do the 4 such add to?
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Is this formula I derived for net acceleration correct? I was thinking about acceleration due to gravity and I thought of deriving a formula that gives the net acceleration due to gravity between two bodies. Now, by net acceleration, I basically mean the effective acceleration. Please have a look :
Let $A$ and $B$ be two objects with masses $m_1$ and $m_2$ respectively
and the distance between them be $d$.
Let $F$ be the force of attraction between them and $g_{m_1}$ be the acceleration of $m_1$ due to the gravitational force of $m_2$ and let $g_{m_2}$ be the acceleration of $m_2$ due to the gravitational force of $m_1$.
Now, according ot Newton's Law of Gravitation: $$F = G \dfrac{m_1m_2}{d^2}$$
We know that: $$F = m_1g_{m_1}$$
$$\text{and}$$
$$F = m_2g_{m_2}$$
This implies that: $$m_1g_{m_1} = G \dfrac {m_1m_2}{d^2} \implies g_{m_1} = G \dfrac {m_2}{d^2}$$
In a similar manner: $$g_{m_2} = G \dfrac {m_1}{d^2}$$
Now, this is the part where I think I'm making a mistake.
What I think is that here both the objects are accelerating towards each other with accelerations of $g_{m_1}$ and $g_{m_2}$ respectively.
So, I think the net acceleration between $A$ and $B$ would be $$g_{m_1}+g_{m_2} = G \dfrac{m_2}{d^2} + G \dfrac{m_1}{d^2}$$
$$\ \ = \dfrac {G}{d^2}(m_2+m_1)$$
Now, I think that if this formula is correctly derived, it gives the net acceleration by which two masses mutually attract each other. And wouldn't this formula imply that the acceleration due to gravity does depend on the mass of the object that is into consideration which is actually not the case.
So, that would imply that this formula is wrong. So, please let me know where the error is
Thanks!
PS : All edits on formatting are welcome :)
| Your formula is just fine. You can read about the "2 body problem" and solve it entirely.
You seem to be upset about that the relative acceleration depends on both masses, but is correct.
Think of $m_2$ >>> $m_1$. You get $g_1$ >>> $g_2$. You can aproximate relative acceleration by $g_1$ and
the acceleration of $m_1$ towards $m_2$ does not depend on $m_1$.
| {
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Acceleration as a function of position and time I know if you have an acceleration as a function of $t$, $a(t)$, to find the velocity you simply integrate $a(t)$ with respect to $t$. Moreover, if the acceleration was a function of position, $a(x)$, you use the fact that $a(x) = v(x) \cdot dv/dx$ and solve for $v(x)$. However, what if the function of acceleration was dependent on both $x$ and $t$, $a(x,t)$. How would you solve for a velocity $v(x,t)$ ?
| the acceleration a is:
$$a(t)=\frac{dv(t)}{dt}$$
thus:
$$v(t)=\int a(t)\,dt+v_0\tag 1$$
if $a(t)\mapsto a(x(t),t)$
you get from equation (1)
$$v(t)=\int a(x(t),t)\,dt+v_0$$
and $x(t)$ is the solution of the differential equation $\ddot{x}=a(x(t),t)$
| {
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Does tangential acceleration change with radius? Do tangential velocity and tangential acceleration change with radius (change of radius on the same object)?
For example consider a spinning disk. Does the equation $$a_t = \alpha R$$ (where $a_t$ is the tangential acceleration, $\alpha$ is the angular acceleration and $R$ is the radius of the disk) give me the tangential acceleration of the centre of mass or of a point on the edge of the disk?
As you go inwards from a point on the edge of the disk , the radius decreases. So doesn’t that mean the tangential acceleration of the centre of mass is zero? I have the same doubt regarding tangential velocity. What is wrong with my reasoning? Does the centre of mass have the highest tangential velocity and acceleration or the lowest of all points on the disk?
| A note on the case you're considering:
A disk is an extended body. This means it's a collection of points: and must be treated as such. To speak of the displacement/velocity/acceleration of a point on the disk and of the disk itself - the center of mass of the disk - are two distinct analyses.
Particles on the disk
Particles of the disk all travel in circles around the axis of rotation. (From your question, it's implied that the axis passes through the center of mass, so we'll continue to use the same convention (call the axis $x$).)
First, any given point is a part of two disjoint sets: it's either part of the disk, or it isn't. Since the disk is rotating, any particle that's on the disk is in circular motion around $x$. If it isn't part of the disk, it's not in circular motion.
Clearly, any particle not on the disk has zero angular velocity. Of the points on the disk, all those with non-zero distance from the axis have some tangential velocity.
$$\vec v = \vec\omega \times\vec R$$
Since $x$ passes through the centre of mass, its tangential velocity is, as correctly stated, zero.
| {
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Why X-ray and radio waves can penetrate walls but light can not? Why can visible light, which lies in the middle between X-ray and radio waves in terms of frequency/energy, not penetrate walls?
| The interaction of photons with matter is complicated. The electromagnetic spectrum covers many orders of magnitude in frequency and photon energy, and there are qualitatively different processes that occur in different regimes. The results depend on the electrical properties of the material, such as conductivity and permittivity. We have materials like glass that are transparent to visible light, and low-energy x-rays that are strongly absorbed.
But speaking very broadly, it is possible to understand the main trends over the whole spectrum. We have a region (1) in the visible spectrum, where the frequency of the light is similar to the frequency of condensed-matter resonances, which in many cases you can think of as resonances of the electrons, as if the electrons are little objects attached to atoms by springs; and region (2) in low-energy x-rays, where the wavelength of the photon is comparable to the wavelength of the electrons in an atom. This splits up the spectrum into three parts.
At low frequencies $f$, below region 1, we have a skin depth, which depends on $f^{-1/2}$. As $f$ gets smaller, the skin depth grows without bound. Hence radio waves tend to be penetrating.
Around region 1, you get strong classical resonant behavior. You can see this if you look at a plot of the index of refraction of glass as a function of frequency. It has a series of spectacular peaks. Each of these peaks has a classic Lorentzian shape, in which the response on the right-hand side of the peak approaches zero. So if you ignore the peaks themselves, which are narrow, then you get a series of stair steps. At frequencies above region 1, you've gone down all the stair steps, and the response approaches zero. This is why, classically, we expect high-frequency electromagnetic radiation to interact with matter very weakly.
But in region 2 you get the photoelectric effect. In first-order perturbation theory, this depends on the extent to which the electric field overlaps with the wavefunction of the electron. When the two wavelengths are similar, you get a strong cross-section. This is why matter strongly absorbs soft x-rays, but not gammas and hard x-rays.
| {
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During particle-antiparticle annihilation, are the photons expelled perfectly perpendicular? During particle-antiparticle annihilation, are the photons expelled perfectly perpendicular to the original direction of the particle-antiparticle pair?
There is very little information on the web about this topic, though I read a book awhile ago that stated that the photons are emitted from an angle, not perfectly perpendicular to the particles. Is this true? If so why?
|
During particle-antiparticle annihilation, are the photons expelled perfectly perpendicular to the original direction of the particle-antiparticle pair?
No.
In the center-of-momentum frame, the two photons can come out back-to-back in various directions, with a probability distribution. Any direction is consistent with energy-momentum conservation.
When non-relativistic electrons and positrons annihilate, all directions are equally likely for the photons; the differential cross section is isotropic. When the electron and positron are ultra-relativistic, the photons strongly tend to be along the direction of motion the annihilating particles. (This is for unpolarized beams.)
The calculation of the differential cross section for any energy may be found here.
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Does putting a thin metal plate beneath a heavy object reduce the pressure it would have applied without it My dad bought an earthen pot and he kept it on our glass table. Worried that the glass could break on filling the pot with water. I kept a metal plate beneath it. At first, it seemed like a good idea , but on further thinking I was unsure if it would actually help in bringing down the pressure on the glass. what if I put three coins beneath the pressure points instead of the plate, would it be any different
than placing the plate (assuming the coins to be nearly as thick as the plate).
and if it won't be any different, am I right to think that I could further keep reducing the area of the coins until they start looking extensions of the stand on which the earthen pot rests.
| I don't think a metal plate is necessary for a thick glass like that, supported by an wood table.
Being transparent, glass looks like more fragile than it actually is.
I have a ceramic floor for example. Each ceramic plate has 60 x 60 cm, and much thinner than the glass of the picture. Moreover, the contact with the concrete behind it is far from perfect. But there are heavy sofas over it anyway. And ceramic is also a brittle material. It must be properly protected during transportation from the store to house, otherwise breaks easily.
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Does the angular velocity of a spinning disk increase if it has a completely inelastic collision with a object with a greater tangential velocity?
A roller of radius 10cm is spinning with a angular velocity of 15 rad/s. It has a completely inelastic collision with a hunk of clay, with mass m moving at 3m/s at it's very bottom edge. Does the angular velocity of the roller (now with stuck clay) increase, decrease or stay the same? (The picture should clarify)
I think it increases because at the moment of the collision there is a torque on the roller, but the moment of inertia also increase so I am not sure. Thanks in advance for any help!
| Angular momentum is conserved in an inelastic collision, so all you have to do is calculate the total angular momentum of the system the moment before the collision and the moment after. Before the collision, the total angular momentum is the sum of the angular momentum of the roller and the angular momentum of the clay (as measured from the axis of the roller) right before it hits the roller. After the collision, you can once again calculate the total angular momentum of the system. The new moment of inertia is different, but it is easy to calculate. All you have to do is add the moment of inertia due to the extra spinning mass of clay. Set the final angular momentum equal to the angular momentum before the collision, and you have your answer.
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Why is the morse potential shifted for two different electron energy levels? I saw this image from the Wikipedia article on the Franck-Condon principle:
But couldn't find an explanation for the shift $q_{01}$ to the right (i.e. increasing the inter-nuclear distance) as the electron energy level goes from $E_0$ to $E_1$. As far as I understand it, the absorbed energy in an electron will push it to an outer-more electron shell, but is this the sole reason for the shift towards right in this graph? Is it simply, the further binding electrons are to the nucleus, the further apart the nuclei binds?
And if so, does this trend continue for each subsequent energy level, $E_2$, $E_3$, etc. or are there other factors at play here?
| There is no reason why equipotential curves should have their minima lined up at the same internuclear distance or why the shapes would be similar.
As an example, when in a single-electron picture an electron is excited from a bonding orbital to an antibonding orbital, if there still is a minimum, it would be at a larger internuclear distance. Or when an electron is excited from a $\sigma$ orbital to a $\pi$ orbital.
I see no obvious reason why excited states could not have a shorter equilibrium distance, for example when an electron is excited from an antibonding orbital to a bonding orbital.
In general, it is all more complicated because of electron correlation.
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What would this inclined force equate to? In the attached problem, i'm curious what effect, if any, the incline would have on the torque generated by the 50 kips force (shown below). Would it lead to a smaller/larger torque and, if so, why? I think the incline wouldn't influence torque necessarily, since the angle between the applied force and the lever arm is 90 degrees, but i'm not 100% sure.
I appreciate any help or guidance anyone can offer on this matter.
| The incline should not affect the torque. The torque of the 50 kips force (F) should be T = hxF, where h is distance AC. To calculate force of 60 kips force (F1) you will need to project the force vector outward so that it can form a line that is perpendicular from A to that point (supposing A is the pivot) if not is there more info for the question?.
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Applying the principle of Occam's Razor to Quantum Mechanics Wolfgang Demtröder writes this in his book on Experimental Physics,
The future destiny of a microparticle is no longer completely determined by its past. First of all, we only know its initial state (location and momentum) within limits set by the uncertainty relations. Furthermore, the final state of the system shows (even for accurate initial conditions) a probability distribution around a value predicted by classical physics.
If the quantum probabilistic distribution always lie near the classical prediction, why do we need quantum mechanics in the first place? According to the Feynman interpretation, if an electron has to go from A to B, it can take all the paths but the weight is more on the path predicted by classical mechanics. We know that it is unlikely that the electron travel through the mars to go from A to B on earth. Then, is not that path through mars is unnecessary? Should not in the spirit of Occam's razor, we exclude such thing in a theory?
| Further to the points discussed, the relevant principle when comparing theories of different empirical success isn't Occam's razor, but the correspondence principle. Originally that refers to quantum mechanics recovering classical mechanics in a certain limit, but more generally it means a new theory is accepted when it explains both the successes and failures of an old theory, so it's consistent with all observations and reduces to the old theory in its domain of validity. (For an illustration of when the principle isn't honoured, as a hallmark of pseudoscience, see this.)
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What's the debate about Newton's bucket argument? I visited some other QA threads about this topic, and I don't understand why people think it's mysterious that the bucket knows about its rotation.
If a non-rotating bucket is all there is in the universe, then, initially, all the parts of the bucket are at rest wrt to each other.
But if we want to rotate that bucket with an angular velocity $\omega$, then we basically require the different parts of it to have relative acceleration wrt each other. Because if we divide the bottom of the bucket into many concentric rings, then each ring would've an acceleration $\omega^2 r$ towards the center, depending on the radius $r$ of ring. This means that the rings have relative acceleration wrt to each other. Laws of physics would take different forms for people standing on different rings. Hence, a rotating bucket is a collection of non-inertial frames having relative acceleration.
But non-inertial frames are supposed to detect acceleration in Newtonian physics. So what am I missing?
| In Newtonian mechanics (and also relativity and quantum mechanics), a hypothetical physicists sitting in the bucket would definitely be able to do an experiment to detect that the bucket is rotating. I'm not sure why that would be mysterious. It should be noted that the velocities of the particles involved (relative to one another) are a fundamental part of the system. You cannot describe a physical situation using only the mass and position of the particles. You need to include their relative velocities. For this reason, a bucket sitting alone in an empty universe is fundamentally different from a rotating bucket sitting alone in an empty universe.
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What does the Problem 14 from Goldstein's book on classical mechanics chapter-7 (special relativity) really mean? I am having difficulty in understanding problem number 14 in Goldstein's Classical Mechanics, 3rd edition, chapter 7 on special relativity. Here is the problem ---
A rocket of length $l_0$ in its rest system is moving with constant
speed along the $z$ axis of an inertial system. An observer at the
origin of this system observes the apparent length of the rocket at
any time by noting the $z$ coordinates that can be seen for the head and
tail of the rocket. How does this apparent length vary as the rocket
moves from the extreme left of the observer w the extreme right? How
do these results compare with measurements in the rest frame of the
observer? (Note: observe, not measure).
How does this differ from the usual length contraction? What is the meaning of the hint given by asking the reader to "observe" not "measure", what is the difference here?
| I'd like to add to what 'PM 2Ring' wrote. The observer will measure the rocket to have a constant length no matter where it is in the observer's frame of reference (assuming it is moving at a constant velocity -- in which case it will be length contracted).
However, the observer will observe the rocket to be longer when it moves towards him and shorter when it moves away from him. This has nothing to do with relativity, just with the fact that there is a path difference between the light coming from each end of the rocket, which can make the rocket appear longer/shorter when it is moving at very high speeds. It may be a bit hard to visualise at first, make two diagrams of the rocket, separated by a small unit of time (in which case the rocket will have moved of course), and compare light pulses from the nose and tail.
It's a nuance of terminology, just have in mind that some people take measure to mean something different to observe. The difference should be explained whenever it makes a difference, which it clearly wasn't in the question.
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Topological Insulator - why does a band have to be isolated to have a Chern number of 1? I'm trying to understand the principle of topological insulator.
Why a band has to be isolated to have a Chern number of 1? More precisely, why, in the case of Haldane Model, all the bands in the valence band don't participate with a Chern number of 1 as in the case of IntQHE?
| There's an intuitive answer and a mathematical answer. Mathematically, the Chern number is defined as $\frac{1}{2\pi i}\int\epsilon^{ij} \langle\partial_i u|\partial_j u\rangle$. Thus, it requires you to have $|u\rangle$ be a locally differentiable function of $k$. But if a band touches another band, $|u\rangle$ can change abruptly across the band touching point. Intuitively, you know that the quantum Hall effect occurs exactly when you have zero longitudinal conductivity, so it only happens to insulators. Thus the bands can't touch.
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Why don't hovercrafts move West relative to the Earth Suppose that there is a hovercraft floating a few centimetres above the Earth's surface. As it is disconnected from the Earth, which is spinning from West to East, shouldn't it appear to move East to West to observers on the ground? Does this happen? If not, why not?
| Compare the case of a plumb line. Anywhere on the planet a plumb line will hang perpendicular to local level surface. This is true both on a planet that is rotating, such as our Earth, and on a planet that is not rotating.
The Earth rotates at a constant angular velocity. For any object that is co-rotating with the Earth the speed is constant. Maintaining this constant speed does not require a force, only acceleration requires a force. Therefore a plumb line hangs perpendicular to the local level surface.
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How does the reversibility of physics interact with nuclear fission? The laws of physics are reversible and quantum information is never destroyed.
Given this, how do I time reverse the $U_{235}$ fission reaction, n which
${}^1_0n + {}^{235}_{92}U \rightarrow {}^{141}_{56}Ba + {}^{92}_{36}Kr + 3 {}^1_0n + \gamma +$ 202.5 MeV (in kinetic energy plus gamma ray energy).
That is, would reversibility require that if we bring together the $Ba$, $Kr$, $3ns$, and the $\gamma$ all at the same instant, a $U$-$235$ and a neutron might just pop out with a certain probability?
(EDIT: Here is a fun video from PBS Space Time explaining why quantum information can never be created or destroyed.)
| The reverse reaction really is
$${}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\ {}^1_0n + \gamma + 202.5\ \mathrm{MeV} \rightarrow {}^1_0n + {}^{235}_{92}\mathrm{U} $$
as you'd expect. Yes, your deduction from time symmetry is absolutely correct (as far as we know) - it has to be. The above will do the job. Yes, the four elements on the left will combine, fuse to form a uranium-235 nucleus and it will pop out a neutron just as going in the forward direction causes all those other things to pop out. No magic; it's entirely possible.
And yes, you're right to suspect it's not easy to make it happen. You'd have to shoot all 4 of those things (think about like 4 particle guns all aiming into one point) into the same tiny little area of space (remember that nuclei are measured in fermis [fm], $10^{-15}\ \mathrm{m}$, and your Uranium nucleus may be on the order of 10 fm, as a proton is about 1 fm), and with just the right energy - if you shoot with too much, you might end up "breaking" the nucleus you are trying to create: per nucleon, the binding energy of $^{235}_{92}\mathrm{U}$ is around 7 or 8 MeV, so if you put in maybe 8 MeV more than that 202.5, you stand a chance to end up with something not 235U, because you had enough energy to kick at least one more nucleon (in addition to the neutron that inevitably pops out) out of it.
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Why do peaks and troughs of a wave cancel each other out? And why peaks and peaks or troughs and troughs add up? I have thought that the cancellation of peaks and troughs is a consequence of Newton's third law of motion that equal and opposite forces cancel each other out.
Or it has something to do with conservation of energy or momentum.But I have never truly understood it correctly.
I believe there is an obvious and clear explanation that I am simply not aware of.
In other words, how does interference work?
| Mathematically it can be seen as a consequence of waves obeying a linear second order differential equation.
Physically, for mechanical waves, it’s the fact that forces are additive that result in the addition (superposition) of waves. This means that if two forces are acting at a point, then the net force that acts is the resultant force.
Similarly for EM waves, the fields themselves are additive in nature and the net outcome is the resultant of all the fields at that point.
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Why doesn't water boil in the oven? I put a pot of water in the oven at $\mathrm{500^\circ F}$ ($\mathrm{260^\circ C}$ , $\mathrm{533 K}$). Over time most of the water evaporated away but it never boiled. Why doesn't it boil?
| It might also be prudent to consider the environment of the oven itself. The "atmosphere" in the oven is already at a temperature > 100°C and this means that water in the "air" is in the gaseous state. As water evaporates at the surface at temperatures nominally > 40°C this would mean that the water vapour is "immediately" absorped into the gaseous state (immediately in the sense that the energy transfer occurs far fast because of the molecular energy) and the remaining water is nominally cooling the surface of the vessel via thermal convection. I believe this is thermo-dynamics in action :)
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What is the torque produced by 2 rotating bodies with a clutch I am trying to simulate a car engine etc, but I have failed to find any equations governing the torque created by $2$ different constant velocity shafts of different angular momenta joining together with some given slip or friction factor. I know $I_1w1 + I_2w_2 = I_3w_3$ which gives me the end angular velocities, but its not giving me the torque acting on each shaft at any moment in time whilst they are joining.
| In a system with two drive shafts connected by a clutch and a system of gears, the power being transmitted by each shaft will be the torque times the angular velocity. (The angular momentum of each shaft is not of concern.) The ratio of the two angular velocities is determined by the system of gears. The clutch allows the second shaft to start at rest and be brought up to speed. In bringing the second shaft up to speed, some energy will be lost to friction in the clutch, and some will be needed to increase the kinetic energy of the second shaft and the gears. After that, (if you have friction-less bearings and gears) all of the power supplied to the first shaft should emerge from the second shaft.
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Why does bottle with long straw get emptied first? Two identical bottles with different straw length are filled with identical liquids (obviously up to the same height as depicted in the picture). Bottle with long straw is emptied first.What can be it's possible reason?
| In a fluid like water or air, the pressure $P$ and velocity $v$ depend on the height $h$ and density $\rho$ in a way described as the Bernoulli equation,
$$
P + \frac12 \rho v^2 + \rho g h = \text{constant}
$$
In your setup, the opening of the straw and the upper liquid surface at the base of the bottle are both open to the air, whose pressure doesn't really change over the size of the apparatus (that is, $\rho_\text{air} g \Delta h \ll \rho_\text{water} g \Delta h$). If the straw is narrow enough that you can neglect the motion of the fluid at the top surface, $v_\text{top}\ll v_\text{bottom}$, this gives
$$
\frac12 \rho v_\text{bottom}^2 = \rho g (h_\text{top} - h_\text{bottom})
$$
The draining time is proportional to the mass flow rate $v$, so you can drain a container about twice as fast if you quadruple the length of the vertical drainpipe underneath it.
(Modulo the usual assumptions about how turbulence is hard, approximations are approximate, et cetera. This is an engineer's approach.)
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Why do we describe probability amplitude rather than probability itself in quantum mechanics? In the quantum mechanics, the dynamics of quantum system are described in terms of probability amplitude. However, we want to calculate the probability in the end which can be measured. Why don't we develop quantum mechanics directly describing the probability instead of probability amplitude? Wouldn't this make the quantum mechanics more interpretable and simple?
| To make sense of interference fringes without probability amplitudes, you would need a messy ad-hoc theoretical framework like the Bohmian pilot wave. Probability amplitudes are used simply because it explains interference in a simple way
$$P_{A+B} = (\langle \Psi_A | + \langle \Psi_B | )(| \Psi_A \rangle + | \Psi_B \rangle) = \langle \Psi_A | \Psi_A \rangle + \langle \Psi_B | \Psi_B \rangle + 2 \langle \Psi_A | \Psi_B \rangle $$
If the $| \Psi_A \rangle$ are eigenstates of energy expressible as $e^{-i\omega_A t}| \phi(x)_A \rangle$, then the above expression becomes:
$$ \langle \phi_A(x) | \phi_A(x) \rangle + \langle \phi_B(x) | \phi_B(x) \rangle + 2 \cos((\omega_A - \omega_B)t) \langle \phi_A(x) | \phi_B(x) \rangle = {P_A + P_B + 2 \cos((\omega_A - \omega_B)t) \langle \phi_A(x) | \phi_B(x) \rangle}, $$
where $P_A$ and $P_B$ are the separated probabilities of each state. Even if you were able to express everything without amplitudes, you would probably need a lot of mathematical contortions to get that last interference term. In fact it is possible since Bohmian pilot wave theory exists and it is equivalent to Quantum Mechanics, but it is not straightforward for calculation purposes
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Why exactly do we feel a shock when we place our hand into a conducting solution? I have a very naive question.
Suppose you have pure water in a flask, and you place two ends of a copper wire (which are connected to a battery) into the water.
If you were to place your hand into the water, you would not feel any shock, as pure water does not conduct electricity.
However, if you add an electrolyte like common salt to the same water, you would probably feel a shock.
Adding salt makes the solution conducting. When the two wires are placed in the solution, the ions are attracted to the end of the wire which has an opposite charge.
However, what does the movement of those ions have to do with whether or not your hand feels a shock? Shouldn't whether you feel a shock just depend on what resistance your hand offers?
| If I may ask, why do you feel a twinge when you step on a nail or a pin pricks you? The same way your nerves make you feel the painful sensation of stepping on a nail is the same way that they react when current flows through your hand. It's not that current flows through your body but how your nervous system reacts to the current flow. When your hand goes across a salt solution connected to a battery, electrons zip through the solution unto your hands and your body makes you react to it by giving you the sensation of a shock.
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Oscillations of Dielectric Slab in Parallel plate capacitor In the book Principles of Physics by Resnick Halliday:
The decrease in Potential energy of a parallel plate capacitor due to a dielectric is because the slab would start to oscillate and the energy would transfer back and forth between the kinetic energy of the moving slab and potential energy stored in the electric field.
First of, can somebody provide me the math of the SHM and could somebody explain why it would become the potential of the capacitor i.e. the electric field and not the potential of the slab?
As the slab is doing SHM, shouldn't the slab(and not the capacitor) have potential energy at one point of time and kinetic at another like a spring? If the SHM is because of the electric field due to the conducting plates in capacitors, won't they always be in one direction (Plus, it's magnitude is not even a function of distance!)?
Could someone be kind enough to provide the math and the reason for Potential and Kinetic Energy transfer question?
| The point is that there are two ways of insertion that is with constant potential difference maintained where it doesn't depend on x so it is periodic not shm but in 2nd case it will perform shm as it is dependent on x and directly proportional
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Does it make sense to say that something is almost infinite? If yes, then why? I remember hearing someone say "almost infinite" in this YouTube video. At 1:23, he says that "almost infinite" pieces of vertical lines are placed along $X$ length.
As someone who hasn't studied very much math, "almost infinite" sounds like nonsense. Either something ends or it doesn't, there really isn't a spectrum of unending-ness.
Why not infinite?
| In physics if a quantity, call it $\lambda$, in a theory was said to be "almost infinite", I would interpret this as stating the effective theory obtained by taking the limit $\lambda \to \infty$ is accurate up until some very long length scale or time scale after which it breaks down.
Crucially this breakdown length/time scale is much greater than the intrinsic scales of the effective theory (at least in some useful regimes), so there is a very small approximation error induced by using the $\lambda \to \infty$ effective theory on its own intrinsic timescales.
I think a lot of the answers here have missed the key point that $\lambda$ is only meaningfully close to $\infty$, if the theory's predictions are close to those of the $\lambda \to \infty$ effective theory.
Obvious examples are
*
*the the speed of light $c$ in classical mechanics
*the inverse planck constant $\hbar^{-1}$ in general relativity
*the Heisenberg time in many body quantum physics
*the stiffness of a billiard ball when playing pool/snooker/billiards
More boringly I would just note that "almost infinite" is just the reciprocal of "almost zero".
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Why is the electric dipole moment parallel/antiparallel to angular momentum? In several textbooks and papers, like this one for example, a claim like the following is made:
The EDM of a system $\vec{d}$ must be parallel (or antiparallel) to the average angular momentum of the system $\hbar\langle\vec{J}\rangle$.
What is the motivation for making such an assertion? It seems not to make much sense if you consider, for example, two opposite electric charges held apart by a short, rigid massless rod (the usual classical picture of an electric dipole). The system can have zero angular momentum, but still has a nonzero electric dipole moment.
| They are probably talking about the quantum mechanical expectation value, which has to be along the angular momentum vector. If the angular momentum is zero, the expectation value of any vector would be zero.
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Yo yo motion and energy conservation I am struggling with balancing energy in a yo-yo. So, we have a yo-yo (massless string wrapped around solid cylinder). It is allowed to fall through a distance $h$ without rotation. The loss in potential energy $= mgh$ will be converted into kinetic energy (KE) and we can find the KE after it has fallen through a distance $h$.
Now, we allow the yo-yo to fall like a yo-yo so that it moves the same distance, but also rotates. So, again, we apply the conservation of energy. Now, the KE of the yo-yo will have 2 parts (i.e. due to the rotation around the center of mass and due to the translation of the center of mass). We find that the final KE in this case is 1.5 times than that of the previous case.
My question is, how come the 2nd case has more final energy when the initial energy in both the cases was $mgh$?
| Obviously, the yo-yo's energy has to be conserved. The potential energy of the yo-yo at the stationary starting point where it starts to move down ($mgh$, $h$ being the length of the rope) is converted in a linear and a rotational kinetic energy part (as you wrote). The linear acceleration downwards though is less than in free fall. This means that the yo-yo's linear kinetic energy when it's rolled off completely is less than $mgh$. The difference with $mgh$ is contained in the yo-yo's rotational energy.
Because you took the kinetic energy to be the same in both cases (free fall vs constrained fall), you took $g$ as the downward acceleration in both cases, which isn't the case.
In fact, the equipartition theorem implies the potential energy is distributed in equal amounts of kinetic energy among the two degrees of freedom of the motion (vertical linear motion and the rotation around one axis).
All this means the KE in the first case (constrained motion with rotation) is equal to the KE in the second case (free fall without rotation) (the kinetic energy you get in the second case becomes half that value in the first case, i.e. $\frac{1}{4} m{v_{cm}}^2$).
So the only mistake you made is assuming the linear kinetic energy at the end of the roll is equal in both cases.
| {
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What is the Topologically Twisted Index? I know that one can take a supersymmetric theory defined on $\mathbb{R}^n$ and topologically twist it by redefining the rotation group of the theory into a mixture of the (spacetime) rotation group and the R-symmetry group. However, what I'm a bit confused about is: what is a topologically twisted index? What does it physically mean? I can't seem to find a definition anywhere. Most papers seem to assume that the reader already knows the definition.
| It would be great that you sharpen your question by asking for a specific case.
But the general intuition is as follows: In a non-twisted theory, an index compute the dimension (possibly the virtual dimension) of the space of solutions of some differential equations that preserve some amount of supersymmetry.
Twisted theories dosen't have propagating degrees of freedom and the path integral (when a lagrangian description is avaliable) of those theories typically localizes to contributions coming from instantons. A twisted index does exactly what a "non-twisted" idex does because they typically come from twisting operators from non-twisted theories. They are indices for twisted theories.
But is important to specify a particular case because twisted indices can be defined for non-twisted theories by composing them with some projector that localizes the space of solutions of some equations to a subset of them that are non-dynamical and preserve some supersymmetry or transform well under some symmetry.
The most general definition that can be stated is that a topologically twisted index for a non-twisted theory is the composition of an index with a possible twist projection operator for the theory (when a twist is avaliable in the theory). What is interesting of this generality is that it makes manifest the fact that a non-twisted theory actually receives non-perturbative contributions from all its possible twists.
In the case of a twisted theory, a topologically twisted theory is just any index.
| {
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One-loop Feynman integral over Euclidean momenta I am trying to perform the following one-loop computation
$$
\int \frac{d^Dq}{(2\pi)^D} \frac{(k+q)^2 q^2}{((k+q)^2+m^2)(q^2+m^2)}
$$
where $k$ is fixed and everything is on the Euclidean setting, so there is no need to perform any Wick rotation.
I can not find the solution anywhere and I am not realizing how to do it by myself. Any help will be appreciated!
| This question is a couple years old, but I ran across it in a search, and it has a fairly simple answer. The trick for dealing with the numerator is
$$\int \frac{d^Dq}{(2\pi)^D} \frac{(k+q)^2 q^2}{((k+q)^2+m^2)(q^2+m^2)}=\int \frac{d^Dq}{(2\pi)^D} \frac{((k+q)^2+m^2-m^2)(q^2+m^2-m^2)}{((k+q)^2+m^2)(q^2+m^2)}$$
$$=\int \frac{d^Dq}{(2\pi)^D} \left[1-2m^2 \frac{1}{q^2+m^2}+m^4 \frac{1}{((k+q)^2+m^2)(q^2+m^2)}\right]$$
The first term in brackets vanishes in dim reg. The second and third terms are standard one loop integrals.
| {
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Describing forces in rolling Consider a wheel on a frictionless horizontal surface. If we apply a horizontal force (parallel to the surface and above the level of the center of mass), what happens to the wheel? Does it roll or slide forward or rotate only or does any other phenomenon happen? Please guide me. Also draw a free body diagram.
Note: This is a thought experiment. If the question is not satisfying, I am sorry for that and please guide me.
| It should roll. Whether it will be rolling with sliding or pure rolling we don't know until we know what height the force was applied at.
Breaking the motion into Translation and Rotation, we can write one force and torque equation each for both respectively as such-
$F=ma$
$ \tau=I\alpha=rF$
We can also calculate the condition on the height at which you must apply the force for pure rolling by equalling the net acceleration of the bottom most contact point to $0$. This point will have two accelerations, one from rotation ($=R\alpha$) and one from translation ($=a$). Notice their directions are opposite, so for net acceleration of zero they must be equal. So, we have
$F=ma$
$a=\frac{F}{m}$
$I\alpha=rF$
$\alpha=\frac{rF}{I}$
For pure rolling, $a=R\alpha$
Substituting and rearranging, we have
$r=\frac{I}{mR}$
| {
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What is happening when magnetic field lines snap or break? In discussions of sun spots and auroras on Earth, magnetic field lines are often described as "snapping" or "breaking", with the result of releasing charged particles very energetically.
My understanding is that field lines are just a visualization tool. I don't understand, intuitively, how a field line could snap or break, or why that would result in a release of energy.
I'm having trouble even framing this question because the concept of a field line breaking just doesn't make sense to me. What is happening when a magnetic field "snaps"?
| I simulated two bar magnets with 4 dipoles each and plotted the field lines and field intensity around them.
Here is what I observed.
I noticed that the dipole chain forms a set of null spots in the field on either side of the chain. As the magnets are pulled apart, two of these null spots (green dots) move away from the point of separation, and this is where the field lines "snap". What is actually happening is that the field lines reform, changing from the field lines surrounding the combined bar magnet to the field lines surrounding the two separate bar magnets.
The field lines "snap" as they pass through the null spot. They are not really snapping, but as the field strength vanishes at the null spot, each field line can reform smoothly into two new field lines surrounding the two separate magnets. Field lines are drawn to follow the field direction, but they do not show field intensity, so they appear to snap when they pass through the null spots.
As material moving along a field line encounters a null spot, it will be free of the field, and if the surrounding field is too weak to recapture it, this material will escape into space.
| {
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Why use virtual displacement to make constraint forces vanish? Why do we use virtual displacement to vanish work done by constraint forces instead of the actual displacement?
| H. Goldstein, Classical mechanics, Chapter $1$ says
"Note that if a particle is constrained to a surface that is itself moving in time, the force of constraint is instantaneously perpendicular to the surface and the work during a virtual displacement is still zero even though the work during an actual displacement in the time $dt$ does not necessarily vanish."
I think that this is just an additive advantage of taking virtual displacement instead of actual displacement and that the main reason leads to the linear independence of the equation
$$\sum_j \left\{\left[\frac{d}{dt}\left(\frac{\partial T}{\partial \dot q_j}\right)-\frac{\partial T}{\partial q_j}\right]-Q_j\right\}\delta q_j=0\tag{1.52}$$
which leads to the Lagrange's equations.
I used actual displacement $$dr_i=\sum_j \frac{\partial r_i}{\partial q_j}dq_j+\frac{\partial r_i}{\partial t}dt$$ instead of virtual displacement $$\delta r_j=\sum_j \frac{\partial r_i}{\partial q_j}\delta q_j\tag{1.47}$$ in the whole derivation of the Lagrange's equations and got an extra term depending on $\displaystyle\frac{\partial r_i}{\partial t}$ on the LHS of the equation $(1.52)$ which did not let me apply the logic of linear independence.
That was two years ago. I can't find that notebook and I am too lazy to do it again. You try it yourself. Comment below, should you face any issues.
PS: Again, this is not from a verified source.
| {
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Instantons in 1+1 dimensional Abelian Higgs model Let's consider the Abelian Higgs model in 1+1 dimensions in Euclidean space-time:
$$L_E=\frac{1}{4e^2}F_{\mu\nu}F_{\mu\nu}+D_\mu\phi^\dagger D_\mu\phi+ \frac{e^2}{4}(|\phi|^2-\zeta)^2$$
where $\zeta>0$ and $D_\mu\phi= (\partial_\mu-iA_\mu)\phi$.
We are looking for finite action field configurations, i.e instantons.
Because in this case $\phi : S^1_{phys}\rightarrow S^1_{gauge}$ and being $\mathbf{\pi}_1(S^1)=Z$,
I expect that for an instanton configuration the action is bounded by its topological number (winding number).
More precisely, the finiteness of the action comes from the following behaviour for large $r$:
$$ 1) \quad\phi(r,\theta)=\sqrt{\zeta}\,g(\theta)$$
$$ 2) \quad A\mu=g\partial_\mu g^{-1}+O(1/r^2)$$
where $g\equiv g(\theta)$ is an element of U(1).
Therefore I would define the winding number $k$ as the one of pure gauge:
$$w=\frac{i}{4\pi}\int d^2x\epsilon_{\mu\nu}F_{\mu\nu}$$
So I would expect, up to numerical factors, something like: $S\ge \zeta w $, where I have inserted $\zeta$ for dimensional reason.
I am interested in demonstrating that inequality and in solutions saturating the bound. Any reference on this topic?
| Advanced Topics in QFT (Shifman) talks about U(1) anomaly in 1+1, Aspects of Symmetry (Coleman) have topics on Higgs 1+1, and probably something about it in David Tong lectures.
| {
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Wick theorem exercise I'm a newbie in QFT and I have some doubts with this simple exercise:
Using the Wick Theorem evaluate
$$\langle0|T(\phi^4(x)\phi^4(y)|0\rangle$$
*
*Use a diagrammatic approach to represent the possible contractions
(how many $\phi$-lines are attached to each vertex).
*Determing numerical factor in front of each diagrams.
| First of all the Wick's theorem states
$$
T(A B C \ldots Y Z)=:\{A B C \ldots Y Z \,+"\text { all contractions" }\}:
$$
(In the case of fermions we have to take care about anticommutation relations,
i.e. every time when we interchange neighboring fermionic operators a minus sign appears.)
In this exercise you shall apply Wick's theorem for bosons.
It is clear that all normal-ordered terms fall off, because their vacuum expectation value is equal to zero. Thus the only remaining terms are those with four contractions.
If you contract one $\phi(x)$ with one $\phi(y)$ four times you shall get $(\langle 0|T(\phi(x) \phi(y))| 0\rangle)^{4} .$ This can be done in $4 !=24$ ways. The next possibility is to make two contractions between fields $\phi(x)$ and $\phi(y) .$ One field $\phi(x)$ can be contracted in 4 ways with one of the $\phi(y)^{\prime}$ s. The next $\phi(x)$ can be contracted in three ways with one of the remaining $\phi(y)^{\prime} \mathrm{s} \cdot$ The obtained result has to be multiplied by $6,$ because this is the number of ways in which two fields $\phi(x)$ can be chosen from the four possible. Thus, there are $4 \cdot 3 \cdot 6=72$ possible contractions of this type. There are three mutual contractions between two fields $\phi(x),$ the similar is obtained for fields $\phi(y),$ so the corresponding coefficient is 9. Thus,
\begin{equation}
\begin{aligned}
\left\langle 0\left|T\left(\phi^{4}(x) \phi^{4}(y)\right)\right| 0\right\rangle &= 24 \,\langle 0|T(\phi(x) \phi(y))| 0\rangle^{4}\\\\
&+72\,\langle 0|T(\phi(x) \phi(x))| 0\rangle\langle 0|T(\phi(y) \phi(y))| 0\rangle(\langle 0|T(\phi(x) \phi(y))| 0\rangle)^{2} \\\\
&+9\,(\langle 0|T(\phi(x) \phi(x))| 0\rangle)^{2}(\langle 0|T(\phi(y) \phi(y))| 0\rangle)^{2} \\\\
&=24\,\left(\mathrm{i} \Delta_{\mathrm{F}}(x-y)\right)^{4}+72\left(\mathrm{i} \Delta_{\mathrm{F}}(x-x)\right) \mathrm{i} \Delta_{\mathrm{F}}(y-y)\left(\mathrm{i} \Delta_{\mathrm{F}}(x-y)\right)^{2} \\\\
&+9\,\left(\mathrm{i} \Delta_{\mathrm{F}}(x-x)\right)^{2}\left(\mathrm{i} \Delta_{\mathrm{F}}(y-y)\right)^{2}
\end{aligned}
\end{equation}
where $\Delta_{\mathrm{F}}(x-y)$ is the Feynman propagator.
The last expression can of course represented by Feynman diagrams.
Now, go through these steps and depict the last expression as diagrams. Good luck and if you have some trouble with the computation/drawing feel free to ask for details.
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Is it possible for two observers to observe different wavefunctions for one electron? Suppose there are 2 scientists who have decided to measure the location of an electron at a same fixed time. Is possible that while one observes the wavepacket localized at (position=x) while the other observes the wavepacket localized at (position=y). The condition however is position x is not equal to y.Please dont confuse about the degree of localization which can be quite varying depending upon which measurement-momentum or position is given priority
:( I have little to no experience with quantum superposition and gaussian wavepackets.......kindly manage with my rough knowledge
| The OP specified that the two observers make their measurements simultaneously. Two simultaneous measurements of the same observable are equivalent to a single measurement, which can have only one result. Therefore, the observers must see the same result.
| {
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Why does the entanglement of quantum fields depend on their distance? When watching Seans Carrol's "A Brief History of Quantum Mechanics", he mentioned around the 50th minute (the video I linked to starts at that point) that
[about quantum fields in vacuum] ... and guess what! The closer they are to each other, the more entangled they are.
Why is it so? I was under the impression that entanglement is not dependent on the distance (two entangled particles getting further from each other are not less entangled).
If this is at all possible I would be grateful for an answer understandable by an arts major - just kidding a bit, I simply would like to avoid an answer which starts with
courtesy of Redorbit
| I expect that Carrol is referring to cluster decomposition, a principle satisfied by many quantum field theories. This principle says that if two quantities are located in spacelike-separated regions very far from one another, then they are going to be uncorrelated. That is, if operators $A$ and $B$ are localized in two spacelike-separated regions a distance $\ell$ apart, we have $\langle A B \rangle \to \langle A \rangle\langle B \rangle$ as $\ell \to \infty$. We can usually give a more detailed statement of how quickly the correlation dies off. For example, in a theory of a field with mass $m$ the bound on the difference $\lvert\langle A B \rangle - \langle A \rangle\langle B \rangle\rvert$ goes like $e^{-m\ell}$, and in a massless theory the difference goes to zero faster than $\ell^{-N}$ for any $N$.
| {
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Cold neutrinos - how are they distributed? Cold or slow neutrinos have non-relativistic velocities and hence very low energies. That makes them very difficult to detect. Answers to Where are all the slow neutrinos? make it clear that they are vastly abundant. Do we have any significant model of how they are distributed and what proportion of the Universal mass/energy they make up?
I do know that current models of cold dark matter discount cold neutrinos as a significant constituent because the ones we know about were not cold when the CDM shaped the galaxies. But in the absence of any strong candidate for CDM, how can we be so sure that there are not even more which we do not yet know about?
| We do have a model which gives the proportion of energy density the neutrinos occupy in the universe, the standard cosmological model or $\Lambda$CDM model. Neutrino number density is given by integrating the Fermi-Dirac distribution over the momentum space:
$$
n_\nu = n_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{e^{p/T_\nu}+1} = \frac{3T_\nu^4}{2\pi^2}n_s\zeta(3),
$$
where $n_s$ is the number of neutrino species (six in the standard model: three flavours and their antiparticles), $\zeta$ is the Riemann zeta function and $T_\nu$ is the neutrino temperature. It is about 1.95 K right now.
Assuming the cosmic neutrinos are nonrelativistic today, their energy density is
$$
\rho_\nu = m_\nu n_\nu.
$$
Latest PLANCK measurements imply the fraction of neutrino energy density of the total energy density of the universe to be
$$
\Omega_\nu \equiv \frac{\rho_\nu}{\rho_\text{total}} \approx \frac{\sum m_\nu}{94h^2 \text{ eV}} \lesssim 0.0025
$$
where $h \equiv H_0$/(100 km/s/Mpc) $\approx 0.68$ is the reduced Hubble constant (dimensionless). Therefore the neutrinos can account at most about 0.25 % of the energy density of the universe.
As for the distribution of the slow neutrinos - one would expect them to be homogeneously and isotropically distributed to high accuracy. Some small anisotropies would be unavoidable though due to gravitational clustering and quantum fluctuations.
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Invariance of Lagrangian under rotations in a constant magnetic field
The Lagrangian for the motion of a particle with mass $m$ and charge $q$ in a constant magnetic field $B$ is given by $$\mathcal{L}(x,v)=\frac{m}{2}\left|v\right|^2-\frac{q}{2c}\left(v\cdot[x\times B]\right).$$
Show that rotations around the $B$-axis leave the Lagrangian invariant, where each rotation is given by $O_{\eta}:=\exp(\eta\,[B\,\times \,.]),\,\eta\in\mathbb{R}$.
I can see that $\left|O_{\eta}(v)\right|^2=\left|v\right|^2$, since rotations are supposed to leave the "length" unchanged but that's about as far as I've gotten with this. I'm guessing that one needs to apply some certain identities here regarding the cross product and the $\exp$ function, which I haven't been able to find on Wikipedia or other websites so far.
| The conservation of the second term seems intuitive, as simply a preservation of volume, formed by vectors $v, x, B$. Nevertheless, I also present a way to deduce this for infinitesimal transformation, which will imply for a finite rotation. Under a small rotation with $\eta \ll 1$, the variation of vector $a$ is $\delta a = \eta [a, B]$. Then:
$$
\delta (v \cdot [x \times B]) = ( \delta v \cdot [x \times B]) + (v \cdot [\delta x \times B]) = \eta [v \times B] \cdot [x \times B] + \eta \ v \cdot [[x \times B] \times B] =
$$
$$
= \eta (v \cdot x \ B \cdot B - v \cdot B \ x \cdot B) +
\eta (v \cdot (-x (B \cdot B) + B (B \cdot x))) = 0
$$
Here we have used following identitites:
$$
(a \times b) \times c = - (c \cdot b) a + (c \cdot a) b \qquad
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) \equiv
(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})
$$
| {
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Why do reflection gratings work? The law of reflection (that the angles of incidence and reflection are equal) can be derived directly from Maxwell's equations, or from Fermat's principle. However, reflection gratings completely defy this law, and from light incident at a fixed angle comes a whole diffraction pattern - light reflected at a range of different angles.
My teachers and textbooks tend to gloss over this point. Why do reflection gratings work, and why is there not a contradiction with the law of reflection?
| A reflection grating is not a mirror. It is an array of reflective grooves in a surface. Light reflected from the bottoms of the grooves is delayed relative to light reflected from the tops of the grooves, just as light transmitted through optically thick portions of a transmission phase grating is delayed relative to light transmitted through the optically thin portions.
| {
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Is a capacitor a dipole? A few more questions about understanding dipoles I recently learned about dipoles, according to its definition I was wondering if a capacitor can be considered also as a dipole?
Also I was wondering what is the physical meaning of the dipole moment $\vec{p}=qd$?
And my last question is what is the motivation of studying dipoles? What is so special about it? From what I have learned it is just 2 equal and opposite charges that are a distance $d$ apart and they create an electric field according to what expected.
| Since we believe that the universe is electrically neutral, any separation of positive and negative charges can be considered as a combination of electric dipoles.
| {
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Why can't primordial fluctuations be super-"horizon" without inflation? I am trying to understand why, in cosmology, it is said that the presence of fluctuations at scales above the Hubble distance would not be expected in the absence of inflation or something like it.
We treat density fluctuations using the density contrast $\delta = (\rho - \bar{\rho})/\bar{\rho}$ and then we Fourier analyse. Let's write $\delta({\bf k}, t)$ for the amplitude of a Fourier component. We then have a distance scale $2\pi/k$ associated with this Fourier component, and also a distance scale $c/H$ (the Hubble distance) associated with the rate of expansion. Also, the particle horizon would be of a similar order of magnitude to that if it were calculated on the assumption that ordinary GR applies all the way down to $a=0$ (a highly questionable assumption of course). Anyway the main point is that it is asserted that one would not expect fluctuations $\delta({\bf k},t)$ for $2\pi/k$ larger than the separation between points that have had no causal contact. At least that is what I understand to assertion to be.
But I think that if independent random processes gave rise to independent fluctuations in separate spatial regions, then when one Fourier analysed the outcome one can get $\delta({\bf k},t) \ne 0$ for values of $k$ of any size. The independent random processes just happen to do this.
But perhaps the assertion is something more like, "if we treat the fluctuations on all scales as independent random processes, then the result is not compatible with what is observed". But is that true?
| You can have field fluctuations on any scale, but in a non-inflationary spacetime they just return to the vacuum. You need inflation to convert the fluctuation to a classical curvature perturbation.
| {
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How can we conclude from Maxwell's wave equation that the speed of light is the same regardless of the state of motion of the observers? I am reading a book titled "Relativity Demystified --- A self-teaching guide by David McMahon".
He explains the derivation of electromagnetic wave equation.
$$
\nabla^2 \, \begin{cases}\vec{E}\\\vec{B}\end{cases} =\mu_0\epsilon_0\,\frac{\partial^2}{\partial t^2}\,\begin{cases}\vec{E}\\\vec{B}\end{cases}
$$
He then compares it with
$$
\nabla^2 \, f =\frac{1}{v^2}\,\frac{\partial^2 f}{\partial t^2}
$$
and finally find
$$
v=\frac{1}{\sqrt{\mu_0\epsilon_0}}=c
$$
where $c$ is nothing more than the speed of light.
The key insight to gain from this derivation is that electromagnetic waves (light) always travel at one and the same speed in vacuum. It does not matter who you are or what your state of motion is, this is the speed you are going to find.
Now it is my confusion. The nabla operator $\nabla$ is defined with respect to a certain coordinate system, for example, $(x,y,z)$. So the result $v=c$ must be the speed with respect to $(x,y,z)$ coordinate system. If another observer attached to $(x',y',z')$ moving uniformly with respect to $(x,y,z)$ then there must be a transformation that relates both coordinate systems. As a result, they must observe different speed of light.
Questions
Let's put aside the null result of Michelson and Morley experiments because they came several decades after Maxwell discovered his electromagnetic wave derivation.
I don't know the history of whether Maxwell also concluded that the speed of light is invariant under inertial frame of reference. If yes, then which part of his derivation was used to base this conclusion?
| If Maxwell's equations have the same form in all frames of reference, then the wave speed is defined by the product of two physical constants, irrespective of coordinate system. i.e. Your book just implicitly assumes that, but of course it requires experimental testing - i.e. Michelson-Morley etc.
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Do atoms absorb the same amount of light? I'm currently working on a project on my own where I'm interested in finding information about an object based on a spectrum. Namely, I want to use the spectrum that I input into my program to be able to analyze what atoms are present in the analyzed object. (I know this is probably hard but it's a fun project). However, when I started to work on this my question arose:
Do atoms that are exposed to the same amount of light absorb the same amount as well? (Albeit different frequencies).
So, when the atoms are exposed to light (uniform over the EM spectrum), will two atoms that absorb different frequencies absorb the same amount of light? And if so, one could infer that the less light of a specific frequency that we can find, (the less compared to the maximum that would be emitted at that frequency) the more there is of the element that absorbs this specific frequency? (Though it would probably be useful to look at more than one "black line" in the spectrum)
|
Do atoms that are exposed to the same amount of light absorb the same
amount as well?
The question is a little fuzzy, so let's try an be more concise.
Take sodium ($\text{Na}$), famous for its strong, yellow doublet emission line.
In identical conditions all sodium atoms emit the same intensity of that yellow doublet (and its other weaker lines)
A constant gas flame of a given sodium concentration would thus emit yellow light, the intensity depending only on the sodium concentration. Similarly, such a flame would also absorb the same amount of light made up of those frequencies (the principle of Atomic Absorption Spectrometry), depending only on sodium concentration.
| {
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Baryon number vs electromagnetic charge, what is the difference? What exactly is a Baryon number? I looked up definition from wikipedia and still struggle to understand this. And how does this differ than the
electromagnetic charge?
My textbook did the following computation:
It is calculating the electromagnetic charge right and not the Baryon number?
| A baryon is any particle held together by the strong force (i.e. a type of hadron) that comprises three quarks. An antibaryon has three antiquarks.
The baryon number $B$ is just the sum of all the quarks $n_q$ minus the sum of all the anti-quarks $n_{\bar q}$ :
$$ B = \frac{1}{3}(n_q-n_{\bar q}).$$
So, a quark has baryon number $B = 1/3$, an antiquark $B = -1/3$, baryons have $B = +1$, antibaryons have $B=-1$, mesons have $B=0$ and so on.
A proton, in your example, has $3$ quarks and hence has a $B=1$. Just $1$, a natural number. No units.
Its quarks are $u$, $u$, and $d$, so if we perform the separate and independent from the above sum of the electric charge we get:
$$ q = q_u + q_u + q_d = 2e/3 + 2e/3 -1e/3 = 1e.$$
So the total electromagnetic charge of a proton is $q = 1e = 1.60 \times 10^{-19}$ C.
The baryon number of a proton is $1$.
| {
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In metals, the conductivity decreases with increasing temperature? I am currently studying Principles of Optics: Electromagnetic Theory of Propagation, Interference and Diffraction of Light, 7th edition, by Max Born and Emil Wolf. Chapter 1.1.2 Material equations says the following:
Metals are very good conductors, but there are other classes of good conducting materials such as ionic solutions in liquids and also in solids. In metals the conductivity decreases with increasing temperature. However, in other classes of materials, known as semiconductors (e.g. germanium), conductivity increases with temperature over a wide range.
An increasing temperature means that, on average, there is greater mobility of the atoms that constitute the metal. And since conductivity is due to the movement of electrons in the material, shouldn't this mean that conductivity increases as temperature increases?
| The characteristic feature of metals is that the valence electrons of the atoms delocalise across the crystal lattice- this is an intrinsically quantum mechanical phenomenon. In essence, the electrons propagate as plane waves and this delocalisation lowers the energy of the electrons. As you increase the temperature, very few electrons are promoted to higher energy plane-wave states, the effect of the lattice vibrations also become quite important. As you increase the temperature of the crystals, you can excite quantised vibrational modes of the crystal lattice- these are called phonons. Electrons can scatter off phonons, resulting in decreased mobility. Since the phonons carry momentum (actually crystal momentum), collisions with electrons will alter the momentum of the electron. This effect reduces the average number of electrons with momentum directed in the direction of an applied electric field, thus decreasing the conductivity.
| {
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Why Ohm’s law doesn’t work in these scenarios (inside ideal battery and in vacuum)? Scenario 1 - Ideal battery
Suppose I have an ideal battery whose electrolyte’s resistance is zero. In the working battery there will be current flowing inside the battery also (due to battery forces) from lower potential to higher potential. Now I choose two points inside the battery, one on the positive terminal and other on the negative terminal now there is current between these two points so I apply Ohm’s law $V=IR$. Now $V$ is non zero(since my points lie on the terminals) and $R$ is zero, which is not in accordance with Ohm’s law.
Scenario 2 - Current in vacuum
Suppose in vacuum there are two pints which are at different potentials and let there be a beam of electrons travelling from the point which is at lower potential to the other. Now since it’s vacuum $R=0$ but $V$ is non zero which is again not in accordance with Ohm’s law.
I may be wrong somewhere in my understanding of Ohm’s law since I’m a new learner. Please correct me so that I can understand it well enough.
| Despite the name, Ohm’s law is not a general law of nature. It is instead a defining characteristic of a small class of materials and devices called resistors (and conductors). Ohm’s law does not apply to other materials and devices, including insulators, capacitors, inductors, switches, transistors, vacuum, voltage sources, current sources, dielectrics, semiconductors, and many others. All of these devices and materials violate Ohm’s law.
All of that is to say that you are correct that both batteries and vacuum violate Ohm’s law.
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Capacitance of ferroelectric capacitor in saturated regime Consider a ferroelectric plate capacitor connected to an AC source in the presence of a strong static external electric field which sets the ferroelectric medium in the saturated regime.
The question:
Does the static polarisation $P$ of the ferroelectric have an influence on the capacitance $C$ of the capacitor?
Two contradicting attempts:
*
*Considering the electrical energy of the system, there is an influence:
The electrical energy $\mathcal{E}$ is given by $\delta\mathcal{E}=E\delta D$ with $E$ the electric field and $D$ the electric displacement which results in $\mathcal{E}=\epsilon_0E^2/2 + \epsilon_0EP$ for the saturated ferroelctric. Electrical energy is also given by $\mathcal{E}=QV$, and thus the capacitance becomes
$$C=\frac{Q}{V} = \frac{\mathcal{E}}{V^2} = \frac{\epsilon_0E^2/2 + \epsilon_0EP}{V^2}\,.$$
Hence, according to this equation, there is an influence of the ferroelectric medium.
*Considering the classical capacitance equation, there is no influence of the ferroelectric medium: In the saturation regime of the ferroelectric, the relative permittivity \epsilon_r is 1. So, according to
$$C=\epsilon_0 \epsilon_r \frac{A}{d} = \epsilon_0 \frac{A}{d}$$
there is no influence of the ferroelectric medium...
Which approach is the correct one?
| Your second approach is correct.
There are a few problems with the first one. First of all, $Q=CV$ no longer applies, because when you have no charge on the capacitor, there is still a non-zero E-field within the ferroelectric, and thus a non-zero potential difference across the capacitor plates. The appropriate definition of the capacitance would be
$$C = \frac{dQ}{dV}.$$
Apart from the fact that you seem to be mixing up electrical energy and electrical energy density, the former is not $QV$, this is not even true in an ordinary dielectric capacitor ($\mathcal{E}=\frac{1}{2}QV$). In general, energy density must be found from the relation $$\delta u=E\delta D,$$ or energy from $$\delta\mathcal{E} = V\delta{q}.$$
With energy density $u$ taken to be zero when $E=0$,
$$u=\int\limits_0^EE'\frac{d}{dE'}(\epsilon_0E'+P)\ dE'=\frac{1}{2}\epsilon_0E^2$$
$$\mathcal{E}=Adu=\frac{1}{2}\epsilon_0AdE^2$$
$$V=Ed$$
$$C = \frac{dQ}{d\mathcal{E}}\frac{d\mathcal{E}}{dE}/\frac{dV}{dE}=\frac{1}{V}(\epsilon_0AdE)/(d)=\frac{\epsilon_0 A}{d}. $$
This is obviously not the most direct way to calculate the capacitance but it follows how your first approach attempts to do it.
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How is the frequency of a wave defined if it propagates on three different directions? Let's consider a wave which propagates on 2 or three directions, like for instance an electromagnetic wave inside a rectangular waveguide totally closed on two ideal conductor surfaces:
The walls of the guide force the wave to assume an integer number of half-wavelenghts along x,y,z:
$$l_{x,y,z} = m_{x,y,z} \cdot \frac{\lambda}{2}$$, with m integer.
When we indicate a certain mode, such as $TM{2,1,1}$ we mean that there are 2 half-wavelength along x, 1 along y and 1 along z. Suppose now $$l_{x,y,z} = l$$ (i.e. all dimensions are equal: the waveguide is a cube).
Obviously lambda will be different for x,y,z:
$$\lambda_x = \frac{2l}{m_x}=l$$
$$\lambda_y = \frac{2l}{m_y}=2$$
$$\lambda_z = \frac{2l}{m_z}=l$$
So, three different wavelenghts. What does it mean? In physics I have always studied that frequency corresponds to wavelength, if the propagation medium is fixed. What is the definition of frequency in this case?
| The frequency is, as always, the number of cycles per second of the oscillations. It is related to the spatial wavelength by $f = \frac{c}{2 \pi} |\vec{k}|$, where $c$ is the speed of propagation of free waves in the medium and
$$
\vec{k} = (k_x, k_y, k_z) = \left( \frac{2 \pi}{\lambda_x}, \frac{2 \pi}{\lambda_y}, \frac{2 \pi}{\lambda_z} \right).
$$
and so
$$
|\vec{k}| = 2 \pi \sqrt{ \lambda_x^{-2} + \lambda_y^{-2} + \lambda_z^{-2}}.
$$
There are an infinite number of possible $\lambda_i$ for each direction, depending on the dimension of the box in that direction and the number of nodes & anti-nodes. Each one can, in principle, give rise to a different frequency.
Note that this relationship between frequency and wavelength is exactly the same as it would be for a free wave with the same wave vector $\vec{k}$. This is because a standing wave solution, such as a wave in a waveguide, can always be expressed as a sum of traveling waves that just happen to interfere at the boundaries of the waveguide. In the 3D case you need to have a sum of waves whose $\vec{k}$ vectors are of the form $\left( \pm \frac{2 \pi}{\lambda_x}, \pm \frac{2 \pi}{\lambda_y}, \pm \frac{2 \pi}{\lambda_z} \right)$; but all eight possibilities have the same magnitude $|\vec{k}|$ and so have the same frequency.
| {
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Pseudo Force and Inertial and Non-Inertial frames In the figure given below is block placed on an incline $\theta$. Now the lift is accelerating upwards with an acceleration $a_0$. Now if we make our measurements from the lift frame we will have to apply a pseudo force $-ma_0$. Which will have two components one in the direction of $Mg\cos\theta$. And other in the direction of $Mg\sin\theta$. Now $Mg\sin\theta+Ma_0\sin\theta=Ma_\text{net}$. Where $a_\text{net}$ is the net acceleration in that direction.
Now let's observe it from the ground or an inertial frame here the object has a net upward acceleration which has a component opposite to $Mg\sin\theta$. Therefore $Mg\sin\theta=- Ma_0\sin\theta$. Now what I thought was that this is not possible and hence there is another force acting opposite to $Mg\sin\theta$, $Ma_\text{net}$. Now this doesn't make any sense to me if there is a force acting opposite to $Mg\sin\theta$, and the net is also in that direction, then won't the object move upwards on the incline. Now that doesn't make any sense. Can someone tell me what Is happening and from where is this $a_\text{net}$ coming from when observing in the inertial frame?
| The acceleration of the mass, a, in the inertial frame is the sum of the acceleration of the elevator, $a_o$ and the acceleration of the mass relative to the incline, a', in the elevator. To avoid using the normal force, I'll chose the +x axis parallel to and up the incline. Then for x components: -mg sin(θ) = m$a_x$ = m($a_o$ sin(θ) + a') giving a' = -(g + $a_o$) sin(θ).
| {
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What do $\ell$ and $A$ precisely mean in the formula for electrical resistance? The formula for resistance is
$$R=\rho\frac{\ell}{A}$$
Generally in most of the textbooks it simply written that $\ell$ is the length of the conductor and $A$ is it’s cross-sectional area. But my question is which length and area do we need to consider as a 3D body has many possible lengths and cross sectional areas. Textbooks simply take an example of a solid cuboid whose opposite faces are supplied with potential difference. But what if I change the faces across which potential difference is applied (for example if I choose two adjacent faces of same cuboid) or I change the shape of the conductor itself (for example a solid sphere whose two faces (across whom potential difference is applied) are opposite semi-hemispherical surfaces.
I’m a beginner in electromagnetism and needs a lot of new learning. So please help.
| In the formula the area(A) is perpendicular to the flow current,
The length (l) is along the flow of current.
Consider an example that will clear you doubt.
Consider a hollow cylinder with inner radius 'a' and outer radius 'b'
and length 'l' .
Case 1- Potential difference is applied along the length 'l' of cylinder.
Here current flows along the length (l) and area perpendicular to it is $$π (b^2-a^2)$$
$$R = \frac{pl}{π(b^2-a^2)}$$
Case 2- Potential is applied across the inner part and outer part of cylinder
Here the the current flows from inner part to outer part of cylinder.
The area perpendicular to current flow is different for different distance from the centre of cylinder. Therefore it will require integration.
Consider a cylinder of radius $\pmb x$ from the centre of hollow cylinder, its AREA=$\pmb {2\pi xl}$( this is perpendicular to current flow)
Consider a width $\pmb {dx}$ along $\pmb x $, this will be along the flow of current hence this will be the length of small elemental part considered .
Now consider infinite such cylinders from $\pmb a \ to \ \pmb b$ each of length $\pmb {dx} $ .
All these cylinders will be in series.
Hence
$$R = \int_a^b dR = \int_a^b\frac{p dx }{2πxl} =\frac{p ln \frac ba}{2πl}$$
Hope it clears your doubt , try using this concept for finding resistance of a cuboid along different edge lengths .
As to your second question - it can be done similarly by considering that the potential difference is applied across diametrically opposite ends of sphere,
The area perpendicular to current can be taken as circular plate having width $\pmb {dr}$,
and then integrating along the diametric length. I leave it upto you to try the integration for this.
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Does tau decay to rho meson + tau neutrino? In the particle data group table: http://pdg.lbl.gov/2020/tables/rpp2020-sum-leptons.pdf ,only $\tau \rightarrow \pi + \nu_{\tau}$ is documented. But does $\tau \rightarrow \rho + \nu_{\tau}$ as well? I don't see any conservation laws that could forbid such decay to happen.
| The $\rho$ has the quantum numbers of $\pi\pi$ coupled to $I=1$. If they only list particles stable under the strong interaction then $\tau^-\to\rho^-\nu$ is the same as $\tau^-\to\pi^-\pi^0\nu$, which indeed has a branching ratio of 25%. The table is more helpful than that: It lists the non-resonant (non $\rho$) part of $\tau\to \pi^-\pi^0\nu$, which is very small. So the branching ratio is about 25%.
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Why is internal resistance of battery considered outside the terminals although it is present between the terminals inside the battery In ideal battery the internal resistance is zero whereas in non-ideal battery there is some internal resistance now this internal resistance is due to the battery material (electrolyte) and is present inside the battery between the terminals then why do we represent and eventually do calculations by considering that internal resistance to be connected with battery terminals externally. I’m totally unable to get the point. Please help
|
Why does one represent the internal resistance of a non-ideal battery as external?
It is just a way of representing a non-ideal battery. Wherever the battery goes, the resistance goes with it, then why worry if one is showing it inside or outside. It will be considered in the Kirchhoff's voltage law anyhow.
An uncanny failure of this representation would be applying loop law to a circuit consisting of a resistor between the electrodes of a battery because the law wouldn't include the internal resistance (that is shown as outside the terminals) then. This can be tackled with extra care to consider the internal resistance in the law expression or simply changing the representation.
| {
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Finding acceleration of center of mass in cart pole problem In this link about finding equations of motion of cart pole problem,
There is an equation about acceleration of center of mass of the pole.
Screenshots of them below.
I don't understand why they have more than two parts about angular acceleration - $\varepsilon \times r_p$ and $\omega \times (\omega \times r_p)$?
If I'm being right, first one is torque, and second one is acceleration of a point in circular movement.
I guess in some part I'm being incorrect, but I don't understand why they put two angular acceleration of it? It's copy of them, aren't they?
| That last term is the centripetal acceleration associated with the rotation of the rod.
| {
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Work Equals Torque? Horsepower, Pulleys While reading one definition of torque, I saw its units are Newton-meter, which is the same as work. But sources usually make it a point to emphasize "even though both work and torque units are the same, they should not be confused, they are very different". One is like an object being pushed with force certain distance, the other force applied to a wrench etc. at certain length, applied around an axis of rotation.
But if we think of the pulley seen below,
Isn't radius and distance related here? If I rotate a wrench of length $r$ with force $F$ from top position to 90 degree position, isn't the same thing as pushing an object with force $F$ at a distance of $r$?
| This is a little bit of a pet peeve of mine, so I’ll chime in with my two cents / rant.
Torque and work don’t actually share the same units, torque actually has units of $\frac{\mathrm{N\,m}}{\mathrm{rad}}$. Dimensionally speaking, the distinction is irrelevant because radians are dimensionless, however most formulas for rotational dynamics fail if you are measuring angles in degrees, rather than radians. Basically measuring torque in $\mathrm{N\,m}$ is akin to measuring angular velocity in $\mathrm{s}^{-1}$ rather than $\frac{\mathrm{rad}}{\mathrm{s}}$.
You already have an example in the answer above: work being $W = \tau\theta$ only works if $\theta$ is in radians, meaning $\tau$ itself is “intrinsically” linked to radians. Another, more long-winded, example: $U = \frac{1}{2} I \omega^2$, in order to get the correct units for energy without “dropping” radians, you would need $I$ to be measured in $\frac{\mathrm{kg\,m^2}}{\mathrm{rad}^2}$. Then you have $\tau = I \alpha$, where you can see only one of the radians cancels out, giving the aforementioned $\frac{\mathrm{N\,m}}{\mathrm{rad}}$.
How do we get this to “match” with the way torque and moment of inertia are defined in terms of force, mass and position vectors? It can be taken as a result of the presence of cross products in their definition. The usual definition is $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ which can be re-written as $\boldsymbol{\tau} = A_\mathbf{r} \mathbf{F}$, using the cross product matrix of $\mathbf{r}$. $$A_\mathbf{r} = \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix}$$
But, you might say, there is no cross products in the definition of $I$. Well, the moment of inertia is actually a tensor quantity and, in tensor terms, is given by $I = m A_\mathbb{r} {A_\mathbb{r}}^\mathsf{T}$, two occurrences of our cross product matrix, meaning two occurrences of radians.
The same is true for angular momentum, which should have units of $\frac{\mathrm{J\,s}}{\mathrm{rad}}$.
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How do balances (not scales) work? More specifically, why does having two objects of different weights make the balance lean to one or the other then stop partway? Why does the balance not just keep tilting until it falls to the side? How can something be heavy enough to tip the balance but not heavy enough to make it tilt the rest of the way?
| Balances work because they only need to tell when things are in balance, not how far out of balance they are. As such, the only behaviors needed are those which are very close to equilibrium.
When one is that close to equilibrium, small things matter. For example, as you tip, you roll the fulcrum ever so slightly to one side (the more knife edged the fulcrum is, the more sensitive it is). That roll does shorten the lever arm of the heavier mass and length the length of the lever arm on the lighter mass. Its a small amount, but when you get very close to equal weights (or, more precisely, equal torques), it's a stabilizing effect that results in the balance settling on a position (rather than moving freely as one might expect if one assumed the fulcrum was an unmoving point)
Naturally, if one has two weights which are within 1/100th of each other's mass, a shift of only 1/100th in the length of the lever arms is enough to bring torques into equilibrium. The balance will settle on the angle which yields that 1/100th of the length shift.
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Newtonian Limit of Schwarzschild metric The Schwarzschild metric describes the gravity of a spherically symmetric mass $M$ in spherical coordinates:
$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2+r^2 \,d\Omega^2 \tag{1}$$
Naively, I would expect the classical Newtonian limit to be $\frac{2GM}{c^2r}\ll1$ (Wikipedia seems to agree), which yields
$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1+\frac{2GM}{c^2r}\right)dr^2+r^2 \,d\Omega^2 \tag{2}$$
However, the correct "Newtonian limit" as can be found for example in Carroll's Lectures, eq.(6.29), is
$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1+\frac{2GM}{c^2r}\right)\left(dr^2+r^2 \,d\Omega^2\right) \tag{3}$$
Question: Why is the first procedure of obtaining the Newtonian limit from the Schwarzschild solution incorrect?
| Consider the change of coordinate
$$r=r'\left(1+{{\cal G}m\over 2r'c^2}\right)^2={r'}^2(1+U)^2$$
where $U={\cal G}m/2r'c^2$. One can check that
$$1-{2{\cal G}m\over rc^2}=1-{2{\cal G}m\over r'c^2(1+U)^2}
={(1-U)^2\over (1+U)^2}$$
Moreover,
$${dr\over dr'}={d\over dr'}\left[r'\left(1+{{\cal G}m\over 2r'c^2}
\right)^2\right]=(1-U)(1+U)$$ so that
$$dr=(1-U)(1+U)dr'$$
The Schwarzschild metric becomes
$$\eqalign{
&ds^2=\!c^2\!\left(1-{2{\cal G}m\over rc^2}\right)dt^2
-\left(1-{2{\cal G}m\over rc^2}\right)^{-1}dr^2
+r^2d\theta^2+r^2\sin^2\theta d\varphi^2 \cr
&=c^2\left(1-{2{\cal G}m\over rc^2}\right)dt^2
-{(1+U)^2\over (1-U)^2}(1-U)^2(1+U)^2dr'^2
-{r'}^2(1+U)^4\left[d\theta^2\!+\!\sin^2\theta d\varphi^2\right]\cr
&=c^2\left(1-{2{\cal G}m\over rc^2}\right)dt^2-(1+U)^4
\left[d{r'}^2+{r'}^2d\theta^2\!+\!{r'}^2\sin^2\theta
d\varphi^2\right] \cr
&=c^2\left(1-{2{\cal G}m\over {r'}c^2}+{\cal O}(U^2)\right)dt^2
-\big(1+4U+{\cal O}(U^2)\big)\left[d{r'}^2+{r'}^2d\theta^2\!
+\!{r'}^2\sin^2\theta d\varphi^2\right] \cr
&=c^2\left(1-{2{\cal G}m\over {r'}c^2}\right)dt^2
-\left(1+{2{\cal G}m\over {r'}c^2}\right)\left[d{r'}^2
+{r'}^2d\theta^2\!+\!{r'}^2\sin^2\theta d\varphi^2\right] \cr
}$$
as expected.
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In what sense are spin coherent states "classical"? Spin coherent states are often introduced as "the most classical states of a finite-dimensional system", or as the analogous of coherent states of light for finite-dimensional systems. See e.g. (Radcliffe 1971) and (Chryssomalakos et al. 2017).
One way to define them (using a notation similar to Radcliffe 1971) is as the states
$$\lvert\mu\rangle=N^{-1/2}\exp(\mu S_-)\lvert S\rangle,$$
where $S_z\lvert S\rangle=S\lvert S\rangle$, $S_-\equiv S_x- i S_y$, and $N$ is a normalisation constant.
While the formal analogy between these states and coherent states of light (a.k.a. Glauber states), $$\lvert\alpha\rangle=\exp(\alpha a^\dagger - \alpha^* a)\lvert0\rangle=e^{-\lvert\alpha\rvert^2/2}\exp(\alpha a^\dagger)\lvert0\rangle,$$
is clear, what I don't find too clear from the references above is why these states should be regarded as "the most classical states", as is stated e.g. in the abstract of (Chryssomalakos 2017).
In the optical case, we justify calling $\lvert\alpha\rangle$ classical observing e.g. that it gives Poissonian photon-counting statistics, and that it cannot produce entangled states using only linear operations.
Is there any similar physical justification in the case of spin coherent states?
| *
*They saturate the "displaced" uncertainty relation. If
\begin{align}
\vert \Omega\rangle=R(\Omega)\vert jj\rangle
\end{align}
is the coherent state, and
\begin{align}
J_k^\prime=R(\Omega)J_kR^{-1}(\Omega)
\end{align}
for any rotation $R(\Omega)$, then
\begin{align}
\Delta J_x^\prime \Delta J_y^\prime=\frac{1}{2}\vert\langle \Omega \vert J_z^\prime\vert\Omega\rangle\vert \, .
\end{align}
*Their Wigner function is localized on the sphere, as for instance the WF of this coherent state with $J=9$ rotated about $\hat y$ by $\beta=2\pi/9$:
*The time-evolution of a coherent state under a Hamiltonian which is linear in the generator of $SU(2)$ is just a continuous rotation on the sphere, without the WF changing its shape.
| {
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Are there any quantum effects which we can see in every day life? I am wondering if there are any natural phenomenon in every-day life that cannot be explained by classical physics but can only be explained by quantum mechanics. By classical physics, I mean Newtonian mechanics and Maxwell's electromagnetic theory.
I know that there are macro-scale quantum phenomena such as superconductivity, but that isn't something that we can see in ordinary life.
| In addition to the other answers, it is also true for ferromagnets, the strong magnetism is explained by the means of exchange interaction. With the machinery of QM one explains the hysteresis and formation of magnetic domains. The weak magnetism, paramagnetisim and diamagnetism can be explained on a classical level.
To add more, the energetical structure of levels in crystals and semiconductors is due to collective quantum phenomena.
| {
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In the generation of X-Rays, why the incoming electron generated from anode knocks out the K shell electron rather than outer shell electrons? When the high energy beam of particles or photon hits the cathode, electrons from $K$ shell are knocked in the generation of characteristic x-rays. Why do inner electrons get knocked out?
| Interactions of electrons with atoms needs a quantum mechanical formulation, classical electrodymics, where bremsstrahlung and synchrotron radiation of X-ray energies can be modeled is not enough to model X-rays from atoms. As Farcher shows in another answer the quantized nature of the energy levels of the electrons have to be taken into account.
In a rough way, an electron has a probability of scattering off the electron orbitals of an atom by exchanging a virtual photon , losing momentum and energy in ejecting the electron. Then that level is filled again giving off the characteristic spectrum of the energy level the bound electron had occupied.
Why do inner electrons get knocked out?
All have a quantum mechanical probability to be knocked out but it is only the inner energy levels that have enough energy to be in the X-ray range.
The same if it is a photon that ejects the K shell electron, it has to have enough energy to do that ( in a type of compton scattering) so that the deexcitation gives off the characteristic -ray.
| {
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Derivation of non-linear Schrödinger equation from many-body QM I hope this (and not MathOverflow) is the right place to post this question. I am a math student taking a methods of mathematical physics course, in which we cover the solution theory the non-linear Schrödinger equation $$i u_t+ \frac{\Delta}{2}u=\lambda|u|^{p-1}u.$$ I understand that this PDE (or some special case of it) may be derived as some mean-field limit from many-body quantum mechanics. How is this done?
| The derivation is in the paper by Pitaevskii posted in the comment above.
| {
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Why are there no operators in classical mechanics? I have been wondering to find the answer of some fundamental questions in quantum mechanics and the answer to the above question will help me to clear doubts of quantum world
| Operators do exist in classical physics. The thing is, we don't use them in the same way.
In mechanics, for example, compute the eigenvalues and eigenvectors of an operator is important to study vibrational modes. You also can find examples of diagonalization of operators in electrodynamics and heat propagation.
Diagonalization of differential operators, Green Functions, and other elements of QFT lore already exist in classical physics.
The interpretation of the spectra of the operators in QM is different, often they are possible states, but the difference lives in the physics, not in the mathematical structure.
| {
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Impulse operator on real wave function The impulse operator in quantum mechanics is given by
\begin{align}
\hat{p} = \frac{\hbar}{i}\nabla
\end{align}
As a Hermitian operator, the expected value of this operator $\langle{p}\rangle = \langle \psi|\hat{p}\psi\rangle$ should be real. However, for a real wave function $\psi(\vec
{r})\in \mathbb{R}$ (a valid solution to the Schrödinger equation) the resulting integral is imaginary:
\begin{align}
\langle{p}\rangle = \frac{\hbar}{i}\int d^3r \cdot \psi \nabla \psi
\end{align}
Is there an error in my thinking or is it impossible to calculate the expected value that way? An alternative approach would be to use the Fourier transform.
| Just to add to Vadim's answer: The integral
$$\int_{-\infty}^{\infty} \psi \partial_x \psi dx= \frac12 \int_{-\infty}^{\infty} \partial_x( \psi^2) dx = [\psi^2]_{-\infty}^{\infty}=0
$$
fo all wavefunctions that vanish at infinity.
| {
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Measuring the Hubble constant in a curved universe In an article from the University of Chicago, July 17, 2020, it is stated that
"Judging cosmic distances from Earth is hard. So instead, scientists measure the angle in the sky between two distant objects, with Earth and the two objects forming a cosmic triangle. If scientists also know the physical separation between those objects, they can use high school geometry to estimate the distance of the objects from Earth."
That seems straightforward, except for the fact that high school geometry only works in flat space where the angles enclosed by a triangle add up to precisely 180 degrees. In a curved universe, a triangle can enclose either more or less than 180 degrees. Unless the curvature is known, triangulation shouldn't work reliably in a curved space.
So my question is: in measurements of the Hubble Constant by the triangulation method, what assumptions are made about curvature of the universe? And, how well-founded are those assumptions?
| I guess you are looking for angular diamater distance. For different curvature the equation takes different forms. See here https://en.wikipedia.org/wiki/Angular_diameter_distance
| {
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If a pendulum of mass $m$ and length $l$ is considered quantum mechanically, what will be the approximate ground state energy? I have seen that the solution of the quantum pendulum is obtained by solving the Mathieu's equation form of the Schrodinger equation and it also depends on a parameter 'q' called the energy barrier.
But can we approximate the energy levels of the quantum pendulum to that of the quantum harmonic oscillator? Under what limit can we do this?
| My best guess would be $\dfrac{h\nu}{2}$. Note that it requires quantum gravity to check my answer .
| {
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Optics: mirrors While drawing ray diagrams for plane and spherical mirrors, what is generally taken as the point of observation?
Eg, If a concave mirror is presented with, say a wire turned into a triangle, placed from focus towards the mirror, the image can be obtained following rules of reflection but where would the eyes need to be for that particular image to be seen? Like if I were to move around, I'd see different parts of an object. What angle is the images drawn from?
| In ray tracing, typically light rays leave a point of an object, go through an optical system, and pass through a point of an image.
Sometimes a detector is placed at the image. Different points of the detector see different points of the image. The outcome may be different pixels of a photograph.
In other cases, the light rays keep going past the image. This might be the case for a microsocpe or telescope with eyepieces instead of a camera. You put your eye in the path of the outgoing light. Your eye is a second optical system.
The image point is where light originates from the point of view of your eye. It is the object. Your lens creates an image on your retina.
This just means that you need to do more ray tracing to find the answer. Perhaps make a bigger system that include the mirror and your eye.
| {
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Is this a kinematics paradox? You consider a shaft which can rotate freely (there will be of course a frame with ball bearing to hold the shaft firm and to allow it to rotate with low friction); fixed on this shaft there are two gears, one with radius $R$ and another with radius $2R$. Note that the two gears rotate together with the shaft since they are welded on it. Now, there is a second identical shaft which is put close to the first one, in order the teeth of the gears can touch each other, the picture should clear the setup:
You suppose you drive the first shaft with an angular speed $\omega$, while the second shaft isn't driven directly. Then the speed of the teeth of the gear with radius $R$ is $v_{a}= \omega R$, the one of the gear with radius $2R$ is $v_{b}= 2\omega R$. It follows that the speed of the teeth of the smallest gear on the second shaft is $v_{b}= 2\omega R = \omega_{2} R$ then $\rightarrow \omega_{2}=2 \omega$; the speed of the teeth of the biggest gear on the second shaft is $v_{a}= \omega R=2 \omega_{2} R$, then $\rightarrow \omega_{2}=0.5 \omega$. The only solution is the trivial one, then $\omega=0$. This means that if you try for example with a crank to rotate the first shaft, it won't move at all. This last fact seems a bit counterintuitive. I don't have the chance to build a model of this setup to verify if the shaft actually can't rotate. Someone has an opinion about it?
PS:
I've written about gears (with teeth) but it's the same if you consider discs with an high friction layer on the edge (in order the edge of the discs don't slip one on the other).
| By examining the gears on the left we can see that for every revolution of 1, 2 must undergo two revolutions. By examining the gears on the right we can see that for every revolution of 1, 2 must undergo half a revolution. This is not possible, so the gears will not turn.
| {
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Maximum velocity of elastic pendulum released from rest Consider the following problem.
Construct a pendulum with a mass $m$ at the end. However, we don't a rigid rod of length $\ell$, we instead use a spring of natural length $\ell$ and spring constant $k$. We lift the mass so that the spring is horizontal and at its natural length (i.e. $\theta = \pi/2$ in the picture). We then release the mass, letting the pendulum swing. How low is the mass at the lowest point of the swing?
What I have so far:
We can find second-order equations of motion, either using Lagrangian mechanics or re-writing $F=ma$ in terms of its angular and radial components. But just like the regular pendulum, these equations are impossible to solve. I need some kind of invariance to solve this problem.
If we use conservation of energy, we get
$$\frac{1}{2}m(\ell + r)^2 \dot\theta^2 + \frac{1}{2}m\dot r^2 + \frac{1}{2}kr^2 - mg(\ell + r)\cos \theta = 0$$
at any time in the swing.
Since the height of the mass is $-(\ell + r)\cos \theta$, at the lowest point of the swing we have
$$ 0 = \frac{d}{dt} ((\ell + r)\cos \theta)$$
or rather,
$$\dot \theta \tan\theta = \frac{\dot r}{\ell + r}$$
And here I'm stuck. The above isn't enough to find the lowest part of the swing, so I need some other invariant. Looking at momentum seems to do no good; the force applied by the spring varies with time. Looking at angular momentum is similarly difficult.
For context: This is problem 76 of chapter 15 in "Physics for Scientists and Engineers" by Jewett and Serway, 8th edition. I looked at this problem some time ago, and it's nagged me ever since.
Also, in general, the elastic pendulum is chaotic, so any solution should make specific use of the given starting conditions.
| You can construct a free-body diagram for the pendulum bob at the lowest point of its motion and balance out the forces in the vertical direction to obtain yet another equation.
The weight of the bob subtracted from the force on the bob due to the extension of the spring should provide the necessary centripetal force for the bob to sustain its motion.
| {
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Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges on them repel each other with a force $F$ Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges on them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is
My question is: Is there any difference between in the answers of conductor and point charge?
| The answers will differ, because the charge distribution on each sphere will not be uniform. Effectively, each sphere will have an induced dipole moment due to the presence of the other, which will change the force between them.
This effect of the charge distribution, however, will generally be smaller than the basic effect from treating the spheres as monopoles (point charges). To within an order of magnitude, these corrections to the force will be smaller than the “point charge” force by the ratio of the spheres’ radii $r$ to their separation $D$. In particular, if the size of the spheres is negligible compared to their separation ($r \ll D$), then the approximation off treating the spheres as point charges will be a good one.
| {
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Will cutting sand paper with scissors make the scissors sharper or duller? This is a little question that I have been wondering when I need to cut sand paper with scissors.
Sand paper can be used to sharpen knives etc. when applied parallel with the blade surface. Also it can be used to dull sharp edges when applied nonparallel with the blade surface.
My assumption is that it should dull the scissors since paper is being cut using the sharp edge and nonparallel with the abrasive material. But I still have doubts about the validity of the assumption.
How is it?
| Sand paper removes material. When used properly, that removal of material can make a blade sharper. However, when cutting the sandpaper, there is no attempt to structure the removal of the material. It will simply dull the scissors. It will remove material in a relatively haphazard manner, taking off the sharp edge.
If you have any questions of this, ask someone who sews for their nice fabric scissors, and let them know you're going to go cut some paper with them. Find out how quickly they respond in an effort to avoid dulling their scissors. Perhaps its not the most scientific approach, but it is a well documented one, and very evidence based! And that's just normal paper!
| {
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Classical Wave theory and the photoelectric effect I read that according to classical wave theory, light is viewed as a wave whose intensity is continuously variable. And for this reason, it is unable to explain the photoelectric effect. My questions are:
*
*What does "continuously variable" mean?
*How does the intensity of light being continuously variable make it unable to explain the photoelectric effect?
| Measurements indicate that in the photoelectric effect, the energy of the electrons which leave the surface depends on the frequency of the incoming light, but not on the intensity of the light. This indicates that the incoming light consist of bundles (or quanta) of energy whose size is determined by the frequency. The number of electrons emitted depends on the intensity of the light (or the density of quanta.)
| {
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Is the entropy of a rotating body largest when the axis of rotation passes through it's centre of mass? I am looking for an answer to the observation that a body always rotates about its centre of mass when freely tossed. It can be explained if the entropy is highest in the case when the axis passes through the com, however, I am unable to prove it.
I am doing this to be able to visualise the motion of a body in space, when struck tangentially.
| This question is based on the false premise that an object has a unique axis of rotation. One can choose any point, whether inside, on, or outside the object, as the center of rotation. Pick some other point that is not on the resulting axis of rotation and you'll get another axis passing through that point that is parallel to the original axis of rotation.
The reason that the center of mass is oftentimes used to describe the motion of an object in space is because the Newton-Euler equations of motion become much more complex when the key point of interest is not the center of mass.
When multiple bodies are connected via joints at specific locations, it is oftentimes preferable, despite the added complexity, to use the full-blown Newton-Euler equations (or their Hamiltonian or Lagrangian equivalents) because doing so makes representations the constraints inherent to those joints feasible. The modeling of a robotic arm rarely uses a center of mass formulation.
| {
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Is energy required in generating magnetic field in simple resistance circuit? Consider a simple resistance circuit with a cell and a resistor. It is stated that energy stored in cell appears as heat in resistance as current flows in ideal circuit (neglecting EM radiation) as whole.
POWER/RATE OF HEAT GENERATION = POWER/RATE OF ENERGY CONSUMPTION in CELL = VI
However we also know that flowing current produces magnetic field.
So my questions are:
*
*Is energy needed to create magnetic field in general?
*Does the energy of cell also appears in the energy of the magnetic field?
*Is there any such thing as "energy of magnetic field"
*Any relevant information.
P.S. I am an undergrad. I do not know Special Relativity but I understand that feeling the effects of magnetic field depends on frame of reference.
| I agree with the answer @Dale provided.
To put things into perspective, the energy stored in the magnetic field of a straight conductor is minuscule compared to the energy dissipated in the resistance of the conductor.
The inductance of a straight copper wire 1 mm in dia and 10 cm long is about 105 nH. The energy stored in the magnetic field of the wire carrying 1 ampere of current is then about 57 nJ ($\frac {Li^2}{2}$). The dc resistance of the same wire is about 2.13 mΩ. The power dissipated in the wire resistance is then about 2.13 mJ per second ($i^{2}R$), or about 74,000 times more joules dissipated every second in the resistance than the total energy stored in the magnetic field.
Hope this helps.
| {
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Trying to understand a visualization of contravariant and covariant bases I was trying to intuitively understand the covariant and contravariant bases for a coordinate system and I came across this image on Wikipedia:
Edit: After reading the first two answers I think I may have not posed my question correctly so I have changed it a bit. I understand that vectors and dual vectors are vastly different objects and occupy different spaces. A proper treatment of them would have to be done solely mathematically. This image seems more of a way to visualize the tangent and cotangent spaces and allow you to visually find the covariant and contravariant components of a vector. I was wondering how to interpret this image and if it was a useful way of visualizing covariant and contravariant vectors.
Based on this image, it seems that the covariant basis vectors $\hat{e}_i$ can be visualized as vectors that point in a direction tangent to the lines of the coordinate grid, while the contravariant basis vectors $\hat{e}^i$ can be visualized as vectors that point in a direction normal to the lines of the coordinate grid.
Is this a correct interpretation of this image and what the covariant vs. contravariant basis vectors are visualized by in it? Even if it is correct if there is a better way to intuitively/graphically understand covariant and contravariant vectors and what their components mean let me know.
Next, if that interpretation is correct, then would an expression like $\dfrac{\partial \hat{e}_i}{\partial x^j}$ be interpreted as the vector displacement of one of the visualized covariant basis vectors $\hat{e}_i$ in this image if you move an infinitesimal distance along the coordinate grid lines of the image in the direction $x^j$?
| It would be more intuitive if you look at the illustrations form Daniel A. Fleisch's book on A Student's Guide to Vectors and Tensors.
Here they are:
The parallel projections represent contravariant components of a given vector $\vec{A}$ and the perpendicular projections represent the covariant components of the given vector $\vec{A}$.
| {
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Are there non-constant potentials that result in eigenstates of the Hamiltonian that are all plane waves? It is commonly known that the eigenstates to the Hamiltonian of a constant potential are plane waves, aka
$$
V(r) = V_0 \Rightarrow H\psi = n \text{ with } \psi = \exp\left(\frac{ip}{\hbar}x\right)\exp\left(-i\frac{E}{\hbar}t\right)
$$
But is there another $V(r)$ for which this holds?
I suppose the question can be simplified to: is a Hamiltonian uniquely defined by its eigenstates, because if this is the case then the answer should be no, there are no other hamiltonians with those eigenstates.
| Plane waves are eigenstates of the momentum operator. Another observable can have the same eigenstates only if it commutes with the momentum operator (see this answer for more details).
Now, the momentum operator generates translations, so the Hamiltonian commutes with it if it is invariant under translations. The kinetic part of the Hamiltonian is invariant (i.e. it commutes with the momentum operator). The potential in general not. The only potential that commutes with the momentum operator is the flat potential.
In summary: if $H$ has to have the same eigenstates as $p$, then you must have that $[H,p]=0$. This is equivalent to $[p^2/2m+V,p]=[V,p]=0,$ which tells you that $V=\mathrm{const}$.
| {
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