Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Suppose infinite 3D space is filled with matter except a spherical cavity. Will a particle in the cavity experience force? Suppose infinite 3D Euclidean space is uniformly filled with dense matter except a spherical cavity. Will a particle in this cavity experience gravitational acceleration towards the nearest wall?
| Well, after reading other answers, I say different. Point me out where I am wrong and I would be glad to understand my mistake.
Suppose the entire region was filled with matter. So, using any kind of symmetry, there would be no force any particle. Now we remove a sphere, which contains that particle that we're observing. Isolating the particle and the sphere, were know that the force due to the sphere is same as in the case of force on particle inside a sphere with uniform mass distributed.
Thus, by removing the sphere, we tend to create some sort of 'negative' mass, with causes a force on the particle on the opposite direction.
Also note that there is symmetry only when the particle we consider is at the center of the sphere, wherein my answer also stands in agreement with the no force result.
Hope this helped!
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Difference between energy and coulombic efficiency of perfect battery I have been trying to reconcile coulombic and energy efficiency for a perfect battery, but failed. Where have I gone wrong?
a coulomb is a unit of charge. Equal to lots of electrons.
a battery has a fixed amount of chemicals which react and need a fixed amount of electrons to complete reactions between empty and full.
therefore an ideal battery needs a fixed number of coulombs
an amp is a coulomb per second
so, charge at 100 amps for 2 hours is 200Ah which is 720000 coulombs
discharge at 50 amps for 4 hours is 200Ah which is 720000 coulombs.
Everybody happy. Ideal battery has coulombic efficiency of 100%. Coulombs is constant.
charge occurs at 13v which is 1300 watts which is 2600 Wh.
discharge occurs at 12v which is 600 watts which is 2400 Wh.
where has the 200Wh of energy gone .............
I think the answer lies in the maths and assumptions but where .....?
I'm guessing either a missing term or a misunderstanding of coulomb.
| An ideal battery has the capability of driving a certain amount of charge around a circuit that being 200 Ah or 720 kC in your example. This represents 12 $\times$ 720,000 = 8.64 MJ of energy stored in the battery.
That is about it for an ideal battery other than actually giving it an infinite capacity, a constant terminal potential difference independent of the current though it and its temperature which implies zero internal resistance.
Compared to a real battery quite a number of assumption have been made.
The chemical reaction is not 100% reversible so the capacity of the battery drops with time and so the battery has a finite lifetime usually stated as number of time that it can be recharged.
Both for charging and discharging of the battery there are lossy elements within the battery which can be lumped together as an "internal resistance" and so energy is converted into heat within the battery.
If a battery is discharged at a greater rate its capacity to supply electrical energy drops.
With time during the discharging process the potential difference across the terminals of the battery drops showing that non-ideal processes are occurring within battery.
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Massless particles in a universe with compact extra-dimensions One common idea behind many extensions to the Standard Model (such as String Theory or Kaluza-Klein Theory) are small or hidden "Extra-Dimensions", that are compactified.
According to my understanding of Quantum Physics, this would result in each particle's wavefunction having a component into the direction of these extra-dimensions, and only discrete energy-states would be allowed (similar to electrons in an atom).
Now imagine a photon, which is considered to be a particle without rest mass in the Standard Model. Its wavefunction would also have components into the direction of the extra-dimensions. Consequently, it would have to occupy one of these energy states. So there would be some energy consisting out of the photon's standing wave in the extra dimension, which - according to my understanding - would behave just like a finite rest mass of the photon.
So how can there be particles without any invariant mass in a theory with compact extra dimensions?
|
The string tension Tstring is the energy per unit length of the string. If the string is wound w times around a circular dimension with radius R, then the energy Ew stored in the tension of the wound string is
The mass of an excited string depends on the number of oscillator modes N and Ñ excited in the two directions of propagation around the closed string, minus the constant vacuum energy. Kaluza-Klein compactification adds the quantized momentum in the compact dimensions, and the tension energy from the string being wrapped w times around the compact dimension, so that the total squared mass becomes
As an experimentalist skimming over theory I am satisfied that zero masses can exist even with compactified dimensions, since there exists a minus sign in the mass formula. :).
This also helps:
The theory gains extra massless particles when the radius R of the compact dimension takes the minimum value possible given the above symmetry of T-duality
One will have to do the maths to really convince oneself of the above statements, but that takes time and effort.
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Determinant as a fermionic path integral I understand that the determinant of a matrix can be written in terms of a fermionic path integral. The expression is:
$$Z = \int D\bar{\psi}D\psi e^{-\iint d^4x' d^4x \bar{\psi}(x')B(x',x)\psi(x)}\tag{1}$$
The proof proceeds by rewriting the complex Grassmann fields in terms of basis functions:
$$\psi(x) = \sum_n c_n\chi_n(x)\qquad \text{and}\qquad \bar{\psi}(x) = \sum_n \bar{c_n}\chi_n^\dagger(x)\tag{2}$$
This induces a change in the measure in the form:
$$D\bar{\psi}D\psi = \prod_nd\bar{c_n}dc_n.\tag{3}$$
I am failing to prove that the measure indeed changes in the above manner under the stated transformation. An explicit calculation would be much appreciated.
| there is a quick and dirty way of proving
$$\mathcal{D}\psi\mathcal{D}\bar{\psi} = \prod_n dc_nd\bar{c}{}_n,$$
although i admit that the method used is still subject of research.
First, we have to see how the integral $\mathcal{D}\psi$ is defined. You can find yourself that,
$$\mathcal{D}\psi = \prod_x d\psi(x)$$
so, consider the change of variables $\psi(x) = \sum_n c_n \chi_n(x)$, then
$$\frac{\partial \psi(x)}{\partial c_n} = \chi_n(x)$$
so, using,
$$d\psi(x) = \frac{\partial \psi(x)}{\partial c_n} dc_n$$
and the fact that volume measures change under a change of variables with the determinant of the Jacobian, we can write the identity,
$$\mathcal{D}\psi\mathcal{D}\bar{\psi} = \det\left (\frac{\partial \psi(x)}{\partial c_n}\right )\det\left (\frac{\partial \bar{\psi}(x)}{\partial \bar{c}_n}\right )\prod_n dc_nd\bar{c}{}_n=\det\left (\chi_m(x)\right )\det\left (\bar{\chi}_n(x)\right )\prod_n dc_nd\bar{c}{}_n$$
since $\chi_n(x)$ satisfies completeness relations,
$$\sum_n \chi_n(x) \bar{\chi}_n(x') = \delta(x-x')$$
$$\int dx \chi_n(x) \bar{\chi}_m(x) = \delta_{nm}$$
we can write that, as a matrix with indices $x$ and $n$, $\chi_n(x)$ satisfies,
$$\chi \cdot \chi^\dagger = \mathbb{1}$$
and so $\det \chi \det \bar{\chi}= 1$.
| {
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Is an electron attracted to one of the magnetic poles ...? I noted the question above had been posted. And I wanted to comment, but nay, it was locked out. However, what of the old 'CRT' tubes, in which magnetic fields are used to steer the electron stream ? Now I know the 'Electron's' do not get attracted, or repulsed directly to the source of the magnetism, but the magnetic field certainly has an effect. So if one were to extend the CRT tube to ... well, rather long, would the electron eventually move to, or away from the 'North' or 'South' pole ?
| The electron has an intrinsic magnetic dipole moment $\mu$ related to its spin. Therefore,in addition to the Lorentz force, it should also experience a net magnetic force in the gradient of a magnetic field and move in or against its direction depending on its spin orientation.Therefore, an electron should indeed be attracted or repelled by a magnetic pole.
| {
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Calculating height of a ball formula based on bounce I'm trying to figure out a formula for the new height of a ball, up until it stops bouncing.
The time iteration between each could be for example one second; and gravity of course would be 9.8
I've tried doing research at different websites such as physics @ illinois.edu and whatnot;
I want it to be regardless of properties of the ball and ground (such as material of the ball and ground).
So assume height starts at 10 meters, after one second it would be x height, then after another second it would bounce and go up to x height, then go down to x height, back up to x height, etc.
Assume it's bouncing straight up and down, so the only plane it moves on is the height plane.
The reasoning behind this is for a programming application; however once I have the formula for the ball bounce I can write the program itself, I just need to figure out the formula.
| Excluding drag the equation for the height at time $t$ is
\begin{equation}h(t) = h_0 - \frac{1}{2}g(t-t_0)^2\end{equation}
This is zero (ie at the ground) when $t = \pm\sqrt{\frac{2h_0}{g}}+t_0$. In this case, $h_0$ is the height from which it is dropped, and $t_0$ is the time when it is dropped. So assume $h_0 = h$ is the initial height and $t_0=0$ is the start time.
It hits the ground for the first time at $t = \sqrt{\frac{2h}{g}}$. Then you need to figure out your $h_1$, the height of the second bounce. This depends on how much energy you're losing, but a reasonable model would probably be $h_{i+1} = ph_i$, where $0<p<1$, although this means you will bounce forever. At some point, some other physics will take over.
You need to figure out $t_1$ (or in general $t_i$). In general it is going to be:
\begin{equation}t_i = t_{i-1}+\sqrt{\frac{2h_{i-1}}{g}}+\sqrt{\frac{2h_{i}}{g}}\end{equation}
This is because $t_i$ is the time at the apex of each bounce, and the $\sqrt{\frac{2h_i}{g}}$ is the time from the apex to the floor or vice versa.
Then use this generalization of the initial equation:
\begin{equation}h(t) = h_i - \frac{1}{2}g(t-t_i)^2\end{equation}
In each case, $t$ will run from $t_i-\sqrt{\frac{2h_i}{g}}$ to $t_i+\sqrt{\frac{2h_i}{g}}$, except for the first one, which runs from $t=t_0(=0)$ (ie from the apex, not the floor).
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Can an object *immediately* start moving at a high velocity? What I mean is, suppose a ball is fired from a cannon. Suppose the ball is moving at 100 m/s in the first second. Would the ball have started from 1m/s to 2m/s and gradually arrived at 100m/s? And is the change so fast that we are not able to conceive it? Or does the ball actually start its motion at 100m/s as soon as the cannon is fired?
Suppose a 40-wheeler is moving at the speed of 100 kmph. And it collides with a car and does not brake (The car is empty in my hypothesis ;-) ). If a body does not reach a certain speed immediately and only increases speed gradually, does that mean that as soon as the truck collides with the car, the truck momentarily comes to rest? Why I ask this is because the car is not allowed to immediately start at the truck's speed. So when both of them collide, the car must start from 0 to 1kmph to 2kmph and finally reach the truck's speed. Does this not mean that the truck must 'restart' as well? Intuition tells me I'm wrong but I do not know how to explain it physically being an amateur.
When is it possible for a body to accelerate immediately? Don't photons travel at the speed of light from the second they exist?
| Nothing propagates instantly
For starters, any real object has a non-zero size. If you apply force to one side of a cannonball, it will start moving before the other side of the cannonball can be affected by that force. The time will be very small ("so fast that we are not able to conceive it") but nonzero. The cannonball will experience some compression, and in general (for reasonable energies) the propagation will be limited by the speed of sound in that material.
So even without even going to gradual acceleration (the main purpose of the cannon barrel - there's a reason why they're so long!), even a simple transfer of momentum by a single impact (e.g one Newtonian ball hitting another) isn't instant.
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Does spin state matters during interaction? I'm asking whether the interaction between a pair of spin-up or spin-down electrons be any different from the interaction between a pair of electrons that comprises of opposite spin state? I think since the dipole moment is a physical property then I can assume spin-up and spin-down electrons are 2 different matter particle despite both come from a common field.
| I guess that the question is about the electromagnetic interaction. For simplicity, forget the magnetic part and consider only the electrostatic interaction. The electrostatic potential energy
$$V(\vec r_1,\vec r_2)={e^2\over 4\pi\varepsilon_0||\vec r_1-\vec r_2||}$$
depends only on the position of the electrons and not on their spins.
However, electrons are undistinguisable particles which implies in quantum mechanics that their wavefunction should be anti-symmetric under the exchange of two electrons. In the absence of interaction, the wavefunction of a free electron is the product of two terms: one depending on its position and one depending on its spin. Therefore, there are four possibilities to construct an anti-symmetric wavefunction for two non-interacting electrons. If the anti-symmetry is imposed to the spin part of the wavefunction, the total spin is equal to one. In contrast, if the anti-symmetry is imposed to the spatial part of the wavefunction, the total spin is zero. When one computes the shift of energy due to the electrostatic interaction for these four wavefunctions, one finds that the result depends on the total spin. The difference is called the exchange energy and is the origin for ferromagnetism.
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Torque of the contact forces about the centre of mass A person leans in the opposite direction when he or she lifts a heavy load in one hand.
I read a reason for the same in a book.
According to the book, if a body is placed on a horizontal surface, the torque of the contact forces about the centre of mass should be zero to maintain equilibrium. This may happen only if the vertical line through the centre of mass cuts the base surface at a point within the contact area or the area bounded by the contact point. That's why the person leans.
But I am unable to understand that how does it explain the leaning of person on lifting a heavy load.
| The person does not have to lean. When carrying a heavy load you should keep the load close to the vertical line through your centre of mass (CM). For stability, the combined CM (you plus load) must be between your feet. For heavy loads (eg a suitcase) you may need to lean slightly to achieve this, even when carrying close to your body.
If the object is bulky and does not have handles, this may prevent you from keeping the load close to your centre of mass. Then you are forced to lean backwards, even when using two hands, to place the combined centre of mass (your own plus that of the load) between your feet.
In all cases, it is the combined CM which must be between your feet.
| {
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Finding intensity/noise measurement of hysteresis loop I'm hoping I could get some help from the experimentalists on some data analysis.
I have a number of hysteresis loops taken at different angles around extinction of a polarizer on a photodiode. I want to find the loop (and by proxy the angle) with the optimal signal/noise ratio but I'm not sure how to go about analyzing the hysteresis loops to get a number that reflects this ratio.
Any help would be appreciated.
| You can use the symmetry properties of hysteresis loops to separate half of the noise from the signal. Let's say your signal is $U$, which is proportional to the magnetization, and you apply a field $H$. Even in the presence of exchange bias there is a point $(H_0,U_0)$ with respect to which the signal is antisymmetric: $U(H_0-\Delta H)=2U_0-U(H_0+\Delta H)$. The point $(H_0,U_0)$ can be determined from the coercive fields and the saturation signals. Now determine the symmetric and the antisymmetric part of the hysteresis loop with respect to that point. Your noise level is the standard deviation of the symmetric part normalized to the maximum signal in the antisymmetric part.
Note that this "noise" also includes quadratic Kerr signals (assuming you are measuring Kerr loops). But normally, you don't want to have them in your signal as well.
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Does a trumpet operate using an open air column or a closed air column Just as the title states. I could not find a coherent answer online.
Many thanks in advance
| The issue is slightly complicated, due to the ambiguity of the terms "open pipe" vs "closed pipe" used to classify wind instruments.
Some authors use it to refer to the boundary conditions on the ends of the instrument. If the pressure of the standing waves in a pipe must be continuous with the outside pressure on both ends of the instrument, then it is considered an open pipe, whereas if the pressure is continuous with environmental pressure on one end and at a maximum on the other end, then it is considered a closed pipe. This is the definition used by this page. By this definition, brass instruments (lip reed instruments) are closed pipes.
However, some authors use it to refer to the qualities of overtone series of an instrument. If an instrument has a full overtone series, then it is considered an open pipe. If an instrument only has odd multiples of the fundamental frequency in its overtone series, then it is considered a closed pipe. By this definition, brass instruments are open pipes.
However, the second definition is not very rigorous, since the shape of the instrument plays a very important role in its overtone series. Consider the clarinet and the saxophone. They have the exact same sounding mechanism (single reed on the closed end), but the clarinet only has odd multiples of the fundamental in its overtone series, while the saxophone has a full overtone series. The reason is that the former has a cynlindrical bore, but the latter has a conical bore. This is also intuitively explained by this UNSW page. Essentially, the wavefront in a conical bore is no longer planar, but spherical, which results in different allowed or forbidden modes.
I suspect the reason why the second definition arises is the misconception that all wind instruments can be approximated by long cylindrical pipes, which would mean that their harmonic series can be classified by their open-ended-ness.
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How to derive the formula for angular velocity in three dimensions? The formula for the velocity of a point $\vec {r}$ with angular velocity $\vec{\omega}$ is given by $\vec{v}=\vec{\omega} \times \vec{r}$.
But, how can we derive the formula $\vec{\omega}=\frac{\vec{r} \times \vec{v}}{|\vec{r}|^2}$ from the above formula, which is known?
| Use the identity (see this)
$$
{\bf v} = \left({\bf v} \bullet \hat{\bf r}\right) \hat{\bf r} + \left(\hat{\bf r} \times {\bf v}\right) \times \hat{\bf r}
$$
and the fact that ${\bf v} \bullet \hat{\bf r} = \left(\omega \times {\bf r}\right) \bullet \hat{\bf r} = 0$.
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Why is nuclear waste more dangerous than the original nuclear fuel? I know the spent fuel is still radioactive. But it has to be more stable than what was put in and thus safer than the uranium that we started with. That is to say, is storage of the waste such a big deal? If I mine the uranium, use it, and then bury the waste back in the mine (or any other hole) should I encounter any problems? Am I not doing the inhabitants of that area a favor as they will have less radiation to deal with than before?
|
But it has to be more stable
That's where you're wrong. Most of the decay products are much more radioactive than the $\rm U^{235}$ that was used in the reactor. Uranium is not very dangerous at all. I have held a uranium rod in my hand. Admittedly it was a) coated in nickel and b) $\rm U^{238}$ which is less radioactive than $\rm U^{235}$.
The energy released in the reactor is not the radioactivity of the $\rm U^{235}$. Instead, the energy is generated by an artificial splitting of the $\rm U^{235}$ nucleus by impact from neutrons. The reaction products have a smaller combined mass than the $\rm U^{235}$ had, and the difference in mass is converted into energy.
These fission products tend to be very unstable, so they decay rapidly and release a lot of radiation in the process. They include isotopes like strontium $\rm Sr^{90}$ and cesium $\rm Cs^{137}$. Both have a half-life of about 30 years. On top of that, $\rm Sr$ is taken up by the body as a replacement for calcium, so all its radiation is released inside the body.
As most of the fission products decay fairly rapidly, their danger also diminishes quickly. However, the word "quick" is relative. For example, 30 years is long in human terms but very fast compared to the half-life of $\rm U^{235}$, 700,000,000 years. Hence, especially the initial containment is crucial but, since other reaction products have half-lives measured in millennia, long-term storage is also very important.
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Why is silver the best conductor of electricity? I've been wondering why silver is the best conductor of electricity for a while now, and I've observed that in Group 11 transition metals where silver is located, copper and gold too are also one the best conductors of electricity (Cu, Ag, Au are all in Group 11). I believe that Cu Ag and Au share some physical similarities that makes them a very good conductor of electricity as they are in the same group. More importantly, why is silver, specifically silver (not gold nor copper) the BEST conductor of electricity?
| Electron-electron scattering is lower in silver. Conduction in the coinage metals is in the $sp$-band that are wide and quite free-electron-like. But there is scattering with the $d$-electrons. This is least important in silver where the $4d$-electrons are about 4 eV away from the Fermi-level (which is also why silver is colorless). But the $3d$ bands in copper and and the $5d$ bands in gold are closer to the Fermi level (absorption in the blue) and they cause more scattering with the $sp$ electrons.
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Shape of water in rotating bucket I need to show that the surface of water in a bucket rotating with constant angular velocity will have parabolic shape. I'm quite confused by this problem, but here's what I did:
$$\vec{F}_{cf} + \vec{F}_{grav} = -\vec{\nabla} U = m(\vec{\Omega}\times\vec{r})-mg\hat{z}$$ where $F_{cf}$ is the centrifugal force, $F_{grav}$ is the force of gravity, $U$ is potential energy, $\vec{r}:=(x,y,z)$.
So $\nabla U(z) = mg-m\Omega^2 z$, hence $$U(z) = gmz-\frac{1}{2}\Omega^2 z^2+C$$
which is a parabola.
In this approach I was trying to use the fact that the surface is equipotential for $F_{cf}+F_{grav}$. But apparently my approach doesn't quite hold any water. Please give some suggestion on approaching this problem, as I have no other idea.
| I'm pretty late to the question but I think I know a simpler way. You can solve this really easily with conservation of energy, the kinetic energy of a circle of water at a constant radius is $\frac{m r^2 \Omega^2}{2}$, in a idealised system the energy has nowhere else to go but increasing the waters gravitational potential energy, so $mgz = \frac{m r^2 \Omega^2}{2}$ or $z = \frac{r^2 \Omega^2}{2g}$.
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Path of light rays in uniform velocity elevator I have been told that light rays seen from inside an accelerated elevator will seen as following a curved path, parabola. Like the image below. But what happens when a pulse of light entered into a non accelerating but uniformly going upward. It seemed to me that like the image below the pulse will entered at the top of the elevator and then after an amount of time light will move forward a bit. Meanwhile the elevator will go a bit upward. so when viewing from inside it will appear that the light rays have again followed a parabolic path! in a non accelerating elevator! I know that can't be true. Where am I going wrong?
| Your picture is used to help visualize a thought experiment where the speed could be constant or accelerating depending on the statement. In reality you would only see a light spot on the right wall. If the elevator was moving at normal speeds, the spot would be directly across from the opening whether moving at constant or accelerated speeds. The faster the elevator moves, the lower the spot will be. When the speed is constant the light spot is stationary. If the speed is accelerating faster then the spot slowly moves down the wall. Keep in mind the light is not curving but the wall of the elevator is rising before the light gets there.
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Mass falling onto a spring from an inclined plane I was studying for my AP Physics test when I came across the following problem, which seemed to confuse me:
The ramp is frictionless, and the spring is initially at it's relaxed length, and the angle is $\theta = 45^{\circ}$.
I worked through this by considering the potential energy lost as the kinetic energy gained just before the mass hits the spring, with the equation
$mgh = \frac{1}{2}mv^2$, and plugging in the known values, getting:
$40\cdot9.8\cdot\frac{2}{\sqrt{2}} = \frac{1}{2} \cdot 40 \cdot v^2$, to solve for $v$, which gives $v = 5.25 \frac{m}{s}$. However, I am unsure whether this is the correct answer, as I am unsure when the maximum speed occurs. If it exists after the compression of the spring, how do I go about calculating it?
Thank you.
| The motion of the box along the slope is dictated by two forces which act parallel to the slope.
The component of the weight of the box down the slope which acts all the time and the force up the slope due to the spring which acts only when the box is in contact with the spring.
As long as there is a net force down the slope the speed of the box will increase.
Once that net force is up the slope the box will slow down .
So at some stage in the motion of the box the net force on the box along the slope will be zero, the acceleration of the box will be zero, the box will be moving with maximum speed.
So make the component of weight down the slope equal in magnitude to the magnitude of the force up the slope due to the spring being compressed an amount $x_o$.
At this point the box has moved $d+x_o$ along the slope and so you can now use energy conservation to find the maximum speed of the box.
| {
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What is the role of pillars in bridges?
As I can see in the picture, there are so many pillars which are holding the bridge. This picture gave a question to me that what are these pillars doing below the bridge?? An appripriate answer could be "these are providing support to bridge".
I tried to get the answer as follows:
In the first image there are two pillars holding a bridge of mass $M$, since gravitaional force is acting downwards thus pillars are bearing a force of $\frac{1}{2}Mg$.
In the second image there are four pillars bearing a force of $\frac{1}{4}Mg$. I'm assuming that mass of bridge is uniformly distributed and each pillar is bearing an equal amount of the load.
Now the question is that since the pillars are bearing the force, so if we make strong enough pillars to bear a large force then there will be no need of so many pillars.
But that is not the case, we see a large number of pillars holding a bridge. What is wrong with the work I did? Shouldn't the number of pillars depend upon the strength of the pillars we make rather than the length of the bridge ??
I shall be thankful if you can provide more information about this topic.
| The number of the pillars does not depend by the load that each one of them can carry.
Mainly the number of the pillars is selected in order to reduce the distance between them and so to minimize the moments and so the stresses produced and act on the beams that holds the bridge's deck, as very nicely @valerio92 answered.
We can see from the photo that you uploaded, that in many cases this is not enough. So there's a little trick: the pilars are extended above the bridges deck, so that we can hang the deck using cables. This way we provide extra supports to the deck . These are the well-known 'Suspensions bridges' .
(see also : https://en.wikipedia.org/wiki/Suspension_bridge)
Of course there are plenty of other dynamic loads such as earthquakes, winds , sea waves etc that may determine the number of pillars but in general this is not the case.
| {
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Hermitian properties of the gamma matrices The gamma matrices $\gamma^{\mu}$ are defined by
$$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}.$$
There exist representations of the gamma matrices such as the Dirac basis and the Weyl basis.
Is it possible to prove the relation
$$(\gamma^{\mu})^{\dagger}\gamma^{0}=\gamma^{0}\gamma^{\mu}$$
without alluding to a specific representation?
| Not exactly, one can choose gamma-matrices that satisfy the anti-commutation conditions, but do not satisfy the hermiticity conditions.
EDIT: (11/19/2016) However, one can argue that the hermiticity condition for gamma-matrices can be derived from hermiticity of the Dirac Hamiltonian (https://arxiv.org/pdf/physics/0703214.pdf). Nevertheless, one can live with gamma-matrices that do not satisfy the condition, but are still related to the gamma-matrices of standard representations via a similarity condition.
| {
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Proton - neutron fusion? In reviewing some problems in an elementary book, I ran across a reference to the reaction $p+n\rightarrow d$ + "energy".
Is that possible? I don't see any reason why not, but I don't find any mention of this reaction at all using Google. It seems to me that the "energy" would have to be a combination of deuteron kinetic energy and a gamma.
| Small mistake in my answer
"This means that 1,4e-6 grams of uranium, 0,05e-6 grams of beryllium and 0.0059e-6 grams of hydrogen"
And i would add, that this cold nuclear fusion induced by radioactive decay will not result in a nuclear catastrophe. It is not a self-reinforcing process. In nuclear power, the more nuclei that are split, the more neutrons are released that potentially split nuclei again. Here, radioactive decay does not accelerate when there is more fusion.
| {
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How does an object undergoing circular motion increase its tangential velocity? I completely understand the concepts behind uniform circular motion. But let's say you are spinning a ball connected by a string to a motor in a horizontal circle. When increasing the angular velocity of the spinning motor, I can't see how the ball connected to the string will have any force that allows it to increase its tangential velocity. How would the string be able to pull it so it accelerates tangentially all while undergoing circular motion? An example would be if you are swinging a ball in a circle above your head and you begin to spin it faster. How is the string causing the object to increase its linear speed? I believe tension can't cause it unless it's working at an angle less than 90 to the tangent because work must be done to increase the kinetic energy.
| There must be tangential acceleration, from a component of the force parallel to the velocity.
Note that if your center-pointing force is different from the magic value $F_\text{centripetal} = mv^2/r$ for uniform circular motion, the radius of your circular motion will change, and the tangential velocity will change as well to conserve angular momentum.
| {
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Wave nature of matter in bonds We say that electrons are probability waves. Then if two or more elements form bonds by overlapping of orbitals then what happens to the wave function of the electrons? Is it possible that during bonding, these electrons still behave as waves and interfere?
|
Is it possible that during bonding, these electrons still behave as waves and interfere?
Of course. This has been confirmed experimentally.
The unequivocal experimental proof of the wave (frequency) nature of matter is the electron image recorded in 2007. This image was recorded by a team of Swedish scientists from the University of Lund. The image turned out to be a soliton image.
Electron movie
If we look at the video frame by frame, we will see interfering waves from others electrons (from the upper and lower sides).
| {
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Exact solution to electron-electron scattering? One of the first results seen in elementary quantum mechanics is the closed form solution to the bound states of the hydrogen atom. In the usual approach, scattering theory is placed on the opposite side of spectrum; exclusively though of as a perturbative process. I am interested in knowing about the scattering states of the hydrogen atom, but more specifically in terms of electron-electron scattering.
My question is regarding the scattering states of electron-eletron scattering, which has the (non-relativistic) Hamiltonian below.
$$H=\left(\frac{\hbar^2}{2m_e}\nabla^2_1+\frac{\hbar^2}{2m_e}\nabla^2_2\right)+\frac{e^2}{|\mathbf{r}_1-\mathbf{r}_2|}$$
Because the problem is so similar to that of the electron-proton system, I would expect a closed form solution for dealing with the scattering states non-perturbatively. However, this problem (Rutherford scattering) is usually treated in textbooks using the Born approximation, with divergences at low deflection angles where the approach breaks down. I am especially interested in this regime.
So my questions are the following:
*
*In the non-relativistic limit, is there a non-perturbative solution to electron-electron scattering? By solution I (naively) am thinking of something akin to an expression for the scattering cross-section.
*When relativity is introduced (either with the Dirac-equation, or the full machinery of QED), at what point is a non-perturbative solution out of reach? Are there any special cases (e.g. total energies are below that needed for pair-production) that are significantly easier to handle?
| To solve this problem, it is better to use the relativistic equation M2.
$$\Delta \Psi -\frac{1}{\hbar^{2}}\left [ \frac{m^{4}c^{6}}{\left ( E-U\left ( \overrightarrow{r}\right ) \right )^{2}}-m^{2}c^{2} \right ]\Psi =0$$
The problem is analogous to the hydrogen atom. It is necessary in the equation for the hydrogen atom to replace the mass of the electron m by the reduced mass $\frac{m}{2}$. And instead of the Coulomb potential of attraction$U\left ( \overrightarrow{r} \right )=-\frac{Ze^{2}}{4\pi \varepsilon _{0}r}$, substitute the Coulomb repulsion potential $U\left ( \overrightarrow{r} \right )=+\frac{Ze^{2}}{4\pi \varepsilon _{0}r}$.
As a result, we obtain the equation: $$\Delta \Psi -\frac{1}{\hbar^{2}}\left [ \frac{\left ( \frac{m}{2} \right )^{4}c^{6}}{\left ( E-\frac{Ze^{2}}{4\pi \varepsilon _{0}r}\right )^{2}}-\left ( \frac{m}{2} \right )^{2}c^{2} \right ]\Psi =0$$
$\left ( Z=1 \right )$ If energy is emitted during the electron-electron interaction, this means the formation of a potential well. In this case, the solutions show the possibility of forming a composite particle of two electrons.
| {
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Effect of Change of Potential Difference applied to an X-Ray tube I am a high school student, so I know only the basics of X-Rays. I simply know about continuous X-rays, cutoff wavelength and threshold wavelength.
Now if I increase the potential applied to the X-ray tube, I am certain that the minimum wavelength of emitted radiation decreases. I am unsure about its intensity. How does that change?
| For a typical x-ray tube used in medical imaging, the radiation output (as measured in mGy/mAs) changes roughly with the square of the ratio of the ${kV_p}$
$$
y_1/y_2 = ({kV_p}_1/{kV_p}_2)^2
$$
where $y_n$ is the radiation output in mGy/mAs at ${kV_p}_n$
So if you measure the output of an imaging tube at 70 kVp and then change to 100 kVp (maintaining all other parameters the same), you can expect the output of the tube to roughly double.
A typical medical x-ray tube normally has about 2-3 mm Al filtration at the output port of the x-ray tube which serves to remove most of the low energy x-rays in the beam that don't contribute to image formation.
The graph below comes from data I've collected over several years of testing radiography units
| {
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Strange quark decay into two down quarks and an anti-down quark I saw that a $\Sigma^+$ can decay into $n+\pi^+$, which means that the $s$-quark must decay into $dd\bar{d}$. However, is there a Feynman diagram to represent this? I cannot find one for either the $\Sigma^+$ decay or the $s$-quark decay. I have only just started learning so I am not sure. Thank you.
| One possibility for this decay is as follows. Note that the initial quark content is $uus$ while the final quark content is $udd+u\bar{d}$.
The $s$-quark can emit a $W^-$ and so turn into a $u$-quark, and this $W^-$ can then be absorbed by one of the initial $u$ quarks turning it into a $d$-quark. We then have quark content $uud$. A gluon (or a photon or a $Z$-boson) can then be emitted (from any of these quarks) and subsequently turn into a $d\bar{d}$ pair. We then have the correct final quark content, and these five quarks can then form into an $n+\pi^+$ final state.
| {
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Fetter & Walecka's derivation of second-quantised kinetic term in many-particle systems On page 9 of Quantum theory of many-particle systems by Alexander L. Fetter and John Dirk Walecka, during the derivation of the second-quantised kinetic term, there is an equality equation below:
\begin{align}
\sum_{k=1}^{N} \sum_{W} & \langle E_k|T|W\rangle C(E_1, ..., E_{k-1}, W, E_{k+1},...,E_N, t)
\\&=
\sum_{k=1}^{N}\sum_{W}\langle E_k|T|W\rangle\times \bar{C}(n_1, n_2,...,n_{E_k}-1, ..., n_{W}+1,...,n_\infty, t)
\end{align}
Why is the number of particles with quantum numbers $n_{E_k}$ decreased by 1 whereas the number of particles with quantum numbers $n_W$ increased by 1?
Anybody know how to get this equality?
| The coefficient $C(E_1, E_2, \dots, E_k, \dots, E_N, t)$ corresponds to the configuration:
$1^{st}$ particle in state $E_1$, $2^{nd}$ particle in state $E_2$, $\dots,$ $k^{th}$ particle in state $E_k$, $\dots,$ $N^{th}$ particle in state $E_N$.
Say, this corresponds to the following occupation numbers:
$n_1$ particles in state $1$, $n_2$ particles in state $2$, $\dots,$ $n_l$ particles in state $l$, $\dots,$ $n_{\infty}$ particles in state ${\infty}$.
where $n_1+n_2+\dots+n_{\infty}=N$. Then, we rewrite the coefficient for this configuration in terms of occupation numbers as $\bar{C}(n_1, n_2, \dots, n_l, \dots, n_{\infty}, t)$.
Now, the new coefficient after the action of the kinetic energy operator is $C(E_1, E_2, \dots, E_{k-1}, W, E_{k+1}, \dots, E_N, t)$, which corresponds to the configuration:
$1^{st}$ particle in state $E_1$, $2^{nd}$ particle in state $E_2$, $\dots,$ $(k-1)^{th}$ particle in state $E_{k-1}$, $k^{th}$ particle in state $W$, $(k+1)^{th}$ particle in state $E_{k+1}$, $\dots,$ $N^{th}$ particle in state $E_N$.
$i.e.,$ the $k^{th}$ particle, which was originally occupying the state $E_k$, is currently in state $W$. Therefore, the number of particles occupying the state corresponding to $E_k$ should be decreased by one, and the number of particles occupying the state corresponding to $W$ should be increased by one. (In other words, the kinetic energy operator has annihilated a particle in the state corresponding to $E_k$, and created a particle in the state corresponding to $W$.)
So, the coefficient for this new configuration in terms of occupation numbers is $\bar{C}(n_1, n_2, \dots, n_{E_k}-1, \dots, n_W+1, \dots, n_{\infty}, t)$.
| {
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Quantizing a complex Klein-Gordon Field: Why are there two types of excitations? In most references I've seen (see, for example, Peskin and Schroeder problem 2.2, or section 2.5 here), one constructs the field operator $\hat{\phi}$ for the complex Klein-Gordon field as follows:
First, you take the Lagrangian density for the classical Klein-Gordon field
$$
\mathcal{L}=\partial_\mu \phi^\dagger\partial^\mu\phi-m^2\phi^\dagger\phi \tag{1}
$$
and find the momentum conjugate to the field $\phi$ via
$$
\pi=\frac{\partial\mathcal L}{\partial \dot\phi}=\dot\phi^\dagger.\tag{2}
$$
Then, one imposes the usual canonical commutation relations on $\hat\phi$ and $\hat\pi$:
$$
[\hat\phi(x),\hat\pi(y)]=i\delta^3(x-y).\tag{3}
$$
So, one needs to find operators $\hat{\phi}$ and $\hat\pi$ such that they obey the above commutation relations, and such that $\hat\pi=\dot\phi^\dagger$. The textbooks then go on to show that defining
$$
\hat{\phi}(x)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2p_0}}[a_p^\dagger e^{-i p_\mu x^\mu}+b_pe^{i p_\mu x^\mu}]\tag{4}
$$
$$
\hat{\pi}(x)=i\int\frac{d^3\vec{p}}{(2\pi)^3}\sqrt{\frac{p_0}{2}}[a_p e^{i p_\mu x^\mu}-b_p^\dagger e^{-i p_\mu x^\mu}]\tag{5}
$$
where $a$ and $b$ are bosonic annihilation operators, satisfies these properties.
My question is: Why do we need two different particle operators to define $\hat\phi$ and $\hat\pi$? It seems to me that one could simply define
$$
\hat{\phi}(x)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2p_0}}a_p e^{-i p_\mu x^\mu}\tag{6}
$$
$$
\hat{\pi}(x)=i\int\frac{d^3\vec{p}}{(2\pi)^3}\sqrt{\frac{p_0}{2}}a_p^\dagger e^{i p_\mu x^\mu}\tag{7}
$$
with $\hat{a}_p$ a single bosonic annihilation operator. Then clearly $\hat{\pi}=\dot{\hat{\phi}}^\dagger$, and also
$$
\begin{array}{rcl}
[\hat\phi(x),\hat\pi(y)]&=&i\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\sqrt{\frac{q_0}{4p_0}}e^{i(q_\mu y^\mu-p_\mu x^\mu)}[a_p,a_q^\dagger]\\
&=&i\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\sqrt{\frac{q_0}{4p_0}}e^{i(q_\mu y^\mu-p_\mu x^\mu)}(2\pi)^3\delta^3(p-q)\\
&=&i\int\frac{d^3p}{(2\pi)^3}\frac{1}{2}e^{ip_\mu (y^\mu-x^\mu)}\\
&=&\frac{i}{2}\delta^3(y-x)\\
\end{array}\tag{8}
$$
which is, up to some details about normalizing the $\hat{a}_p$, correct. We would then have a Klein-Gordon field with just one kind of excitation, the $\hat{a}_p$ excitation. Why do all textbooks claim we need two separate bosonic excitations, $\hat{a}_p$ and $\hat{b}_p$?
| There is a conceptually simple (but fiddly) way of relating this expansion to the usual Fourier expansion. TL;DR: Requiring $\phi$ to satisfy the Klein-Gordon equation divides the nonzero Fourier components into two classes, corresponding to particles and antiparticles.
For a general complex scalar field defined on spacetime,
$$ \phi(x) = \int\frac{d^4 p}{(2\pi)^4} \hat{\phi}(p) e^{-ip \cdot x}. \label{fourier}\tag{1} $$
This is the ordinary four-dimensional Fourier expansion. Now, let us impose the Klein-Gordon equation
$$ 0 = (\partial^2 + m^2) \phi = \int \frac{d^4 p}{(2\pi)^4} (-p^2 + m^2) \hat{\phi}(p) e^{-ip \cdot x}. $$
A general field satisfying this must consist of only modes where $p^2 = m^2$, i.e. on-shell modes. Thus,
$$ \hat{\phi}(p) = 2\pi \delta(p^2 - m^2) f(p) $$
for some function $f$ (the factor $2\pi$ is convenient for comparing with the standard expansion). Writing $p = (p^0, \mathbf{p})$,
$$ \delta(p^2 - m^2) = \delta\left((p^0)^2 - (\mathbf{p}^2 + m^2)\right). $$
The argument of the $\delta$-function has two zeros $p^0 = \pm \sqrt{\mathbf{p}^2 + m^2}$ (for fixed $\mathbf{p}$), so we use the general rule
$$ \delta(f(x)) = \sum_{f(x_i) = 0} \frac{1}{|f'(x_i)|} \delta(x - x_i) $$
to find
$$ \delta(p^2 - m^2) = \frac{1}{|2 p^0|} \left[ \delta\left(p^0 - \sqrt{\mathbf{p}^2 + m^2}\right) + \delta\left(p^0 + \sqrt{\mathbf{p}^2 + m^2}\right) \right]. $$
Putting this back into (\ref{fourier}) and performing the $p^0$ integral (abbreviating $\sqrt{\mathbf{p}^2 + m^2} = E_\mathbf{p}$):
$$ \begin{align} \phi(x) &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[
f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x}
+ f(-E_\mathbf{p}, \mathbf{p}) e^{-i(-E_\mathbf{p}, \mathbf{p}) \cdot x}
\right] \\
&= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[
f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x}
+ f(-E_\mathbf{p}, \mathbf{p}) e^{i(E_\mathbf{p}, -\mathbf{p}) \cdot x}
\right] \\
&= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[
f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x}
+ f(-E_\mathbf{p}, -\mathbf{p}) e^{i(E_\mathbf{p}, \mathbf{p}) \cdot x}
\right] \\
\end{align} $$
where we have swapped $\mathbf{p} \mapsto -\mathbf{p}$ in the second term.
We identify the usual (Heisenberg picture) expansion
$$ \phi(x) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_\mathbf{p}}} \left[
a_\mathbf{p} e^{-ip \cdot x} + b^\dagger_\mathbf{p} e^{ip \cdot x} \right] $$
with
$$ \begin{align}
a_\mathbf{p} &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(E_\mathbf{p}, \mathbf{p}) \\
b_\mathbf{p}^\dagger &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(-E_\mathbf{p}, -\mathbf{p})
\end{align} $$
($f(E_\mathbf{p}, \mathbf{p})$ and $f(-E_\mathbf{p}, -\mathbf{p})$ are the relativistically normalized annihilation and creation operators).
If $\phi$ is real, we know from Fourier analysis that $\hat{\phi}(p) = \hat{\phi}(-p)^\dagger$, which immediately translates to $a_\mathbf{p} = b_\mathbf{p}$.
Up until now, the field is entirely classical ($a_\mathbf{p}$ and $b_\mathbf{p}$ are simply complex numbers, and $\dagger$ is complex conjugation). Thus even the classical field has two types of excitations: Positive-frequency solutions (with coefficients $a_\mathbf{p}$) and negative-frequency solutions (with coefficients $b_\mathbf{p}^\dagger$). After quantizing, they correspond to particles and antiparticles.
| {
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"Unaccounted for" vertices in certain Feynman diagrams? I'm looking to understand where certain "unexplained" vertices in some Feynman diagrams come from, in a physical sense. For example, in the top figure (Aaij et al. 2015), in diagram (b), there is an "unaccounted for" $u\bar{u}$ pair that seems to come from nowhere.
Are these quarks similar to the spectator quarks from the initial $\Lambda_0^b$, or do they arise from the virtual quark/gluon sea somehow? If they come from the quark sea, how can they come to comprise real, observable particles like the kaon and $P_c^+$?
Image source: Observation of $J/\psi p$ resonances consistent with pentaquark states in $Λ^0_b→J/\psi K^−p$ decays, LHCb collaboration. Phys. Rev. Lett. 115, 072001 (2015), arXiv:1507.03414.
Image source: Hidden-charm molecular pentaquarks and their charm-strange partners, R. Chen et al. Nucl. Phys. A 954, 406-421 (2016), arXiv:1601.03233.
| Once you have quarks in a Feynman diagram, the strong interaction enters, and note, the strong, it means Quantum Chronodynamics . It means from that point on the usual perturbative expansion with the nice fixed rules for calculating the total integral with electromagnetic and weak interactions can only be used with approximations for the QCD part, which are highly divergent using the usual tools. QCD calculations use different tools, like lattice QCD.
In the diagram with the u u_bar vertex, what is implied is a gluon connecting the two quark lines, (a number of possible combinations) as for example shown in this link. The fact that the authors do not use a gluon line is because that line is not really a single particle exchange simply under the integrals, but a complicated approximation on strong force exchanges that are allowed within the kinematics. A gluon giving the vertex would make it simple, but would not stress the real complexity of the mathematical situation.
In the top diagrams, the spectator approximation is implied. As far as stron interactions go there should be innumerable gluon-quark vertices between the quarks, keeping them within a hadron.
| {
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Can we tell that we're in a flat space time? Or just in reference to other objects? If you are sitting in space, far from any planet or large gravitational object, can you be sure that you're in a flat spacetime? Is it possible that some very distant object is heavily warping spacetime, but because it's so far away, it's acting uniformly on the area around you, so there is no way to detect that you are in warped spacetime?
Put another way, how do you know how warped spacetime is around you? Can you only measure it based on how other objects gravitational are accelerating towards you?
Put yet another way, is the curvature of spacetime absolute with some points in space having 0 spacetime curvature, or is the curvature relative to other areas?
| Not an expert, but if think we are comfortable with 3d space. It is easier to visualize flat space time. Please correct my notion if at all possible and not a violation of the rules. I am just here to learn
| {
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How are we stardust? I am a layperson and certainly not a physics student, though I'm sure that is palpably obvious. I have a strong appreciation for people like Carl Sagan, Lawrence Krauss, and Sean Carrol.
I'm often challenged by creationists over the claim that we are all "starstuff" or "stardust". I can understand that most or all of the atoms and things needed for life to evolve on earth came from dead starts, but how are new humans, babies, etc... the stuff of stars?
How did those star elements get into living things? I would have thought they were only necessarily present for the early chemical-into-biological evolution. Is there an infinite amount? As the human population rises, do we ALL possess material from long dead stars? Any deeper explanation and clarification would be most appreciative. I apologize in advance for my ignorance, I hope I get some real answers, because it's a serious question.
| Well, people are made out of a bunch of elements, none of which are created on Earth: they all got here from somewhere else.
There are two sources for elements:
*
*Hydrogen, some Helium, some isotopes of them and I think some Lithium were made by 'primordial nucleosynthesis' within a few minutes after the big bang;
*everything else is made in stars.
So, for instance, Carbon, which is extremely important to life, was made in stars, as was Oxygen and so on.
| {
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What is the definition of "Complexity" in physics? Is it quantifiable? I don't know much about the discipline of "Complex systems studies" but I know in the field of "Statistical mechanics" there is much talk about the "Complexity of the system". Like "...the state of this system is more complex..." or "...as we see the complexity of the system is arising..." and so on.
My question is:
*
*How "Complexity" is actually defined in physics?
*Is "Complexity" a quantifiable property of the system? I mean can we define a quantity like $\mathfrak{C}$ representing the measure of complexity of the system?
| As others have said, complexity is still a notion that does not have one and only one definition. However, you did ask for a quantitative definition, so one example of complexity of a quantum field (or even a quantum lattice system) I've seen in the context of quantum gravity is:
"the minimum number of quantum gates needed to take you from the ground state of the system (or really any distinguished state) to the state in question."
This definition is due to Suskind (pg 7) to the best of my knowledge. It's not the only way to do it, but generally, quantum information theory is interested in complexity of a quantum system and its relationship to entanglement.
| {
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One-loop Higgs mass in the Standard Model I cannot reproduce the one-loop Higgs mass expression with a cutoff regularization.
The standard result found in literature is (which leads to the so-called Veltman condition):
\begin{equation}\delta m_h^2 = \frac{3\Lambda^2}{16\pi^2 v^2}(m_h^2+m_Z^2+2m_W^2-4m_t^2)\end{equation} (ignoring contribution from lighter fermions), where $v=246$ GeV.
I manage to recover the expression for the gauge bosons and fermions contributions but get an additional factor $1/2$ for the Higgs contribution.
Defining $V_{Higgs}= -\mu^2|H|^2 + \lambda|H|^4 \subset -\frac{\lambda}{4} h^4$, where $H=(0, \frac{v+h}{\sqrt 2})^T$, the tree-level Higgs mass is $m_h^2=2\mu^2=2\lambda v^2$. The relevant self-energy function is (neglecting the mass in the propagator):
\begin{equation}
-i \Pi_h = \frac{1}{2}\left(\frac{-i4!\lambda}{4}\right)\int \frac{d^4q}{16\pi^4} \frac{i}{q^2}=3\lambda (-i)\frac{\pi^2}{16\pi^4}\int_0^{\Lambda^2}dq^2_E = (-i) \frac{3\lambda}{16\pi^2}\Lambda^2=(-i) \frac{3\Lambda^2}{16\pi^2 v^2}\frac{m_h^2}{2}
\end{equation}
where the first $1/2$ factor is a symmetry factor (which can easily be recovered from Wick's theorem). Therefore my result differs by a factor 1/2 as compared to the literature.
For the top contribution for instance, I have no problem:
\begin{equation}
-i\Pi_t = -3\left(\frac{iy_t}{\sqrt 2}\right)^2\int \frac{d^4q}{16\pi^4}\text{Tr} \frac{i q_\mu\gamma^\mu}{q^2}\frac{i q_\nu\gamma^\nu}{q^2} = 3\frac{y_t^2}{2}(-4)\int \frac{d^4q}{16\pi^4}\frac{1}{q^2}=-
(-i)\frac{3\Lambda^2}{16\pi^2}\frac{4m_t^2}{v^2} \end{equation}
which indeed leads to the right result.
Do you see where the problem is ?
| One should actually (obviously..) add the Goldstone contributions. The Feynman rules are (consistent with my previous conventions):
\begin{equation}
hh\phi_+\phi_-: -2i\lambda , \ \ hh\phi_0 \phi_0: -2i\lambda
\end{equation}
Since the diagram involving the neutral Goldstone has a symmetry factor of 1/2, the total contribution to the Higgs self-energy from the Goldstones is (Landau gauge):
\begin{equation}
-i\Pi_\phi=-i\lambda(2+\frac{2}{2})\int \frac{d^4q}{16\pi^4} \frac{i}{q^2} = (-i)\frac{3\lambda}{16\pi^2}\Lambda^2
\end{equation}
which is the same as the Higgs contribution, hence the problem is solved.
| {
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Rubbing of a fur - can the electrons in a fur be ever ran out? As we all know, if we rub ebonite rod with a fur, the rod becomes negatively charged(electrons get passed from fur to rod). But what happens if we will do that many times with the same fur but each time with another ebonite rub? I mean, is it possible that once(at one point) all electrons in this fur will run out? The fur without electrons? And, how can we to restore a previous state of this fur (with balance of protons and electrons)?
| The ebonite rod is a good conductor, so electrons will be floating on every part of it's orbit having sufficient electrons on it's highest energy level. The fur is just like the carbon atom that is a solid but a non metal, chemically the carbon atom has 2 inner electrons and 4 valence outer electrons which it really doesn't give away. So electrons are donated by the ebonite rod, and the ebonite rod becomes negatively charged.
| {
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Why is bench pressing your bodyweight harder than doing a pushup? Why does bench pressing your own bodyweight feel so much harder than doing a push-up?
I have my own theories about the weight being distributed over multiple points (like in a push-up) but would just like to get a definite answer.
| Consider leverage. Assume as in @Michael's comment that the centre of mass is somewhere near the middle. Further assume that the toes are fixed to the floor. If the torso and legs are rigid the centre of mass does not lift as far as the shoulders (at which the pressing force is exerted), so you've got leverage in your favour. This probably reduces the lift to about 2/3 or 3/4 of what it would be if you didn't put any weight on your feet.
| {
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spacelike and timelike event or interval? Just a short question:
Is it valid to say that some event is space/time-like or is this term reserved for the distance between two events (the space-time interval) and four vectors?
| The terms spacelike, timelike, and lightlike refer to space time intervals only. They are used when discussing the sign of the distance between two different events. Mathematically, this comes from the sign of the metric tensor acting on two different four-vectors, $g(\mathbf{x},\mathbf{y}) = g_{ij}x^i y^j$. The metric tensor applied to a single four vector (event) will not return a number, but rather a covector. Hence, it does not make sense to talk about the sign of the metric tensor applied to a single vector.
| {
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Dark matter annihilation cross section What is the meaning of upper limit of annihilation cross section? Sometimes I have seen that $\langle\sigma v\rangle = 3.0 \times 10^{-26}\,{\rm cm}^{3}\,{\rm s}^{-1}$ or $2.0 \times 10^{-25}\,{\rm cm}^{3}\,{\rm s}^{-1}$.
Which is the better value and why?
| By "which is better" I assume that you mean "which is most constraining".
First, $\langle\sigma v\rangle$ refers to the cross section per unit velocity. This is important because it is conjectured that dark matter scattering with a velocity independent cross section has several problems that it seems to be possible to resolve by assuming the cross section changes with velocity. This explains the units of $\langle\sigma v\rangle$: it relates how big a target the particle represents (${\rm cm}^{2}$) and how fast it is traveling (${\rm cm}\,{\rm s}^{-1}$).
The scattering/annihilation cross section of DM (work on both types of interactions uses similar language) has never been measured, which is a bit of a given since DM scattering/annihilation have never been detected (well, there are claims, but none generally accepted at the moment). A smaller upper limit represents a more constraining non-detection. Basically, the limit says "if the scattering/annihilation cross section were larger than $<{\rm value}>$, we would have seen it in our detector, so it must be smaller". The limit only applies to the types of interactions the detector is designed to be sensitive to, and the design is motivated by some theoretical candidate particle, such as a WIMP.
| {
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Symmetry factor via Wick's theorem Consider the lagrangian of the real scalar field given by $$\mathcal L = \frac{1}{2} (\partial \phi)^2 - \frac{1}{2} m^2 \phi^2 - \frac{\lambda}{4!} \phi^4$$
Disregarding snail contributions, the only diagram contributing to $ \langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle$ at one loop order is the so called dinosaur:
To argue the symmetry factor $S$ of this diagram, I say that there are 4 choice for a $\phi_y$ field to be contracted with one of the final states and then 3 choices for another $\phi_y$ field to be contracted with the remaining final state. Same arguments for the $\phi_x$ fields and their contractions with the initial states. This leaves 2! permutations of the propagators between $x$ and $y$. Two vertices => have factor $(1/4!)^2$ and such a diagram would be generated at second order in the Dyson expansion => have factor $1/2$. Putting this all together I get
$$S^{-1} = \frac{4 \cdot 3 \cdot 4 \cdot 3 \cdot 2!}{4! \cdot 4! \cdot 2} = \frac{1}{4}$$ I think the answer should be $1/2$ so can someone help in seeing where I lost a factor of $2$?
I could also evaluate $$\langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle = \langle p_4 p_3 | : \phi(y)^4 \phi(x)^4 : | p_1 p_2 \rangle + \dots + (\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 :| p_1 p_2 \rangle + \dots $$ where dots indicate diagrams generated via this correlator that do not contribute at one loop. (I don't know the latex for the Wick contraction symbol so I just write contract). Is there a way to find out the symmetry factor from computing the term $(\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2: | p_1 p_2 \rangle?$
| The diagram of interest (the $s$ channel dinosaur) is generated at second order in the Dyson expansion (alongside the $t$ and $u$ channel dinosaur diagrams and the snail $1 \text{PI}$ reducible diagrams) within the correlator $\langle p_4 p_3 | T( \mathcal L(y) \mathcal L(x)) | p_1 p_2 \rangle$. Using Wick's theorem, we can write this explicitly as $$\langle p_4 p_3 | T( \mathcal L(y) \mathcal L(x)) | p_1 p_2 \rangle = \langle p_4 p_3 | : \phi(y)^4 \phi(x)^4 : | p_1 p_2 \rangle + \dots + X(\text{C}(\phi (y) \phi(x)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 : | p_1 p_2 \rangle + \dots $$ where dots indicate other diagrams not of interest, $\mathcal L(x) = -\frac{\lambda}{4} \phi(x)^4$ and $X$ is the number of ways the contraction $(\text{C}(\phi(y) \phi(x)))^2$ can be formed.
There are four possible $\phi(x)$ fields that can be contracted with four possible $\phi(y)$ fields. This means a single contraction shows up $4 \cdot 4$ times. There are then three $\phi(x)$ fields to contract with three $\phi(y)$ fields. This gives a factor $3 \cdot 3$. To avoid an overcount, we divide by $2!$. Therefore, $$X = \frac{4 \cdot 4 \cdot 3 \cdot 3}{2!}$$
Now, we may write the remaining correlator in which the fields are contracted with the in and out external states in terms of field creation and annihilation operators to yield $$ \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 : | p_1 p_2 \rangle = 4 \langle p_4 p_3 |(\phi(y)^- \phi(x)^- \phi(y)^+\phi(x)^+ )| p_1 p_2 \rangle.$$ where we explicitly normal ordered. The computation leads to $$ \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 : | p_1 p_2 \rangle = 4 \left(e^{-i(p_1 +p_2)y} e^{i (p_3+p_4)x} + \begin{cases} p_1 \rightarrow -p_4 \\ p_2 \rightarrow p_1 \\ p_3 \rightarrow p_3 \\ p_4 \rightarrow -p_2 \end{cases} + \begin{cases} p_1 \rightarrow -p_4 \\ p_2 \rightarrow p_2 \\ p_3 \rightarrow p_3 \\ p_4 \rightarrow -p_1 \end{cases} + \left\{ y \leftrightarrow x \right\} \right) $$ where the permutations of the momenta within the braces give rise to the $t$ and $u$ channel dinosaur diagrams. Focusing only on the $s$ channel contribution and inserting back into the $\mathcal T$ matrix element we obtain the expression $$\mathcal A = \frac{(- \lambda)^2}{(4!)^2} \int \text{d}^4 y \,\,\text{d}^4 x \frac{\mathrm{i}^2}{2!} \left(\frac{4 \cdot 4 \cdot 3 \cdot 3}{2!}\right) (\mathrm{i} \Delta_F(y-x))^2 4(e^{-i(p_1 +p_2)y} e^{i (p_3+p_4)x} + (y \leftrightarrow x)), $$ with $(C(\phi(y) \phi(x)))^2 \equiv (\mathrm{i} \Delta_F (y-x))^2$, the Feynman propagator. The $( y \leftrightarrow x)$ term yields the same contribution as that displayed, readily seen by relabelling the indices on the space time measures. Collecting all numerical factors in front of the integration we arrive at $$\mathcal A = \underbrace{\left(\frac{4 \cdot 4 \cdot 3 \cdot 3 \cdot 4 \cdot 2}{4!\cdot 4! \cdot 2! \cdot 2!}\right)}_{1/2} (-\mathrm{i}\lambda)^2 \int \text{d}^4 y \,\,\text{d}^4 x \,\,(\mathrm{i} \Delta_F(y-x))^2 e^{-i(p_1 +p_2)y} e^{i (p_3+p_4)x} $$ which is the coordinate space representation (easily recovered using Feynman's rules in position space) of the diagram in question with symmetry factor $S = 2$.
| {
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Lorentz force (equation) i have a question
When we consider the motion of a particle, then write
ordinary Lorentz force
Why do not we consider a charged particle's own field, because
it is known, is always much larger than the external field and the even formation
It becomes infinite at the point where the particle is?
| You can't consider particle's own field because its module will be infinite and it is impossible to state in which direction it is directed. For that field, expressed in spherical coordinates for example, the point r=0 is a singular point and even the force will be infinite and without a definite direction.
| {
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Why faraday rotation is not effective in case of circular polariation? As in case of linear polarization the plane of polarization gets rotated. Why it's not happening with circular polarization?
| Circular polarization is rotated just as much as linear polarization.
The catch is that the circularly polarized wave is itself “rotating” about the same axis, so the only effect is to change the phase of the wave. It's like taking a spinning wheel and rotating it by 90° — it's still spinning, and you need some notion of where it was at a given moment in time to be able to say that a change was made.
Therefore, in order to detect the change you need to use a technique which is sensitive to phase, i.e. interferometry. Any measurement that is not comparing the rotated wave to an un-rotated wave cannot detect a difference.
| {
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Human body calories consumption estimate based on physics approximation I was looking for references online about equations that relates human kinetics and calories consumption, and it seems there are no many engineers interested in weight lifting (go figure...). I think under the right assumptions it cannot be to complicated.
If I lift a box of 30 Kg, for 0.5 meters in one seconds, following a perpendicular trajectory to earth, I can say:
WORK = FORCE * DISPLACEMENT
FORCE = m * a = 30Kg*9.8m/s^2 = 294Kg*m/s^2 = 294N
WORK = 294N * 0.5m = 147J
POWER = WORK / TIME
POWER = 147J / 1s
POWER = 147W
1W = 0.24Cal/s
147W = 35.11Cal/s in 1 seconds = 35.11 calories
So assuming the human body is a perfect machine with no losses, is it correct to say that I just used (at least) 35.11 Calories?
| The CN Tower in Toronto holds annual stair climbs for charity.
The record for climbing the $1776$ stairs is a little under $8$ minutes.
Data for physics calculations is surprisingly hard to find, but assuming an $80$ kg climber going up $1776$ steps, each $0.18$ m high in $480$ seconds, we come out with a power output of $522$ Watts
This compares well with the functional threshold power of $400$ Watts for elite cyclists working indefinitely (not just 8 minutes) in @Pirx's answer...
| {
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Sensation of atmospheric pressure Pressure is force divided by area, and force is mass times acceleration.
Now, the newton [N] is the force needed to accelerate 1 kg by 1 m/s, and the kilogram-force [kgf] is the force needed to accelerate 1 kg by g m/s, where g is the standard gravity.
The standard atmospheric pressure is set at 101325 Pa, which would translate to ~10333 kgf/m^2.
So why don't we all implode? What is missing from the picture?
| Missing from the picture is that all macroscopic bodies in fluid or solid state do not like being compressed. These macroscopic bodies can be thought of as made up of molecules or atoms which below some distances will always start to repel each other. This causes them to exert a force against being compressed, i.e., brought more close to each other. So pressure is balanced by the resistance of bodies to be compressed beyond a certain point.
| {
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Can atomic orbitals of an isolated atom rotate relative to the nucleus? I am a beginner at orbitals. It seemed interesting to me while studying p orbitals that the texts don't suggest that the orbitals are rigid relative to the nucleus.
But I thought mathematically these rotations would not be possible to know because the wave function of the orbitals only tell about the probability. Until this:(Simple rotation of an atomic orbital wavefunction)( I didn't understand it as I don't know the mathematics; I am trying)
So, is it possible for orbitals to rotate relative to the nucleus? If yes, what validates it; if no, what disallows it?
| Atomic p orbitals have angular momentum. The angular part of the wavefunction is described by the spherical harmonics $Y_{\ell,m}$ where $\ell=1$ and $m=-1,0,1$. When one multiplies this with the time-dependent part $e^{iEt/\hbar}$, the result is a rotating phase.
The $p_x$ and $p_y$ orbitals of chemical compounds are different linear combinations of $Y_{1,1}$ and $Y_{1,-1}$ without angular momentum along the z-axis.
Absorption of linearly polarized light may transfer an electron from an $s$ orbital to a $p_x$ orbital, but it will not remain oriented because of spin-orbit coupling.
| {
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Why should the perturbation be small and in what sense? In time-independent perturbation theory, one writes $$\hat{H}=\hat{H}_0+\lambda \hat{H}^\prime$$ where $\lambda H^\prime$ is a "small" perturbation.
*
*Why should the perturbation be small for perturbation theory to work?
*Both $\hat{H}_0$ and $\hat{H}^\prime$ are operators. Therefore, what does it mean to say the perturbation is "small"? I think, saying $\lambda \hat{H}^\prime\ll \hat{H}_0$ is meaningless.
*Is it that the matrix elements of $\lambda\hat{H}^\prime$ much smaller than that of $\hat{H}_0$ in the eigenbasis of $\hat{H}_0$? If yes, why is such a mathematical requirement necessary? In other words, what if the matrix elements of $\lambda\hat{H}^\prime$ are comparable to that of $\hat{H}_0$?
| When one assumes the solution to the perturbed system is of the form
$$|\psi\rangle=\sum_{n=0}^\infty \lambda^n|\psi_n\rangle$$
where $|\psi_0\rangle$ is an eigenstate of $H_0$, one hopes that the expression is meaningful and that only the first few terms are significant which is to say that $|\psi\rangle$ is in some sense close to $|\psi_0\rangle$, i.e., that the unperturbed Hamiltonian is in some sense dominant.
| {
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Car Power/Torque to Weight -- non linear? ( Hello, my first question ... )
Assume two cars, each with exactly the same body, but one a scaled down version of the other (hence, aerodynamics approach equality):
"junior car" - has an engine that delivers 200 ft-lbs of torque and weighs 3000 pounds
"base car" - has an engine that delivers 400 ft-lbs of torque and weighs 6000 pounds
My question is this: isn't the effect of "mass" non-linear? Doesn't the junior engine have less than half the work to do to accelerate than the base car because momentum, friction, drive train length, drive train mass, turbo charger mass, gearing all drain power in a non-linear way; the larger a vehicle you make? I've done the G-search thing and haven't found any answer to this type of "real world" condition.
(Or, asking the same thing in a different way: would the two vehicles above accelerate with identical track times?)
Thanks for responding.
.R.
| In the real world, at constant speeds, coefficient of drag matters more than the weight of the vehicle, at least within the normal range of vehicle weights, else a vehicle weighing 2500 lbs. would get twice the gas mileage of one weighing 5000 lbs., and this isn't true. Loss from drivetrain friction and other areas will be a constant percentage. In other words, no. The smaller vehicle would perform better under the conditions you stated if horsepower is equal.
| {
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How to vary the current using batteries? I am making a door bell as a school assignment. It works by having a solenoid produce a magnetic field which attracts a pice of iron attached on a conductor. when the iron is attracted towards the solenoid the current is broken so it falls back, inducing the magnetic field again. This will produce a frequency. The investigation is based on the influence the current has on the frequency.
I have tried to power the circuit by ten 4.5 volt battery packs connected in parallel as well as in series. But however I connect the batteries to the circuit Current will stay constant, even when I disconect battery packs.
In series I get a constant current of 3 amps. In parallel I get a current of 7 amps maximum.
What should I do to vary the current?
| You can use potentiometer in series with the battery to vary the current.
| {
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Regarding Energy and Momentum in QM I've been trying to learn Quantum Mechanics for a few months now, and there's a pretty fundamental thing I never quite understand: What is energy and momentum in quantum mechanics?
I've been following MIT's QM course, and the instructor talked about the energy and momentum of light/wave-like objects, and said that De Broglie proposed that $E=\ \hslash \omega$ and "it turned to be true".
My question is, what is/was the QM definition of energy before de Broglie, and how did $E=\hslash \omega$ "turn out to be true"? Same question goes for momentum (the only definition of momentum I know is $\mathbf p = m\mathbf{\dot x}$).
| This is actually more of a comment than an answer but I don't have enough reputation to post a comment yet. Anywho, it's important to distinguish between classical dynamical definitions of dynamical properties and quantum mechanical definitions. First of all the definition you gave for momentum is that of linear momentum $\mathbf{p}=m\mathbf{\dot{x}}$. Furthermore, what you are discussing are the deBroglie relations for matter waves. For e.g. the energy of a photon would be given by $E=hf$ where $h$ is plancks constant and $f$ is the frequency so the momentum can be obtained by the deBroglie relation $\lambda = h/p$ where $\lambda$ is the wavelength. The description of energy (and the beginning of the quantum era) pre-deBroglie is given by Einstein and Planck. Perhaps reading the wiki article will help, given here. then of course, we have the classical $E=T+V$ relation in classical conserved systems.
P.S. with regards to the course you are taking, I think you will find that in terms of mathematics, the book by Griffiths is good and very doable the whole way through. Good luck!
| {
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use of static vs kinetic friction when orthogonal forces are applied Suppose I have a mass sitting on a surface with friction. Now I start pushing on the mass in one direction (call this direction the x direction). To get this mass accelerating, I have to push the mass with a force greater than $\mu_{s}mg$. Assuming the force is large enough such that the mass moves until there is a significant x component of velocity.
Now here's where my question arises. While this mass is moving in the x direction suppose I decide to push it along a direction orthogonal to it's motion (call this direction y). Is the minimum force required to get the mass to move in the y direction going to be $\mu_{s}mg$ or is it $\mu_{k}mg$ since the object is already moving?
How would I think of this problem in terms of a microscopic model of friction?
| There doesn't seem to be a clear and obvious answer from first principles, but if the object is in motion, kinetic friction should be applied. It might be useful to consider the microscopic model - that the static form of friction is applied when there are molecular attractions between the object and the floor.
However, kinetic friction deals with dynamic forces that are random and time dependent. This can be due to how the floor is not perfectly uniform, or that there are imperfections in the object. Then, kinetic friction is somewhat an average of many different electromagnetic interactions between the object and the floor.
What you are describing seems to be much more kinetic than static. The object is not at some equilibrium with the floor - it is passing through many different patches. Therefore, it is interacting with the floor in a different way than static friction implies.
| {
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What does it mean to say that two optical systems are equivalent to each other? I am studying about Michelson interferometer from Pedrotti and they transformed a Michelson interferometer into an equivalent system consisting of light rays originating from two virtual images of the sources.
When can such conversions into equivalent systems attempted and why ?
Relevant image :
Excerpt :
The actual interferometer in Figure 11-la possesses two optical axes at right
angles to one another. A simpler but equivalent optical system, having a single
optical axis, can be displayed by working with virtual images of source 5 and mirror
M1 via reflection in the BS mirror. These positions are most simply found by
regarding the assembly including S, M1, and beams 1 and 3 of Figure 11-la as rotated
counterclockwise by 90° about the point of intersection of the beams with the BS
mirror. The resulting geometry is shown in Figure 11-lb. The new position of the
source plane is S', and the new position of the mirror Ml is MГ. Light from a
point Q on the source plane S' then effectively reflects from both mirrors M 2 and
Ml', shown parallel and with an optical path difference of d. The two reflected
beams appear to come from the two virtual images, Q'l and Qî, of object point Q.
| The notion of "equivalent" in physics generally has many different uses and associated definitions. It is often a question of input/output. Two systems would be said to be equivalent if they have the same output for the same input. This is how the concept is used in circuit theory, for example.
In optics it is easy to see that the light reflected from a mirror is moving just as if it had come from a source behind the mirror and so there is a kind of equivalence. This is the idea being used in your Michelson example. It is a useful idea as long as we place the "equivalent" source correctly and precisely. With that done, we may then find it easier to trace where the waves will go, and how they accumulate different amounts of phase, depending on the optical path length.
| {
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Inconsistency in Vertex Factor in Scalar QCD I'm trying to find the vertex factor for the scalar-scalar-gluon-gluon interaction. The diagram is
This corresponds to the term $g^2 \phi^{\dagger}A_{\mu}^a\left(T^aT^b\right)_{ij}A_{\mu}^b\phi_j$ in the Lagrangian density. Naively, I expect the factor to be
$$
-ig^2\left(T^aT^b\right)_{ij}g^{\mu\nu},
$$
which is what appears in Schwartz's book. However, other sources have
$$
-ig^2\left(T^aT^b + T^bT^a\right)_{ij}g^{\mu\nu}.
$$
Which is correct? This makes a difference in calculation I'm doing (finding the beta function in scalar Yang-Mills theory), in which I have the following corrections to the three-point vertex:
If I use Schwartz's convention, the color factor is $T^bT^aT^b$, which can be rewritten as $\left(C(R) -\frac{1}{2}T(A)\right)T^a$. If I use the other convention, I have an extra factor of $C(R)T^a$.
| Since $A^a_\mu A^b_\nu g^{\mu\nu}$ is symmetric in $a$ and $b$, only symmetric part of $T_aT_b$, i.e. $\frac{1}{2}(T_aT_b + T_bT_a)$ should contribute to the vertex. Also, since the term is quadratic in gauge potential, you get a combinatoric factor of 2.
You can also compute functional derivative of the action, $\frac{\partial S}{\partial \phi^\dagger_i \partial\phi_j \partial A^a_\mu \partial A^b_\nu}$, to obtain the correct expression for the vertex.
| {
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Why do electrons move from amber to wool during static electricity charging? We know our word for electron comes from the Greek word for amber, since the ancient Greeks experimented with static electricity by rubbing amber on wool.
We know that a static charge is built up by the electrons in the amber moving to the wool.
But why do the electrons move? Aren't they happy orbiting the protons of their own atoms? (Or happily caught in the bonds of their molecules?)
Why would an electron go travelling simply by rubbing? (And stay there?)
My question is: Why do electrons move from amber to wool during static electricity charging?
| Electrons, like gas molecules, will diffuse. So, a material with
a high concentration of surface electrons will lose some to any material
nearby that has a lower concentration.
The diffusion halts when buildup of
charge creates sufficient voltage drop to reach equilibrium between
ohmic (field-driven) current and diffusion-driven current.
There may only be a fraction of a volt in such a contact potential, but
the very short distance between objects in contact means high
capacitance. Capacitance equals charge divided by voltage, so the charge
can be significant.
Of course, when you pull nonconducting objects
apart, the gap between the net-positive and net-negatively-charged objects
is increased, and separation does work on the charges, i.e. it raises the voltage.
If those objects
are conductive, pulling them apart doesn't create much electric separation,
because the charges can conduct to the last few square microns of contact area,
and the resulting smaller capacitance effectively neutralizes the objects even
with the same small contact voltage present.
| {
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Orthogonality of the particle in a box model I found that for the particle in the box model, since the solutions represent the wave functions $\psi_n=A\sin{\frac{n\pi x}{L}}$ and $\psi_{n+k}=A\sin{\frac{\left(n+k\right)\pi x}{L}}$, hence the integral shown below $$\int_{-\infty}^{+\infty}A^2\sin{\frac{n\pi x}{L}}\sin{\frac{\left(n+k\right)\pi x}{L}}\mathrm{d}x=\frac{A^2}{2}\int_{-\infty}^{+\infty}\left(\cos{\frac{k\pi x}{L}}-\cos{\frac{\left(2n+k\right)\pi x}{L}}\right)\mathrm{d}x=0$$
From this I imply that $$\langle\psi_n|\psi_{n+k}\rangle =\int_{-\infty}^{+\infty}{\psi_n}^*\psi_{n+k}\mathrm{d}x=0.$$ Is this valid, only for the particle in the box model or for all potential shapes as well? But, this means that all the $\psi$ waves are vectors orthogonal to each other and electronic transition from one eigen value level $E_n$ to the other $E_{n+k}$ would not be deemed as a possibility. Yet electronic transitions do exist; am I doing something wrong or am I going wrong somewhere?
| To calculate electronic transition you need to have some time dependent potential then at least to first order the probability of transition between eigenstates is $|\int dt <\psi_k|V(t)|\psi_{k'}>e^{iE_k -E_{k'}t}|^2$ this is very different from $\int \psi_k^*\psi_{k'}dx$.
| {
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Do unstable equilibria lead to a violation of Liouville's theorem? Liouville's theorem says that the flow in phase space is like an incompressible fluid. One implication of this is that if two systems start at different points in phase space their phase-space trajectories cannot merge. But for a potential with an unstable equilibrium, I think I've found a counterexample.
Consider the potential below (excuse bad graphic design skills).
A particle starting at rest at point A, $(q,p) = (x_A,0)$ at $t = 0$, would accelerate down the potential towards the left. Because it has the amount of energy indicated by the purple line, it would come to rest at the local maximum B at $t = T$, an unstable equilibrium $(q,p) = (x_B,0)$. However any particle started at rest at the top of the local maximum B at $t = 0$ would also stay that way forever, including up to $t = T$. Thus there appears to be two trajectories that merge together in violation of Liouville's thorem.
| A few comments on the question you are asking:
*
*The Liouville equation describes not a single particle, but an ensemble of states that evolve according to the laws of mechanics. The Liouville equation says that the entropy of the system stays constant at all times, which is to say that this equation describes isentropic processes and nothing else. The approach to equilibrium is accompanied by entropy increase, something that the Liouville equation cannot handle.
*The statement that "two systems start at different points in phase space their phase-space trajectories cannot merge" is correct, but it is not a consequence of the Liouville equation. It follows from the fact that the differential equations of motion have a unique solution for given initial condition. Merging would imply that two different initial conditions produce the same state at some later time, which is not possible.
*Your example does not violate the uniqueness of trajectories. To violate it you need to find two different trajectories in the same potential, with both trajectories starting at $A$ and both ending at $B$.
| {
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Is the universe non-linear? First of all, I've read this other question Is the universe linear? If so, why? and I'm aiming at a different kind of answer.
Theories like General Relativity or QFT, which are believed to be quite fundamental, are strongly non-linear. However, in the end, both theories must be just low energy limits of an unified theory. So this question arises: Could this unified theory be linear? I'm looking for both a mathematical and a physical answer. In other words, I'd like to know if 1) it is possible to make a linear theory that has non-linear low energy aproximations and if 2) a non-linear universe would make any physical sense in view of the superposition principle.
| The obvious example is hydrodynamics.
The interactions in a fluid all originate from the interactions between atoms and molecules that are described by quantum mechnics, and QM is as far as we know linear. However the Navier-Stokes equations are (scarily) non-linear and produce all sorts of weird behaviour.
It's an interesting question whether general relativity is a fundamental theory or whether it's an emergent theory based on some more fundamental interactions. The simple answer is that we don't know as there is no evidence either way. The work on this idea is being led by Erik Verlinde.
| {
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If the ground's normal force cancels gravity, how does a person keep rotating with the Earth? When I am on earth, the weight of my body is countered by the reaction of the ground. So, there is no net force acting on me.
But I am spinning with earth. But if there is no centripetal force then why am I spinning? And the equal air pressure on both side of my body won't be enough for me to stay in the same angular velocity as the earth.
Is it just conservation of angular momentum?
| It is easier to consider you standing on the Equator.
Assume that the gravitational field strength at the Equator is $g$. This would be the acceleration of free fall at the Equator with no air resistance if the Earth was not spinning.
If the reaction of the Earth is $N$ then assuming down is positive and using N2L, $mg-N=0$ if your mass is $m$.
If the Earth of radius $R$ is spinning with angular speed $\omega$ then using N2L one gets $mg-N'=mR\omega^2$.
So the reaction force due to the Earth $N'$ has decreased.
The acceleration of free fall would also decrease to $g-R\omega^2\;(\approx 0.03 \rm ms^{-2})$ as would your apparent weight $m(g-R\omega^2)$.
So measuring your "weight" at the Equator using a spring balance would yield a smaller value than that at the geographic poles where you your centripetal acceleration would be zero.
If it so happened that the period of rotation of the Earth was 84.5 minutes you would find that there was no reaction force due to the Earth and the acceleration of free fall would be zero.
Objects which you let go of would not fall closer to the Earth.
This would be a state of weightlessness.
It so happens that 84.5 minutes is the theoretical speed of a satellite of the Earth whose circular orbit had a radius equal to that of the Earth.
All this has ignored the effect of air resistance and the fact that if the Earth was made to spin that fast it would disintegrate due to the brittle crust not being very good at sustaining tensile stresses.
| {
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Perturbation Theory for a ring in an Electric Field
A particle of mass $m$ move on a circular ring of radius $a$. The only variable of the system is the azimuthal angle, which we will call $\varphi$. The state of the system is described by a wave function $\psi(\varphi)$ that must be periodic,
$\psi(\varphi + 2\pi) = \psi(\varphi)$ and normalized.
Now assume that the particle has a charge $q$ and that it is placed in a uniform electric field $ε$ in the $x$-direction. We must therefore add to the Hamiltonian the perturbation
$$\delta H = −q\epsilon a \cos \varphi$$
Calculate the new wave function of the ground state to first order in $ε$. Use this wave function to evaluate the induced electric dipole moment in the $x$-direction: $\langle\psi|q_x|\psi\rangle $. Determine the proportionality constant between the dipole moment and the applied field $ε$. This proportionality constant is called the “polarizability” of the system.
My problem is when I'm trying to estimate the first correction
$$E_1=\langle \psi_0|−q\epsilon a \cos \varphi|\psi_0\rangle.$$
It is coming out zero so I don't understand why the wave function of the ground state will change?
| This is a simple exercise in properly applying time-independent perturbation theory to the eigenvalue equation
$$
\frac{d^2\psi_n}{d\varphi^2} +\frac{2E_n ma^2}{\hbar^2} \psi_n(\varphi) =-\epsilon\frac{2 m a^3 q}{\hbar^2} (\cos \varphi ) \psi_n(\varphi).
$$
All you have to do is substitute 1st order approximations to the eigenfunction and eigenvalue, something like $\psi_n(\varphi) = \psi_n^{(0)}(\varphi) + \epsilon \psi_n^{(1)}(\varphi)$ and $E_n = E_n^{(0)} + \epsilon E_n^{(1)}$, and then separate terms corresponding to different powers of $\epsilon$.
If you do this correctly you'll find that the equation for the 1st order correction to the eigenfunction turns out to be
$$
\frac{d^2\psi_n^{(1)}}{d\varphi^2} + \frac{2E^{(0)}_n ma^2}{\hbar^2} \psi_n^{(1)} =- \frac{2ma^2}{\hbar^2} \left( a q \cos \varphi + E^{(1)}_n\right) \psi_n^{(0)}
$$
It's not hard to see that even if both $E_n^{(0)}$ and $E_n^{(1)}$ happen to be zero (and I am not saying that they are), the $\psi_n^{(1)}$ correction does not vanish. As a rule, both $\psi_n^{(1)}$ and $E_n^{(1)}$ are determined by the perturbation, but this does not mean that $\psi_n^{(1)}$ is "proportional" to $E_n^{(1)}$.
| {
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How varying length of loop affects induced emf
*
*What is happening when delta x of this loop increases? Give me a theoretical idea and how is emf increasing? I know that flux is changing but I think that the rails on which conductor rod is moving is not contributing that much to induced emf
*What is induced emf actually in this case is it the energy with which charges are moving or the increasing electric field in conductor because of the charge accumulation at the corners of conductor creating a region acting as source for energy
| The induced emf blv is called motional emf. thus we are able to produce induced emf by moving a conductor instead of varying magnetic field that is, by changing the magnetic flux enclosed by the circuit.
it is possible to explain the motional emf expression by invokin the lorentz force actin on the free charge carries of conductor.Consider any arbitrary charge q in the conductor. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field B. The lorrentz force on this charge is qvb in magnitude, and its direction, irrespective of their position in the rod.
It is not obvious how an emf is induced when a conductor is stationary and the magnetic field is changing-a fact which Faraday verified by numerous experiments. In the case of a stationary conductor, theforce on its charges given by
F =q(E + v×B)=q E
since v=0 , thus any force on the charge must arise from the electric field term E alone . Therefore, to explain the existence of induced emf or induced current, we must assume that a time-varying magnetic field generates an electric field . However we hasten to add that electric field produced by static electric charges have properties different from those produced by time varying magnetic fields. we know that charges in motion(current) can exert force/torque on a statiionary magnet. conversely, a bar magnet in motion(or more generally, a changing magnetic field) can exert a force on the stationary charge. this is the fundamental significance of the Faraday's discovery. Electricity and magnetism are related.
| {
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Can we measure the exact value of the Fermi Level in semiconductor? Or is it always measured relatively to the Conduction/Valence Band energy level? From the books that I read, the discussion and the formulas related to the Fermi Level are always relative to the energy level of Conduction/Valence Band, or Fermi Level in intrinsic semiconductor.
Let's assume that the measurement is done at exact temperature.
| Fermi level characterizes the filling of the energy levels, e.g., the concentration of electrons is given by
$$n = \int dE D(E)\frac{1}{1 + e^{-\beta(E-\mu)}},$$
where $D(E)$ is the density-of-states and $\mu$ is the Fermi level. As you see from this equation, shifting the Fermi level will make the occupations of all the energy states change and this will change the electron concentration and other parameters. The only way to have everything remaining consistent is to treat the Fermi level as any other energy, i.e. it is measured in respect to the same origin as all other energies.
What might be a possible source of confusion here is that Fermi level is not the same thing as Fermi energy, which is the energy distance between the bottom of the conduction band and the last occupied state, as I discussed here.
| {
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Is there a standard, coded table of physical units? I am a programmer in the medical/biosignals area, and I want to represent physical units in a database table. Ideally, I would like to have a code that uniquely identifies a given physical quantity.
As an example, ISO defines strings to represent languages, such as "en-US", or "pt-BR". I am looking for similar standard (ideally from ISO or similar) that identifies electric potential, force, angle, acceleration, etc.
| That would be the International System of Units by the Bureau International des Poids et Mesures, as seen here.
| {
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Concept of negative mass in solving problems I have searched a lot on the internet regarding negative mass. And in the meantime, I came across this question.
The question states:
Consider two spherical empty regions (C1 and C2) in an otherwise
uniform and essentially infinite intergalactic gas cloud of density
$\rho$ as shown in the figure. As a result of gravitational
effects, describe the movement of the empty regions.
One very obvious approach is to consider the spherical empty regions as a superposition of two spherical regions of densities $\rho$ and $-\rho$. From this one can easily deduce that the spherical regions will move towards each other. However, I am not satisfied with this because the theory of negative mass is not defined in reality. Also, what would happen if the density varied according to some law? Will the same approach work?
| As my previous answer was not understood and the wrong one was chosen as correct, I'll try to present a clearer answer.
The integrals in forky's answer are fine, but the bubbles repel!
Imagine that in stead of completely empty bubbles you have balls of styrofoam, with $\rho'= (1/10000)\rho$, or any other density, as small as you like. Now things get easier, the balls still have a practical density of almost 0 compared with the rest of the bulk, but now they also have a well defined gravitational and inertial mass.
Now if you do the integrals, you will see that the gravitational force on $C_1$ will be: $$F=\frac{\rho'V_{C_1}(\rho'-\rho)V_{C_2}\hat{R_{12}}}{R_{12}^2}=-9999\frac{\rho'^2V_{C_1}V_{C_2}\hat{R_{12}}}{R_{12}^2}=\rho'V_{C_1}a$$
$$a=-9999\frac{\rho'V_{C_2}\hat{R_{12}}}{R_{12}^2}=-\frac{9999}{10000}\frac{\rho V_{C_2}\hat{R_{12}}}{R_{12}^2}$$
The acceleration is in the direction of $-\vec{R}$, meaning the balls repel. As we take $\rho'\rightarrow 0$, the acceleration will approach $a\rightarrow-\frac{\rho V_{C_2}\hat{R_{12}}}{R_{12}^2}$, but is will still be in the "out" direction.
| {
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Thermodynamics: What is the entropy change of an electrical resistance taken as system? Consider a system consisting only of an electrical resistor through which a constant current is flowing. Simultaneously, it is rejecting heat such that its temperature is constant throughout the process. With this much of information, what can we conclusively say about the entropy change of the system?
Thanks in advance.
| (I am assuming that temperature is constant in the resistor. If you want, think about it as a quasi-1D object)
If we assume the system to be stationary, then all the thermodynamic parameters in the resistance will be constant. Therefore, since entropy is a state function, the entropy change in any given time interval will be $0$: the system is increasing its entropy by absorbing energy from the battery, but since we are assuming that $T$ is constant it means that it is simultaneously releasing the same quantity of energy in the environment, thus decreasing its entropy.
The medium to which the resistance is giving off its heat, on the other hand, will heat up, and its entropy will increase. Using the formula for Joule heating, we have
$$dS = \frac {\delta Q} T = \frac{I^2 R dt}{T} \to \frac{dS}{dt}= \frac{I^2 R} T$$
| {
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What is the upper limit on gravitational time dilation? First, please forgive me: I'm not versed enough in mathematics to even approach working this out for myself.
Now, the question: how much "slower" could time pass for a person around a supermassive black hole (such as S5 0014+81)? Would this increase indefinitely with the mass of the hole, or is there something about maximum density or whatnot that places a limit on this?
I'd imagine that the increasing volume of a hole would somewhat cancel the effects of increasing mass, but I'm such a layperson that I'm merely speculating at this level of knowledge.
| The time dilation becomes infinite at the event horizon. That is, an observer watching from far away would see the rate that a clock runs slow to zero as the clock approaches the event horizon. This is not directly related to the mass of the black hole. The mass determines where the event horizon is, but it's the distance relative to the event horizon that determines the time dilation.
For a static black hole the time dilation is given by:
$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{r_s}{r}} \tag{1} $$
where $r_s$ is the event horizon radius and $r$ is the distance from the black hole. For interest, the event horizon radius is given by:
$$ r_s = \frac{2GM}{c^2} $$
Anyhow, when you are far from the black hole $r \gg r_s$ so $r_s/r \ll 1$ and the time dilation is small. As $r$ decreases towards $r_s$ the time dilation gets greater until when $r = r_s$ equation (1) gives us:
$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{r_s}{r_s}} = \sqrt{1 - 1} = 0 $$
so at $r = r_s$ time stops completely.
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How to choose origin in rotational problems to calculate torque? We know that $\text{Torque} = r \times F$ and $r$ is the position vector. But the position vector depends on the choice of the coordinate system and in turn on the choice of origin. So, where should we take the origin?
Also, do torque, angular velocity and angular acceleration point of out the plane of rotation for 2D objects because otherwise they wouldn't have constant direction?
Many sources (including my textbook) seem to say that the origin should lie on the axis and that it wouldn't make a difference where it is on the axis.. but I don't get why it shouldn't, since position vector would be different from different origins and so the torque, according to me, might come out to be different.
| It says the torque about the axis is the same irrespective of where you take the origin. Because the moment arm will always be the same.
Another way view it is $rsin\theta$. The more away you take the origin, the greater the $r$ and hence smaller the $\theta$. However you take it the $risn\theta$ will result same regardless of the values of $r$ and $\theta$ about the axis.
But when choose the origin differently the torque gets inclined and the torque around the axis is a component of the torque about the origin. The other component produces torque perpendicular to axis and parallel to centripetal force.
And about angular velocity, angular velocity about the axis always parallel to the axis. But that doesn't mean there aren't any other angular velocity perpendicular to the axis.
And for this reason, they put bearings on the axis so that it can counterbalance that torque and the angular velocity.
| {
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Reversible process in General Physics Usually, for christmas , I have lunch with my family and a couple of other families. Most of the people got a Phd on chemistry, or molecular biology, and are high academics (they're in they 50-70). On the other side, i'm the only one who studies physics there, and 2 years ago, in the same situation, they told me that physics only study reversible proccess (In the context that we're like, noobs). At that moment I was very young and have litle idea about physics. What's your opinion?
| Never-the-less, there is the law of entropy, which chemical re-actions follow, which has an irreversibility of direction to it. probably that is behind the comment. A chemist might not have picked up how that law applies in other areas (or maybe I should say at other scales/levels) if they have specialised, I suppose.
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Solution of Dirac equation-Positive and Negative energy For particles defined with positive energy, we use $$\phi=
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
$$ or $$
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}
$$
while for particles defined with negative energy, we use this instead
$$\chi=
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
$$ or $$
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}
$$
where ∅ is the upper component while $\chi$ is the lower component of the bispinor in Dirac equation.
Can we do it the other way round or
$$\phi_p=
\begin{pmatrix}
1 \\
0 \\
0 \\
0 \\
\end{pmatrix}
....$$ instead? I don't understand is why is the 4-spinor split into two components, one for the positive energy and the other for the negative energy. It did not say anything here.
| I don't understand is why is the 4-spinor split into two components, one for the positive energy and the other for the negative energy. It did not say anything here.
You are misinterpreting the large and small components of the Dirac 4-spinor for the positive and negative energy solutions (a common mistake). Each Dirac spinor (whether a positive or negative energy solution) has four components. For a positive energy solution the upper 2-spinor is called the large component because the lower 2-spinor is much smaller (of order $\frac {v}{c}$ in comparison, $v$ is the average velocity of the particle described). The same applies to the negative energy solutions except that the relative size of the upper and lower components is reversed.
Your link describes the situation correctly as does the answer by @probably_someone.
| {
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Why can't the Klein-Gordon equation explain the hydrogen atom but the Dirac equation does? Why can't the Klein-Gordon equation with a Couloumb potential describe the hydrogen atom?
Why can the first order Dirac equation explain it?
What are the failures?
| The application of the Klein-Gordon equation for the hydrogen atom leads to logical contradictions.
Allowance for relativism in the Klein-Gordon equation should lead to more accurate solutions for the hydrogen atom compared with the Schrodinger equation. However, on the contrary, solutions became increasingly worse with an increase in the nuclear charge. Finally, for the element Z = 69 the solution of the Klein-Gordon equation terminates.
The Klein-Gordon equation has other drawbacks. Therefore, we can say that the Klein-Gordon equation is erroneous.
| {
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After measuring momentum, it seems like the particle's position could be literally anywhere? Once measuring momentum, the wavefunction "collapses" into something that looks like this
If you were to then measure the position, couldn't it be literally anywhere? What am I missing? Is it even possible to measure momentum perfectly?
| Each observable corresponds to a mathematical operator in Hilbert Space. There are pairs of observables which are called conjugate variables, these cannot both be known accurately at the same time. The measurement of one immediately makes the measurment of the other impossible. Position and momentum are such a pair. It is possible to measure one to a lesser precision, and then measure the corresponding conjugate variable in a similar manner. The relationship between the accuracy of the two measurments is given by the Heisenburg uncertainty.
| {
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Deviation in refraction related question Why in refraction incident ray remain undeviated when incident angle is 0°? Please give the molecular interpretation.
| Refraction is explained by the wave nature of light. What happens is that the incident light wave wavefront is slowed down by the optically denser material (glass slab, lens etc.). If the angle of incidence is more than 0 degrees, the parts of wavefront are slowed down as soon as they interact with the denser material but since the parts away from the material are still moving at a faster speed(original speed) , the light deviates from its path . In the case where the angle of incidence is 0 degrees, the light does slow down but it keeps moving at the same path as before and there is no deviation.
This is kind of hard thing to explain in words, so I would suggest that you watch some videos on refraction.
Why light slows down is because the way the light(Electromagnetic Radiation) interacts with the molecules of the material. I don't know much about how exactly it happens, but I can tell you that it has something to do with the electric field of the electromagnetic radiation and the charges inside the material(charges have their own electric fields). Search on Google if you are interested.
I am sorry if I couldn't clear your doubt.
| {
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Thermodynamics Question: Does measuring the temperature of an object change its temperature? Suppose that I want to measure the temperature of an object, such as a pot of hot water. When I stick the thermometer into the pot, I know that the temperature measured by the thermometer is its own temperature when it reaches thermal equilibrium, which, according to the Zeroth Law of Thermodynamics, is equal to the temperature of the object (the pot of hot water) at thermal equilibrium. Does this imply that the temperature of the object that I am now measuring is different than the initial temperature, and if so, is the change significant? Also, can I somehow use the information of the system at thermal equilibrium to find the initial temperature of the object?
| Yes the thermometer affects the temperature of the body that you are measuring. The change in temperature can be found doing some simple calorimetry (so you have to know the temperature, mass, and specific heat of the thermometer before you use it). And the change in temperature might be significant (but typically isn't at all), it just depends. Suppose you are measuring the temperature of a very small amount of a low specific heat substance and your thermometer is particularly cold (or hot, relative to the material of interest); here the change in temperature might be significant. Again, understanding how much the temperature will change just boils down to calorimetry.
| {
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Projectile Motion - $V_f = V_i + at$ - Divide by zero If I have a projectile that is thrown at some horizontal velocity at some height, and horizontal acceleration is zero, can't I use the equation $v = v_0 + at$? The problem is when I use it since $v$ will equal $v_0$ (acceleration is zero so velocity won't change) I get $\frac{0}{0}$ which is undefined and definitely not the answer.
| Resolve the vertical component until it h=0, it hits the ground. Use this time t in the horizontal $s=ut+(1/2)at^2$ where a is 0. Because horizontal acceleration is zero.
| {
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Why is the circumference of a circle drawn on a sphere less than the circumference of the circle of same radius drawn on a flat surface? Why is the circumference of a circle drawn on a sphere less than the circumference of the circle of same radius drawn on a flat surface? This is an extract from the book "The Elegant Universe" but I couldn't figure out why this actually happens. How can one figure out that circumference is going to decrease or increase if a circle is drawn on different types of curved surfaces?
| I think the crucial ingredient to this is that the difference arises because one assumes that this is from the perspective of "sphere-people" who live confined to the surface of the sphere. They cannot see, hear or reach points interior or exterior to the sphere. The surface of the sphere is their whole world and existence. When these "sphere people" draw a circle, and then walk from the edge to the center of the circle, they don't travel in what we (the "3D" people) would observe to be a straight line. They travel along a curve (the surface of the sphere). So, they (the "sphere-people") measure the radius of their circle to be longer than we would measure it to be. This is because what they see as a "straight line", we see as a curved line, and what we see as a "straight line", they can't see at all.
| {
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How does a small object move with constant velocity when drag force is equal to its weight? When drag force ($bV$) equals to object's weight (mg) then upward and downward force becomes equal. As a result the object comes to rest. If this is true, how is a body moving with constant velocity?
| When opposing forces are equal the object does come to arrest. Just try standing on a scale. As for a car traveling down the highway it matches the wind and road resistance with more throttle until an equilibrium is reached at whatever speed you desire.
| {
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Light clock with 1 m ticks A standard light clock has two mirrors, say one metre apart, and a light pulse. When the clock is stationary the light path is perpendicular. When moving, diagonal. The clock ticks over each time the light hits the mirrors. The light takes longer to travel the diagonal path, so the clock takes longer to tick over, so we get time dilation.
What if the clock ticked over each time the light travelled one metre? The light would still travel the same paths, perpendicular or diagonal,except those paths would be divided up into one metre lengths. Is one metre the same in both frames? It has to be, it is the distance the mirrors are apart in both frames. If you are having trouble visualising this,draw a light clock and just mark the light's path in one metre lengths, as opposed to where the light hits the mirrors.With one metre ( or whatever the mirrors perpendicular distance apart is ) the clock will tick over at the same rate in both frames. This contradicts time dilation. Can you explain why?
Thanks.
DAC.
|
Is one metre the same in both frames? It has to be, it is the distance the mirrors are apart in both frames.
No, a one meter light path is most assuredly not the same in both frames. If the mirrors are 1 m apart then in one frame the pulse of light travels 1 m but in another frame the same pulse of light travels more than 1 m. The point where the light has traveled 1 m in the “moving” frame will be less than 1 m in the stationary frame.
Note that the relevant distance is the length of the light path, not the distance between the mirrors
| {
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What are Hamilton's equations with respect to a nonstandard symplectic form? Hamilton's equations for a Hamiltonian $H(q,p)$ w.r.t. to a standard symplectic from $\omega = dq \wedge dp$ are $$\dot{q} = \partial H_{p}, \quad \dot{p} = - \partial H_{q}$$
How do Hamilton's equations write w.r.t. a nonstandard symplectic form $F(q,p) dq \wedge dp$, where $F(q,p)$ is some smooth function?
| A Hamiltonian $H:M\rightarrow \mathbb{R}$ defines a vector field $X_H$ through the equation
\begin{equation}
\omega(X_H,\cdot)=dH.
\end{equation}
For $\omega=F(q,p)dq\wedge dp$ and substituting the components $X_H=X_{Hq}\partial_q+X_{Hp}\partial_p$ we get
\begin{equation}
F(q,p)(X_{Hq}dp-X_{Hp}dq)=(\partial_qH)dq+(\partial_pH)dp.
\end{equation}
The integral curves $t\mapsto(q(t),p(t))$ of the vector field $X_H$ represent the Hamiltonian flow of the system. Therefore, we have
\begin{align}
\dot{q}=\frac{\partial_qH}{F(q,p)};\;\;\;
\dot{p}=-\frac{\partial_pH}{F(q,p)};
\end{align}
| {
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Apparent weight in circular motion Here's a diagram below. From my understanding, the apparent weight is the weight that you feel > which is the upward force on you > which is the normal force.
I can clearly see how you'd weigh more in the bottom of the circle, because the net force is directed up, and so that would mean normal force gets increased. Thus saying, you'd feel like you weight more.
From that same reasoning, I'm having trouble seeing why you'd weigh less at the top of the loop, just from looking at the free body diagram.
I have seen physical demonstrations of situations of this with scale readings, and I know you have to weigh less on the top of the loop, but the free body diagram doesn't seem like it shows that, or am I wrong?
What I don't understand is that the length of the normal force vector looks the same as the weight vector. And isn't it true that at the top of the loop, you're upside down so your upward force would be directed down? In that case, this would make your apparent weight equal to your actual weight but that's wrong. Why? I'm having lots of trouble understanding the free body diagram.
| It would be helpful if the meaning of the vector symbols n & w can be defined. it looks like w is gravity, and N is centrifugal acceleration. In the case N will be smaller at the top as the v is lower.
F seems to be confusing, it seems to be labelled in two places as the overall resultant force, but it would be different in each place.
At the bottom gravity and centrifugal force (high speed) combine to produce a greater force.
At the top gravity works in the opposite direction from centrifugal force (which is anyway less as the speed is lower). The result is a reduced net force.
| {
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Scattering, 4 point correlator, #distinct Feynman diagrams
In order to compute the scattering probability that two particles of type 1 (associated to $\phi_1(x)$) which come from the far past with the momenta $p_1$ and $p_2$, to scatter and evolve into two particles of type 2 (associated to $\phi_2(x)$) with the momenta $p_3$ and $p_4$, I am going to apply the momentum space scattering rules, I'm just first of all trying to establish the number of distinct Feynman diagrams of $$\langle0|T( \phi_1(x_1)\phi_1(x_2)\phi_2(x_3)\phi_2(x_4))|0\rangle,$$ where the interacting Hamiltonian is $$H_\mathrm{int}= \frac g4 \phi_1^{2}(x)\phi_2^{2}(x)$$
QUESTION:
The solution lists these two Feynman diagrams only, see attachment 'dia 1'
But I am also getting this diagram, see attachment 'dia 2'
Where solid lines are associated to $\phi_1(x)$, dotted to $\phi_2(x)$, and $z$ and $w$ are the internal variables I am integrating over.
Is there a reason this should be excluded? I know that 'bubble/vacuum' diagrams are excluded - disconnected diagrams with no connection to external points, but I'm unsure here since it's connected...
On another note, my lecture notes describe the contribution from the $g^{0}$ term as 'trivial' since it describes non-interacting particles. so yes in this case it's $x_1, x_2$ and $x_3, x_4$ contracted since you can not contract different fields, however, what is meant by 'trivial' in this sense? Because I know that diagrams that have no connections to external points are not included since they are vacuum contributions and will just cancel out anyway, however here it is a diagram solely between external points, so whilst it doesn't describe any interaction, it still needs to be included right?
| Without knowing the text you are referencing, it's possible that they are only considering amputated diagrams, since the third diagram is a self-energy correction to the 4-point contact diagram. If you consider amputated diagrams, you only look at the diagrams close to the interaction vertex, and you effectively discard the self energy contribution, since it could happen arbitrarily far in the future of particle $x_2$ and not around the interaction point.
| {
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Why is the speed of light said to be constant when we know it slows down in a medium? If we ask someone what the speed of light is, they will say that it is constant and its value is $3\times 10^{8}\ \mathrm{m/s}$.
But if we recall refraction of light, we say that when the light travels from a rarer medium to a denser medium then it slows down.
But we still say that its speed is constant. Why?
| Speed of light is a constant. That is $3\times10^8$ m/s.
Now the question is why it is thought to be a fundamental constant.
Let's say there is a ball moving with 3 m/s speed and you are moving towards ball with speed 2 m/s, what is the relative velocity of ball for you , that is 5 m/s.
Now apply same logic for light. Light is coming towards you with speed c and you are moving towards light source with velocity v what is the speed of light relative to you c+v? No! That will remain c.
Several times the experiments were performed (look for Michaelson Morley experiment) to check this result but it was found that the light speed do not change with the relative speed of observer with respect to the light source.
Due to this reason the speed of light in vacuum is taken as constant.
| {
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Using Delta Dirac function as a mathematical tool in Green's functions So, I was studying green's functions and in general I understood that if I have an operator $\mathscr{O}$ that acts of a function $h_1(\vec{r})$ such that
$$\mathscr{O}h_1(\vec{r})=h_2(\vec{r})$$ Then all I need to do is to find the function, $g(\vec{r})$, on which the operator acts to yield the delta function.
Then I can write, $$h_1(\vec{r})=\int h_2(\vec{\tau})g(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}$$
Reason being
$$\mathscr{O}h_1(\vec{r})=\mathscr{O}\int h_2(\vec{\tau})g(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}=\int h_2(\vec{\tau})\delta(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}$$
So far, so good, but then in an effort to solve the Poisson's equation, writes $$V(\vec{r})=\frac{1}{4\pi}\int \frac{\rho(\vec{\tau})}{\epsilon}\frac{1}{|\vec{\tau}-\vec{r}|}\mathrm{d}^3\vec{\tau} $$
Because (and I'm back-calculating)
$$-\nabla ^2\left(\frac{1}{4\pi|\vec{r}|}\right)=\delta(\vec{r}).$$
I am unable to understand this move. Is there some mathematical basis for this or just to equate the preconceived notion of potential of point charges does this equation hold good?
| *
*Calculate the Laplacian of $1/r$ using spherical coordinates, you get that it is zero where $r \neq 0$.
*Use Green's Theorem to calculate the volume integral of $\nabla^2(1/r)$ in a sphere (of arbitrary radius) about 0, the value of this integral is $-4\pi$ (regardless of the radius of the sphere chosen).
Since the volume integral is nonzero and there is only one point (0) where the function is nonzero, we have a $\delta$-function.
*edited to be less cryptic :)
| {
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Why do we get tired after walking While we were studying pure rolling of bodies in the chapter rotation we were told that energy conservation holds if a body is purely rolling ( The point of contact between the rolling body and ground is stationary w.r.t. ground ). Considering a person to be system. Whenever the person walks his point of contact with the ground does not move. So apparently the only force(s) which acts on him are $(m_{man} + m_{clothes}) \cdot g,$ static friction and normal due to ground. Now energy can be conserved because normal and $mg$ while cancelling out each other are also perpendicular to the direction of walking. And since the force acting is static friction energy can be conserved. However if this is the case a person should not tire while walking becuase his energy is conserved. But practical experiences deny this. So what part of the work am I missing.
| You are, unfortunately, missing out on two crucial mechanisms of energy dissipation: kinetic friction in our joints (and other internal and external parts, like clothes) and the inefficiency of converting chemical energy into motion. As we walk, our muscles exert forces constantly in various directions; in order to do this, they have to do work against the kinetic friction that inevitably occurs at various points in the body. This work is dissipated as heat.
In addition, in a simplified sense, muscles receive chemical fuel in order to do work; however, the conversion from chemical bond energy to mechanical work is nowhere near 100% efficient. Once again, we lose this energy, as heat. So, as a person keeps walking, he or she gradually exhausts this finite chemical reserve, waste products build up in the muscles, and they feel tired.
| {
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Would it be possible to use light to transfer physical object (nanoparticle) from one place to another? I know that light has momentum and it has been used to pick up nanoparticles. But I am thinking, is it possible for us make a pouch of light to carry a very small object from one place to another. The transfer should cover a large distance.
| I do not understand what you mean by a "pouch" of light. However, yes it is possible to move very small objects using light.
Light can be used to manipulate micron-sized particles by a technique known as Optical Tweezers.
Light is refracted as it passes through a transparent or translucent object. Light carries momentum so there is a change of momentum which results in a force on the particle in the opposite direction. Particles are deflected towards the most intense part of the beam of light, and can be trapped at a focus. Moving the focus of the beam moves the particle.
Very small reflective particles can also be pushed by bouncing light off them.
| {
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In deriving the heat transfer equation, why do we use heat capacity at constant pressure? I have seen many derivations of the heat transfer equation. It always has a form something like the following:
$$\rho C_{P} \frac{\partial T}{\partial t}-\nabla\cdot(k\nabla T)=\dot{q}_{V}$$
No matter how you write it, there is always a $C_{P}$ term, for specific heat capacity at constant pressure even though you are not necessarily considering an example that is at constant pressure. None of the derivations explain why they choose $C_{P}$ specifically either (they just say that one should use heat capacity, but why not $C_{V}$, for instance). Where does the constant pressure part come into play?
| In the derivations you are referring to, the fluid is assumed to be incompressible. in the limit of an incompressible fluid, the heat capacity and internal energy are functions only of temperature. So it doesn't matter whether you call it Cp or Cv or just C.
| {
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Magnetization currents (amperian currents): how to show that they are always zero in total from the definition? For magnetic field in matter the two following amperian current densities are defined:
*
*Surface currents: u.m. $[\frac{A}{m}]$
$$J_{A,s}= \bf{M} \times \hat{n}$$
*Volume currents: u.m. $[\frac{A}{m^2}]$
$$J_{A,V}= \nabla \times \bf{M} $$
Where $\bf{M}$ is the magnetization and $\hat{n}$ is the normal outgoing from the object considered.
My question is : how to see, from the definition that the total amperian current is zero? I.e., why is
$$I_{A,S}+I_{A,V}=0 \,\,\, \mathrm{everywhere}$$
?
| If you're comfortable with delta functions etc., you can prove that the surface current formula is a special case of the volume current formula (the special case where M goes sharply to zero across a boundary). So we really only need the volume current case.
Kelvin-Stokes theorem says
$$\oint_\Gamma \mathbf{M}\, \cdot\, d{\mathbf{\Gamma}} = \iint_S (\nabla\times\mathbf{M})\, \cdot\, d\mathbf{S}$$
Let's prove the z component of $J$ is zero. (x and y are obviously the same argument.)
Say we have some object of finite extent surrounded by air (air has magnetization zero). For each number $z_0$, we draw a loop in the $z=z_0$ plane, entirely outside the object, and then apply Kelvin-Stokes. The LHS is zero, so the formula says that the integrated z-component of $J_{A,V}$ in this $z=z_0$ slice of the object is zero. If the total in each $z=z_0$ slice is zero, then (by integration) the total in the whole object is also zero.
| {
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Getting thermal comfort by cooling just one point of body? Can I get thermal comfort by cooling or heating just one point of body like wrist or foot?
As far as I know, Wrist and foot are the most thermally sensitive points of body. I saw that wrapping a wet cloth in the wrist can give someone thermal comfort during summer. Is it really effective? If it is really effective, where should I wrap wet cloth between foot and wrist to get more cooling effect?
| yes, there are products such as wrist bracelets that use the fact that your blood vessels are close to the surface on the wrist.
Search for "wrist cooling device"
They seem to be active, using e.g. Peltier effect.
There was an article about a device that uses a closed water circuit and was at least tested for tennis players. The method is apparently effective, I am not sure how safe it is though.
| {
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Phase Transitions and Bubble Nucleation The potential for a first order phase transition is shown below
The phase transition occurs from the spontaneous formation of bubbles. Inside the bubbles the field value is at the "true vacuum" and outside the bubble the field value is at the "false vacuum". In many texts, a second order phase transition is described to occur in a smooth fashion.
My question is can bubble nucleation occur in a second order phase transition? Or is a first order phase transition necessary?
| I think nucleation is only relevant to first order phase transitions. This is because only a first-order phase transition has an entropy curve $S(U)$ which leads to the possibility of metastable phases (e.g. supercooled vapour or superheated liquid).
| {
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Why does torque produce a force on the axis of rotation? If a door is rotated about its fixed axis in (outer) space, a force parallel to the door on the hinges will arise due to centripetal force on the centre of mass and conservation of momentum (Newton's third law).
But any torque on the door will create a force on the hinges which is equal to $t/r$ or torque divided by radius. I'm looking both an intuitive and mathematically based explanation for this fact. I can sort of 'see' why, but my understanding is vague and uncertain.
| It is only due to the hinges that that when the door is pushed ,that it doesn't translate along the line of action of force applied and the hinges changes the velocity of the door and thus a force is necessary for that.This force is variable , changing its direction every instant
| {
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Is there a difference between observation and entanglement? I have very basic knowledge about quantum mechanic. I started wondering if there is any difference theoretically between observation and entanglement between subject and object.
I.e. if we simplify an observer state to q-bit and initial state of second is $\alpha\left|0\right> + \beta\left|0\right>$ then observation can be roughly represented as:
$$\left|0\right> \otimes \left(\alpha\left|0\right> + \beta\left|1\right>\right) \to \alpha\left|00\right> + \beta\left|11\right>$$
(Unless I'm missing something this is unitary operation)
That should behave from 'POV' of left q-bit as-if the measurement was $\left|0\right> $ or $\left|1\right>$ respectively (sort of like in multiple worlds instead of splitting the words the observer becomes in quantum state). [Of course most scientist are represented by slightly more complex state then one q-bit ;) ]
Am I missing something or would it be a valid interpretation of quantum physics?
| The observer you model here is Wigner's friend as seen by Wigner: he may have observed something, but for Wigner he is now in a superposed state.
You are right that entanglement is the proper relationship between a quantum system and an observer, when both are described as a larger quantum system, but upon measurement (not a well defined theoretical notion unfortunately, but a very sound experimental one) entanglement is broken: measurement yields only one of the possible outcomes, and the quantum state of the observed system has to be updated accordingly.
| {
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What did the big bang "look like"? I've been reading here for a while now and something I always see is people saying "the big bang happened everywhere" or "the center of the universe is where you are", explaning that the big bang didn't happen from a single point, but everywhere at once.
The problem is that I am unable to get an "image" of what that might look like in my head. What does it mean when the universe expands everywhere at once? I know that this might make sense from a mathematical point of view, but what would it actually look like?
| It probably is not possible for humans to get a physical picture of that time, because to do so would involve comparing it to something in the world around you today. But the Big Bang was almost certainly so completely different in it's "appearance", that we do not have words, past experiences or even intuition to help us describe it.
If it helps, we cannot describe an electron in any definitive physical way either, yet we are surrounded by, and composed of, elementary particles.
In both cases, the best we can do is describe the Big Bang and electrons mathematically, which has worked out very well for us, and accept that trying to get a physical picture of either of the above is futile.
If someone claims that he/she has achieved an accurate mental picture of the Big Bang, or an electron, that they can relate to the "real" world, I doubt that very many physicists would believe them.
| {
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Relationship between Coordinate Time and Proper Time While I was reading Ta-Pei Cheng's book on relativity, I was unable to derive the correct relationship between coordinate time $dt$ (the book defined it as the time measured by a clock located at $r=\infty$ from the source of gravity) and proper time $d\tau$ from the definition of metric.
The book states that for a weak and static gravitational field, $g_{00}(r)=-\left(1+\frac{2\Phi(r)}{c^2}\right)$ (with the metric signature $(-1,1,1,1)$ and $\Phi(r)$ is the gravitational potential) and the proper time $d\tau=\sqrt{-g_{00}}\,dt$.
From the gravitational redshift result I know that the above result is correct (in a more unambiguous form $d\tau=\sqrt{-g_{00}(r_\tau)}\,dt$).
However, if I simply use the formula for spacetime interval $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ (assuming two clocks that measure proper time and coordinate time are at rest relative to each other), I have
$$ ds^2=g_{00}(r_\tau)c^2d\tau^2=g_{00}(r_t)c^2dt^2=-c^2dt^2\\ \implies \sqrt{-g_{00}(r_\tau)}\,d\tau=dt$$
This suggests that time flows faster with a lower gravitational potential which is incorrect.
I'm not sure why the above method lead to a wrong conclusion, did I misunderstood the the definition of proper time, coordinate time or spacetime interval?
Update:
*
*One mistake I've made is letting $ds^2=g_{00}(r_\tau)c^2d\tau^2$,
which should be $ds^2=-c^2d\tau^2$ by definition. However, I'm
confused about two definitions of $ds^2$ now.
$ds^2=-c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$, this suggests that
$g_{00}$ is always $-1$ for the frame that measures proper time, but
in my problem $g_{00}$ is a function of $r$ which is only equals to
$-1$ if $r=\infty$, how could two both be true at the same time?
*Assuming $ds^2=-c^2d\tau^2$ is true, as all the answers pointed out that $d\tau=\sqrt{-g_{00}}\,dt$. But by the definition of $g_{00}$ and $ds^2$ the $g_{00}$ used here must be $-(1+2\Phi(r_t)/c^2)=-1$, but I want $g_{00}$ here to be $-(1+2\Phi(r_\tau)/c^2)$ so that
$$d\tau=\sqrt{-g_{00}}\,dt=\sqrt{1+2\Phi(r_\tau)/c^2}\,dt\approx (1+\Phi(r_\tau)/c^2)\,dt\\
\implies \frac{d\tau-dt}{dt}=\frac{\Phi(r_\tau)}{c^2}=\frac{\Phi(r_\tau)-\Phi(r_t)}{c^2}$$
Please correct me if I've made any mistakes!
| The answer provided by Rumplestillskin is correct
$$d\tau = \sqrt{1-\frac{r_g}{r}} dt$$
but it is only valid for the clock statically residing in the gravity at coordinate distance $r$ from the center. For moving clocks in the Schwarzschild coordinates, for example in radial free fall, the relation between the proper time and coordinate time is different
$$d\tau=dt+\sqrt{\frac{r_g}{r}}\left(1-\frac{r_g}{r}\right)$$
| {
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How is the isospin quantum number calculated? $$\Lambda^+_c$$ and $$\Sigma_c^+$$ are both made up of the same quarks, and have the same I_3, but have different isospin quantum numbers. How is the isospin quantum number determined? Also is the isospin quantum number indicated by the greek letter used?
| From Wikipedia (excluding the anti-quarks for now to keep it simple):
In the modern formulation, isospin (I) is defined as a vector quantity
in which up and down quarks have a value of $I = 1/2$, with the
3rd-component ($I_3$) being 1/2 for up quarks, and -1/2 for down
quarks, while all other quarks have $I=0$. In general, for hadrons,
therefore \begin{eqnarray} I_3=\frac {1}{2}(n_u-n_d) \end{eqnarray}
where $n_u$ and $n_d$ are the numbers of up and down quarks
respectively.
Or otherwise it can be stated in form of $|{I, I_3}>$ that, $u =|{1/2, 1/2}>$, $d=|{1/2, -1/2}>$ and other quarks as $|{0, 0}>$.
Thus for the hyperon with quark content $uds$, it can have its composite isospin $I = |I_u - I_d|, I_u + I_d$. i.e. 0, 1. Thus, it can be expressed as two states: $|{0,0}>$ and $|{1,0}>$.
This is similar to your example, $\Lambda^+_c$ and $\Sigma_c^+$, which have quark content $udc$ as the $strange$ has been replaced with a $charm$. Then their $| I, I_3>$ should be $|1/2 - 1/2, 1/2-1/2>$ and $|1/2+1/2, 1/2-1/2>$, $i.e.$ $|0,0>$ and $|1,0>$
| {
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Why don't humans burn up while parachuting, whereas rockets do on reentry? I guess it has something to do with their being both a high horizontal and a vertical velocity components during re-entry. But again, wouldn that mean there is a better reentry maneuver that the one in use?
| Re-entry velocity from LEO is $~7,800 \frac m s$, from lunar space it is as high as $~11,000 \frac m s$ [1].
Different books give the terminal velocity of a skydiver as about $56 \frac m s$ or $75 \frac m s$ [2, 3]. The exact value isn't material, but the fact that it is two powers of ten smaller then re-entry velocity, is.
The difference between skydiving and re-entry is that in order to orbit, you need to go very fast sideways. You essentially fall so fast sideways that you miss the ground when falling towards it (see related xkcd). A skydiver, whether they dive from a plane or balloon, has only marginal horizontal velocity and almost zero vertical velocity to start with. The skydiver then accelerates to terminal velocity, which is quite slow compared to re-entry velocity (see above).
In comparison, the Apollo capsule had a terminal velocity of $150 \frac m s$ at $7,300 \ m$ altitude. It is from that moment on that Apollo behaves like a skydiver. Drogue chutes are pulled that slow down the craft to $80 \frac m s$, and then finally the main chutes that slow down the craft to $8.5\frac m s$ [4].
But that is only the very last phase of the flight. You need to somehow slow down from $7,800 \frac m s$ to $150 \frac m s$ first and descent from space deep into the atmosphere. Creating chutes that can both withstand that and are big enough to slow the craft down enough that high up in the atmosphere is simply not feasible from an engineering point of view, and even if it were, it would probably prohibitive from a weight/delta-v point of view.
The Falcon 9 first stage does not have problems with re-entry heating, although it also reaches space. But that stage does not achieve orbital velocity. It only goes about $2,000\frac m s$ at separation, which is slow enough that heating is not an issue when it comes down (see this question on Space StackExchange).
1: Atmospheric Entry. Wikipedia, the free encyclopedia.
2: Tipler, Paul A. College Physics. New York: Worth, 1987: 105.
3: Bueche, Fredrick. Principles of Physics. New York: McGraw Hill, 1977: 64.
4: W. David Woods. How Apollo Flew to the Moon. Springer, 2008: 371.
| {
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Where does all the heat go during winter? I do not understand where actually the heat in our surroundings go during the winter season. Is it radiated out into space? I know it cannot coz global warming would not be a issue then. It might get absorbed but where? I tried figuring it myself but couldn't please help.
|
Where does all the heat go during winter?
There is less energy coming from the sun in the form of electromagnetic radiation impinging on the land during winter.
Depending on the latitude, in regions where there is winter , the difference is large.
The closer to the equator the smaller the effect of "winter".
So it is not where the energy goes, but why it does not fall , and this is explained to first order by the inclination and the distance to the sun during the orbit of the earth.
In general , a body in space radiates energy away the rate depending on various conditions, like green house gases, cloud cover, convection , albedo ...the numbers change . It is the continuous radiation from the sun that keeps replenishing the energy so that the earth does not freeze. During winter at high lattitutes , less energy comes and cold settles.
| {
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How force exerted by spring is always opposite to the direction of displacement in Hooke's law Suppose a spring lying on a horizontal table, displaced from its equilibrium length by an external agent. The external agent is removed, the spring will head back to its equilibrium length. Here, the direction of spring force and displacement will be same.
But according to Hooke's law,
$$\mathbf{F}=-k\Delta\mathbf{x}$$
The minus sign tells us that the force exerted by spring is always opposite to the direction of displacement.
How is this? Please explain the reason for the minus sign.
Thanks.
| Imagine a spring which has a force $\vec F_{\rm sy}$ applied on it by you and this produces an extension $\vec x$.
You then have $F_{\rm sy}=k\vec x$
However it is usual to be interested in the force the spring exerts on you $F_{\rm ys}$.
Using Newton's third law $\vec F_{\rm sy}=-\vec F_{\rm ys }$ so $\vec F_{\rm ys}= - k \vec x$.
Introducing a unit vector in the positive x-direction $\hat i$ and let the magnitude of the forces $F_{\rm ys}$ and $F_{\rm sy}$ be $F$.
$\vec F_{\rm ys}= - k \vec x$ becomes $F \hat i = - kx \hat i \rightarrow F=-kx$ in terms of components in the positive x direction.
| {
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Does folding a paper towel help dry your hands faster by creating interstitial forces? The question is based on a TEDx video, where the speaker claims that folding a paper towel before using it creates interstitial forces which help dry your hands faster.
The question:
Does this effect actually occur? If it does, then why, and how significant is it?
In addition, would there be a significant difference between folding the towel once (i.e. in half), versus folding it twice?
| The speaker actually claims it works by "interstitial suspension". I think he is referring to capillary action, which you are aware is involved here. He does not claim this method is faster, only that it is more environment-friendly. Shaking before wiping is a crucial factor.
Folding has 2 advantages : it reduces excess towel at the edges which is not used, and it makes the towel thicker so that the paper between the hands does not become saturated, when it will have no further drying capacity.
Yes, folding twice might be even more effective, depending on the size, thickness and quality of the towel. However, excessive folding leaves paper in the folds usused (ie dry) while the outer surfaces are overused and become saturated. So a single fold might be optimal.
Why not do some personal research and report your findings? (Be quick, or Floris will beat you to it!)
| {
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Pauli matrix for triplet state? Question is, what would be the result of applying the operator $\hat A = [3I + \vec\sigma_1 . \vec\sigma_2]$ on the |singlet$\rangle$ and |triplet$\rangle$ states ($\vec\sigma_1$ acts on the 1st particle and $\vec\sigma_2$ acts on the second particle ONLY), ie, $$\hat A|singlet\rangle=?|singlet\rangle$$
and $$\hat A|triplet\rangle=?|triplet\rangle$$
I am stuck at the triplet part of the question.
For a system of 2 spin half particles, where $\vec\sigma_1$ acts on the 1st particle and $\vec\sigma_2$ acts on the second particle ONLY, (like adding angular momentum of two electrons) $$\vec\sigma=\vec\sigma_1+\vec\sigma_2$$
squaring both sides, $$\vec\sigma^2=(\vec\sigma_1+\vec\sigma_2)^2$$
from which we have$$\vec\sigma_1 . \vec\sigma_2 = (\sigma^2 - \sigma_1^{2} - \sigma_2^{2})/2$$
Now, $\sigma_1^{2}=\sigma_{1x}^{2}+\sigma_{1y}^{2}+\sigma_{1z}^{2}=3I$ and similarly, $\sigma_2^{2}=3I$.
and that for the singlet state, the value of $\sigma^2=0$, (which i gathered from the total spin being $0$ for the singlet state) which gives $$\vec\sigma_1 . \vec\sigma_2 = (0 - 3I - 3I)/2=-3I$$
I dont know what the value of $\sigma^2$ is for the triplet state (i do know that the total spin $S$ is $\sqrt2\hbar$)?
I am not able to relate the total spin with the $\vec\sigma$ properly
| As @rob asked you to, you are meant to simply write down
$$
\hat{B}\equiv\vec{\sigma}_1\cdot\vec{\sigma}_2 = {\sigma}_1^x {\sigma}_2^x +{\sigma}_1^y {\sigma}_2^y+{\sigma}_1^z {\sigma}_2^z \\= ({\sigma}_1^x+i {\sigma}_1^y)({\sigma}_2^x -i{\sigma}_2^y )/2 +({\sigma}_1^x-i{\sigma}_1^y ) ({\sigma}_2^y +i{\sigma}_2^y)/2+{\sigma}_1^z {\sigma}_2^z\\
\equiv {\sigma}_1^+ {\sigma}_2^- +{\sigma}_1^- {\sigma}_2^+ +{\sigma}_1^z {\sigma}_2^z ~,
$$
where $\sigma^+ \uparrow=0$, and $\sigma^+ \downarrow=\uparrow \sqrt{2}$, etc... for both 1 and 2. Recall
$$
\sigma^+ = \sqrt{2}
\begin{pmatrix}
0&1\\
0&0
\end{pmatrix} .
$$
Acting on the singlet, $\uparrow \downarrow- \downarrow \uparrow$ ,
this $\hat B$ has the obvious eigenvalue -3.
The triplet is $\uparrow \uparrow$; $(\uparrow\downarrow+\downarrow\uparrow)/\sqrt{2}$; $\downarrow \downarrow$, and so it obviously has eigenvalue 1 under the action of $\hat{B}$.
Your $\hat A= 3 1\!\!1 +\hat{B}$ has eigenvalues 0 and 4 respectively, given my normalizations. This is to say, of course, that, for the triplet,
$\sigma^2/4=2=(1+1)1$, as expected.
| {
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Confusing working of lens Why do lens don't splits light into its seven constituent colors, like Prism?
*
*Why is lens left is correct, not right one?
*How does lens came to know that rays are coming from infinity or are at Focus and converge/diverge them at different point accordingly?
| Real lenses suffer from chromatic aberration due to the refractive index of glass varying with the wavelength of light.
All a lens does is to refract incoming rays of light.
It so happens that if rays of light close to and parallel to the principal axis of a lens cross at a point after refraction.
| {
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Why does water flow slower on butter? Last night I was in my bathroom conducting an experiment about the speed of water on butter.
The experiment went something like this:
*
*Cover half a sheet pan with butter and leave the other half untouched.
*Incline the sheet pan at an angle of roughly 50°.
*Start the stopwatch, and with an oral syringe, put a drop of water on the highest part of the sheet pan on the buttery side.
*Stop the stopwatch when the drop of water reaches the bottom of the sheet pan.
*Repeat steps 3 and 4 on the clean side of the sheet pan.
For some reason the water ran alot slower on the buttery side. The water droplets for both sides started with close to no speed.
Observations
BUTTERY SIDE:
1st try: more than 15 minutes. The water started out very slow and lost speed quickly. The water was about 1/2 the way to the bottom after 15 minutes.
2nd try: more than 15 minutes. The water appeared to stop completely at one point. It was about 1/2 the way to the bottom after 15 minutes
3rd try: more than 15 minutes. The water appeared to stop completely at one point, around 1/4 of the way there. The water was around 3/4 the way to the bottom after 15 minutes.
NON-BUTTERY SIDE:
1st try: 28.32 seconds. The water flowed slowly at the beginning but quickly accelerated and flowed quickly.
2nd try: 35.76 seconds. The water did not accelerate as much as in the first try. It appeared to stop completely at one point and then accelerated again.
3rd try: 32.26 seconds. There is water all over the bathroom floor.
Assuming these times are accurate, why did the water on the buttery side flow SO slowly? I had initially expected it to flow faster because of the oil in butter and also because butter is hydrophobic.
Thanks!
| adhesive forces of water particle are different with base of sheet and butter molecule
From your experiment results we can conclude water-butter molecule adhesive force is higher
pl. support your experiment with video if possible (- providing YouTube link ?)
if you pore honey instate of water it will end up in different result
Also the surrounding temperature have influence on the experiment !
| {
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Can I use Root-Mean-Square Speed to measure the average speed of particles in solids and liquids? Can I use the equation $v_{rms}= \sqrt{\frac{3RT}{M}}$ of Root Mean Square Speed to measure speed of particles in liquids and solids as well? I am doing a Javascript animation about molecular motion of solids, liquids and gases and I can't find anywhere how to measure speeds of particles of solids and liquids and it's getting really frustating.
I will really appreciate any help and if I can't use the equation for Root Mean Square Speed of gases, I will be thankful for suggestions how to measure speed of liquids and solids in other ways with other equations and formulas.
| Due to equipartition theorem, average kinetic energy of particles in a system of classical particles is
$$\langle K\rangle=\frac32k_BT=\frac{m\langle v\rangle^2}2.$$
This is true even if there are some interactions: for derivation see this answer. Thus we have
$$\langle v\rangle^2=\frac1m 3k_BT=\frac{N_A}M 3k_BT=\frac{3RT}M,$$
and after taking square root we get your formula.
Now, as I mentioned, this only applies to classical systems of particles. In actual liquids and solids there're electrons in each atom, and these electrons are in a highly quantum regime. But if you restrict your attention to motion of atoms themselves, rather than taking into account motion of electrons, then you can safely use the classical formula. Atomic nuclei are quite heavy, so at typical energies their motion in liquids and even in solids can be considered classical (i.e. their energy spectrum is very dense).
So to summarize, the answer is yes: for motion of atoms in liquids and solids at not too low temperatures you can safely use your formula.
| {
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In determining sound beam steering angle, where does a constant of 0.514 come from? I've read in some sources (like this, among some other) that for determining sound beam steering angle in a phased-shift array system, it is using a formula of:
$$\theta = \arcsin (0.514 \lambda\mkern-6mu^{_-}\mkern-12mu^{_-} / e)$$
Where does this constant of 0.514 come from? Is it related in some way to the analogy of diffraction grating?
| It comes from point spread function theory in optics and 0.514 is the factor used for specifically considering the part of your beam that encircles half the power (-6 dB of peak). It is derived from the Airy Function
| {
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Scaling argument for friction force on a chain in shear flow In her paper entitled "Deformations of One Tethered Chain in Strong Flows", the author makes the argument that given a shear flow velocity $v_x(y)=sy$ in a solution of viscosity $\eta$, and a tethered particle in the flow of size $y$, then the friction force felt by the particle is $f=\eta sy^2$ (ignoring numerical pre-factors). If we have a chain of such particles tethered together in a line of varying sizes $y_i$, then the total force on the chain is $
\sum_i \eta sy_i^2$. So far so good.
Now the author transitions to a continuum, so that our blob sizes vary continuously as $y(x)$, She then says that $\sum_i \eta sy_i^2\approx \int \eta s y(x) dx$. Can someone shed some light on how that transition is made? I'm sure it's simple, but I'm not seeing it.
EDIT: is it just that the line density of particles is $y^{-1}$, so we divide by the density when moving to the integral?
| In the continuum limit the body (i.e. aggregate of blobs) seems to become a cylinder, but with increasing radius along X direction. Area of elemental circular strip at distance $x$ will be (circumference of cylinder at $x)\times~dx$. But circumference is proportional to diameter $y(x)$. Therefore area of the elementary strip at $x$ is $\sim y(x)~dx$. Shear force on this cylindrical strip is product of shear stress ($\sim \eta s$) and its area.
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Friction of a Car in Motion It is commonly known that friction opposes motion. For example, if a block is sliding down a wooden surface with no forces other than friction acting on it, friction acts in the direction opposite to the block's velocity.
Let's say a car is driving at constant speed along a perfectly circular loop road. Centripetal acceleration works toward the center of the circle, but the car is moving forward along the loop, perpendicularly to the car's acceleration. As we have already established, friction should be in the opposite direction of the car's motion. But no- apparently it acts as the centripetal force that is perpendicular to the velocity.
What is going on with this discrepancy?
Edit: new comment
Edit: Furthermore, the book I am using says that, when the loop is banked, friction ceases to act centripetally and the centripetal force is provided entirely by the horizontal component of the normal force (neither friction nor, to my surprise, parallel force.). Why is that? Thanks!
|
It is commonly known that friction opposes motion.
If that were so you would not be able to move from rest.
Friction opposes or tries to oppose the relative movement between two surfaces.
If you have two blocks on top of one another and at rest and apply a force on the bottom block what will happen?
The bottom block will start to increase its velocity and if the static frictional force is large enough the top block will also increase its velocity by the same amount.
Here static friction is producing the motion of the top block whilst trying to have no relative movement between the surfaces which are in contact.
If the static frictional force is not large enough and kinetic friction kicks in, the kinetic frictional force will still try and increase the velocity of the top block to reduce the relative movement between the two surfaces in contact.
With the car going around the corner the static friction force tries to have a situation that at the point of contact between the tyre and the road there is no slipping.
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"url": "https://physics.stackexchange.com/questions/306151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Can an accelerating frame of reference be inertial? In physics problems, the earth is usually considered to be an inertial frame. The earth has a gravitational field and the second postulate of the general theory of relativity says:
In the vicinity of any point, a gravitational field is equivalent to
an accelerated frame of reference in gravity-free space (the principle
of equivalence).
Does this mean that accelerating frames of reference can be inertial?
| Accelerating frames are never truly inertial; however, in many situations the acceleration is sufficiently small that we can assume the accelerating frame to be inertial. It largely depends on the scale relevant to the problem.
For example, for purposes of projectile motion, we can consider the Earth to be an inertial reference frame and still model the projectile's path accurately. However, in orbital mechanics, we definitely cannot consider the Earth to be an inertial frame, since it constantly accelerates in its orbit around the Sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/306303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
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