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Basics of centripetal force Suppose an object is moving in a circular path. We know that the net force that is working on that object is the centripetal force towards the center. But the object should have gone closer towards the center in that case due to the radially inward force working on it, but it doesn't. Why does the object remain on the circular path instead of going closer towards the center?
For people who would be introducing centrifugal force in this case, i have a doubt on this too. Centrifugal is a pseudo force that only works when we are in the frame of the rotating object meaning we experience a pseudo force that pushes us radially outward. When we are in this frame, does centripetal and centrifugal both work on us?
But let us stay in ground frame as of now. Then what is the cause of the object not being pushed radially inward due to the effect of centripetal force? I am asking this question to clear out my doubts for strengthening my basic concept of physics. Hope the physics lovers will find this question relevant.
| When going in a circle at constant speed, velocity changes. Velocity is speed and direction. Direction changing, is velocity changing. Changing velocity is acceleration. So, it takes acceleration to move in a circle. By $F=ma$, that takes force. That’s called centripetal force, the force to keep something revolving. In direction of velocity change: toward the center.
If you’re in the thing revolving, you feel a force pulling you away from the center, as if extra gravity. Thing circling is an “inertial reference frame” and is accelerating. The apparent force felt by things in the revolving frame is called centrifugal force and it is directly outward.
| {
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How much energy was used to charge the inductor? Electrical systems question I am a math major, and I am taking a mandatory class that looks more like physics and the professor gave us an assignment with a question that I don't have any idea on how to solve it. The question is originally in portuguese, so, I will try my best to translate.
In t = 0, a battery is connected to an inductor with inductance L. At the instant T, the current in the inductor is constant. How much energy was used to charge the inductor?
The question is general, without any explicit number. If anyone can help me, I would be grateful. I feel like it's simple, but I don't even know how to start.
The professor is using the following book: Modern Control Engineering by Katsuhiko Ogata (this problem is not here).
| The battery delivers a constant voltage $V$. The voltage of an inductor follows the differential equation: $$V = L\frac{dI}{dt}$$ where $L$ is the inductance and $I$ is the current. And the energy is the integral of the power $E = \int{VI dt}$
| {
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Free space motion I was going through Kleppner and got this interesting doubt. At page 138 there is an article named rocket in free space.So if there is no external force. The fuel can expand rapidly or slowly without affecting final velocity of rocket. Or we can say the velocity of rocket change by same amount whether we burn all the fuel in one go. or slowly. But at page 148, Question no. 3.14 if men jump from railway flatcar all at same time vs men jump one by one the speed don't remain same but this is similar case as the rocket one there is no friction involved.
| The difference is that in one case, the mass being accelerated is varying, but in the other case the force being applied is varying.
So in the case of the rocket, the amount of energy input will be the same total, regardless of how we do it through time. Same energy in. Also, same mass being accelerated by that energy. Accelerating a mass takes energy. $E=\tfrac{1}{2}mv^2$. If we have a single value for $m$, and we end-up adding the same overall energy $E$, then the final velocity $v$ must be the same regardless.
In the case of the train-car, we are not varying the total energy put in. We are varying what masses get accelerated. So if a man stays awhile and gets accelerated, then that takes up some of the energy. But if he leaves early, then that energy will not go into accelerating him; it will all go into accelerating the car. We could think of two possibilities: 1. One guy jumps right at the beginning. Or 2. This one guy waits until the very end to jump. Either way engine has the same power output over time, and so it has the same total energy to put into accelerating things to velocity.
*
*Total energy $E$ goes into speeding up the train and appears as kinetic energy: $$E= \tfrac{1}{2}~m_{train}~v^2$$
*Total energy $E$ goes into speeding up the train and man and appears as kinetic energy: $$E= \tfrac{1}{2}~m_{train+man}~v^2$$
If the $E$ is the same in both 1 and 2 above, then the $v$ will be lower in number 2 since $m$ is higher.
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What's the best answer to "how many bounces until the ball stops"? A problem question that has been coming has the form:
*
*How many complete trips between two plates moving towards each other can you make? (Given that you are moving between them at a constant velocity, until they both collide, and thus the space between them becomes 0)
*How many bounces does a ball reaching a height 0.6 of its previous one make until it stops? (Dropped from rest... at an initial height...)
There are similar questions such as how much TIME it takes until it stops are simpler to answer since a geometric series can be formed. Is there a way to deal with number of travels/bounces/etc.?
Thanks for any input you may provide!
Edit: I am aware of Zeno's Paradox. Is there a mathematical work-around for it?
| This is the simulation of bouncing ball with the co-efficient of the restitution $~\epsilon=0.9$
you start from the height $~h_0~$ with zero velocity .
Section 0
$$y(t)=h_0-\frac{g\,t^2}{2}\\
v=-g\,t\\
y(t)=0~\Rightarrow~,t_0^2=\frac{2\,h_0}{g}\\ v_{01}^2=-2\,g\,h_0$$
when the ball hit the floor it reach the velocity $~v_{01}$
for the next section the start velocity will be $~v_{01}\mapsto -\epsilon\,v_{01}~$ where $\epsilon~$ is the co-efficient of the restitution.
Section 1
$$y(t)=v_{01}\,t-\frac{g\,t^2}{2}\\
\frac{dy}{dt}=0~\Rightarrow~,t_m=\frac{v_{01}}{g}\\
\text{the max. height is}\\
h_1=y(t_m)=\frac 12 \frac{v_{01}^2}{g}\\
\text{with}~ v_{01}^2\mapsto 2\,g\,h_0\,\epsilon^2\\
h_1=h_0\,\epsilon^2$$
Section 2
you start from $h_1~$ with velocity zero, thus you can use the calculation of section 0, and so on, you obtain the next maximal height $y_2=h_0\,\epsilon^4$
the n'th max height is:
$$h_n=h_0\,\epsilon^{2\,n}~,n=1\ldots$$
from here
$$n=\frac 12 \frac{\ln\left(\frac{h_n}{h_0}\right)}{\ln(\epsilon)} +1$$
plus one because we have to count the "section 0"
Verification
$h_n=20~,~\frac{h_n}{h_0}=\frac{20}{100}$
$\epsilon=0.9$
$n=\frac 12 \frac{\ln\left(\frac{20}{100}\right)}{\ln(0.9)}+1=8.6=8$
this is also what you see in the simulation, the ball bounces eight times until it reaches 20% of the start height
| {
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Is it possible to create a magnetic environment where it pushes and pulls at the same time, making the target “levitate”? To better understand why I ask this, the backstory is I’m getting more and more annoyed by rolling office chairs rotating, hitting my ankles, hard to switch directions when wheels having ~90 degrees differences.
Imagine a ball, and put a bowl on it (upside down). The ball is metal, and the bowl is a magnet that pushes the ball, so it doesn’t touch it (because of the shape), and pulls it at the same time, so if you lift the bowl up the ball goes with it. This used as wheelchair castors could prevent my, and possibly others’ ankles to be hit.
Is it possible to create this environment that the magnets are in perfect balance in every direction, both pushing and pulling?
| You can create a spatial array of magnets which generates a specific spatial force profile against a corresponding magnet array.
Check out polymagnets:
https://vimeo.com/107166551
Spring Polymagnets attract until they pass through a defined transition point, passed which they will repel. These Polymagnets will come to rest at an equilibrium distance.
I don't know whether it'd work for your specific spherical application though. I think it'd be easier to just redesign the normal wheelchair.
| {
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Does relativity mean that the crew of a relativistic rocket would experience less acceleration than in our frame of reference? I have been told regarding a 1 g rocket that "the amount you accelerate would be less due to relativity".
Does that mean that from the crew's time dilated perspective, they would experience less acceleration than we observe in our frame of reference? Could a ship be accelerated at say, 10 g from our frame of reference on Earth, while the crew of the ship only experiences 1 g of acceleration in theirs?
If this were possible, how far can we take this, and how quickly?
|
Does that mean that from the crew's time dilated perspective, they would experience less acceleration than we observe in our frame of reference?
No, the crew’s proper acceleration (the acceleration they feel) is greater than the coordinate acceleration (the derivative of their velocity in our frame) that we observe. They can continue at 1 g proper acceleration indefinitely, whereas we see their coordinate acceleration approach 0 g as they approach c.
| {
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Why have we not found an interior Kerr solution? The Schwarzschild interior solution was found not so long after the exterior solution was found. I understand that Kerr solution is significantly more complicated and there are more conditions at the boundaries but is there anything deep or profound about the interior that makes it so difficult to find a solution that describes it?
| Birkhoff's theorem guarantees that any spherically symmetric interior solution you write down will have the Schwarzschild geometry as its exterior field, so all you have to do is ensure that the interior stress-energy looks reasonable for matter.
In the rotating case, a no-hair theorem guarantees that uncharged black holes have the Kerr form, but there's no result analogous to Birkhoff's theorem. Most rotating objects in fact don't have a Kerr exterior. So you either have to find one of the rare (perhaps nonexistent) objects that does, or find a vacuum solution more general than Kerr's to glue your interior solution to.
I'm not sure that this problem is actually unsolved, though. "Interior solution for the Kerr metric" by Hernandez-Pastora and Herrera (2017) seems to claim to solve it, and is published in Phys. Rev. D.
| {
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Non-Perturbative Effects Of Soliton in Quantum Field Theory I am reading Quantum Field Theory in a Nutshell by A.Zee. In Chapter 5 Section 6, Under the subtitle A nonperturbative phenomenon, He commented
"That the mass of the kink comes out inversely proportional to the coupling is a clear sign that field theorists could have done perturbation theory till they were blue in the face without ever discovering the kink."
I don't quite get the meaning of the sentence, why this is a clear sign? Why does it imply non-perturbative effects? What is the non-perturbative effects he was referring to?
| More generally, if one is doing perturbation theory in some coupling constant $g$, the result is a power series in $g$, i.e. non-negative powers of $g$. To get negative/inverse powers of $g$, one must include non-perturbative effects.
| {
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Simplest exactly solved model displaying a phase transition? The classical example of an exactly solved model which displays a phase transition is the 2D Ising model. However, all the proofs I've seen of this have been very long and complicated.
So, I wanted to know whether there were any other exactly solved models with phase transition, which were easier to solve, or that the 2D Ising model is the simplest such model that we know of.
| The simplest model demonstrating a phase transition is probably the Ising model with an interaction constant that is the same for all spin pairs:
$H=-J\sum_{i,j}S_i S_j$.
I will try to find a reference later.
EDIT (9/6/2021): https://homepages.spa.umn.edu/~vinals/tspot_files/phys5201/2015/hwk8.pdf
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Damped harmonic motion initial conditions I was reading Halliday's section on Damped Simple Harmonic Motion, which stated that this equation:
$$-b\dot{x} - kx = m \ddot{x}$$
Is the differential equation that dictates the displacement of the object, and $b$ is the damping constant of the system.
The author claims that the solution of the equation is:
$$
x(t)=x_m \mathrm e^{-bt/2m}\cos(\omega't+\phi)
$$
Suppose we want to set initial conditions $x(0) = x_0$ and $x'(0) = 0$. Giving the equations
$$
x_m \cos(\phi) = x_0
$$
$$
-\frac{x_m b}{2m} \cos(\phi) - x_m \omega' \sin(\phi) = 0 \implies \frac{b}{2m} \cos(\phi)+\omega' \sin(\phi) = 0
$$
Physically, I would love to say that the initial conditions imposed are, a priori, the maximum amplitude and the initial velocity. If that is correct, then the first equation would imply that $x_m = x_0$ and $\cos(\phi) = 1$ (and therefore $\sin(\phi) = 0$)
However, that does not makes sense, since if we substitute these values into the second equation, we obtain
$$
\frac{b}{2m} = 0
$$
But both $m$(mass) and $b$ (damping constant) are non-zero. What is wrong with my intuition?
Also, if we begin with the second equation and solve for $\phi$, we obtain $\phi = \tan^{-1}\left(\frac{-b}{2m\omega'}\right)$. And then we would be bound to accept that $x_m = \frac{x_0}{\cos(\phi)}$. If correct, then what does $x_m$ represent, if not just a mathematical constant?
| Answers that came before me perfectly OK. I will just add one significant intuitive picture of what's going on.
Picture in your mind the oscillator starting at $x_m$ but with initial velocity running away from equilibrium position.
Why would you identify $x\left(0\right)$ with the amplitude?
Interesting exercise: Fix initial velocity at zero and see what happens --as @147875 suggests.
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When a car accelerates relative to earth, why can't we say earth accelerates relative to car? When a car moves away from a standstill, why do we say that the car has accelerated? Isn't it equally correct to say that the earth has accelerated in the reference frame of the car? What breaks the symmetry here? Do the forces applied to the car have special significance in determining which frame is inertial and which one is not?
Please explain in simple terms.
| The second Newton's law is valid for inertial frames of reference.
If we are for example in a airplane that is braking after landing, any loose object will accelerate forwards, without any force that can be identified. On the other hand, if we hold the object, we do a force on it and it is at rest in the plane's frame. In the first case, an acceleration without a force, and in the second one, a force without acceleration.
It is the criteria to know for sure that we are in an accelerated frame of reference.
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Is it okay to sleep with the wifi router next to me? Where I sleep I have my wifi router right above me. I don't know if this is safe for my health since it emits radiation.
Some people say that it should be safe since the radiation emitted is non-ionising.
But other sources claim that even non-ionising radiation can have harmful effects biologically.
Therefore, I'm quite confused whether it is safe or not.
These articles claim that it is not safe:
https://educateemf.com/is-it-safe-to-sleep-near-a-wireless-router/
https://emfadvice.com/sleep-near-wifi-router/
https://techwellness.com/blogs/expertise/is-wifi-safe-distance-is-key-emf-protection
https://ictbuz.com/wifi-router-in-bedroom/
Some insight would be much appreciated.
| The wifi router, as well as cell phones and microwave ovens all produce non ionizing radiation. To date, the major concern for non ionizing radiation is its heating (thermal) effect on tissue. It is my understanding that the magnitude of the non ionizing radiation of wifi and cell phones is limited by regulations to be well below harmful in terms of heating of tissue. The leakage limits for microwave ovens is likewise regulated to be well below harmful in terms of heating of tissue.
Insofar as the potential harmful non-thermal effects of these sources are concerned, research continues. Clearly, as the number of these sources continue to increase, more data will be available to determine whether or not there are cumulative harmful non thermal effects of non ionizing radiation at the levels being produced.
In the meantime, in my opinion only, though it would seem prudent to limit ones exposure as much as practicable, it should not be because of being alarmed by non scientific information one may encounter on the internet. Rather it should be in recognition of the fact that the long term effects of non ionizing radiation continue to be studied.
Hope this helps.
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What does it mean that uncertainty can be calculated by the "smallest division on a scale"? I have been reading everywhere a little bit to understand the concept of uncertainty but I cannot understand exactly how to find the uncertainty value. On this explanation it says that the uncertainty can be calculated by the "smallest division on a scale". What does this exactly mean?
Thanks.
| It might be easier to appreciate if you imagine using digital devices, such as weighing scales. Imagine you had one in your kitchen that displayed weight to the nearest gram, and another industrial scale, for measuring the weight of lorries say, that displayed weight to the nearest kilogram. Clearly the absolute uncertainty of weights on the first scale seems smaller than the absolute uncertainty of weights on the second scale.
In each case, the scale is rounding the measurement to the nearest whole unit. An apple shown as weighing 233 grams on your kitchen scale could really have a weight anywhere between 232.5 and 233.5 grams, so there is an uncertainty of a gram.
That said, it all depends on an assumption that both scales are equally well calibrated. If your kitchen scale has a fault that randomly adds kilograms to every measurement, then the industrial scale will give you a lower uncertainty for the weight of your apple!
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How to understand Black-Body Curves and get useful information from them?
Here I have got a black-body curve. It shows the amount of radiation released as the wavelength varies for different temperatures. But I don't seem to understand it much.
*
*How do we properly study a black-body curve (like this one) and draw different conclusions about the body ?
*Why after some wavelength, radiation decreases slowly (the low slope area) and why at lower wavelength the radiation decreases quickly (the high slope area) ?
*Also, how the shape of the graph would change as the temperature changes ?
| *
*The main thing to notice is that the peak moves to the left as the temperature increases, the peak of the $3000K$ curve is above 0.9 micrometers, but the peak of the $5000K$ curve is above about 0.6 micrometers, so by looking at the position of the peak the temperature of the body can be deduced.
*You should look up the formula for the 'Planck black body distribution', but the derivation involves quantum theory and is complicated.
*The shape is similar as the temperature changes, but the change mentioned in 1. happens and also the total power emitted is higher as temperature increases, so the peak (and the whole curve) becomes higher.
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Why won't all voltage be used up on first resistor in series? tldr: I am having trouble conceptually understanding voltage between resistors in series, even though I know how to calculate it using Ohm's law. How do the electrons "know" there are more resistors after going through the first one? Why doesn't it use up all the 10V so it's 10V -> resistor 1 -> 0V?
Case A: If a circuit has 1 resistor, the voltage on each side will be completely used up. For example, with a 10V battery and $5\Omega$ resistor, the voltage will be 10V -> resistor -> 0V, with a current of 2A.
Case B: However, if we attach another $5\Omega$ resistor, then the voltage will be 10V -> resistor 1 -> 5V -> resistor 2 -> 0, with a current of 1A.
From the perspective of the electrons, they don't know that there is another resistor down the line.
So why does the voltage drop across the first resistor 10V in Case A but only 5V in Case B?
I'm probably not understanding something very simple. This is my first circuits class and it's a lot to wrap my mind around! Thanks in advance.
|
From the perspective of the electrons, they don't know that there is another resistor down the line.
The current is a moving ensemble of electrons, with a group drift velocity, no single electron goes from the positive to the negative pole of a battery. Electrons interact with the lattice and the molecules of the medium they pass through.
If they find a lot of resistance the drift velocity becomes lower, if the resistance is low the drift velocity becomes higher.No need for knowledge of the future path.
It is important to understand Ohm' law
Since both the momentum and the current density are proportional to the drift velocity, the current density becomes proportional to the applied electric field; this leads to Ohm's law.
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Fire caused by friction with water Friction causes heat and heat causes fire, but could something catch in fire because of friction with a high speed water stream?
If so, what material would it be and how fast would the water speed need to be flowing?
| Ordinary water is a neutron moderator, so we can run water through a nuclear reactor with enriched uranium and get as much heat as we want. One could object that it would not be "friction" with water that causes heat, but at least it is "interaction" with water.
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How come black body have more emissivity and more absorbtivity(a) at same time? We have two definitions to look at
Absorbtivity(a): the ratio of absorbed energy and incident energy on a body
$a_{BlackBody} = 1$
so if i have a tourch light that gives red light, in a dark room I point this tourch light at this black body, then according to above definition, I should see black color
Good absorbers are good emitters from kirchoffs law, and also stefans law
$$\frac{d\theta}{dt} = \sigma AeT^4$$
Now according to that statement, if we reconduct same experiment, when I point my red torch light towards black body it should now emmit red color
Aren't both definitions contradictory to each other or did I misinterpret something
| The red light, with power $P$, will be absorbed. The blackbody will be heated to equilibrium when it's temperature $T$ satisfies:
$$ \sigma T^4 A = P $$
Thus, it all depends on $P$. If $P$ is a 1 TW red laser and $A$ is a pellet, it could get to really hot...mega kelvins. If it's a 10 W light, and $A$ is larger, the radiation will peak in microwave or mm.
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Conservation of angular momentum in an inelastic collision
I have a question about the second method used to solve the problem above.
The moment of inertia with respect to the stick's midpoint after the collision is $ml^2/12 + ml^2/4$ or $ml^2/3$ so the angular momentum with respect to the stick's center after the collision is $ml^2/3*w$. Therefore, equation 8.56 becomes $mv_0l/2 = ml^2/3*w$ but this doesn't give the same w value. Can someone explain why?
| You can calculate moments of inertia about any axis you want. But if the system isn't actually rotating around that axis, then you can't use it to (directly) calculate the rotation speed.
Instead of all the parts moving with speed proportional to $r$ as the distance from your axis, the parts are moving with various speeds as it spins around the center of mass.
(8.56) above always uses the center of mass of the post-collision system as the axis to calculate the rotation speed. The relocated origin is only used for the angular momentum of the linear motion of the center of mass, not as a changed axis for the moment of inertia of the spinning stick.
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What can I assume about Mandelstam variables in four-photon scattering? In the case of four-photon scattering, the Mandelstam variables must satisfy
$s+t+u=0$
Where:
$
s=-(p_1+p_2)^2, t=-(p_1+p_3)^2, u=-(p_1+p_4)^2
$
And we know that $s>0$ is true because it is the total energy squared. What can I assume for $t$ and $u$? Are they free to be one negative and the other positive and viceversa? Or is there a restriction over their possible values?
| If I remember correctly Peskin & Schroeder derive some more general properties:
*
*$\qquad s > 0 \qquad, \qquad t \le 0 \qquad \text{and} \qquad u\le 0 $
*$\qquad t=0$ for $\theta =0$ and $u=0$ for $\theta = \pi\;$ ($\theta$ being the scattering angle)
*$\qquad s\ge |t|$
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Does the frequency of light have any effect on gravitational lensing? General relativity considers gravity as the curvature of space-time instead of a force. Therefore, what is bent around an astronomical object is the space-time itself. The light just follows the path as regular.
In classical physics however, I think we would interpret gravitational lensing as the light's being pulled by a massive object (correct me if I am wrong). This implies some centripetal acceleration on light.
The question is does the frequency of light have any effect on the amount of bending in gravitational lensing? Is it possible to separate the light into its components by gravity, like a glass prism would do?
| Do you know that any mass with the the velocity of earth would take the same orbit?
so since "mass" for light in the more classical approach ist E/c^2 different light has different mass but since all have speed c the same path.
| {
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What happens when a helium white dwarf accretes enough matter? What happens when a helium white dwarf accretes enough matter?
Say you have a white dwarf made of helium, what happens when it's core becomes dense enough to fuse helium and how massive would it have to be for this to happen?
| If the white dwarf (WD) is accreting matter, whether or not it is composed of helium, it will undergo a type 1a supernova or form a cataclysmic variable.
The Chandrashekar limit (maximum mass for a stable white dwarf) is independent of the composition of the WD, what matters is the degeneracy pressure and the mass of the WD. So, when the accreted mass exceeds the Chandrashekar limit, it will either give rise to a type 1a supernova or form a cataclysmic variable.
If the accretion rate is high, then the helium white dwarf will undergo a type 1a supernova as a result of a runaway helium fusion. If the accretion rate is not high enough, then the white dwarf will occasionally give rise to novae, forming a cataclysmic variable.
| {
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Provided a unit vector and Force, how can I calculate it's components? Say I have a $F=kQ_{1}Q_{2}/r^{2}$ and a direction vector $(x, y, z).$ How can I find the component forces $F_{x}$, $F_{y}$, and $F_{z}$?
| To find the components of any vector $\bf F$ using unit vectors,
you can use the dot product between the vector and each unit vector.
So the x-component of $\bf F$ is $\bf F\cdot \hat i$ the y-component is $\bf F\cdot \hat j$ and the z-component is $\bf F\cdot \hat k$
If you have a "direction vector" $\bf u=(x,y,z)$ then its unit vector would be $$\bf \hat u=\frac{u}{\mid u\mid}$$ so that $$\bf F\cdot \hat u$$ is the projection of $\bf F$ in the direction of the unit vector $\bf \hat u$ or the component of $\bf F$ along $\bf u$.
| {
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Rotation in 3D space Suppose a body that rotates around an axis. But the axis also rotates. Now, the axis of the rotation of the axis also rotates, and so on. How far we can go with it? Can it be extended to infinity like a power series?
|
Can it be extended to infinity like a power series?
Yes it can. This is possible not just in 3D space, but in 2D space as well. These motions are collectively called epicycles, and will be familiar to anyone who has produced patterns with a spirograph. Before Johannes Kepler realised that planetary orbits were ellipses, astronomers used systems of epicycles to approximate the motion of the planets. According to Wikipedia:
Any path—periodic or not, closed or open—can be represented with an infinite number of epicycles. This is because epicycles can be represented as a complex Fourier series; so, with a large number of epicycles, very complicated paths can be represented in the complex plane.
| {
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Does a single atom vibrate? Thermal vibration looks impossible, but does a single atom in vacuum vibrate due to the electrons' or subatomic particles' actions? If yes, how much is that vibration and is there a way to stop it?
|
Vibration is a mechanical phenomenon whereby oscillations occur about an equilibrium point. The word comes from Latin vibrationem ("shaking, brandishing"). The oscillations may be periodic, such as the motion of a pendulum—or random, such as the movement of a tire on a gravel road.
Vibration is a classical mechanics phenomenon. Atoms may be considered classical mechanics points when in bulk and a model with mechanical vibrations used, but a single atom can only be described quantum mechanically about its center of mass.
The quantum mechanical wavefunction has electrons in orbitals and nuclei within the nucleus in energy levels depending on the quantum mechanical model used. All the periodic functions describing the atom have to do with the probability of measurement, which means many atoms must be measured in order to see the phase space of the possible positions, and any periodic function has to do with the frequencies displayed in the probability distributions, not space distributions for one atom.
Within the envelope of the Heisenberg uncertainty, one cannot know exactly the position of the single atom, because then the uncertainty in its momentum would be very large, but that does not mean it vibrates, it just means that the probability of its location and momentum are correlated.
| {
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How to understand the time reversal symmetry of position operator? How to understand the fact that position operator is symmetric under time reversal? I can visualize the momentum and magnetic field being odd under time reversal.
Got the same doubt for Electric field as well.
| You can easily find out which quantities are even and which
are odd upon time-reversal by considering a video showing
physical processes and then comparing it to the time-reversed
video showing the same processes backwards.
See also "T-symmetry - Effect of time reversal on some variable of
classical physics" for lists of even and odd physical quantities.
Even quantities
Obviously the spatial position $\vec{x}$ doesn't change sign
in the time-reversed video. (A body being on the right side in the
original video is still on the right side in the time-reversed video,
not on the left side.)
Also the acceleration $\vec{a}$ doesn't change sign.
(In the original video a ball thrown up and then falling down
is accelerated downwards. Viewed in the reversed video, the
ball is again accelerated downwards, not upwards.)
From Newton's $\vec{F}=m\vec{a}$ you can conclude, force
$\vec{F}$ also doesn't change sign.
And then from the definition of electric field ($\vec{F}=q\vec{E}$),
you can further conclude, the electric field $\vec{E}$ doesn't
change sign.
Odd quantities
Obviously in the reversed video time $t$ and velocity $\vec{v}$
have the opposite sign as compared to the original video.
The magnetic field $\vec{B}$ is defined by the Lorentz force
($\vec{F}=q\vec{v}\times\vec{B}$). Because force $\vec{F}$ is even
and velocity $\vec{v}$ is odd, you can conclude the magnetic field
$\vec{B}$ must be odd.
| {
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Do electromagnetic waves contain electrons? I understand that EM waves are oscillating electric and magnetic fields. But doesn't this mean that the wave itself contains charged particles that generate the fields?
| No. Electromagnetic Waves do not actually 'contain' anything.
Classically, you can describe them as perturbances of the Electric and Magnetic Fields in space that are in a certain relation ( such as: they satisfy Maxwell's equations and D'Alembert equation, aka the wave equation).
An electromagnetic wave can be described as one ( or more ) photons in a quantum approach.
Either way, they do not contain electrons.
They, however, interact with electrons: the electric field of the wave causes a force on electrons , and if they are moving they will experience a Lorentz force because of the wave's magnetic field.
| {
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Understanding Space-time intervals and its types I am taking Introduction to Modern Physics class. There, we were studying spacetime intervals as a subsection of Lorentz's transformation. My professor said that $\Delta x^2-c^2\Delta t^2$ is invariant, and then he said us that it is a lightlike event if $\Delta x^2-c^2\Delta t^2=0$, timeline if $\Delta x^2-c^2\Delta t^2<0$ and spacelike if $\Delta x^2-c^2\Delta t^2>0$. I understand that a spacelike event is when different reference frames do not agree on the order of the events. That is because we used Lorentz's transformation to show that, if say A and B are space-like events, A and B happen simultaneously in one frame, A happens before B in one reference frame, and B can happen before A in another reference frame. These are all mathematical notions for these events. I do not understand when these events can occur in reality. Can someone give me some examples so that I can understand them?
| An interval between two events is space like if it is not possible for light leaving one of the events to arrive at the location of the other before it happens.
Let's take an example. The Moon is about 1.3 light seconds away. Suppose I synchronise my watch with two friends, Como and Zaquette, on the moon. If I sneeze at exactly noon and Como sneezes one second later, then the interval between our two sneezes is space like, because his sneeze happened 0.3 seconds before light from my sneeze could have reached him. If Zaquette sneezes a second later, then the interval between her sneezes and mine is timeline, because light from my sneeze would have reached her 0.7 seconds before she sneezed.
Everyone will agree that my sneeze happened before Zaquette's. However, in some frames of reference moving relative to mine, Como's sneeze will seem to have happened before mine.
| {
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How to simplify this complex circuit? I am new to circuit solving and I tried to simplify this circuit, but I am unable to do so. I can't figure out which resistances are in series and which resistances are in parallel.
Do I have to use star-delta conversion here or the circuit can be solved without conversion?
I just need a simplified circuit diagram and the rest I can solve. Thanks for any help!
| When you have such seemingly complex circuits, the best way to approach the problem is to name each node and see which elements are in series and parallel. Keep on finding the equivalent resistances of groups of elements like the two resistors in parallel on the extreme right of the circuit. When you hit a hurdle, use star-delta transformations to simplify the configurations.
| {
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Would a pressurized container move by itself if opposite edges have different size surface area? If inside a closed container there is gas with higher pressure than outside the container, and one edge of the container has a larger surface area than the opposite side, would the container move by itself? Wouldn't there be a net force in one direction since one surface area is bigger than the opposite side while the pressure is the same?
| The world would be a lot more interesting if that were true!
No. If you sum the normal forces over all sides of the shape, you'll see that the components in all directions still cancel to zero.
| {
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How is Newton's third law working here? I understand that if I were, for example, to throw a bowling ball in outer space then the ball would move away from me by the force I generated. But the ball would also exert a force on me in the opposite direction and move me away in the opposite direction of the bowling ball.
I don't understand how Newton's third law works in the next scenario though: I saw on YouTube an MIT professor let the gas out of a fire extinguisher on the back of his bicycle which then propelled him forward on the bicycle. What's exactly causing that motion? How is Newton's third law working here?
The gas molecules are pushing on neighboring gas molecules which is ultimately causing gas to escape from the nozzle. The gas molecules are pushing on one another like my bowling ball in space example, right? But where is the gas pushing on the fire extinguisher? Is there more pressure on one side of the fire extinguisher?
| The walls of the chamber that houses the compressed gas is pushed upon by the molecules inside the chamber. The actual gas molecules escaping through the nozzle do not push on the walls of the extinguisher though, but since there is pressure throughout the chamber, the escaping molecules near the nozzle push on the molecules inside the container that also push on those that are near, and at, the walls of the chamber.
In terms of Newton's third law, it may be more intuitive to think about it in terms of momentum conservation. If you have mass exiting in one direction from an object, then the object must move in the other direction with opposite momentum or $$m_gv_g+m_fv_f=0 \\
\rightarrow m_gv_g=-m_fv_f \\
p_g=-p_f$$
where $m_g$, $v_g$ is the mass and velocity of the gas, and $m_f$,$v_f$ is the mass and velocity of the fire extinguisher.
| {
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Is the coordinate speed of light in vacuum in a point direction dependent? Suppose we could control gravitation inside a wire frame cube with a side length of $l$ at will (I assume some yet unknown technology). First we simulate empty space inside and when measuring how long a light beam takes to go through it is $t_0=l/c$.
Now we create a homogeneous gravitational field inside the cube. My hunch is that this would be a constant stress energy tensor field inside the cube. Again we measure the time the light beam takes through the cube and as far as I understand that would be $t_g = b\cdot l/c$ for $b>1$, meaning it takes longer than before, i.e. in coordinate time the light is slowed down.
Question: Will the $b$ inevitably be the same between any two pair of sides of the cube or could there be some directional dependence despite the constant stress energy tensor field?
| The coordinate speed of light is indeed directionally dependent in general. However, that has nothing to do with gravity or stress energy tensors. It is all about the coordinates chosen.
For example, in flat Minkowski spacetime you can use Anderson’s coordinate system described on p 105 here https://www.sciencedirect.com/science/article/abs/pii/S0370157397000513?via%3Dihub
For this coordinate system the metric is $ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$ and the coordinate speed of light is $1/(1-\kappa)$ in the $+x$ direction and $1/(1+\kappa)$ in the $-x$ direction.
Conversely, in a spherically symmetric vacuum spacetime with spacetime curvature, you can use isotropic coordinates $$ds^2=-\left( \frac{1-m/2r}{1+m/2r} \right)^2 dt^2 - \left( 1+m/2r \right)^4 \left( dx^2 + dy^2 + dz^2 \right)$$ where the coordinate speed of light is isotropic.
So whether or not the coordinate speed of light is isotropic or not is not dependent on curvature or anything. It is simply an artifact of the chosen coordinate system.
| {
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How do electrons absorb energy? I'm trying to understand how lamps work.
I have read that the way this happens is when voltage is applied. The free electrons move due to the electromotive force (EMF), and when they collide with the lattice filament inside the lamp, they push the electrons in the lattice's atoms to a higher energetic state.
My question is how do electrons absorb energy to transition to a higher energetic state (by getting repelled by other free electrons) or is there something happening under the hood that I'm still unaware of?
| The EMF, $\varepsilon$, corresponds to an electric field $\mathbf{E}$ within the lamp circuit. Neglecting induction,
$$ \varepsilon = \int_A^B \mathbf{E} \cdot d \mathbf{l} .$$
This field does work on electrons as they travel from terminal B through the lamp to terminal A. Within the filament the kinetic energy of the electrons is dissipated through resistivity (corresponding to collisions). The filament is therefore heated and emits light.
The potential energy of the electrons is raised in the EMF source, typically through induction or chemistry.
| {
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Why did Rutherford's atomic model predict a continuous emission spectrum for hydrogen? When researching the flaws of Rutherford's atomic model, I get that one of them is that it predicts the electrons would spiral and collapse into the nucleus.
However, I don't understand the second flaw which says it can't explain the discrete emission spectrum of excited gases like hydrogen. Why doesn't Rutherford's model explain the discrete emission spectrum? Moreover, why does it predict a continuous spectrum instead?
| You have sort of answered your own question.
That the electrons would spiral in toward the nucleus was shown by Maxwell, in that accelerating charged particles would emit radiation loosing their energy. That is the first problem, as you have pointed out.
But this is also related to the second problem, in that this would lead to a continuous spectrum.
If an accelerating charged particle emits radiation continuously, then the detected emitted energy spectrum (of electrons accelerating around and toward the nucleus) would also have to be continuous. There are no discontinuous "jumps" on the way. The fixed energy orbits were later theorized (by Niels Bohr).
And experiment showed at the time a discrete spectrum for Hydrogen (and other elements), and this is why the Rutherford model could not explain discrete spectra, nor could it explain why electron orbits did not completely decay if they are continuously emitting radiation.
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Graph for Coulomb Force vs $1/r$ My teacher told me that the graph for the coulomb force $F$ vs $1/r$ where $r$ is the distance between the 2 charges should be parabolic but I can't seem to understand why. I am aware that equations of the form $y^2=4ax$ are parabolic but why should $F$ vs $1/r$ graph be parabolic?
| The Coulomb force magnitude is given by
$$F=\frac{kQq}{r^2}$$
Pulling out the $(1/r)^2$ gives us
$$F=kQq\cdot\left(\frac1r\right)^2$$
So then F has a quadratic dependence on $1/r$, which is a parabola if you were to graph $F$ vs $1/r$.
If you still cannot see it, then call $1/r$ something else, like $x=1/r$. Then $F(x)=kQqx^2$
| {
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Can particles smaller than Planck Length be detected? It is my understanding, that anything smaller than a Planck Length cannot interact with photons, because photons with such small wavelength are massive enough to collapse into a black hole (source). Particle of such size would not interact with electromagnetic forces, but what about the three other fundamental forces? If this particle existed and was abundant in the universe but not uniformly distributed, could we be able to observe it's effect or could we build a device that could detect those particles?
I'm interested in this question, because dark matter famously doesn't interact with electromagnetic force and have a gravitational pull.
| I will make the comment into an answer:
Note that the standard model of elementary particle interactions is a quantum field theory where the particles in the table
are axiomatically assumed to be point particles (photon included) .
In the abstract of a book :
Quantum Field Theory of Point Particles and Strings (Frontiers in Physics)
.....
Part I of this book follows the development of quantum field theory for point particles,
A point particle has zero length by construction.
Considering that the model fits the preponderance of experimental data with these particles, and is very predictive, one can consider their "length" measured,and equal to zero ! The particles are surely detected in experiments.
| {
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What is wrong with the high-school definition of a vector? Why is the high-school definition of a vector as "a quantity with a magnitude and a direction" incomplete? For example, Griffiths Introduction to Electrodynamics book says:
The definition of a vector as "a quantity with a magnitude and direction" is not altogether satisfactory.
(section 1.1.5 "How vectors transform")
However, I am not very satisfied with his chain of arguments.
| In addition to this answer , here is another simple argument why definition of vector is not satisfactory. Electric current in simple circuits has both magnitude (Amps in SI units) and direction (positive to negative terminal or from high to low potential).
But electric current does not satisfy either the triangle law of vector addition nor the parallelogram law of vector addition. Thus it cannot be regarded as a vector.
Similarly there are many quantities (Higher dimensional) which have magnitude and direction but don't follow the usual triangle law of vector addition.
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Why can basis vectors change direction? I thought that basis vectors were of magnitude one and located at the origin and were each linearly independent, so how in things like polar coordinates can the basis vectors be moving?
| The short answer is: Because there is a separate vector space with a separate basis at every point of space and you can choose a basis for each of theses vector spaces independently.
To understand this properly you need the theory of manifolds (as explained in Dale's answer).
It is just an "accident" of Euclidean space that vectors don't change when you transport them around. We can identify all the vector spaces at all points and choose a single basis for them in Cartesian coordinates because of this accident.
Acutally, you can also work in polar coordinates for the position, and still use the global Cartesian basis vectors to express your velocity/electric field/.... It is just very convenient to match the basis vectors to your coordinates, as it simplifies calculations.
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Can Vera Rubin's findings be explained by a distribution of charge? Vera Rubin found that the rotational velocity of galaxies is much greater than expected at greater distances from the center. Gravity from an invisible mass is assumed to account for this measurement. Can we not account for such a force by a distribution of charge? For example consider the sun and the inner planets positively charged. Perhaps there is even a gradient of decreasing charge the further away from the center. And now consider that the outer planets are oppositely charged. Even perhaps increasing charge as we travel towards the outer edge of the galaxy. Since each solar system is 'insulated' from every other by empty space, the charge separation is maintained. Since solar systems are small compared to galaxies, a uniform charge on all parts of the solar system would not be detectable.
| Unfortunately (or fortunately), some back-of-the-envelope calculations show it is unlikely that flat rotation curves are caused by charge separation. The acceleration of the Sun when moving around the center of the galaxy is around $1.6\cdot 10^{-10}\mathrm{m/s^2}$, so, lets say, the acceleration caused by forces from electric field is of the order $a=10^{-10}\mathrm{m/s^2}$. The charge of the Sun is about $Q=77 \textrm{C}$, according to this paper, and therefore the electric field needed to give the Sun needed acceleration would be
$$
E=\frac{M a}{Q}\approx 2\cdot 10^{18} \textrm{V/m},
$$
which is impossibly high (for comparison, an electric field needed to rip an electron from a hydrogen atom is about $10^{11} \textrm{V/m}$)
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Basic question about orbital speed I was reading a Sci-Fi book recently and had a weird thought:
I know that objects closer to a gravitational well need to move faster to stay in orbit and objects further away move slower. But if you want to increase your orbit/escape the gravitational well you have to speed up while if you want to lower your orbit you have to decrease your speed.
In my mind this seems like a paradox. I'm sure I'm just thinking about it the wrong way but I can't figure out how to solve this. Can someone explain it to me, please?
| I will try to explain it mathematically. Our orbital velocity is given by formula:
$$v_{\text{orbital}}=\sqrt{\frac{GM_{\text{star}}}{R_{\text{orbit}}}}$$
It’s obvious that numerator terms are not going to change. So, if the orbital velocity changes, the radius of the orbit changes. Since they are inversely proportional, if one increases, the other decreases and vice versa. So you get the result you are looking for.
In an elliptical orbit itself, the velocity changes as the celestial body revolves around the star (present at the ellipse's focus). Here, also:
$$L=mv_{\text{orbital}}a=\text{constant}$$
And you can see that perpendicular distance ($a$) and orbital velocity are inversely proportional.
Or think another way. We are stuck to our planet's gravitational field. But if we increase our velocity, to escape velocity, we can leave the planet altogether.
Calculating the new radius by altering velocity can be easily found by conservation of energy.
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Current density operator derivation in Fourier space Free particle Hamiltonian is $H_0 = \int d\mathbf{r} \frac{\hbar^2}{2m}(\nabla\Psi^\dagger)(\nabla\Psi)$. The Fourier transform representation of $\Psi^\dagger(\mathbf{r})$ is
$$
\Psi^\dagger(\mathbf{r}) = \sum_ke^{-i\mathbf{k}\cdot \mathbf{r}} c_\mathbf{k}^\dagger
$$
So,
$$
\boxed{H_0 = \frac{\hbar^2}{2m}\sum_k k^2 c_\mathbf{k}^\dagger c_\mathbf{k}}
$$
And the density operator $\rho(\mathbf{r})=\Psi^\dagger(\mathbf{r})\Psi(\mathbf{r})$ is
$$
\rho(\mathbf{r})=\sum_{q_1q_2}e^{-i\mathbf{q_1}\cdot\mathbf{r}}e^{i\mathbf{q_2}\cdot\mathbf{r}}c^\dagger_\mathbf{q_1}c_\mathbf{q_2}
$$
$$
\boxed{\rho(\mathbf{r})=\sum_{q_1q}e^{i\mathbf{q}\cdot\mathbf{r}}c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q}}
$$
where $\mathbf{q=q_2-q_1}$.
The current density operator is calculated by
$$
\nabla\cdot \mathbf{J(r)} = -\frac{i}{\hbar}[H_0,\rho(\mathbf{r})]
$$
$$
\nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\sum_{k,q,q_1}k^2[ c_\mathbf{k}^\dagger c_\mathbf{k},c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q}]e^{i\mathbf{q}\cdot\mathbf{r}}
$$
$$
\nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\sum_{k,q,q_1}k^2\bigg\{ c_\mathbf{k}^\dagger[ c_\mathbf{k},c^\dagger_\mathbf{q_1}]c_\mathbf{q_1+q}+c^\dagger_\mathbf{q_1}[ c_\mathbf{k}^\dagger ,c_\mathbf{q_1+q}]c_\mathbf{k}\bigg\}e^{i\mathbf{q}\cdot\mathbf{r}}
$$
$$
\nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\sum_{k,q,q_1}k^2\bigg\{ c_\mathbf{k}^\dagger c_\mathbf{q_1+q} [ \delta_\mathbf{{k,q_1}}] -c^\dagger_\mathbf{q_1}c_\mathbf{k} [ \delta_\mathbf{{k,q_1+q}}] \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}}
$$
$$
\nabla\cdot \mathbf{J(r)} = -\frac{i\hbar}{2m}\bigg\{\sum_{k,q,q_1}k^2 c_\mathbf{k}^\dagger c_\mathbf{k+q}-\sum_{k,q,q_1}(q_1+q)^2 c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q} \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}}
$$
If we include the Fourier transform of LHS as $f(x)=\sum_q e^{ikx}f_k$
$$
\boxed{\sum_\mathbf{q} (i\mathbf{q}) \mathbf{J_q} e^{i\mathbf{q}\cdot \mathbf{r}} = -\frac{i\hbar}{2m}\bigg\{\sum_{k,q}k^2 c_\mathbf{k}^\dagger c_\mathbf{k+q}-\sum_{q,q_1}(q_1+q)^2 c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q} \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}}} \tag{1}
$$
I know the result for $J_q$ should is
$$
\mathbf{J_q} = \frac{\hbar}{m}\sum_{k}(\mathbf{k}+\frac{\mathbf{q}}{2}) c_\mathbf{k}^\dagger c_\mathbf{k+q} \tag{2}
$$
Question:
How to reach at Eq (2) from Eq (1)?
| After a help from @Jakob, I solved it:
$$
\sum_\mathbf{q} (i\mathbf{q}) \mathbf{J_q} e^{i\mathbf{q}\cdot \mathbf{r}} = -\frac{i\hbar}{2m}\bigg\{\sum_{k,q}k^2 c_\mathbf{k}^\dagger c_\mathbf{k+q}-\sum_{q,q_1}(q_1+q)^2 c^\dagger_\mathbf{q_1}c_\mathbf{q_1+q} \bigg\}e^{i\mathbf{q}\cdot\mathbf{r}}
$$
Change dummy variable $q_1\to k$
$$
\sum_\mathbf{q} (i\mathbf{q}) \mathbf{J_q} e^{i\mathbf{q}\cdot \mathbf{r}} = -\frac{i\hbar}{2m}\sum_{k,q}\bigg\{k^2 -k^2 -q^2-2\mathbf{k}\cdot \mathbf{q} \bigg\}c^\dagger_\mathbf{k}c_\mathbf{k+q}
e^{i\mathbf{q}\cdot\mathbf{r}}
$$
$$
\mathbf{J_q} = \frac{\hbar}{m}\sum_{k}\big\{ \frac{\mathbf{q}}{2}+\mathbf{k} \big\}c^\dagger_\mathbf{k}c_\mathbf{k+q}
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/671287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The effect black holes have on light We all know that light loses its energy when it is moving through expanding space and time. And sense a black hole can be summed up to a super compressed space time, shouldn't that mean that a photon gains energy when it enters a black hole because it is experiencing the opposite effect of the expansion of spacetime
| Basically yes it does lose energy and it can not get it back. For your second qwestion well, we don’t know what happens in a black hole but we have general ideas of what happens in there and basically it’s just contrubiting energy into the black hole BUT a black hole could just be a incredibly different thing like a worm hole so you would have to be more specific with your qwestion
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If our solar system and galaxy are moving why do we not see differences in speed of light depending on direction? May be a silly and simple question, but I've been wondering if:
The speed of light is constant, and
*
*When we're moving in the same direction (where both the emitter and the receiver move with the light direction) we would be making it take more time for the light to reach the other end.
*Conversely when moving in the opposite direction we'd be shortening the time it takes.
Why do we not see a non-uniform speed of light caused by solar/galactic movement?
Or do we?
| The speed of a wave is constant regardless of the source velocity. Only the frequency of the wave is affected by the source velocity. See this answer for sound waves: Does sound waves pick up the speed of its source?
| {
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Hydrogen line spectrum lately I have come across the hydrogen line spectrum, I quite understand it as I could easily solve all numerical questions. However, conceptually I feel stuck at two points.
*
*If an electron absorbs the em radiation of a particular wavelength to jump to a higher shell say $n=1$ to $n=3$, then when it returns to the ground state (from $n=3$ to $n=1$) will it emit the radiation of the same wavelength or of a different wavelength. Kindly explain what exactly happens?
*In the absorption spectrum of hydrogen, em wavelengths of 434 nm, 486 nm, and 656 nm are absorbed because they correspond to the energy difference between the shells so they are not present on the photographic plate but I don't understand that if excited electrons (from the absorbed light) jump back to their ground state in nanoseconds releasing the same radiation with the same wavelength then why does it not appear in the spectrum? Also, how do the electrons figure out which light to be absorbed (I know it sounds dumb but I just can't understand how do electrons figure out that a particular wavelength does not suffice the energy requirement to jump?)
P.S.: Please consider my doubt and excuse it if they appear dumb as I have taken these questions to various platforms and have never received any satisfactory response.
| 1.) An electron falling down from level n=3 can decay into the ground state in two ways, either directly into the ground state n=1, or first into n=2 and then into n=1. These decay paths have certain statistical probabilities, so from many atoms you will see three lines in this case with a relative intensity given by these probabilities.
2.) The absorption lines in a spectrum are caused by the fact that a beam of light travelling in a given direction will be re-emitted again in all directions, so light is lost from the beam and re-appears in other directions instead. See the following graphic for an illustration
from https://courses.lumenlearning.com/astronomy/chapter/formation-of-spectral-lines/)
Classically, you can explain the physical mechanism here by a forced harmonic oscillator. Light will only be absorbed by the atom at the resonant frequencies, which correspond to the spectral lines.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Particle in a one-sided infinite step function plus a delta-function attractive well Given the following potential:
$$
\begin{cases}
\infty & x\leq -d \\
-V_0 \delta(x) & x > -d
\end{cases}
$$
with d > 0
I would like to compute the condition that has to be verified in order to have at least one bound state.
From $\hat{H} \psi(x) = E_n \psi(x)$ and assuming we have a bound state (E < 0), I get as general solution for $\psi(x):$
$$
\begin{cases}
0 & x < -d \\
A(-e^{-2kd} e^{kx} + e^{-k x}) & -d < x < 0 \\
Ce^{-kx} & x > 0
\end{cases}
$$
where $k = \sqrt{\frac{2m|E|}{\hbar^2}} $
By imposing the boundary conditions in $x = 0$: $\Delta \frac{d \psi(x)}{dx} = -\frac{2 m V_o}{\hbar^2} \psi(0), \hspace{0.5 cm} \psi(0^-) = \psi(0^+)$ $\hspace{0.5 cm}$ and $\hspace{0.5 cm}$ $\psi(-d) =0$,
I conclude that the quantization of the energy of the system is given by:
$$
\begin{equation}
\frac{1+e^{-2kd}}{1-e^{-2kd}} = 1 - \frac{2 m V_0}{\hbar^2 k}
\end{equation}
$$
Which doesn't make any sense due to the fact that $kd > 0$ and the left member of this equation is bigger than 1 in this region.
Could someone explain me what is wrong?
| My guess would be to say that you are right which would mean that there are no bound states for this potential.
I think this is the case since for a simple delta-function attractive well there is only one bound state, so if we add another condition (in this case the infinit step) it could remove that unique allowed state, which would result in no allowed bound state.
EDIT: If your exercise states that there is a bound state, my guess must be wrong but I still don't see a mistake in your calculations...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can a material generally score or cut itself by hand? I'm wondering if a given solid material can, in general, score or cut the same material, when applied by (at most) human muscular strength.
I've tried searching for this online, but it seems like a difficult-to-express search target. For example, the site Answers.com has a brief entry:
Q: What mineral can scratch diamonds? | The material quality you're looking for is hardness. The Mohs Hardness Scale is an empirical table of minerals sorted by hardness. Two materials are said to be the same harness if each can scratch the other. A homogeneous or mostly homogeneous solid, like diamond, has an identifiable hardness. A nonhomogeneous solid, like concrete, does not.
A solid-liquid mixture, like a tomato, or a solid-gas mixture, like aerogel, does not. Nonhomogeneous Solid mixtures like concrete will scratch like their hardest component and be scratched like their least hard component. Solid-liquid and solid-gas mixtures may also crush, tear, or burst, unrelated to scratching.
So: if an object has a well-defined hardness, it can scratch itself because it has the same hardness as itself. And an object has a well-defined hardness if it is a homogenous solid.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/672466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Matrix element and Dirac notation If
$$
T=
\left[
\begin{array}{cccc}
e^{\beta J} & e^{-\beta J} \\
e^{-\beta J} & e^{\beta J} \\
\end{array} \right]
$$
and
$$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \exp{\beta J(\vec{S_1}\vec{S_2}+\vec{S_2}\vec{S_3}+...+\vec{S_{N-1}}\vec{S_N}+\vec{S_N}\vec{S_1})}
$$
Then why can we say that
$$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \langle S_1|T|S_2\rangle\langle S_2|T|S_3\rangle...\langle S_N|T|S_1\rangle ?
$$
| $\newcommand{\e}{\boldsymbol=}$
$\newcommand{\p}{\boldsymbol+}$
$\newcommand{\m}{\boldsymbol-}$
$\newcommand{\gr}{\boldsymbol>}$
$\newcommand{\les}{\boldsymbol<}$
$\newcommand{\greq}{\boldsymbol\ge}$
$\newcommand{\leseq}{\boldsymbol\le}$
$\newcommand{\plr}[1]{\left(#1\right)}$
$\newcommand{\blr}[1]{\left[#1\right]}$
$\newcommand{\lara}[1]{\langle#1\rangle}$
$\newcommand{\lav}[1]{\langle#1|}$
$\newcommand{\vra}[1]{|#1\rangle}$
$\newcommand{\lavra}[2]{\langle#1|#2\rangle}$
$\newcommand{\lavvra}[3]{\langle#1|\,#2\,|#3\rangle}$
$\newcommand{\x}{\boldsymbol\times}$
$\newcommand{\qqlraqq}{\qquad\boldsymbol{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad}$
Consider two complex $n\m$vectors expressed also as kets
\begin{equation}
\mathbf x\e
\begin{bmatrix}
x_1 \vphantom{\dfrac{a}{b}}\\
x_2 \vphantom{\dfrac{a}{b}}\\
\vdots \vphantom{\dfrac{a}{b}}\\
x_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}\e \vra{\mathbf x}\qquad \texttt{and} \qquad
\mathbf y\e
\begin{bmatrix}
y_1 \vphantom{\dfrac{a}{b}}\\
y_2 \vphantom{\dfrac{a}{b}}\\
\vdots \vphantom{\dfrac{a}{b}}\\
y_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}\e \vra{\mathbf y}\quad \in \mathbb C^n
\tag{01}\label{01}
\end{equation}
Complex conjugating and transposing these one-column matrices we obtain the bras
\begin{equation}
\mathbf x^{\boldsymbol*}\e
\begin{bmatrix}
\overline x_1 & \overline x_2 & \cdots & \overline x_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}\e \lav{\mathbf x}\quad \texttt{and} \quad
\mathbf y^{\boldsymbol*}\e
\begin{bmatrix}
\overline y_1 & \overline y_2 & \cdots & \overline y_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}\e \lav{\mathbf y}
\tag{02}\label{02}
\end{equation}
Their usual inner product in $\,\mathbb C^n\,$ is
\begin{equation}
\overline x_1\,y_1\p\overline x_2\,y_2\p\cdots\overline x_n\,y_n\e
\begin{bmatrix}
\overline x_1 & \overline x_2 & \cdots & \overline x_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}
\begin{bmatrix}
y_1 \vphantom{\dfrac{a}{b}}\\
y_2 \vphantom{\dfrac{a}{b}}\\
\vdots \vphantom{\dfrac{a}{b}}\\
y_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}\e \lavra{\mathbf x}{\mathbf y}
\tag{03}\label{03}
\end{equation}
Given a $\,n\times n\,$ complex matrix $\,\mathrm A\,$
\begin{equation}
\mathrm A\e
\begin{bmatrix}
a_{11} & a_{11} & \cdots & a_{1n} \vphantom{\dfrac{a}{b}}\\
a_{21} & a_{22} & \cdots & a_{2n} \vphantom{\dfrac{a}{b}}\\
\vdots & \vdots & \vdots & \vdots\vphantom{\dfrac{a}{b}}\\
a_{n1} & a_{n2} & \cdots & a_{nn}\vphantom{\dfrac{a}{b}}\\
\end{bmatrix}
\tag{04}\label{04}
\end{equation}
the notation $\,\lavvra{\mathbf x}{\mathrm A}{\mathbf y}\,$ is the inner product of the vectors $\,\mathbf x\,$ and $\,\mathrm A\mathbf y\,$ expressed by matrices as
\begin{equation}
\lavvra{\mathbf x}{\mathrm A}{\mathbf y}\e
\begin{bmatrix}
\overline x_1 & \overline x_2 & \cdots & \overline x_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}
\begin{bmatrix}
a_{11} & a_{11} & \cdots & a_{1n} \vphantom{\dfrac{a}{b}}\\
a_{21} & a_{22} & \cdots & a_{2n} \vphantom{\dfrac{a}{b}}\\
\vdots & \vdots & \vdots & \vdots\vphantom{\dfrac{a}{b}}\\
a_{n1} & a_{n2} & \cdots & a_{nn}\vphantom{\dfrac{a}{b}}\\
\end{bmatrix}
\begin{bmatrix}
y_1 \vphantom{\dfrac{a}{b}}\\
y_2 \vphantom{\dfrac{a}{b}}\\
\vdots \vphantom{\dfrac{a}{b}}\\
y_n \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}
\tag{05}\label{05}
\end{equation}
Under this spirit you could look at the $\,S_k\,$ as $2\times 1$ matrices and more precisely
\begin{equation}
S_k\e
\left.
\begin{cases}
\begin{bmatrix}
1 \vphantom{\dfrac{a}{b}}\\
0 \vphantom{\dfrac{a}{b}}\\
\end{bmatrix} \texttt{for} \p 1\\
\\
\begin{bmatrix}
0 \vphantom{\dfrac{a}{b}}\\
1 \vphantom{\dfrac{a}{b}}\\
\end{bmatrix} \texttt{for} \m 1
\end{cases}\right\}
\tag{06}\label{06}
\end{equation}
(note : this reminds us the up and down states of a spin-1/2 particle or the up and down quarks of isospin-1/2 particle).
So if for example $\,S_3\e\m 1\,$ and $\,S_8\e\p 1\,$ then
\begin{equation}
\lavvra{S_3}{\mathrm T}{S_8}\e
\begin{bmatrix}
0 & 1 \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}
\begin{bmatrix}
t_{11} & t_{11} \vphantom{\dfrac{a}{b}}\\
t_{21} & t_{22} \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\begin{bmatrix}
1 \vphantom{\dfrac{a}{b}}\\
0 \vphantom{\dfrac{a}{b}}
\end{bmatrix}\e
\begin{bmatrix}
0 & 1 \vphantom{\dfrac{a}{b}}\\
\end{bmatrix}
\begin{bmatrix}
t_{11} \vphantom{\dfrac{a}{b}}\\
t_{21} \vphantom{\dfrac{a}{b}}
\end{bmatrix}\e t_{21}
\tag{07}\label{07}
\end{equation}
For the rest look in the other till now two answers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can the degree of entanglement be an order parameter in a phase transition? In most continuous phase transitions, there is a well-defined order parameter $\langle \psi \rangle$ of some observable that is zero above the transition temperature, and continously grows below the transition. In the cases that I am familiar with, this observable is usually the expectation value of some physical quantity (spin, density, momentum, position, number) that is generally a local observable. On the other hand, it has become increasingly clear that there is a lot of non-local physics captured in quantities like entanglement, so perhaps more complex types of non-local order parameters may exist.
My question is, can entanglement entropy, or a similar entanglement witness, ever be an order parameter in a phase transition? If yes, what class of phase transitions do they appear in? If not, can you explain why?
I am interested to hear about both the trivial answer, where an entanglement measure and conventional observable can both be the order parameter, and the non-trivial case where only entanglement can be treated as the order parameter. I suppose at some level this whole discussion must involve quantum phase transitions, not just classical phase transitions.
| Entanglement over an ensemble can be used to tell measurement induced phase transitions. The idea arises in a quantum circuit where measurements and unitary gates take place randomly over time. The ratio between measurements and gates determines whether the system is driven to a "more entangled" or "more disentangled" phase.
An example can be found here. You can see in figure 3 that the entanglement entropy is obtained for chains of different sizes, and the phase transition is obtained from a scaling law. The entanglement is, however, defined over an ensemble of quantum circuits run under the same ratios of gates/measurements.
This is a pretty recent field of research. More information can be found here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Adding angular velocity vectors We know we can add two angular velocity vectors to get a total angular velocity. Whereas I more of less understand the basic principle and the mathematical formulation, I have problems in visualizing this addition. Example picture (from my old Resnick and Haliday book) is attached. However, once the base rotated with $\omega_1$ the $\omega_2$ vector should constantly change its position, thus the sum should also rotate. Are there any visualisations, animations or better explanation on why all vectors in such addition remain constant in space and/or how the body rotates along such axes (along components $\omega_1$, $\omega_2$, and the sum $\omega_1+\omega_2$). I could not find any. With many thanks.
|
I see it like this:
you obtain the angular velocity components from the
rotation matrix
the disk is rotating with the angle $~\varphi_z~$ about the z-axes the rotation matrix is:
$$\mathbf R_d= \left[ \begin {array}{ccc} \cos \left( \varphi _{{z}} \right) &-\sin
\left( \varphi _{{z}} \right) &0\\ \sin \left(
\varphi _{{z}} \right) &\cos \left( \varphi _{{z}} \right) &0
\\ 0&0&1\end {array} \right]\quad
\Rightarrow\quad
\vec\omega_d=\left[ \begin {array}{c} 0\\ 0\\
\dot\varphi _{{z}}\end {array} \right]
$$
the wheel is rotating relative to the disk with the angle $~\varphi_y~$ hence the transformation matrix between the wheel and inertial system is:
$$\mathbf R_w=\left[ \begin {array}{ccc} \cos \left( \varphi _{{y}} \right) &0&\sin
\left( \varphi _{{y}} \right) \\ 0&1&0
\\-\sin \left( \varphi _{{y}} \right) &0&\cos
\left( \varphi _{{y}} \right) \end {array} \right]
\,
\left[ \begin {array}{ccc} \cos \left( \varphi _{{z}} \right) &-\sin
\left( \varphi _{{z}} \right) &0\\ \sin \left(
\varphi _{{z}} \right) &\cos \left( \varphi _{{z}} \right) &0
\\ 0&0&1\end {array} \right]
\quad\Rightarrow\\\\
\vec\omega_w=\begin{bmatrix}
\dot\varphi_y\,\sin(\varphi_z) \\
\dot\varphi_y\ \\
\dot\varphi_z\,\cos(\varphi_z) \\
\end{bmatrix}
$$
you can now add the angular velocities components
$$\vec\omega=\vec\omega_d+\vec\omega_w$$
| {
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Proof that you can't disentangle two parties if you only operate on one Let $A$ and $B$ be two entangled systems. Can someone prove or sketch a proof of why you cant unentangle $A$ and $B$ by only acting on $A$ or $B$ alone? i.e. by only applying $\mathbb{I}_A\otimes U_B$, with $U_B$ unitary.
| For a "positive" proof (as in, a proof not by contradiction), write the Schmidt decomposition of a generic bipartite pure state $|\Psi\rangle\in\mathcal H_A\otimes\mathcal H_B$ as
$$|\Psi\rangle=\sum_k \sqrt{p_k}(|u_k\rangle\otimes|v_k\rangle),$$
for some positive reals $p_k\ge0$ summing to the identity, and orthonormal bases $|u_k\rangle$ and $|v_k\rangle$. Note that you can always do it: this decomposition amounts to the SVD of $|\Psi\rangle$ when thought of as an operator $\mathcal H_B\to\mathcal H_A$.
The entanglement of $|\Psi\rangle$ is then encoded in the Schmidt coefficients $(\sqrt{p_k})_k$. This statement can be made precise e.g. via the theory of majorization and its relation to entanglement. What we care about here is simply that separable states (which are equal to product ones for pure states) are all and only those with Schmidt coefficients equal to $(1,0,0,...)$, up to permutation of the elements.
Now, observe that local unitary operations do not affect the Schmidt coefficients. This is trivially seen writing
$$(U\otimes V)|\Psi\rangle =
\sum_k \sqrt{p_k} ((U|u_k\rangle)\otimes (V|v_k\rangle)),$$
and remembering that $\{|u_k\rangle\}_k$ is an orthonormal basis if and only if $\{U|u_k\rangle\}_k$ is, for any unitary $U$.
So, not only local unitary operations cannot disentangle states: they cannot affect the entanglement in any way.
It is worth noting that this is not the case for generic local operations: non-unitary local operations can absolutely degrade the entanglement. The standard example of this being entanglement-breaking channels. An entanglement-breaking channel $\Phi$ is such that
$(\Phi\otimes I)\rho$ is separable, for any (possibly entangled) $\rho$. See this answer for a proof of this.
| {
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"question_score": "8",
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Age of a black hole Is there a way to measure the age of a black hole by find Hawking radiation or calculating the stable orbits around the black hole?
| No. The No-hair theorem tells us that an isolated black hole is completely described by its mass, angular momentum, and net electric charge. In fact, the charge is expected to be basically zero, so it's basically just the mass and angular momentum. Hawking radiation and orbits only depend on those parameters as well, so they wouldn't tell us any extra information.
For realistic astrophysical black holes, there could be indirect evidence that gives us some hints, though. For example, if the black hole was formed as a result of a supernova, it may be possible to observe evidence of the time since that supernova. There are numerous examples of supernova remnants that can be directly imaged. If you analyze those images — especially multiple images taken over a period of many years — you could figure out when the supernovae occurred, thus suggesting when the corresponding black hole may have formed. (Though maybe the supernova was triggered by interaction with an existing black hole...)
There are other scenarios that I can imagine might be able to give us some hints about how long a black hole has been hanging around influencing nearby matter. In principle, you might also be able to say something about the statistics of a population of black holes. But indirect evidence of these sorts are really the only possibilities in accepted physics. A black hole on its own gives us no hint about its age.
| {
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"timestamp": "2023-03-29T00:00:00",
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In the Aharanov-Bohm experiment, how are we sure that the electrons do not ever enter regions with $\vec B\neq 0$? In the Aharanov-Bohm experiment, we say that the electron passes through a region where the magnetic field $\vec B=0$ but the vector potential $\vec A\neq 0$. The electron never passes through the region where $\vec B\neq 0$ i.e. through the interior of the solenoid. But how can we be sure of that? Because the electron wavefunctions are smeared out. How can we be sure that the tails of the electron wavefunction do not penetrate the interior of the solenoid?
| We can hypothesize an infinite potential barrier around the solenoid. The Aharonov Bohm effect can be shown theoretically all the same.
We can also hypothesize the solenoid to be infinitely thin. The area of the solenoid does not affect the Aharonov Bohm effect.
| {
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} |
How do stars produce energy if fusion reactions are not viable for us? From what I've learned, fusion reactions are not currently economically viable as of right now because the energy required to start the reaction is more than the energy actually released. However, in stars they have immense pressures and temperatures which are able to allow these reactions to take place. However, if these reactions are considered endothermic for us, how are they exothermic in stars? i.e. how are stars able to release energy?
Moreover, why are such fusion reactions for us endothermic in the first place? Given we are fusing elements smaller than iron, wouldn't the binding energy per nucleons products be higher and hence shouldn't energy be released?
| It is the fact that fusion reactions are very exothermic that makes them so hard to control.
Coal releases its chemical energy so slowly that a coal fire does not need any confinement - it does not blow itself apart. Refined hydrocarbons, such as petrol, release energy more violently, but the walls of a metal cylinder are strong enough to contain the explosion (e.g. in an internal combustion engine) or at least direct the explosion products in a useful direction (jet engines, rocket engines).
We have various ways of achieving nuclear fusion (see this Wikipedia article for an overview) but they all either blow themselves up (thermonuclear devices), require too much power to achieve confinement (magnetic and inertial confinement devices) or produce a low density of reactions (colliding beam devices).
| {
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"url": "https://physics.stackexchange.com/questions/674089",
"timestamp": "2023-03-29T00:00:00",
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Concept of Gravitational potential energy Change in Potential energy corresponding to a conservative force is defined as $$\Delta U = U_f - U_i=-W_f$$ and gravitational potential energy is $$\Delta U = U_f-U_i = -W_g $$ Suppose a mass $m_1$ is kept at a fixed point $A$ and a second mass $m_2$ is displaced from point $B$ to point $C$ such that $AB = r_1$ and $AC = r_2$.
$\therefore$ , $$\Delta U = -W_g = \int{\frac{Gm_1m_2}{r^2}}dr$$ $$U(r_2)-U(r_1) = Gm_1m_2\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$$ Now I am free to choose any reference point thus if I take potential energy at $U(r_1) = 0$ and $r_2 = \infty$ Then I will get potential energy at infinity as $$U(\infty) = \frac{Gm_1m_2}{r_1}$$ which I think is wrong as a reference point at $r_1$ the potential energy at infinity should be infinite.
So where I am wrong, is my concept of gravitational potential energy wrong itself.
|
I think is wrong as a reference point at r1 the potential energy at infinity should be infinite.
The potential energy at infinity is only infinite if it takes an infinite amount of work to get to infinity. However, because the gravitational force decreases rapidly with distance, a projectile rapidly reaches a space where the force of gravity is not strong enough to reverse its velocity. Potential energy keeps increasing, but there is a maximum value that is approached asymptotically.
We have actually built space probes that have escaped Earth's gravity and the Sun's gravity. Voyager 1 and 2 have exited the Solar System and are never coming back. If they don't run into anything, they will reach arbitrarily far distances. The only thing stopping them from actually reaching infinity is the infinite time it would take to get there. To reach such a state, it only took the energy in the rocket fuel and some kinetic energy stolen from planets during planetary slingshots.
There are systems with a potential energy that reaches infinity at infinite distance. The spring potential energy $U = \frac{1}{2}kx^2$ is an example. When $x=\infty$, $U=\infty$ because the spring force keeps getting stronger as the spring is stretched.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why can a very large body of water not store summer heat? On this page, it states "The key disadvantage of using a very large body of water to achieve heat exchange with a relatively constant temperature is that you are not able to store summer heat in that body of water – to have the benefit of retrieving those higher temperatures in winter."
Why is it so? Is it because a very large body of water would have more heat exchange with the air and hence would lose the heat gathered in summer?
But "heat exchange with a relatively constant temperature" also points in the direction of having a large body of water, so I am a bit confused.
| I find the statement wrong. Living in Greece and, when younger, swimming in the sea from end of May to beginning of November, I know that the large body of water , Corinth gulf , is much warmer in November than the small body of water of a connected sea lake. So large bodies of water store heat energy of the summer much better than small bodies.
It is interesting to see the temperature of the sea water in the weather reports all winter, the sea cools much less than air in the winter. The opposite is true in the summer. The variations is seen in these average sea temperature per month color scaled charts.
| {
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Is an expression of a quadrupole as an expansion of dipoles possible? Would it be possible to express a quadrupole as an expansion of dipoles? Because a possible definition of a quadrupole seems to be: an electric field equivalent to that produced by two electric dipoles.
| You seem to mix up two things here: Which charge distributions generate what multipole fields and how multipole-expansion works.
Yes, you can reach a pure quadrupole field by a limiting procedure involving two dipoles (just take two anti-parallel dipole sources and let their distance go to zero from above).
But that doesn't mean you can expand the quadrupole field into dipoles in the sense of a multipole expansion. One natural way to think about multipole expansion is in the sense of a far-field approximation: If you have an arbitrary configuration of charges restricted in a radius $R$ around the origin, then the field for $r \gg R$ will be given by a multipole expansion, the leading term will be the lowest non-vanishing multipole contribution.
If you have a quadrupole (constructed as above), you will see that the dipole moment is zero (because the dipole moments of the two dipoles cancel out). In a multipole expansion there is no way to put two dipoles "next to each other", all the multipole sources in the expansion are exactly located at the origin.
| {
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What is faster? Pure rolling or rolling with slipping? So we know that a ball will slide down an incline when there is no frictional force. Once you say there is static friction it causes a angular momentum which in turn causes a torque and makes the ball roll. Lets say the ball number one does pure rolling, where the static friction is large enough to keep it rolling pure all the way down. Now lets say ball 2 rolls as well, but slips some ways down the incline. What would reach the ground first? The ball rolling and slipping, or the ball purely rolling? If so, why?
|
you have those two equations:
$$m\,\ddot s=m\,g\sin(\alpha)-F_r\\
I_\theta\,\ddot\theta=F_r\,r$$
in case of pure rolling is $~\ddot s=r\,\ddot\theta~$ and in case of partial rolling $~F_r=\mu\,m\,g\cos(\alpha)~$
pure rolling
$$\ddot s=\frac{m\,g\,r^2\,\sin(\alpha)}{m\,r^2+I_\theta}
\quad \Rightarrow\\
s_1(t)=\frac{m\,g\,r^2\,\sin(\alpha)}{m\,r^2+I_\theta}\,t^2\quad (1)$$
partial rolling
$$\ddot s=g\,(\sin(\alpha)-\mu\,\cos(\alpha))
\quad \Rightarrow\\
s_2(t)=g\,(\sin(\alpha)-\mu\,\cos(\alpha))\,t^2\quad (2)$$
hence (with $~I_\theta=m\,r^2$)
$$\frac{\dot s_1}{\dot s_2}= \frac 12\,\frac{\sin(\alpha)}{\sin(\alpha)-\mu\,\cos(\alpha)}< 1$$
so partial rolling is faster then pure rolling
| {
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Second Kepler's law explanation What is the explanation for the second Kepler's law? Why is the law valid?
Is it that the total energy of a planet equals to the kinetic energy plus the potential energy?
| Kepler's second law
according to Newton's second law
$$m\,\mathbf{\ddot{r}}=F(r)\,\frac{\mathbf r}{r}$$
where $~F(r)~$ is central force
from here
$$\mathbf r\times m\,\mathbf{\ddot{r}}=\frac{F(r)}{r}\left(\mathbf r\times \mathbf r\right)=\mathbf 0$$
with
$$\frac{d}{dt}\left(\mathbf r\times m\,\mathbf{\dot{r}}\right)=
\mathbf{\dot{r}}\times m\,\mathbf{\dot{r}}+\mathbf r\times m\,\mathbf{\ddot{r}}=
\mathbf 0$$
hence
$$ \mathbf r\times m\,\mathbf{\dot{r}}=\mathbf L=\textbf{const.}\tag 1$$
this is the conservation of the angular momentum
additional from equation (1) you obtain that $\mathbf r\cdot \mathbf L=\mathbf 0~$ this means that $~\mathbf r\perp\mathbf L~$ so the mass point m has a planner motion
from equation (1) $~\mathbf r\times \mathbf{\dot{r}}=\frac{\textbf{const.}}{m}$ with polar coordinate you obtain
$$\mathbf r=r\,\begin{bmatrix}
\cos(\varphi) \\
\sin(\varphi) \\
0 \\
\end{bmatrix}\\
\mathbf{\dot{r}}=\left[ \begin {array}{c} \cos \left( \varphi \right) {\dot r}-r\sin
\left( \varphi \right) \dot\varphi \\ \sin \left(
\varphi \right) {\dot r}+r\cos \left( \varphi \right) \dot\varphi
\\ 0\end {array} \right]\quad \Rightarrow\\
|\mathbf r\times \mathbf{\dot{r}}|=r^2\,\dot\varphi
$$
or $$\frac{d A}{dt}=\frac 12 r^2\,\dot\varphi\\
\boxed{~A=\frac 12 r^2\,\varphi}$$
Kepler's second law
| {
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Understanding conservation of energy in a pulley problem I am trying to check if my understanding of conservation of energy is correct. Imagine a pulley problem like so with $m_2$ heavier than $m_1$ and the pulley is ideal (in the original problem I borrowed this from $v$, $m_1$, and $m_2$ are given:
and our goal is to find $h$.
Splitting the energy up we have at rest, all energy must balance so:
$E_i = KE + PE = m_2gh$
Taking the ground as the zero potential plane. Since $m_1$ is at rest at the $0$ potential plane there's no energy there. Since nothing is moving, there is also no kinetic energy.
Now, looking at the energy situation in the final period when we release the system and $m_2$ hits the ground:
$E_f = KE + PE = \frac{m_2v^2}{2} + \frac{m_1v^2}{2} + m_1gh$.
Of course, setting these equal (since energy is conserved) and solving for $h$ gives the answer.
I want to reason out why the kinetic energy is the way it is in the final configuration. Initially I had thought that since the system is once again at rest, the only kinetic energy that would matter would be the kinetic energy of $m_2$. But it seems that I must also consider the kinetic energy used to lift $m_1$ to $h$ as well.
Is this because $m_2$ strikes the ground with the sum of the kinetic energies of $m_1$ and $m_2$ due to conservation, or am I interpreting this result incorrectly?
| The Energy is:
$$E=\underbrace{\frac 12 m_1\,v^2+\frac 12 m_2\,v^2}_{K_E}-\underbrace{\left(m_1\,g\,h_1-m_2\,g\,h_2\right)}_{P_E}=0$$
at $~t=0~$ you obtain the initial energy
$$E_i=E(v=0~,h_1=0)=m_2\,g\,h_2$$
with the conservation of the energy $~E=\text{const.}~$ you can obtain the final velocity when $~m_2~$ collides with the floor.
$$E=E_i=\text{const}\quad \Rightarrow\\
v={\frac {\sqrt {2}\sqrt { \left( m_{{1}}+m_{{2}} \right) m_{{1}}\,g\,{\it
h_1}}}{m_{{1}}+m_{{2}}}}\bigg|_{h_1=h_2=h}
$$
| {
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Doppler Effect and Relativity paradox Let, Alice is moving towards Bob at very high speed. Therefore, events in Bob's frame will appear to happen slowly in Alice's frame due to time dilation. Since velocity is relative, the same is also true for Bob.
Now, since Alice is moving towards Bob, Bob's light will be blueshifted towards him. Special relativity will cause it to blueshift even greater. That means, time between two light pulses, therefore two events on Bob's actually decreases as seen by Alice. Not to mention, the same thing will also happen for Bob.
So Alice sees the events at faster than normal speed even though they are happening at slower than normal speed. How is this patadox resolved?
| "That means, time between two light pulses, therefore two events on Bob's actually decreases as seen by Alice."
No.
c=λf=λ/Τ
c in vaccum space must be fixed at all times. The frequency will increase (blueshif) but the wavelength will proportionally decrease to keep c fixed. Meaning the distance between B and A will contract to compensate for the faster pace of time.
A blinking white light for example with a 1 sec pause period will be measured for both B and A to blink at the same rate with the only difference that both B and A will see it now blueshifted in color.
You confuse SR time dilation and length contraction with some kind slow motion video effect. This is a common misconception of SR.
What SR says about your light perception (image) of a relative moving light source towards a target observer, is at the time light has reached you the only effect you will observe will be a color change (Doppler shift). This is now because the increased momentum thus energy of a moving light source does not affect the propagation speed of light which remains fixed but only its frequency E=hf.
| {
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Intuition for the trace-free energy-momentum tensor condition in CFTs It is a textbook exercise to show that
\begin{equation}T^{\mu}_{\,\,\,\mu}=0
\end{equation}
is a sufficient condition for there to be a conserved current associated with a dilation symmetry. This condition is very important in CFTs so my question is,
Question: is there a nice intuitive way to visualize the trace-free condition, or is there some physical example (like a fluid?) where one can better understand what the trace-free condition corresponds to physically?
As an example of the type of intuition I am looking for: the $T^{00}$ component is the energy density, whereas the diagonal components $T^{ii}$ have the interpretation of a pressure (the flux of $p^i$ momentum in the $i^{th}$ direction). So the trace-free condition is somehow equating the energy density with the pressure. Making this more precise or providing a physical example along these lines would be helpful.
| I will elaborate the answer I gave in a comment.
As mentioned the trace of the stress tensor in a QFT is related to the $\beta$ function by $$T_\mu^\mu \sim \beta(g).$$ The $\beta$ function indicates how couplings change with a change of scale but in a CFT there is no scale so the $\beta$ function vanishes; CFTs are fixed points of the RG flow.
There is more to the story. The stress tensor is usually defined at the quantum level by point-splitting:
$$\lim_{\delta \rightarrow 0}\left[\partial_z \phi(z + \delta/2)\partial_z\phi(z - \delta/2) -\frac{1}{2 \delta^2}\right].$$
This definition fails to commute with conformal mappings because it introduces a scale $1/2\delta^2$ that is not mapped to $1/2(|f'(z)|\delta)^2$ in the image. An anomalous term appears in the transformation law because the regularization procedure does not respect the conformal symmetry:
$$T(z) = f'(z)^2 T(f(z)) -\frac{c}{12}\{f,z\}$$
where $c$ is the famous central charge and $\{f,z\}$ is the Schwartzian derivative of $f$ with respect to $z$.
| {
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Time averages of complex quantities If an electric field $E$ oscillates as $E_0\sin(ωt)$ then the average value of $E^2$ over one period of oscillation will be
$$E_0^2\left< \sin^2(ωt)\right>=E_0^2/2$$ since the average value of $\sin^2(ωt)$ is well known to be $1/2$.
However if we write $E$ using complex numbers as $E_0e^{iωt}$ and then take real parts, as is often the case, then we have
$$\left< E^2 \right> = E_0^2\left< (e^{iωt})^2 \right> = E_0^2\left< e^{2iωt} \right> $$
$$=E_0^2 \left< \cos(2ωt) + i\sin(2ωt)\right>=0$$ as the average value of an unsquared $\sin$ or $\cos$ will be equal to $0$ over a single period of oscillation, and the real and imaginary parts (should be) independent of one another.
What's gone wrong here? Is there a fault in my above assumption?
| So the complex numbers that we use in Classical Electromagnetism are more of a mathematical trick. The electromagnetic waves are in fact "real" quantities, i.e. they are represented by sines and cosines. However, it is often easier to work with complex exponentials, and so we cheat a little: since any linear combination of the sines and cosines is mathematically allowed, we choose the combination with a complex number $$\cos(\omega t) + i \sin(\omega t) = e^{i\omega t}.$$
The understanding is then that we will work with such "complex" waves, but at the end we will always return the "real" part. The imaginary part is just along for the ride, to make calculations simpler. (The definition of what is the "real" part is also not set in stone, you could work with either $\text{Re}(e^{i\omega t})$ or $\text{Im}(e^{i\omega t})$, and as long as you're consistent, that's perfectly acceptable.)
However, this argument only works so long as the operations you perform are linear operations. Happily for us, Classical EM is linear, since Maxwell's Equations are linear. Thus, this "complex" wave satisfies the same equations that a "real" wave would.
Thus, you cannot use the complex representation of the wave when you are working with physical quantities that are non-linear. In particular, when calculating physical quantities that depend non-linearly on the wave, like the power, intensity, or even the Poynting vector, one should always take the real part. Thus, your first approach (using $\sin^2(\omega t)$) is the right one.
It turns out that you can use the complex representation if you redefine what you mean by the average, using a complex conjugate representation. (See these notes for a review of phasor notation.) Thus, the average of the product of two waves $\mathbf{A}$ and $\mathbf{B}$ is given by: $$\text{average of $\mathbf{A}\mathbf{B}$} = \frac{1}{2}\text{Re}(\mathbf{A}\mathbf{B}^*)$$ So, for example, you can see why the average of $\mathbf{E}^2$ is: $$\text{average of $\mathbf{E}^2$} = \frac{1}{2}\langle \mathbf{E} \mathbf{E}^* \rangle = \frac{1}{2}.$$ Similarly, it is possible to define the "Poynting vector" for a complex wave as (ignoring some constants): $$\mathbf{S} = \mathbf{E} \times \mathbf{H}^*,$$ and so on. But when in doubt, always go back to the real notation when performing non-linear operations on a complex wave.
| {
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Why are there both antinodes at both ends of the tube? I learned stationary/standing waves the other day. For stationary waves in open tubes, the textbook says both ends must have an antinode. Can anyone tell me why? (shown as figure)
And also, when playing with the instruments like guitar, what's the number of harmonic on the string, i.e. how many antinodes and nodes are there on the string?
| It's a pressure node as the pressure at the open ends of tube is atmospheric pressure. It's then automatically a displacement antinode because the equation of motion
$$
\rho_o \partial_t v = - \partial_x P
$$
means that a standing wave
$$
P(x,t)= A\sin(kx) \sin (\omega t)
$$
with a pressure node at $x=0$ makes the velocity obey
$$
\rho_0 \omega^2 v(x,t)= \partial^2_{xt} P= A k\omega \cos(kx) \cos (\omega t),
$$
which has a velocity (and hence a displacement) maximum at $x=0$.
| {
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Passive transformation, pseudo vectors and cross product Let's consider the passive transformation i.e. inversion only of the basis vectors (coordinate axes) and all other vectors remaining the same and check if the cross product is a pseudo vector.
After the inversion using a passive transformation a vector $a$ remains $a$, $b$ remains $b$, $a \times b$ remains $a \times b$ and any other vector remains the same. So in this case $a \times b$ behaves like any other vector, it remains the same geometrically. Where is the 'pseudo-vector' in this case?
Do we have to use a left-handed screw rule in the inverted coordinate system to find the cross product?
Why should we change our rule on how to find the cross product?
This is very confusing to me. Can anyone please help me.
| with
$$\vec a=a_1\hat g_1+a_2\hat g_2+a_3\,\hat g_3\\
\vec b=b_1\hat g_1+b_2\hat g_2+b_3\hat g_3$$
where $~a_i~,b_i~$ are the vectors components and $~\hat g_i~$ are the basis vectors
with $~\hat g_i\cdot\hat g_j=1~,i=j~$ and $~\hat g_i\cdot\hat g_j=0~,i\ne j$
the cross product
$$\vec c=\vec a\times \vec b$$
now $~\hat g_i\mapsto -\hat g_i~$
$$\vec a\mapsto -a_1\hat g_1-a_2\hat g_2-a_3\hat g_3=-\vec a\\
\vec b\mapsto -b_1\hat g_1-b_2\hat g_2-b_3\hat g_3=-\vec b$$
the cross product
$$\vec c\mapsto -\vec a\times (-\vec b)=\vec a\times \vec b$$
coordinate axes are left -handed.
$$\hat g_1\times \hat g_3=\hat g_2\\
\hat g_2\times \hat g_1=\hat g_3\\
\hat g_3\times \hat g_2=\hat g_1$$
coordinate axes are right -handed.
$$\hat g_1\times \hat g_2=\hat g_3\\
\hat g_2\times \hat g_3=\hat g_1\\
\hat g_3\times \hat g_1=\hat g_2$$
| {
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High school physics problem - having trouble understanding This is a fairly straightforward problem which doesn't require the usage of more than one or two formula but I find it hard to grasp the concept behind this.
Let's say we have two trains, one which moves at the speed of $45 \frac{km}{h}$ and the other at the speed of $60\frac{km}{h}$. Now, let the first train start moving, and let the second one start moving an hour after the first one. The question is after how many hours will the second train catch up to the first one.
I have always had trouble visualizing these kind of problems. I know that the second train starts with a delay of $1$ hour and that during that time the first train passes $45$km. But how do I calculate this?
I know that $v_2 - v_1 = 15\frac{km}{h}$ which is the relative speed of the second train with respect to the first one. This probably means that in such a frame of reference, the $v_1$ is zero so we can imagine it as being static, under the condition that the new $v_2=15 \frac{km}{h}$.
But how do I calculate this? $t=\frac{s}{v}$, thus I need a length in order to calculate this. I can't simply plug in the $45$ km from above because that would be the time in which the second train got to the $45$km mark, but the first train would have moved away from that point. Could anyone explain?
| You want to visualize something that takes time to travel a distance. Its hard to imagine actual trains moving at actual speeds, so you would likely be better off with a substitute.
How about a Monopoly game board?
Two trains, two pieces.
Train 1 is 45km/h. Train 2 is 60km/h. The difference is 15km/h. Luckily, all three are divisible by 15. That gives us Train 1 at 3km/h, Train 2 at 4km/h, and the difference at 1km/h.
Monopoly boards don't have kilometers of space to use, but you can use those reduced numbers to associate with number of spaces on the board.
Now each hour of time can be replaced with a "turn" in the game. On turn 1, Train 1 moves 3 spaces and Train 2 stays put. Turn two Train 1 moves 3 more spaces, Train 2 moves 4 spaces. Repeat until you land on the same space at the end of the turn.
In this manner, you're not calculating everything at once. Here you visually and physically calculate each hour as a turn until arriving at the compiled answer.
| {
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Averaging over spin phase-space for a cross section In Peskin and Schroeder the Dirac equation is solved in the rest frame for solutions with positive frequency:
$$\psi(x) = u(p) e^{-ip\cdot x}$$
$$u(p_0) = \sqrt{m} \begin{pmatrix} \xi \\ \xi \end{pmatrix},$$
for any numerical two-component spinor $\xi.$ Boosting to any other frame yields the solution:
$$u(p) = \begin{pmatrix} \sqrt{p \cdot \sigma}\ \xi \\ \sqrt{p\cdot \bar{\sigma}}\ \xi \end{pmatrix},$$
where in taking the square root of a matrix, we take the positive root of each eigenvalue.
Then, they summarize:
The general solution of the Dirac equation can be written as a linear combination of plane waves. The positive frequency waves are of the form
$$\psi(x) = u(p)e^{-ip\cdot x}, \ \ \ p^2 = m^2, \ \ \ p^0 >0.$$
And there are two linearly independent solutions for $u(p),$
$$u^s (p) = \begin{pmatrix} \sqrt{p \cdot \sigma}\ \xi^s \\ \sqrt{p\cdot \bar{\sigma}}\ \xi^s \end{pmatrix}, \ \ \ s=1, 2, $$
which are normalized: $$\bar{u}^r (p) u^s (p) = 2m \delta^{rs}.$$
Next, we can consider the unpolarized cross section for $e^+e^{-} \to \mu^+ \mu^-$ to lowest order. The amplitude is given by:
$$\bar{v}^{s'} \left(p'\right) \left(-ie\gamma^{\mu}\right)u^s\left(p\right)\left(\frac{-ig_{\mu \nu}}{q^2}\right)\bar{u}^{r} \left(k\right) \left(-ie\gamma^{\nu}\right)v^{r'}\left(k'\right)$$
Then, I quote
In most experiments the electron and positron beams are unpolarized, so the measured cross section is an average over the electron and positron spins $s$ and $s'$. Muon detectors are normally blind to polarization, so the measured cross section is a sum over the muon spins $r$ and $r'.$
...
We want to compute
$$\frac{1}{2}\sum_s \frac{1}{2} \sum_{s'} \sum_r \sum_{r'}|M(s, s' \to r, r')|^2.$$
Why, in order to take the average and the sum, do we only need to sum, rather than integrate, over the spin phase space? Doesn't each incoming particle have an infinite number of spinors? Assuming it is unpolarized, the probability distribution will be uniform, but in principle it still seems like we should integrate over some $\theta$ for spinors $\xi = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}$. Instead, what has been done is to assume each incoming particle is in a definite state of "spin-up" or "spin-down" and assign a $50/50$ probability to each.
| The best analogy I can give is that the incoming and outgoing spins are in a mixed state. In a mixed state that is 50% $+z$ and 50% $-z$ for example, expectation values are equal to
\begin{equation}
\left < A \right > = \mathrm{Tr} [\rho A] = \frac{1}{2} \left < + \right | A \left | + \right > + \frac{1}{2} \left < - \right | A \left | - \right >
\end{equation}
and it is not correct to integrate over the possible relative coefficients between $\left | + \right >$ and $\left | - \right >$. Note that choosing this basis was merely a matter of convenience. Something which is 50% $+x$ and 50% $-x$ is the exact same mixed state.
Edit
Just to be clear, we can take
\begin{equation}
\rho_1 = \frac{1}{2} \left | +x \right > \left < +x \right | + \frac{1}{2} \left | -x \right > \left < -x \right |
\end{equation}
and plug in $\left | \pm x \right > = \frac{1}{\sqrt{2}} \left ( \left | +z \right > \pm \left | -z \right > \right )$ to get
\begin{equation}
\rho_2 = \frac{1}{2} \left | +z \right > \left < +z \right | + \frac{1}{2} \left | -z \right > \left < -z \right |.
\end{equation}
Therefore, when we compute expectations with $\rho_2$, we are not missing the contributions of $\rho_1$ but rather doing something equivalent to if we had used $\rho_1$.
| {
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Is there time reversal symmetry for delta potentials? I was looking at problem $4.6$ of Gasiorowicz's Quantum Physics, where he asks to prove that the scattering matrix of a potential of the form
$$V(x)=\frac{\hbar^2}{2m}\frac{\lambda}{a}\delta(x-b)$$
is a certain unitary ($S^\dagger S=I$) but not symmetric matrix. I thought, as is referenced in the Wikipedia page on the $S$-matrix, that for any real potential $V(x)$, there is time reversal symmetry, meaning the $S$-matrix is symmetric. Where is my mistake?
Also, can we say that in the case where $V(x)$ is even then doing $x\to -x$ implies $$\psi_{in}=S\psi_{out}\implies S^2=Id$$
EDIT: For reference:
| Tragically, the author of the book you reference uses a different convention for the $S$-matrix:
$$\tag{1}
\left(\matrix{C \\B}\right)=S_G\left(\matrix{A \\D}\right)
$$
Instead of
$$\tag{2}
\left(\matrix{B \\C}\right)=S_{RW}\left(\matrix{A \\D}\right)
$$
Where the subscript indicates "rest of world". The letters $A,B\dots$ and notation here follow the wiki page. On conjugating $\psi_L$ and $\psi_R$ we find
$$\tag{3}
\left(\matrix{D^* \\A^*}\right)=S_G\left(\matrix{B^* \\C^*}\right)
$$
As you can see, this is no longer in the same form as (1), the components of the in and out states have been flipped. We cannot continue to the next step of the proof on the wiki page because we cannot substitute the conjugate of (3) into (1). Instead we have
$$\tag{4}
\left(\matrix{D \\A}\right)=S_G^*\left(\matrix{B \\C}\right)
$$
On introducing the matrix $X=\left(\matrix{0 \ 1 \\1 \ 0}\right)$ we can write this as
$$\tag{5}
X \left(\matrix{A \\D}\right)=S_G^* \ X \left(\matrix{C \\B}\right)
$$
Into which may be substituted (1) to yield
$$\tag{6}
(XS_G)^{-1}=(XS_G)^*
$$
Where I have used $X^2=\mathbb{I}$. In this convention, the $S$-matrix does not generally satisfy $S^{-1}=S^*$ for real potentials; instead, (6) is the analogous statement about $S_G$.
The expression in the question is correct using the author's definition.
If the potential is symmetric, you may use the same method above (applying parity rather than complex conjugation, and relating the coefficients $A,B,...$) to show that
$$ \tag{7}
S_{RW}=XS_{RW}X \implies S^{-1}_{RW}S_{RW}=\mathbb{I}
$$
Furthermore, if the potential is even, the energy eigenstates may be chosen to be parity eigenstates. You can read more about that in eg. the notes by Tong or in chapter 16 of Quantum mechanics: a modern development by Ballentine
| {
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How to solve the Biot-Savart Law? I've currently trying to learn electromagnetism in depth and I tried to solve the Biot-Savart law, for a magnetic field generated by a current.
$$
\vec{B} = \frac{\mu_0}{4 \pi} \int{\frac{I \; \hat{r} \times \mathrm{d} \vec{\ell}}{r^2}}
$$
When I looked up for information on how to solve the equation, there are always simplifications, like:
$$
B = \frac{\mu_0}{4 \pi} \int{\frac{I \; \mathrm{d} \ell \sin{\theta}}{r^2}} \rightarrow B = \frac{\mu_0 I}{2 \pi R}
$$
Edit:
It is possible to solve this integral in a vectorial form and without the need of simplification for the magnetic field in a straight wire carrying a current? Would you be able to arrive to a solution similar to the one of the electric field generated by an electric current in a straight wire $\frac{\lambda}{2 \pi \varepsilon_0 r} \hat{r}$ (which you arrive using Gauss's law)?
| What about:
$$d \vec{B} = \frac{\mu_0 I d \vec{\ell} \times \vec{r}}{4 \pi r^3}$$
or
$$d \vec{B} = \frac{\mu_0 I d \vec{\ell} \times \hat{r}}{4 \pi r^2}$$
$B$ = magnetic flux density; $I$ = current in wire; $\ell$ = length of wire; $r$ = distance of a point in the field to a segment of the wire; $\mu_0$ = magnetic permeability of free space.
| {
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Conservation of mass from material derivative Let the mass be $m=\rho \text{Vol}$, where $\text{Vol}$ is the volume of the domain and the velocity is $u$.
Applying the material derivative, then
$$\frac{Dm}{Dt}=\frac{\partial (\rho \text{Vol})}{\partial t}+ u \cdot \nabla (\rho \text{Vol})=0$$
Since the volume is constant, then it reads
$$\frac{\partial \rho}{\partial t}+ u \cdot \nabla \rho=0$$
However, the conservation of mass (the correct one as far as I know)
$$\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho u)=0$$
It is not clear What's wrong I am doing here.
| Both forms are perfectly consistent in the particular case that the divergence of the 3-D vector field is zero. Recall, that the mass continuity equation can also be written as:
$$\frac{D\rho}{Dt}=-\rho\cdot\operatorname{div}V.$$
| {
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Why do we need insulation material between two walls? Consider a slab made of two walls separated by air. Why do we need insulation material between the two walls. Air thermal conductivity is lower than most thermal conductivities of insulating material and convection cannot be an issue in the enclosed volume: hot air rises, so what? it won't go any further than the top of the cavity.
| Complementing the excellent answers on convection; some heat can also transfer through radiation, if the internal surfaces of the two walls are not treated to counter this. Both the warm and the cold wall will radiate, but at different rates, and this results in an equalizing heat flow.
A quick calculation using [1] gives that a wall at 25 C will radiate 85 W/m2 to a wall at 0 C, if the walls have an arbitrary emissivity of 0.64.
Putting some IR-blocking material in-between the walls decreases this mode of heat transfer.
[1] https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html
| {
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Protons and QED I have been researching quantum electrodynamics recently and I have found out that when electrons repel each other, they constantly exchange photons with each other. When two protons repel, do they also exchange photons? If not, what do they exchange (if at all anything)?
| This is the lowest Feynman diagram showing the exchange of virtual photons in electron electron scattering
|
Depending on the problem one would be studying with proton-proton elastic scattering, a similar diagram with "p" instead of "e" would be used for the electromagnetic interaction.
Considering that the protons are composed of charged quarks, for high enough energies of the proton scattering strong interactions will become important, even the quark structure..
| {
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Can I apply Kirchhoff's current law to two separate nodes which are connected only by an ideal wire?
Suppose there are 2 nodes A and B in the above circuit. If A has 2 incoming currents $I_1$, $I_2$, and B has 1 incoming currents $I_3$, $I_4$, is it correct to write the equation of whole nodes and circuit as $I_1+I_2-I_3+I_4=0$?
| In your drawing A and B are not separate nodes, they are different points on the same node. All points connected by a single contiguous conductor are a single node.
If you tried to treat them as separate nodes then you would have to consider the conductor between them to be a 0 ohm resistor. This would do nothing but add some additional equations which would be more complicated but would eventually boil down to $V_A=V_B$, which we already knew by inspection of the circuit diagram.
| {
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Stefan–Boltzmann law applied to the human body The average person consumes 2000 kcal a day, which is equal to ~100 W. Furthermore, if one uses the Stefan–Boltzmann law to calculate how much someone loses heat due to radiation, it can be seen that it equals
$$Q=\sigma T^4 \varepsilon A$$
$$Q\approx1000\ W$$
Considering a surface area of ~2 m², an emissivity of 0.98 and a temperature of 36.5 °C.
However, this is clearly much greater than the maximum possible heat output of a human body, and that doesn't even consider convection and conduction, which would make heat loss even greater. So what is wrong with this analysis?
| Building on the answers from Thomas Fritsch and fraxinus: Alternatively, we could find the ambient temperature at which the human body would not need any heat-control measures such as clothing, unusual exertion, or perspiration. Working the equation in reverse, we can solve for $T_\text{environment} $, given $Q_\text{absorbed} = (1000 \;\text{W} - 100 \;\text{W})$. This yields $T_\text{environment} = 27 °\text{C}$, which seems reasonable to me.
However, as fraxnius points out, the normal temperature of limbs is considerably lower than 36.5°C, so the true answer is more complicated.
| {
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How small would a neutron star be to see the entirety of it? How small in Schwarzschild radii would a neutron star need to be for its gravity to be strong enough to deflect light emitted from one side toward an observer on the opposite side? I know the figure is above $1.5R_s$.
| For a Schwarzschild spacetime outside the neutron star (i.e. spherically symmetric and non-rotating), the neutron star surface would need to be at a radial coordinate $\leq 1.76 r_s$ (e.g. Pechenik et al. 1983). This corresponds to an apparent radius at infinity of $\leq 2.68r_s$.
The derivation (and it is a numerical problem) is to just figure out the closest approach to a Schwarzschild object at which light from infinity is bent through 90 degrees. This means that light emitted tangential to the surface from this minimum radial coordinate at the back of the neutron star will bend through 90 degrees and reach an observer at infinity on the opposite side.
EDIT: I attach a plot from a script I've written to calculate the total deflection angle either as a function of the impact parameter of light ($b$) or the closest approach ($r_{tp}$). The plot shows that a total deflection angle of 180 degrees (corresponding to a 90 degree bend for light emitted tangentially from the centre of the opposite side of the neutron star) occurs for $r_{tp}=1.76r_s$ and for an impact parameter of $2.68r_s$. In other words, the apparent radius of the neutron star at infinity would be $2.68r_s$, with the centre of the rear-side of the neutron star forming the outermost ring of the apparent image.
| {
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Peskin and Schroeder equation 3.136 (book edition 1995) I'm studying Peskin and Schroeder's QFT and I'm confused by equation 3.136 on page 68:
$\textbf{Previously, on page 48, equation 3.62 says:}$
$\textbf{My question is: how do we deduce 3.136 from 3.62?}$ $\textbf{It doesn't seem to me that} \boldsymbol{\xi^s}\textbf{ and }\boldsymbol{\eta^s}\textbf{ are related to each other?}$
| You must have an old edition. In my edition (3.136)has the $\sigma$ as well. Also, $\xi$ and $\eta$ are arbitrary spinors.
| {
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Can current flow in a simple circuit if I enclose the battery in a faraday cage? So suppose I have a regular circuit with a battery connected to a resistor and a lightbulb.
Suppose now somehow the battery is inside a metal box (faraday cage) but the rest of the circuit is outside of it so the wire is maybe poked through a tiny hole in the box.
Since energy flow through a circuit is due to the electromagnetic field as described by the Poynting vector, since the field cannot penetrate through the faraday cage, will current flow through the circuit?
| Good question, related to a controversial Veritasium video. My answer is yes, current will still flow and the bulb will still light. While the region outside the box is shielded from the field inside, there is no reason the portion of the wire outside the box can't generate its own E & B fields.
BTW, while I believe that Veritasium (Derek) is correct in spirit, I disagree with his answer to the multiple choice question. I believe that the answer to his question is none of the above. The current won't ramp up appreciably until about one RL-time constant elapses.
| {
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Vector Addition of velocity
In the arrangement shown in figure, the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with what speed?
I tried to solve this problem with the following way-
Let the upward speed of the mass M be x.
Then,
$$x=\sqrt{u^{2}+u^{2}+2u^{2}\cos 2\theta}$$
$$\implies x=u\sqrt{2(1+\cos 2\theta)}$$
$$\implies x=u\sqrt{4\cos^{2}\theta}$$
$$\implies x=2u\cos \theta$$
Hence, the upward speed of mass M is $2u\cos \theta$.
But the corrcet answer given is $\dfrac{u}{\cos\theta}$.
Can anyone please tell me that where i am doing mistake.
| You are working this problem backwards. Starting from the velocity $u$ and resolving it on $M$ in the vertical direction and adding the two sides up. This is incorrect since velocities do not add up like forces.
While a body might have multiple forces acting on it, there is always only one motion state (velocity + rotation).
So start with the velocity of $M$ upwards with $x$ and find how fast the length $MB$ is changing. This speed is the same as $u$ on the other side of the pulley. To match the vertical component of $u$ with $x$ you need.
$$ u \cos \theta = x $$
| {
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Scalar in compactification in several dimension In sec-$8.4$ $($String theory vol $1$$)$ Polchinski states that
With more than one compact dimension, the anti-symmetric tensor also has scalar components $B_{mn}$
I am not understanding why the scalar is identified to the element $B_{mn}$. The scalar should come from decomposing symmetric-traceless, trace part of metric of the coordinates which are compactified? Like the scalar $g_{25,25}$ when we did KK reduction of $g_{\mu\nu}$. Most probably I’m missing some part?
| First of all, when I hear "the trace part" of a metric in $D$ dimensions, I think of $g_{\mu\nu} g^{\mu\nu}$ which is just a constant.
Polchinski is talking about tensors which have a scalar as one of their restricted representations. In other words, exactly what you were doing by choosing indices that point in a compact direction. The observation is that for a symmetric tensor in $D$ dimensions, you only need to compactify one dimension to get a scalar. This would be $g_{25, 25}$. On the other hand, you would need to compactify at least two dimensions to get a scalar from the $B$ field because it is anti-symmetric. That means $B_{25,25} = 0$. You need to be able to write something like $B_{24,25}$.
| {
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Landau levels degeneracy in symmetric gauge I'm reading David Tong's lecture notes on the Quantum Hall Effect.
When symmetric gauge taken, a basis of the lowest landau level wave functions is
$$\psi_{LLL,m}\sim\left(\frac{z}{l_B}\right)^m e^{-|z|^2/4l_B^2},$$
where $z=x-iy$,
and we have
$$J_z\psi_{LLL,m}=\hbar m \psi_{LLL,m}.$$
On page 25, it says that
the profiles of the wavefunctions form concentric rings around the origin. The higher the angular momentum $m$, the further out the ring.
The wavefunction with angular momentum $m$ is peaked on a ring of radius $r=\sqrt{2m}l_B$. This means that in a disc shaped region of area $A=\pi R^2$, the number of states is roughly (the integer part of)
$$N=R^2/2l_B^2=A/2\pi l_B^2$$
I can't understand these two statements. I think the profile of $e^{-|z|^2/4l_B^2}$ does form concentric rings around the origin, but does not when multiplied by $(\frac{z}{l_B})^m$. And why $r_{max}=\sqrt{2m}l_B$?
For the second statement, my understanding is that it divide the area in real space by the area "a wave function occupies", but if this is the case, shouldn’t there be a $m$ in the denominator?
| *
*$r_\text{max}$ is the location where $|\psi|^2$ is maximized. Even after multiplying the wavefunction by $z^m$, $|\psi|^2$ is still symmetric under rotation around the origin (only a function of $|z|$), so in this sense the wavefunctions still represents concentric rings.
*Remember the integer $m$ actually labels different eigenstates. The area of the annulus between two neighboring states is $\pi r_{m+1}^2-\pi r_m^2=2\pi l_B^2$, so the number of states is roughly $\frac{A}{2\pi l_B^2}$. Alternatively, as $r_{\text{max}}$ grows with $m$, the maximal value of $m$ before the wavefunction completely goes out of the disc is determined by $\sqrt{2m}l_B\leq R$, which gives $m\leq \frac{R^2}{2l_B^2}$. So that's roughly how many eigenstates there are in the disc.
| {
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Does capacitance between two point charges lead to a paradox? Is it possible to have a capacitance in a system of two point charges? Since there is a potential energy between them and they both have charges then we can divide the charge by the potential and get capacitance.
However, capacitance is supposed to depend only on geometry so should therefore be zero. How does one resolve this paradox?
| Last page in this link:
Capacitance of an Isolated Sphere
We obtain the capacitance of a single conducting sphere
by taking our result for a spherical capacitor and moving
the outer spherical conductor infinitely far away
Result:
$C=4 π ε_0 R$
where R is the radius of the sphere.
As you are talking of point particles, $R=0$ .
One could extend the logic to two point particles, two spheres, and again $R=0$ for each will give zero capacitance.
| {
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Seeing gauge-covariance of the wavefunction for a uniform magnetic field Consider the Hamiltonian
$$H=\frac1{2m}(\mathbf p-q\mathbf A)^2+q\phi$$
and let $\psi$ be a solution to the Schrödinger equation
$$i\hbar\frac{\partial\psi}{\partial t}=H\psi$$
Then if we gauge transform
$$\phi\rightarrow\phi'=\phi-\frac{\partial\phi}{\partial t} \quad\quad \mathbf A\rightarrow\mathbf A'=\mathbf A+\nabla\Lambda$$
for any scalar field $\Lambda$, the corresponding Hamiltonian has a solution
$$\psi'=e^{iq\Lambda/\hbar}\psi$$
This is the gauge-covariance of the Schrödinger equation for the minimally-coupled Hamiltonian, which I buy. I am trying to see this in practice for the case of a uniform magnetic field $\mathbf B=B\hat{\mathbf z}$. If I choose a gauge $\mathbf A=-By\hat{\mathbf x}$, I get the solutions
$$\psi_n(\mathbf x) = e^{i(k_xx+k_zz)}e^{-\rho^2/2}H_n(\rho)$$
where
$$\rho=\sqrt{\frac{qB}\hbar}\left(y+\frac{\hbar k_x}{qB}\right)$$
and the $H_n$ are Hermite polynomials. Now, this is manifestly gauge-dependent, which we can see in the asymmetry between the $x$ and $y$ directions. Moreover, if I had chosen instead the gauge $\mathbf A=Bx\hat{\mathbf y}$, I would have gotten the solutions
$$\psi_n'(\mathbf x) = e^{i(k_yy+k_zz)}e^{-\rho'^2/2}H_n(\rho')$$
where now
$$\rho'=\sqrt{\frac{qB}\hbar}\left(x-\frac{\hbar k_y}{qB}\right)$$
The gauge transformation taking me to this second case is $\Lambda=Bxy$, but clearly,
$$\psi'_n(\mathbf x)\neq e^{iqBxy/\hbar}\psi_n(\mathbf x)$$
Perhaps more to the point, we learn as we solve that in the first gauge we have a harmonic oscillator in the $y$-direction, and in the second gauge it is in the $x$-direction. How can these be equivalent up to a gauge transformation? What am I missing?
| Take the gauge $\mathbf{A}=Bx\mathbf{y}$ first. In this case, eigenstates are labeled not just by $n$, but also by $k_x$. In other words, one should really write $\psi_{n, k_x}(\mathbf{x})$. For each $n$, there is actually an infinite number (if the system is infinitely large) of degenerate eigenstates, all with energy $\hbar \omega_c(n+1/2)$.
Similarly, in the other gauge $\mathbf{A}=-By\mathbf{x}$, the eigenstates should be labeled as $\psi'_{n, k_y}(\mathbf{x})$. Again, fixing $n$ there is an infinite number of them.
The statement is then the gauge transformed wavefunction $e^{iqBxy/\hbar}\psi_{n,k_x}(\mathbf{x})$ is an energy eigenstate of the Hamiltonian in the other gauge, but it doesn't have to be exactly one of those $\psi'_{n,k_y}$. In fact, it must be an superposition of different $k_y$'s to create something localized in $y$.
| {
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Why light radiates out of the bend in dielectric waveguides?
Why EM wave radiates out of the waveguide. Is this similar to centrifugal force?
| The confinement of wlectric field intensity within a dielectric waveguide depends upon refractive index, wavelength, and geometry of waveguide.
The total internal reflection is mostly used in ray model to diescribe the acceptance angles and propogation of light in optical fiber. The bending of optical fiber changes the incident angle of light making it smaller than critical angle.
For electromagnetic radiating outside of waveguide, it's better to take wavelength and refractive index in consideration.
If wavelength is higher than waveguide geometry, higher modes will leak out.
If index contrast of waveguide is low, radiation will leak out easily for higher wavelength.
There are still some waveguide with really small radius and still don't radiate the light because of high-index contrast.e.g.
Resonators are made up of SOI waveguides having silicon core (n=3.47) surrounded by a silicon oxide bottom cladding (n=1.44) leads to very strong confinement allowing the light guiding in bends with very small radii without radiation losses.
| {
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If you were invisible, would you also be cold? If you were invisible, would you also be cold? (Since light passes through you, so should thermal radiation.)
Additionally, I'd like to know if you were wearing invisible clothes, would they keep you warm? In my understanding, the heat radiation from the body would pass through the cloth.
Is it even necessary to be permeable for heat radiation in order to be invisible? Could there be a form of invisibility (hypothetically speaking, of course) that makes you permeable for light in the visible spectrum, but not for heat radiation? Can those two things be separated?
| Warm dark matter is not what you may think, but rather a hypothetical component of dark matter, which may consist of, e.g., sterile neutrinos.
The point here is that temperature is essentially the average kinetic energy (at least for a gas), and it does not fundamentally matter whether the particles interacat with the electromagnetic field. Hence, a gas of neutrinos has a well-defined temperature, despite being invisible (to all wavelengths).
Note that while neutrinos and similar particles interact very little and so would probably not form molecules or complex structures, there not fundamental obstacle to imagine another sector of matter which has significant interactions in itself, but is decoupled from standard model matter.
| {
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Landau and Lifshitz - collisions between particles In the book mechanics from Landau & Lifshitz, section 17 collisions between particles there are those two equations in page 46:
$$\tan \theta_1 = \frac{m_2 \sin \chi}{m_1+m_2\cos\chi}, \quad \quad \theta_2 = \frac{1}{2}(\pi-\chi).$$
How did they derive these equations?
The vectors in the image are:
$$\textbf{p}_1' = m \mathcal{v} \textbf{n}_0 + m_1\frac{\textbf{p}_1+\textbf{p}_2}{m_1+m_2}$$
$$ \textbf{p}_2' = -m \mathcal{v} \textbf{n}_0 + m_2 \frac{\textbf{p}_1+\textbf{p}_2}{m_1+m_2},$$
and from these we get
$$\vec{AO} = \frac{m_1}{m_1+m_2} (\textbf{p}_{1} + \textbf{p}_{2})$$
$$\vec{OB} = \frac{m_2}{m_1+m_2} (\textbf{p}_{1} + \textbf{p}_{2})$$
$$\vec{OC} = m \mathcal{v},$$
given that $m$ is the reduced mass
$$m = \frac{m_1 m_2}{m_1+m_2}.$$
A link to the book:
https://cimec.org.ar/foswiki/pub/Main/Cimec/MecanicaRacional/84178116-Vol-1-Landau-Lifshitz-Mechanics-3Rd-Edition-197P.pdf
| You got so close to an answer.
The authors are considering a situation when $|\vec p_1|= m_1v$ and $|\vec p_2|= 0$
This means that $|\vec{AO}| = \dfrac{m_1^2\,v}{m_1+m_2}$ and $|\vec{OB}| = \dfrac{m_2\,m_1\,v}{m_1+m_2} =|\vec OC|$
Taking the common factor, $\dfrac{m_1\,v}{m_1+m_2}$ out of each of the lengths results in the following diagram.
Noting that $OB=OC$ for the second relationship the required equation,
$\tan \theta_1 = \dfrac{m_2 \sin \chi}{m_1+m_2\cos\chi}$ and $\theta_2 = \dfrac{1}{2}(\pi-\chi),$
then follow.
Although they do not answer your question you might find these two articles of interest?
Diagrammatic Approach for Investigating Two Dimensional Elastic Collisions in Momentum Space I: Newtonian Mechanics
Diagrammatic Approach for Investigating Two Dimensional Elastic Collisions in Momentum Space II: Special Relativity
| {
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When a block slides down an incline, why does the incline move back, if the work is done by an external force (gravity)? If I put a point mass on the top of the incline, and if all surfaces are frictionless, I heard, that the incline is going to move back a little (depending on the mass difference), because momentum is conserved and velocity of the center of mass should remain constant. How is this possible though when, it's the gravity (an external force) which is acting on the mass?
| As @Triatticus rightly mentioned, there is no external force in the horizontal direction. Hence the COM's x coordinate (assuming an x-y plane with x axis along the horizontal direction) should stay constant. Now as the point mass is moving away from, the incline plane has to compensate for that and hence it moves in the opposite direction. Depending on the mass differences, how far both the bodies move can be calculated.
| {
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Spacetime effects on human scale objects? For a human standing upright on the earth, gravity would have a different value at the feet than at the head, and gravity influences the flow of time. Does the difference in the flow of time cause any effects?
I was toying with the idea that gravitational acceleration is just nature trying to compensate for time flowing at different speeds with a preference for moving towards slower timeflow.
Highschool level question.
| The time dilation depends on the factor $\sqrt{1-\frac{2GM}{rc^2}}$ see for example gravitational time dilation
From this formula the ratio of the flow of time for two points with a height difference of $h$ is about
$$1+\frac{GMh}{r^2c^2}$$
and using data for the earth and $h=2$, the time difference that would pass over a human lifetime for someone's head and feet is about $3 \times 10^{-7}$ seconds.
| {
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What does GR get right that QFT gets wrong, and vice versa? I wondering what precisely it was, in terms of predictions of observations, that General Relativity gets right, that QFT cannot explain. And what QFT gets right, that GR cannot explain.
I'm assuming GR cannot predict quantum effects, like wave-particle duality, but is there anything else? Or a more thorough list?
| GR is not intended as a model of particle physics, and makes no predictions in that field, and QFT is not intended as a model of spacetime curvature and makes no predictions in that field either. They were formulated to solve completely different classes of problems and it is not particularly surprising that there are problems in one field that can't be solved with the methods of the other.
| {
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If $\mathbf{F}_{net} = m\mathbf{a}$ then how is $m\mathbf{a}$ not a force? The question is in the title: If Newton's second law says that the sum of the forces acting on a body in a given direction is the same as the mass of the object times its acceleration in that direction, then how is $m\mathbf{a}$ not a force? Every book I have read on physics (all basic) says that $m\mathbf{a}$ is not a force. Are forces not "closed" under addition? Is this somehow a loose version of equality?
$$$$
"University Physics" by Young and Freedman says...
Acceleration is a result of a nonzero net force; it is not a force itself.
I guess that's just not enough explanation for me. How can a force be equal to something that is not a force?
| "I'm so hungry, I'd be willing to pay \$100 for a hamburger right now. In other words, my willingness to pay is equal to \$100."
"But a dollar is not the same thing as a willingness! Dollars are green pieces of paper, or balances in checking accounts, but a willingness is a psychological state!"
"Maybe so. But you still understood me, didn't you?"
| {
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Where is quantum probability in macroscopic world? How can macroscopic objects in real world have always-true cause-effect relationships when underlying quantum world is probabilistic? How does it not ever produce results different than what is predicted by Newtonian physics, except for borderline cases?
| Expectation values (of position, for example) in QM track the classical analogues. For large objects, position is localized and so is momentum (given the relative smallness of planck’s constant) - which keeps the position localized. In turn, this means that the expectation values of position closely match what we actually observe.
| {
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Can a rotating neutron star collapse to a black hole through a reduction in rotation? It is well known that non-rotating neutron stars cannot grow without bound, since an increase in mass causes a proportional increase in density, and accretion beyond a critical limit would cause it to collapse to a black hole.
However, it seems that a rotating neutron star could support a supercritical mass, since the additional centrifugal force of rotation would balance its gravitational force. Of course, the star could not be arbitrarily massive since an increase in mass must be accompanied by an increase in rotation, and there is a limit to how fast a star can rotate before breaking apart.
Regardless, we know that neutron stars can rotate as fast as 716 revolutions per second, and that many have extremely strong magnetic fields produced by magnethohydrodynamic dynamos, fields which can persist due to currents in the proton-superconductor phase of matter that exist at intermediate depths within the neutron star.
Therefore, I wonder if it possible that neutron stars could, in effect, lose energy and angular momentum through the production of magnetic fields associated with rotation, and thus slow until centrifugal forces diminish enough that the star collapses to a black hole?
If this is theoretically possible, what might the signature for such an event be? On what timescale would it occur? How common could it be?
| Rotation periods of the order of a millisecond can sustain a maximum mass that is only a few per cent larger than the maximum mass predicted by the TOV equations. Differential rotation can increase the effect, see this paper for some (very) extreme (and probably unphysical) examples. More details are given in the references of the paper (just take a look at the introduction).
Hypermassive neutron stars can be the remnant of neutron star mergers. Their differential rotation is killed by viscosity, and the star then collapses to a black hole. Otherways to spin-down a hypermassive neutron star id radiation emission (electromagnetic and gravitational). However, for the post-merger remnant, the most efficient mechanism is thought to be viscosity.
| {
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Does light have mass or not? We know light is made of photons and so it should not have mass, but light is a form of energy (light has energy) and has velocity ($c$), so according to $E=mc^2$, light should have mass... So what is correct?
| The value $E/c^2$ has sometimes been called relativistic mass, but we make an effort nowadays to discourage that term because it refers to a frame-dependent quantity, and it's more helpful to focus on the invariant mass. A free particle has energy-momentum relation $E^2=m_0^2c^4+p^2c^2$ (see here for a generalization to a non-free particle, but we'll overlook that for now), with $m_0$ a symbol we have to use for invariant mass, instead of the preferred $m$, if someone's already claimed $m$ for $E/c^2$ in the discussion. I'll stick with $m_0$ for now, since we're already "in this deep".
With that preamble out of the way, a photon in the vacuum has $m_0=0,\,E=|p|c>0$. The photon has zero invariant mass, and no rest frame. This makes "rest mass", another term that's been used for invariant mass, unnecessarily confusing! (It's less of a problem for particles with $m_0>0$, which do have a rest frame; indeed, light in a medium with refractive index greater than $1$ has been modelled in terms of $m_0>0$ photons.) But any photon, regardless of the context, has a frame-dependent $E/c^2>0$, which was discussed above.
| {
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Do Einstein's two postulates allow for handling acceleration in special relativity or is something else needed? When I was taught special relativity, we started with Einstein's two postulates and worked from there. However we were also taught that a proper resolution of the twin paradox required general relativity - because one twin accelerates. Apparently this was Einstein's opinion as well.
However modern texts, such as M,T&W's Gravitation, state that special relativity can handle the paradox. Specifically they state that when a uniformly accelerating observer momentarily passes a non-accelerating observer travelling at the same velocity, they will agree that their clocks are running at the same speed. With that statement, if accepted as part of special relativity, the twin paradox can be resolved.
However, I do not see how this last statement follows from Einstein's two postulates. Does it? Or is special relativity, as understood now-a-days, reliant on more than the two postulates?
| I would guess you are referring to the clock hypothesis. Annoyingly I cannot find a web page that neatly describes this, but it is discussed in detail in the question What is the history of adding the Clock Hypothesis to Special Relativity? on the History of Science and Mathematics Stack Exchange.
As far as I know Einstein never stated this and did not include it in his postulates. However Minkowski's formulation of special relativity as a metric theory implicitly assumes it is true.
The way we approach special relativity these days owes more to Minkowski than Einstein, and if we use the geometric approach handling acceleration is straightforward. For example the chapter in MTW (chapter 6 in the first edition) derives the coordinate transformations straightforwardly from the fact the four-velocity and four-acceleration have to be normal to each other.
If you are interested, the resolution to the twin paradox using the geometric approach is described in What is the proper way to explain the twin paradox? though this will be challenging for the novice.
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Why does mass bend the temporal dimension more than the spatial dimensions of spacetime? From my (limited) understanding of general relativity, most of what we experience as gravity is a result of the distortion of the temporal dimension, and not the spatial dimensions. Therefore, most of the spacetime curvature caused by the earth (and most astronomic objects, with the exception of maybe black holes) occurs along the temporal dimension, with very little on the spatial dimensions. This is why the bent sheet analogy is misleading, if I am not mistaken. Why is this so? Why aren't all four dimensions distorted equally, or the spatial dimensions distorted more than the temporal?
| It is the result of our choice of time and spatial coordinates. For practical reasons, we use seconds and meters, so the accelerations of gravity in the earth and solar system are far from negligible.
Suppose that our perception of time were such that we used microseconds instead of seconds, while keeping meters for spatial distances. Everything would seem to be almost at rest, because our daily objects barely moves on a scale of meters and micoseconds. And if things were viewed as at rest, the situation would be similar to a flat Minkowskian spacetime both in temporal and spatial dimension.
| {
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Is the electron a pointlike particle? And if yes, how is that possible, because the energy then would diverge, wouldn't it? My problem is that I read (besides others in this post Why are electrons and quarks 0-dimensional?) that the electron is a point-like particle. My question is on the one hand whether that is true and on the other hand if the electrostatic energy of the electron would not diverge if it was a point-like particle?
| I think you are referring to the diverge of the electrostatic energy of a point-like particle. In classical electrodynamics that is the energy of the electric field which is associated to the charge of the electron. As you are saying, the $1/r^2$ energy of the field diverges at the r=0 position and even its integral from 0 to infinity would diverge.
The solution to this apparent paradox lies in what it means to be point-like. In quantum field theories, such as quantum electro-dynamics, as r approaches zero you need to consider quantum fluctuations, which, among other things add new charged particle in a cloud around the electron.
Therefore, when $r\sim1/m_e$ it is likely to see not just a naked electron, but an electron surrounded by a pair of particles (total 3 particles!): the original electron, one "virtual" positron and one "virtual" electron, which can exists only for times and separations comparable to $m_e$.
This makes the energy of the "point-like" electron much better behaved at short distance. Still, it diverges as the logarithm of the shortest distance you consider. In the end you have to consider renormalization of the theory to get finite answers.
| {
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Why bi-telecentric lens works with a finite focal length while it should be afocal system？ I bought a bi-telecentric lens for experiment. From all I've learned before, it accepts only axis-parelle incoming beams and exits the same parelle light, so it should be an afocal system with relatively constant imaging performance regardless of working distance.
But it's not.
It blurs image when nearer or further from a fixed working distance, just like a regular lens. Why is that? I've looked up many explanatory articles and descriptive drawings, still no clue. Customer service man is of no help, unfortunately.
Really appreciate any help from you guys!
| A telecentric lens is not necessarily afocal.
An a focal system is a system which accepts light from infinity (collimated light) and 'sends' it to infinity at the output (image is at infinity). The most typical example is a laser beam expander.
a telecentric lens is a lens where the aperture coincides with the back\front focal plane of the lens (or both, in the case of bi-telecentric lens), but this does not mean that the system is afocal, rather it means that all chief-rays (or light cones) are parallel to the optical axis (the image below should help illustrate). It also means that the pupil of the system is at infinity, but this is less intuitive. An added benefit of this fact is that for object-space telecentric lenses, perspective error (magnification vs. distance) are reduced in the resulting image.
| {
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A coupled nonlinear dynamical system in four dimensional phase space I have come across a coupled nonlinear dynamical system given below
$$ r\, \ddot{x} + \dot{x} = \sin y~,$$
$$ r\, \ddot{y} + \dot{y} = \sin x~,$$
where $r$ is some real number and $\dot{x}$ denotes $\frac{d\, x(t)}{d\,t}$. Despite the fact that the system is simple looking, numerical study (by finding the large time Lyapunov exponents) shows that the system exhibits periodic (limit cycles) and chaotic behaviors for the parameter $r = \mathcal{O}(1)$ at large time limit. Depending on the initial conditions $(x(0),y(0),\dot{x}(0),\dot{y}(0))$, for a particular value of the parameter $r \sim 1$, the system exhibits these interesting dynamical properties. However, theoretically, I am not able to explain why this happens. One resembling system I have seen is the Standard map. However, that too does not explain why my system shows these strange behaviours. Any insight will be helpful.
| Chaos is very common — regular behavior is actually the special case.
Your system is nonlinear and has more than 3 dimensions, that is in principle enough for chaos to be possible or even likely — see, e.g., 1, 2 or the books suggested here, but, in a nutshell, 3 dimensions give you enough "room" for the continuous trajectories to be complicated, and nonlinearity is needed for the stretch and fold mechanism prototypical of chaos.
| {
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Does Newtonian mechanics work in polar coordinates? Our teacher suggested that Newtonian Mechanics only applies in cartesian coordinates. Is this true?
He gave this example.
Suppose there a train moving with constant velocity $\vec{v}=v_0\hat{x}$, with initial position vector $\vec{r}=(0, y_0)$, where $v_0,y_0$ are constants. He argued that Newton's second law would not hold in polar coordinates. Any ideas?
(We can assume 2D or 3D cases as well, so spherical or polar, it doesn't really matter)
| The advantage of cartesian coordinates is that we can take vectors as simply indexed functions. But there is an additional feature that can not be forgotten when changing variables: coordinate basis.
The velocity vector of the OP example can be expressed in its complete form as:
$v_1(X^1, X^2) = V^1B_{11} + V^2B_{21}$
$v_2(X^1, X^2) = V^1B_{12} + V^2B_{22}$
In cartesian coordinates, $X^1 = x$, $X^2 = y$, $V^1 = v_x$, $V^2 = v_y$, $B_{11} = 1$, $B_{12} = 0$, $B_{21} = 0$, $B_{22} = 1$
That way:
$v_1(x,y) = V^1$ and $v_2(x,y) = V^2$ are the familiar cartesian components of the indexed function (and vector) $v(x,y)$.
When transforming to polar coordinates, it is possible to derive $B_{ab}$ and $V^a$ such that $v_b$ don't change. Here:
$B_{11} = cos(\theta)$, $B_{12} = sin(\theta)$, $B_{21} = -Rsin(\theta)$, $B_{22} = Rcos(\theta)$, $X^1 = R$, $X^2 = \theta$, $V^1 = v_R$, $V^2 = v_{\theta}$
So:
$v_1(R, \theta) = v_Rcos(\theta) - v_{\theta}Rsin(\theta)$
$v_2(R, \theta) = v_Rsin(\theta) + v_{\theta}Rcos(\theta)$
In the example, $v_1 = v_0$ and $v_2 = 0$. It is easy to realize that to keep the same values:
$$v_R = v_0cos(\theta)$$ and $$v_\theta = -v_0\frac{sin(\theta)}{R}$$
In polar coordinates, the components change with time for this vector be constant with time as desired. The vector equations of the Newton laws are valid, but the notion of what is a vector must be closely understood.
| {
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NMR/EPR with an RF electric field Is it possible to perform NMR/EPR spin alignment with an oscillating electric field instead of a magnetic field (so with a sample inside the RF electric field of capacitive plates rather than a RF magnetic field of a coil)? In other words, can an electric field align nuclei or does it just polarize the atoms without having that effect?
| No this will not work. The electron does not have an electrical dipole or higher moment. Note that an oscillating electric field implies an oscillating magnetic field not far away.
| {
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In the Double Slit Experiment, what is the longest time $T$ recorded between shots of particles, electrons, photons, etc.? How can we be certain that electrons fired do not leave some kind of residual interference for the next shot?
What is the longest time recorded between shots?
| You assume that spacetime is plastic, like skiers carving paths through the snow. I was told that this is not the case, in terms of current theory.
On the other hand, assume that the insides of the double-slit device are evacuated. Shoot a photon, let it collide with the sensor array. Then, flush the device with gas (like air), and re-evacuate before shooting the next photon. Does this affect the pattern on the sensor array over time? It was worth thinking about anyway.
| {
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Has anyone charged an object with 1 coulomb? Why was such a ridiculously large charge chosen as the unit of charge? The fact that two balls charged with 1 coulomb each would repel/attract each other from a distance of 1 metre with a force sufficient to lift the Seawise Giant would suggest me otherwise, but has anyone ever charged an object with 1 coulomb of net charge?
Why was such a ridiculously large charge chosen as the unit of charge? Or better, why did we give the Coulomb constant such a big value instead of using a value in the same order of magnitude of the Newton constant ($10^{-11}$)?
EDIT
For the historical reasons that explain why the coulomb was chosen as the unit of charge please refer to the good answers given to this question.
After a bit of research I have found that the highest voltage ever created is $32\,\mathrm{MV}$ at the Oak Ridge National Laboratory. With such a voltage the best we can do is charging a copper sphere the size of a basketball with around 424 microcoulombs:
$$Q = 32 \times 10^6\,\mathrm{V} \times 4\pi\epsilon_0 \times 0.119\,\mathrm{m} = 4.237 \times 10^{-4}\,\mathrm{C}$$
Such a sphere, when placed at a distance of $1\,\mathrm{m}$ from the surface of a similarly charged sphere, would experience a repulsion of $1052\,\mathrm{N}$ (the force needed to lift $107\,\mathrm{kg}$).
If the maximum voltage we can access is $32\,\mathrm{MV}$ and we want to charge a sphere with $1\,\mathrm{C}$, all we need is a sphere $561.7\,\mathrm{m}$ in diameter. It might often snow on the top.
| Actually the ampere (SI unit for electric current) was defined first
(in 1881, see Wikipedia: Ampere - History).
They chose this size for $1$ ampere, probably because at this time
such a current could be produced with a decent electrochemical battery,
and was easily measurable with a galvanometer by its magnetic effect.
The natural consequence of this is: A flowing charge of $1$ coulomb
(i.e. a current of $1$ ampere flowing for $1$ second) is also
a convenient unit, neither ridiculously large nor ridiculously small.
The fact, that a static charge of $1$ coulomb is a really big
thing, is an entirely different story, which has to do with
the large electric force between charges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why in the first Friedmann equation quantity $ρ$ is directly proportional to Hubble's constant despite the fact that gravity counteracts expansion? Here is the first Friedmann equation:
$$H^2 = \left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}$$
We know that matter and energy through gravity slow down or reverse any expansion in the fabric of spacetime. Yet in some context and specially here with this equation I encounter the fact that matter and energy content of universe is increasing the expansion rate instead of the opposite, as if there's an anti gravitational force in effect. How so?
| Your intuition that $\rho$ ought to slow the expansion is correct, but that intuition concerns $\ddot{a}$ not $\dot{a}$ (and indeed in the other Friedman equation you do see a minus sign).
The $(\dot{a})^2$ term comes from the expression for the curvature when the metric is of FLRW form. The Einstein field equation says there has to be a relationship between this curvature and $\rho$, and that is what you are seeing in the first Friedman equation. By comparing with a Newtonian description, it can be roughly interpreted as a statement about energy. $\dot{a}^2$ is related to kinetic energy, and $\rho$ relates to gravitational potential energy.
| {
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