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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 $\hat{H}_0$?
| Answer to (2); saying that the perturbation is "small"...all that means is that $\lambda << 1$.
As a silly example: if $\hat{H}_{0} = \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$ and $\hat{H}^{\prime} = \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$, and if you've got a tiny $\lambda = 0.000001$, then the system you're looking at:
$$
\hat{H} = \left[ \begin{matrix} 1 & 0.000001 \\ 0.000001 & 1 \end{matrix} \right]
$$
Can you see how the overall system is "almost" the same as $\hat{H}_{0}$? This is the sense in which the perturbation is small. You can then safely do expansions in terms of the parameter $\lambda$.
(As somebody else mentioned, you can often set $\lambda = 1$ after you've down all the calculations.)
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Why is lens flare doubled if odd number of blades but equal to number of blades if even? An article in Picture Correct describes how the number of blades in a particular lens correlates to the number of starburst points associated with lens flare. The number of starbursts is double the number of aperture blades if there are an odd number of blades and equal to the number of blades if the lens contains an even number of blades. Here's a jpeg of the article should the above link ever break:
What property of light could cause this phenomenon? This seems more appropriate here on the Physics page than it does on the Photography SE community, since the question is about the physics behind this and not whether or not it's true.)
| A straight edge causes diffraction in the direction perpendicular to that edge. When two straight edges are parallel to each other their diffraction patterns will overlap (point along the same line) - making it look like there is just one.
That happens when there is an even number of blades.
Note - I believe the diagram for the six pointed blade is wrong - it shows the flare aligned with the corners of the hexagon. I am pretty sure they align with the centers of the blades instead. (Note - for an odd number of blades the distinction cannot be made).
| {
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How do EM Wave Boundary Conditions Comply with Conservation of Energy? One of the boundary conditions of an EM wave crossing a boundary (dielectric materials, wave is TE polarized), where part of the wave is reflected and part is refracted, is
$$E+E'=E''$$
where E is the amplitude of the oscillating electric field of the incident wave, E' is that of the reflected wave, and E'' is that of the refracted or transmitted wave.
My concern is, how does this comply with conservation of energy? Especially in the case of going from high index of refraction to low index of refraction, so E' is in the same direction as E (no phase shift), it seems that an incident wave with some E is resulting in two new waves with E' and E'', where E'>0, and E''>E.
Doesn't this violate conservation of energy?
(Note: I have seen the proof of this boundary condition from Faraday's law with the shrinking loop and it makes sense to me, but I haven't been able to reconcile it with this problem.)
(I also considered that the transmitted wave might be moving slower than the incident wave, which would mean it has a smaller energy flux despite the higher E, but this is not the case when going from high to low index of refraction).
| The boundary conditions for electromagnetic field of constant frequency $\omega>0$ at an interface of two media are that the tangential components (the components in the plane of the interface) of electric and magnetic fields $\overrightarrow{E}$ and $\overrightarrow{H}$ are continuous. If these conditions are satisfied, the normal component of the Poynting vector (which vector equals $\overrightarrow{E}\times\overrightarrow{H}$ up to a constant factor) is continuous at the interface, so the energy flux is the same from both sides of the interface.
| {
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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?
| I had a similar problem and the best thing to use would be an Arduino. This is because firstly, Arduino is small - to fit in the project. Secondly, you can use resistors and program them to release certain volts of current. You can set the parameters. This worked for me!
| {
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Why is the pressure gradient zero at a wall? It's accepted to impose a zero pressure gradient normal to a wall when solving the Navier-Stokes equation. Is there any mathematical reasoning for that? Which pressure (static pressure, total pressure...) is actually meant by that?
| This usually only applies to a wall bounded flow and is normally restricted to incompressible fluids. This result usually manifests in boundary layer theory and can be obtained through order of magnitude analysis of the Navier-Stokes equations. The steady, incompressible, and constant property momentum equation in the $y$ direction takes the form,
$$ u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} = -\frac{1}{\rho} \frac{\partial p}{\partial y} + \nu \left( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}\right)$$
The order of magnitude of each term inside the boundary layer goes as follows,
$$ O\left[u \frac{\partial v}{\partial x}\right] = O\left[\frac{\delta}{L^2} U_e^2\right] $$
$$ O\left[v \frac{\partial v}{\partial y}\right] = O\left[\frac{\delta}{L^2} U_e^2\right] $$
$$ O\left[\frac{1}{\rho} \frac{\partial p}{\partial y}\right] = O\left[\frac{\delta}{L^2} U_e^2\right] \text{(at most)}$$
$$ O\left[\nu \frac{\partial^2 v}{\partial x^2}\right] = O\left[\frac{\delta^2}{L}\frac{\delta}{L^2} U_e^2\right] $$
$$ O\left[\nu \frac{\partial^2 v}{\partial y^2}\right] = O\left[\frac{\delta}{L^2} U_e^2\right] $$
Where $\delta$ is the boundary layer height, $L$ is the characteristic length of the body, and $U_e$ is the external flow velocity at the edge of the boundary layer. An additional constraint is that $\delta/L \ll 1$. Notice that each term has an order of magnitude comparable to $(\delta/L^2) U_e^2$, except for the normal pressure gradient term and the viscous term from the $x$ direction. We first recognize that $\delta^2/L$ is a very small quantity and essentially removes the $\nu \partial^2 v/ \partial x^2$ term from the equation. Similarly, only at the edge of the boundary layer where the viscous forces become negligible (i.e. high Reynolds number) does the pressure gradient term order of magnitude approach $(\delta/L^2) U_e^2$. This was first observed by Prandtl, to which he deduced across the boundary layer we can write,
$$ \frac{1}{\rho} \frac{\partial p}{\partial y} \approx 0$$
or more conventionally,
$$ \frac{\partial p}{\partial y} \approx 0 $$
As for your second question, this only applies to the static pressure. Also, all of this assumes the flow is attached to the wall.
| {
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High(very high) frequency sound waves = heat? I had this question in mind and did not quite found the answer on google. If heat is a vibration of particles in matter, and sound is pressure wave moving using particles (causing them to fluctuate), could a very high frequency sound wave raise the temperature of matter which it is travelling in?
| Sound waves can definitely be used for heating of matter. High intensity ultrasound waves are (among other applications) used for internal heating of parts of the human body for medical therapy of a number of conditions. See Therapeutic Ultrasound. The extreme heating of gas bubbles in water by high-intensity ultarsound has also led to the discovery of the baffling phenomenon of sonoluminescence.
| {
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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?
| If you are applying two orthogonal forces their resultant (Which is always greater than both the constituent forces) should be greater than static friction or:
Lets say your forces are $ F_x $ and $ F_y$
The magnitude of these forces' resultant is $ \sqrt{(F_x)^2 + (F_y)^2} $
For the block to move the resultant must be greater than friction
$ \sqrt{(F_x)^2 + (F_y)^2} > \mu_smg$
However in your example since the block is already moving any amount of force you produce on the y axis will always increase the magnitude of the resultant force. and hence all the while you will use $\mu_k$ mathematically:
$ F_x > \mu_smg $
$ (F_x)^2 > (\mu_smg)^2 $
$ F_x^2 + F_y^2 > (\mu_smg)^2$
and therefore
$ \sqrt{(F_x)^2 + (F_y)^2} > \mu_smg$
And since $\mu_s > \mu_k $
$ \sqrt{(F_x)^2 + (F_y)^2} > \mu_kmg$
And notice $F_x$ itself is sufficient to get the block move and hence you use $\mu_k$ So while we work with the resultant which is obviously greater than $F_x$ this time too we use $ \mu_k$ because in this case too friction proves insufficient tto stop motion of block
| {
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Which form of Einstein's stress-energy tensor $T$ is physical, the covariant or the contravariant? Using Einstein's Field Equations for a specific metric I am interested in, I have calculated the first diagonal term of the stress energy tensor $T$, which is the energy density. In my calculations, the covariant density ($T$ with lower indices 00) does not show a singularity, but the contravariant density ($T$ with upper indices 00) does show a singularity. I need to know whether this singularity is physical. Which form of the stress energy tensor represents actual physical density, the covariant or the contravariant?
| Generally speaking the contravariant form $T^{\alpha\beta}$ is the one most naturally related to what we think of as physical observables. My favourite way of thinking about this is to start with the stress-energy tensor of a point particle, and we can understand the more complicated forms as made up from point particles.
But we should note Bob Bee's comments. When you write down $\mathbf T$ you have to choose a coordinate system to do so, and the individual components of the tensor will depend on the coordinates you choose. If your choice of coordinates results in coordinate singularity then you may well find your stress-energy tensor behaves strangely at that point.
It's hard to comment without knowing the details of the geometry you are studying, but if $T_{\alpha\beta}$ isn't singular this suggests very strongly that you have a coordinate singularity rather than a real one. Try calculating the trace of $\mathbf T$ to see if it diverges.
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Why is it not possible to measure the individual neutrino mass from $\beta-$decay? Why do we have to rely only on neutrino oscillations, to measure the mass squared differences of neutrinos? Why is it not possible to measure the neutrino masses directly, say, from $\beta-$decay?
*
*When Pauli hypothesized the existence of neutrinos to save the conservation of energy and angular momentum in $\beta-$decay, why did he assumed the neutrinos to be massless?
*How did experimentalists at that time concluded that neutrinos were massless?
*What about the measurement of energy in the 3-body decay $n\rightarrow p+e^-+\bar{\nu}_e$, in the sophisticated experiments today, and determination of the individual neutrino masses from that?
One has $$E_n=E_p+E_e+E_\nu\Rightarrow m_nc^2+T_n=m_pc^2+T_p+m_ec^2+T_e+m_\nu c^2+T_\nu.$$ $m_p,m_n,m_e$ are known. Therefore, by measuring the kinetic energies $T_n,T_p,T_e$ and $T_\nu$ one can determine the mass of $\nu_e$.
Why is this not possible experimentally?
Is this because $T_\nu$ is not measurable? Or is this because $\nu_e$ being a flavour state, does not have a definite mass?
| There cannot be enough accuracy in the momentum and energy measurements of the end products of beta decay, only limits can be determined because of measurement errors.
See this link for a recent review.
The paper reviews recent experiments on tritium β-spectroscopy searching for the absolute value of the electron neutrino mass m(ν_e). By use of dedicated electrostatic filters with high acceptance and resolution, the uncertainty on the observable m^2(ν_e) has been pushed down to about 3 eV^2. The new upper limit of the mass is m(ν_e) < 2 eV at 95% C.L. In view of erroneous and unphysical mass results obtained by some earlier experiments in β-decay, particular attention is paid to systematic effects. The mass limit is discussed in the context of current neutrino research in particle- and astrophysics. A preview is given of the next generation of β-spectroscopy experiments currently under development and construction; they aim at lowering the m^2(ν_e)-uncertainty by another factor of
100, reaching a sensitivity limit m(ν_e) < 0.2 eV.
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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.
| Simply, Liouville's theorem implicates that the two particles can not take the same state at t=T because their state trajectories behave as currents in incomprisible fluid.
Physically, when the first particle come to B, it will push the second particle and throw it to the left side.
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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.
| First things first, there is no proof that the universe is either. An outstanding question in philosophy is the ontological question of whether the universe is defined by mathematics, or if we created mathematics to understand the universe. Your question only makes sense in the former.
With sufficient feedback, you can create remarkable approximations of non-linear systems with linear systems. Likewise, linearity can be seen in many non-linear systems which operate on a manifold if one looks at small scale.
As for superposition, it only is useful for linear systems, or systems which have reasonable linear approximations. An excellent example appears in electrical engineering with "small signal" models. These are models of how a non-linear system (lots of transistors amplifying currents) can look linear within a small region around a chosen point. Many times that is more than sufficient for what we need.
| {
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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?
| Say you stand on a scale on the surface of Earth, and that it shows your weight $\vec{W}=m\vec{g}$. It is precisely balanced by the normal force $\vec{N}$. The local gravitational constant, little $g\approx 9.8 ~\mathrm{m/s^2}$, is not just due to gravity, despite the name. It is actually a vector sum $\vec{g}=\vec{g}_{gr}+\vec{g}_{cf}$ of gravitational acceleration $\vec{g}_{gr}$ and centrifugal acceleration $\vec{g}_{cf}$. Therefore, you have already implicitly accounted for the fictitious force, the centrifugal force $m\vec{g}_{cf}$, which is present in the accelerated reference system of you & the scale.
| {
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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?
| Let $\epsilon$ point towards positive x.
Then potential energy according to electrostatic force is proportional to $(a \cos \varphi)$ only.
Schrödinger's equation states $$
-\frac{\hbar^2}{2m} \nabla^2 \psi -(a \cos \varphi) \epsilon q \psi(\varphi) =E_n \psi.
$$ According to your problem, we choose $\varphi$ to be the (only) canonical variable. And replace Laplacian with cylindrical expression, keeping azimuthal term only, for others all vanish. $$
-\frac{\hbar^2}{2m} \cdot \frac{1}{a^2} \frac{d^2}{d\varphi^2} \psi(\varphi) -(a \cos \varphi) \epsilon q \psi(\varphi) =E_n \psi(\varphi).
$$ Or, $$
\psi'' +\frac{2E_n ma^2}{\hbar^2} \psi =-\frac{2\epsilon m a^3 q}{\hbar^2} (\cos \varphi ) \psi(\varphi).
$$ The homogeneous part is a sinusoical with frequency $$
\frac{a \sqrt{2E_n m}}{\hbar}
$$ This should equal to $n$, as is the usual "particle in the box", but only now circular. So $$
E_n =\frac{n^2\hbar^2}{2ma^2}.
$$ Resulting unperturbed eigenfunctions $$
\psi_{n,s}^{(0)} (\varphi) =\sin n\varphi \\
\psi_{n,c}^{(0)} (\varphi) =\cos n\varphi
$$ Since we want to perturb $$
\frac{2E_n ma^2}{\hbar^2} \leftarrow \frac{2E_n ma^2}{\hbar^2} +\frac{2\epsilon m a^3 q}{\hbar^2} \cos \varphi
$$ Or, speaking proportionally, $$
E_n \leftarrow E_n +\epsilon a q \cos\varphi
$$ Now, $\sqrt{E_n} \propto n$, and alteration added to $E_n$ has half effect on $n$ by expansion. We have $$
\psi_{n,s}^{(1)} (\varphi) =\sin \left( 1 +\frac{\epsilon a q}{2E_n} \cos\varphi \right) n \varphi \\
\psi_{n,c}^{(1)} (\varphi) =\cos \left( 1 +\frac{\epsilon a q}{2E_n} \cos\varphi \right) n \varphi
$$ For brevity, $$
A :=\frac{\epsilon a q}{2E_n}
$$ Consider $$ \begin{align}
&[\sin^2 (1+A \cos \varphi) n\varphi ] \cdot \cos \varphi \\
=&[\sin n\varphi \cdot \cos (An\varphi \cos \varphi) +\cos n\varphi \cdot \sin (An\varphi \cos \varphi) ]^2 \cdot \cos \varphi \\
\approx& [\sin n\varphi \cdot 1 +\cos n\varphi \cdot An\varphi \cos \varphi]^2 \cdot \cos \varphi \\
\end{align} $$ Where the fact that $\epsilon \ll 1$ is assumed, and only expanded to the 1st order as asked.
If you take $n=0$, then this is 0. And $\langle \psi_{0,s}^{(0)} \mid \delta H \mid \psi_{0,s}^{(0)} \rangle =0$, and similarly $\langle \psi_{0,c}^{(0)} \mid \delta H \mid \psi_{0,s}^{(0)} \rangle$. Really?
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Van der waals equation derivation? He assumed that the intermolecular forces result in a reduced pressure on the walls of the container which has a real gas in it. Also that the molecules are finite in size which means they do not have the entire volume of the container to themselves; something less than that. So when he accounted for the reduced volume by $V-nb$, why did he not do $P-\frac{an^2}{v^2}$ and instead, did the below:
He accounted for the reduced volume first with $V-nb$, then he used $$P(V-nb) = nRT$$ and then $$P=\frac{nRT}{V-nb},$$ then said that the real pressure is less than the ideal gas pressure by an amount $\frac{an^2}{V^2}$ from which follows the below $$P_{real}=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$$and therefore $$(P_{real}+\frac{an^2}{V^2})(V-nb)=nRT.$$
My question is: what is the logic behind this? What if he did the other way around? meaning corrected for the reduced pressure first and then corrected for the reduced volume which would have given the following steps
Correction for the pressure FIRST (reducing the ideal pressure by an amount$\frac{an^2}{V^2}$) $$V=\frac{nRT}{(P_{ideal}-\frac{an^2}{V^2})}$$
Then correcting the volume by reducing it by an amount $nb$, giving
$$V=\frac{nRT}{(P_{ideal}-\frac{an^2}{V^2})}-nb$$
giving
$$(P_{ideal}-\frac{an^2}{V^2})(V+nb)=nRT$$
Should the equation of state be $$P_{real}V_{real}=nRT$$ or $$P_{ideal}V_{real}=nRT$$ or $$P_{real}V_{ideal}=nRT$$
??
| The more formal derivation of the van der Waals equation of state utilises the partition function. If we have an interaction $U(r_{ij})$ between particles $i$ and $j$, then we can expand in the Mayer function,
$$f_{ij}= e^{-\beta U(r_{ij})} -1$$
the partition function of the system, which for $N$ indistinguishable particles is given by,
$$\mathcal Z = \frac{1}{N! \lambda^{3N}} \int \prod_i d^3 r_i \left( 1 + \sum_{j>k}f_{jk} + \sum_{j>k,l>m} f_{jk}f_{lm} + \dots\right)$$
where $\lambda$ is a convenient constant, the de Broglie thermal wavelength and this expansion is simply obtained by the Taylor series of the exponential. The first term $\int \prod_i d^3 r_i$ simply gives $V^N$, and the first correction is simply the same sum each time, contributing,
$$V^{N-1}\int d^3 r \, f(r).$$
The free energy can be derived from the partition function, which allows us to approximate the pressure of the system as,
$$p = \frac{Nk_B T}{V} \left( 1-\frac{N}{2V} \int d^3r \, f(r) + \dots\right).$$
If we use the van der Waals interaction,
$$U(r) = \left\{\begin{matrix}
\infty & r < r_0\\
-U_0 \left( \frac{r_0}{r}\right)^6 & r \geq r_0
\end{matrix}\right.$$
and evaluate the integral, we find,
$$\frac{pV}{Nk_B T} = 1 - \frac{N}{V} \left( \frac{a}{k_B T}-b\right)$$
where $a = \frac23 \pi r_0^3 U_0$ and $b = \frac23 \pi r_0^3$ which is directly related to the excluded volume $\Omega = 2b$.
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Theta Vacuum of Yang-Mills theory and Baryon number violation Background 1. In classical SU(N) Yang-Mills theories, there are a countably infinite number of homotopically inequivalent gauge field configurations of zero energy labelled by a winding number $n\in \mathbb{Z}$. In the corresponding quantum theory, the states $|n\rangle$ labelled by the quantum number $n$ are not the true vacua of the theory. This is because due to instanton effects, there is a nonzero tunnelling amplitude between the states $|n\rangle$. The true vacua of such theories are not the states $|n\rangle$ but given by a superposition $$|\theta\rangle=\sum\limits_{n=-\infty}^{\infty}e^{in\theta}|n\rangle\tag{1}$$ called the $\theta-$vacua. These states do not have a definite winding number.
Background 2. In the Standard Model, the baryon current $J^\mu_B$ is anomalous $$\partial_\mu J^\mu_B=\frac{N_f}{32\pi^2}(g^2W_{\mu\nu}^a \tilde{W}^{\mu\nu a}-g^{\prime 2} B_{\mu\nu}\tilde{B}^{\mu\nu})\tag{2}$$ where $N_f$ is the number of fermion flavours, $W_{\mu\nu}^a,B_{\mu\nu}^a$ are the SU(2) and U(1) field strengths.
One says that when there is a transition from a state of $n=1$ to $n=2$ (say, for example) there is a baryon number violation $\Delta B$ is given by $$\Delta B=2N_f(2-1)=2N_f.\tag{3}$$
This implies that, initially, the Universe is in a state with a definite winding number and finally it can make a transition to another state with definite (but different) winding number. This suggests to me that one considers states $|n\rangle$ as the vacua of the theory (instead the $\theta$ vacua) and somehow baryon numbers are related to the winding number. The states $|n\rangle$ can be considered as vacua only when the tunneling amplitudes are negligible. Another possibility is that, one considers the theory to be classical. Classical vacua are gauge field configurations of zero energy separated by finte energy barriers and having definite $n$.
Questions
$\bullet$ Therefore, my question is why are we calling $|n\rangle$ to be the vacua in the Standard model rather than $|\theta\rangle$?
$\bullet$ How is the winding number related to the baryon number?
| There is a definite fermion number only before and after the transition, as one can see from your equation for divergence of baryon current. Analogously, the system is in the definite vacuum state only at $t= \pm \infty$.
| {
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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.
| Experimentally, the most straightforward way of measuring this is photoemission spectroscopy. But it is often difficult to measure the position of the Fermi level (chemical potential of the electrons) in an intrinsic semiconductor. The charge carrier distribution is small, and the Fermi level is often pinned by surface states. Also, low conductivity may cause charging.
So most of the comparisons of Fermi levels in doped and intrinsic semiconductors are theoretical.
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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.
| As far as I know, no standard provides a list of unique identifiers for each quantity. However, a fairly complete list of quantities with recommended names and symbols is provided by IUPAC in the so called Green Book.
Elecrical quantities are also defined by IEC in its online Electropedia, which provides also translations of terms in different languages.
However, take into account that there are variations in terminology between fields and that the recommended names are not always used.
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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?
| They would actually repel. Yes, the force will act towards the middle but as $F=ma$ and you're looking at negative masses, it will cause repulsion.
How can we see this without negative mass voodoo? Easy, just use test particles.
Put a test particle on the left side of $C_1$, there are 2 holes on the right and 0 from the left, the total gravitational force will be to the left!
Now put a particle to the right of $C_1$, the hole on its left is much closer, so it will be pulled to the right! The forces try to stretch the bubble! (counter-intuitive, but yes gravity like to clump matter together and make the vacuum bigger).
But notice that the force to the left is stronger than the one to the right!
Now if you imagine the the bubbles are contained in a thin hard shell of some small mass, you will see that $C_1$ will move to the left and $C_2$ to the right, without using negative mass. Of course, to really calculate the force on the shell one needs to perform an integral over all the points on its surface, but I don't believe you'd get a different result.
| {
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Confusion about the calculation of 1PI effective action using path-integrals The bare Lagrangian of the $\phi^4$-theory can be written in terms of bare parameters as $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi_0)^2-\frac{1}{2}m_0^2\phi_0^2+\frac{\lambda_0}{4!}\phi_0^4\tag{1}.$$ The same bare Lagrangian, in terms of renormalized parameters and counterterms as $$
\mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2+\frac{\lambda}{4!}\phi_r^4+\frac{\delta_Z}{2}(\partial_\mu\phi_r)^2-\frac{\delta_m}{2}\phi_r^2+\frac{\delta_\lambda}{4!}\phi_r^4 $$
$$=\mathcal{L}_{renorm}+\mathcal{L}_{counterterm}\tag{2}.
$$
I'm interested in evaluating the 1PI effective action for this theory. For this, I have to compute the integral $$Z[j]=e^{iW[j]}=\int D\phi \exp[{i\int d^4x( \mathcal{L} +j\phi)}]$$
I think, it is the total Lagrangian $\mathcal{L}$, in either forms (1) or (2), can be used to evaluate $Z$.
But if I understand it correct, A. Zee's book on Quantum Field Theory in Nutshell, calculates Z using $\mathcal{L}_{renorm}$ part of $\mathcal{L}$ and not using full $\mathcal{L}$. See Chapter IV.3 Eqn. (1), (11). He uses, $A,B,C$ for counterterms.
Why is it that only the renormalized part of the Lagrangian used for the calculation of effective action? Am I missing something?
| A.Z. uses the full Lagrangian $\mathcal L$, not only $\mathcal L_\mathrm{renorm}$. He omits the counter-terms at first to keep the notation as simple as possible, but he includes them back later on: see equation $(15)$ (it seems odd to me that you decided to stop reading at equation $(11)$). You can repeat the calculations that led to equation $(11)$ but including the counter-terms as well, and you will end up with equation $(15)$ (but note that this is not really necessary: after all, the counter-terms have the same structure as the renormalised Lagrangian, and so it suffices to redefine $\mu^2\to\mu^2+B$ and $\lambda\to\lambda+C$ to get the correct expression).
As ACM mentions in the comments, Zee is not particularly rigorous in his book. For an alternative derivation of the Coleman-Weinberg effective potential, see Itzykson & Zuber's Quantum field theory, section 9-2-2 (in particular, page 454). See also Coleman's Aspects of symmetry, chapter 5, section 3.3 (in particular, page 138).
| {
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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?
| Any inelastic deformations and ferromagnetic cycles are always irreversible processes.
| {
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Change in pressure due to dissociation of gas Let's say I have a container with fixed volume, containing Hydrogen Gas at some predefined temperature. Now I pass an electric spark in the whole container such that this reaction takes place
$$H_2 \rightarrow 2H$$
Now actually the number of moles of Hydrogen(atoms) remain constant, the volume is also constant, would there be an increase in pressure assuming the whole situation to be ideal. I just want to know the contribution in increment due to dissociation and not due to increase in temperature of the gas due to the electric spark. Will the pressure increase, decrease or remain constant?
| We use the ideal gas law:
$$PV=nRT$$
Here, $n$ is not the number of moles of hydrogen, but rather the number of moles of gas particles, whatever those may be. Before the spark, there are $n$ moles of hydrogen molecules. After the spark, each molecule dissociates into two hydrogen atoms, making $2n$ moles of hydrogen atoms. So, all else being constant, pressure in the chamber should double.
Of course, this is assuming that you can actually keep the gas dissociated into neutral atoms for long enough to reach steady-state conditions. In practice, this is difficult for any case except very dilute gases, as collisions between the neutral hydrogen atoms favor recombination into $H_2$.
| {
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Conduction of semiconductors at atomic levels Speaking in terms of energy levels:
At 0K (absolute zero) all states below valence energy band of electrons are occupied and all states above the conduction energy band are empty.
At higher temperatures some valence electrons are excited from their parent atom.
My question is: Are the valence electrons which are teared away from their parent atom moved to conduction energy band?
| This is correct! At finite temperature, the valence electrons acquire a specific probability, given by the Fermi-distribution, to enter the conduction band of the semiconductor.
| {
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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?
| Real measuring devices all have a granularity. Your experiment will never tell you that a particle has momentum exactly $4.03752\,\mathrm{MeV}/c$; it will tell you (assuming it is very precises indeed) that it has momentum $(4.037 \pm .014)\,\mathrm{MeV}/c$ which is completely compatible with the particle still being found withing the apparatus at the end of the measurement.
| {
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Nature of metallic bonding in solid state What is the reason behind attraction of metal kernels & free electrons in electron sea model?
| It's mechanically not that different from dropping a lot of small magnets into a sea of iron filings (or better, magnetite sand). The positively charged metal ions attract the diffuse fermi sea of delocalized electrons, and vice versa. It's the ability of the electrons to delocalize that enables this, and that gives metals unique properties and the ability (usually) to bend and flex in interesting ways.
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Lowering the type-I seesaw scale without fine-tuning The type-I seesaw mechanism yields the effective neutrino mass $$m_\nu\sim -M_DM_N^{-1}M_D^T.$$ Here, $M_D$ is the Dirac neutrino mass coming from Yukawa coupling of the left-chiral neutrino with the standard model (SM) Higgs and $M_R$ is the mass of the right-chiral neutrino whose scale is not known a priori. Without fine-tuning the Yukawa couplings $Y$ in $M_D=Yv$ (where $v=246$ GeV, the VEV of SM higgs) one has $M_D\sim 100$ GeV. Now, if one requires a neutrino mass of the $0.1$ eV, then one must choose $M_N\sim 10^{12}$ GeV.
However, this paper, below eqn. 2.2, says that
However, theoretical arguments based on the naturalness of the SM Higgs mass of 125 GeV against radiative effects induced by the neutrino loop suggest the seesaw scale to be below $\sim 10^7$ GeV.
Is this scale of $M_N$ viable without fine-tuning the Yukawa couplings of $M_D$?
| Small neutrino masses from a rather small right-handed neutrino mass scale always imply small Yukawa couplings. Note, however that $M_D \sim 100$ GeV assumes Dirac neutrino masses of the order of the top quark, by far the largest Yukawa coupling in the Standard Model. If you assume that the neutrino Dirac masses are of the scale of the lepton Yukawa couplings, you obtain (for the heaviest state) $M_D \sim 1$ GeV, two orders of magnitude smaller than before.
Since the Dirac mass enters the Seesaw formula quadratically, this lowers the realistic right-handed mass scale by four orders of magnitude. This takes your $10^{12}$GeV down to $10^8$ GeV, very close to the limit you quote.
| {
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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?
| If you use a temperature measuring device that has mass, the answer is "yes", you slightly changed the temperature of the pot of water. If you use a temperature measuring device that doesn't have mass, such as an infra-red thermometer, the answer would be "no", since you didn't contact the water with anything that has any thermal inertia. Regarding how much the temperature changed in the first case, you have to use calorimetry to solve for the final temperature. Examples of calorimetry can be found at
http://www.physicsclassroom.com/class/thermalP/Lesson-2/Calorimeters-and-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.
| I was just thinking about it and I think I realize why that equation didn't work as expected. That equation will always fail to calculate time correctly when acceleration is 0 as it depends on there being a change in velocity between the initial state and the final state. If acceleration is 0 then velocity will always be constant and then time equals all real numbers and the equation will not be able to give a concrete value.
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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?
| An object in such a state has reached what is called a dynamic equilibrium where a constant velocity, free of acceleration is occurring. This, opposed to what is called static equilibrium where the object has reached a position with zero velocity.
For a falling object that experiences body forces due to gravity ($mg$) and opposing drag force ($bV$), and where acceleration has reached zero, the dynamic equilibrium state is often called terminal velocity
| {
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What do the results of a quantum fourier transform describe? The classical discrete fourier transform takes a sequence of values and outputs another sequence of values that describe a set of coefficients for complex sinusoids which can be used to reconstruct (or approximate) the original input.
In contrast, what exactly does a quantum fourier transform do? And how is it said to be the "analogue" of the discrete fourier transform?
More specifically, since all the input qubits are in superposition with each other, is there still the notion of an ordered sequence as there is in DFT (since the order of the inputs of DFT clearly affects the output)?
And how does one interpret the output of QFT? Say the output of a QFT is a superposition of $a |00\rangle +b|01\rangle + c|10\rangle + d|11\rangle$, what does this tell me about the original superposition?
| The quantum Fourier transform applies a discrete Fourier transform to the amplitudes in a superposition. This is, if your input is
$$
\sum a_n |n\rangle\ ,
$$
then the output is
$$
\sum \hat a_n |n\rangle\ ,
$$
where the vector $(\hat a_n)$ is the discrete Fourier transform of the vector $(a_n)$.
The point is that if $|n\rangle$ is encoded in binary (i.e., in a $N$-qubit register, with $n=2^N$), this can be carried out very efficiently, namely in time $O(N^2) = O((\log n)^2$), which is exponentially faster than classically (where $O(n\log n)$ operations are required).
| {
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Is current density (J) in Ampere's law derivable? In Ampere's law:
$$
\nabla\times\mathbf{B}=\mu_0\mathbf{J}
+\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}
$$
the current density is listed explicitly as a separate term from the change in electric field. My understanding of the history (perhaps completely wrong), is that the $J$ term was determined first, and then the $E$ term was added later (by Maxwell?) to account for displacement current.
As the $J$ term is physically a set of moving charges, which each produce a time-varying electric field, why isn't the $E$ term sufficient to calculate the magnetic field? That is, to determine the magnetic field from a set of moving charges, couldn't you determine the magnetic field of a single moving charge from:
$$
\nabla\times\mathbf{B}=\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}
$$
and then the total magnetic field of a current would be the sum of magnetic fields from many moving charges?
| Nothing in Maxwell's equations depends on the fact that the charges that we encounter in the real world are all tied to individual particles.
So we can imagine that we have an infinite rod of charge in space and it's moving with uniform velocity along its axis, and Maxwell's equations will still apply to this situation. And here we have moving charge, but with no net change in the configuration of the charge, so no change in the electric field associated with that charge.
In the real world, conduction in a wire, for example, involves the motion of quadrillions of electrons, mostly moving randomly, but with a slight bias in one direction or the other. Even though there is a current through our hypothetical wire, there's no net change in the configuration of the charge over time, except at the microscopic level, and so no significant change in the electric field resulting from the motion of the charged particles.
| {
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Confusion with partial derivatives as basis vectors So I have seen that the directional derivative can be written as
$$ \frac{df}{d\lambda} = \frac{dx^i}{d\lambda}\frac{df}{dx^i} $$
And we can identify $ \frac{d}{dx^i} $ as basis vectors and $ \frac{dx^i}{d\lambda} $ as components. What I don't understand is why is $\frac{df}{d\lambda} $ considered a vector? It's a derivative of a function w.r.t. a parameter and surely that's not a vector?
I.e. In vector notation the directional derivative is given by a dot product
$$ \frac{df}{d\lambda} = \hat{n} \cdot \nabla f $$ which is a scalar but in tensor notation that seems to not be the case?
| I think the physicspages author is just confused. $df/d\lambda$ is a scalar, not a vector. It's the scalar product of the covector $\nabla f$ with the vector $d\mathbf{x}/d\lambda$. They say, "Regarding the partial derivatives as basis vectors, ..." and go on as if $\partial f/\partial x$ and $\partial f/\partial y$ were the basis vectors. This is wrong. In the notational convention they have in mind, it's the operators $\partial/\partial x$ and $\partial/\partial y$ that are used as basis vectors.
In this notational convention, the partial derivative operators are never actually applied to anything. They never have anything written to the right of them. The convention is a notational trick that exploits an isomorphism between vectors and derivative operators, but it doesn't involve actually taking the derivative of anything.
The thought that they're probably trying to express is that in their example, their paraboloid is embedded in a higher-dimensional space (which would not normally be the case in general relativity). They're being sloppy/confused with their notation, because they're using the symbol $f$ to mean a scalar field defined on the $(x,y)$ plane, but they're also treating $f$ as if it were a position vector in $(x,y,z)$ space.
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Why does stacking polarizers of the same angle still block more and more light? I have some sheets of polarization film. They came in a big box, all stacked at the same angle. I noticed that the entire stack of them lets almost no light through, even though they're all at the same angle.
I pulled out two, and those two also block more light than just one.
Why?
Is this because I have low-grade polarizers? Or because lining them up at EXACTLY the same angle is impossible? Or because the light that gets through the first one is not really polarized exactly to its angle — it's just that less of it is polarized away from its angle than before?
If it's because these are low-grade polarizers, can anyone recommend a linear polarizer that I can stack several of in a row at the same angle and still have a 100% probability of the light getting through?
I feel like I'm probably just misunderstanding polarization theory so please correct me.
| A high-quality supplier of polarizers and other optical equipment would be able to offer you data on the transmission characteristics of even their cheapest polarizers:
I interpret this plot to mean that if you bought two of these devices, aligned their axes parallel to each other, and shone unpolarized $\lambda=550\rm\,nm$ green light on them, you'd only get 40% of your intensity out of the first polarizer, and of that fraction something like $10^{-4}$ still has the "wrong" polarization. From the second polarizer you'd only get $80\% \cdot 40\% = 32\%$ of the original intensity, and another transmission factor 0.8 from any subsequent, also-aligned polarizers. It might be possible to improve the polarization by having multiple parallel filters, at the cost of this lost overall intensity, but you might also run into sneaky laboratory issues.
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Understanding thermodynamic equilibrium for a relativistic system The conventional books on thermodynamics that I know, do not talk about the thermodynamics of relativistic systems. But in Cosmology, the thermodynamic concepts are often applied to the whole universe.
My question is about the concept of thermodynamic equilibrium for a relativistic system consisting of electrons, positrons and photons. In this system, there are processes of the type $$e^++e^-\leftrightarrow \gamma+\gamma.$$ However, for chemical equilibrium, the forward reaction must proceed at the same rate as the backward reaction. Now, if the temperature (of the environment) falls below a certain value, photons will not have enough energy required for pair production. In particular, if the temperature is such that the average photon energy falls below the rest energy $2m_ec^2$ the backward reaction will stop.
*
*Does it mean that the system fails to be in equilibrium below that critical temperature?
*Since the forward reaction continues to occur, shouldn't the system ultimately equilibrate with only photons (that obey the blackbody distribution?).
| 1) Thermodynamics for relativistic systems is pretty straightforward, and discussed in many text books (Landau, Greiner, $\ldots$). The only difference is that in relativistic systems the total number of particles is not conserved (and the associated chemical potential is zero), only the total charge (electric, baryon, $\ldots$) is conserved.
2) The fact that certain equilibration reactions become very slow as $T\to 0$ has nothing to do with relativity, it happens in non-relativistic systems as well. This just means that equilibration takes a long time.
3) In the particular case of electrons and photons there is no issue, because equilibration can take place via Bremsstrahlung $e^-+e^-\to e^-+e^-+\gamma$.
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Light reflected in a cylinder I was cooking and noticed a funny pattern appearing when i was looking in on of my pans. When light fell into a pan with high edges, it seemed to reflect into a hart-shaped pattern. Can anyone explain how the light gets warped into this shape, and does anyone know why the light gets reflected into a sharp line, rather then a hart-shaped plane?
I have added some pictures to make it more clear.
| You want to look up Caustics in optics.
The specifics of your example boil down to the way the geometry works out. In this case the simplest model would be something like a Nephroid, where you have a circular shape which ( in terms of illumination ) is similar to a half circle, and that makes that particular caustic.
| {
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Grashof number as a ratio of buoyant and viscous forces The Grashof number is supposed to be a ratio of buoyant forces to viscous forces.
I find this hard to believe, since if
$$F_b=\beta g \rho \Delta T$$
is the buoyancy force, the definition of the Grashof number,
$$\text{Gr}=\frac{\beta g\Delta T L^3}{\nu^2},$$
implies that the viscous force is something like $\frac{\rho}{L^3}\nu^2$, instead of something linear in $\nu$. How is this supposed to be the viscous force?
| I don't agree with @Pirx that it is to be understood as vague metaphors although I admit it sometimes is a little bit difficult to understand exactly how they are ratio of scales as you have clearly found out.
What makes it a bit difficult is that dimensionless numbers are sometimes themselves ratios of other dimensionless numbers. For example the definition of $Gr$ can be rewritten as:
$$Gr=\frac{\beta g\Delta TL^{2}}{\nu U}\frac{UL}{\nu}$$
clearly we see a role for the Reynolds number here:
$$Re=\frac{UL}{\nu}=\frac{\rho U^2/L}{\mu U/L^2}=\frac{inertial}{viscous}$$
The other term in $Gr$ is easily decomposed:
$$\frac{\beta g\Delta TL^{2}}{\nu U}=\frac{\beta \rho g\Delta T}{\mu U/L^2}=\frac{bouyancy}{viscous}$$
| {
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How are water vapors not visible? This site says that water vapor isn't visible.
However, take a look at this picture:
Isn't that water vapor?
| Water vapour is a clear and colourless gas, so it can't be seen by the naked eye.
What you see in the photo in your second link is (partially) condensed water vapour, i.e. fog (or mist). Fog contains tiny, discrete water droplets and light bounces off their surface in random directions, causing the visibility.
Water vapour by contrast only contain free molecules, too small for light to bounce off, so pure water vapour (without any condensate) is invisible, like most gases (some gases are clear but coloured like chlorine gas).
| {
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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 the most accurate answer is that we probably do not know. I wrote the following paper which looked at inflationary cosmology as a quantum critical version of Landau's tri-critical point. If this is right, or some variant similar to this is right, the tri-critical point falls below the valued of the Landau parameter for type-II phase transitions. The would tend to make sense with the generation of matter and radiation at reheating. That is an abrupt transition with a latent heat that is more of a type-I phase transition.
| {
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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.
| An object can be in rotational non-equilibrium while simultaneously being in translational equilibrium depending upon the arrangement of the forces. In other words, an object can undergo an accelerating spin without translating. You can add rotational energy to an object without adding translational energy.
What I wrote above was meant as a hopefully helpful hint. I'll clarify below.
This is a very simple structural engineering problem. A hinge is a structural component that, by definition, can only have a resultant x-direction force and a resultant y-direction force.
You apply a force to the outer region of the door, setting up a torque acceleration of the door. You can mathematically slide the torque force to the hinge. That force now acts on the door at the hinge location. You've also mathematically noted the torque about the hinge as a circular arrow in the proper direction. The one reaction component at the door hinge exactly counters your applied force so that translation in that direction is impossible because the forces are balanced. The OTHER reaction at the hinge directly balances the centripetal acceleration of the rotationally accelerating door. It's much like whirling a ball on a string.
| {
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How can I handle divergence that appears in many physical problem? I came across with the following type of integration with singularity.
$$\int_{s_2=0}^{s_2=\infty}\int_{s_1=0}^{s_1=s_2}\left(\frac{1}{s_2-s_1}\right)^{3/2} \,ds_1\,ds_2 \, .$$
How can I solve it?
| As suggested I have expanded my comment into an answer.
There's no general prescription for dealing with divergent integrals in physics. Typically when an integral like this shows up it means is that the integral is not the full story, but the missing pieces of the puzzle depend on exactly what it is you're trying to do.
For example in UV divergences in quantum field theory require remormalization, one says you are missing a separate infinite comtribution from the 'counter-terms', when that is added the result is finite. There are also IR divergences which are associated with massless particles, but which go away whenever you ask a truly physical question (e.g., account for the finite resolution of your detector). In electromagnetism it is possible for the electric potential to be divergent, but for the force (which is actually observable) to be finite since it's the derivative of the potential. In other contexts sometimes the argument is that your model is breaking down. A famous example is the singularity at the center of a black hole, which is thought by many to be a signal that general relativity does not apply near the black hole's center, rather than a problem that can be fixed within general relativity.
In any case, often a good first step is to regulate the integral: write the divergent integral as a limit of a sequence of converging integrals. This can help to diagnose where the divergence comes from and how strong it is. In your exaple you could regulate by taking the $s1$ integral from 0 to $s2-\epsilon$ then sending $\epsilon\rightarrow 0$.
| {
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Can a ship float in a (big) bathtub? I am confused.
Some sources say it is possible at least theoretically ( http://www.wiskit.com/marilyn/battleship.jpeg ) and some say it is not true ( http://blog.knowinghumans.net/2012/09/a-battleship-would-not-float-in-bathtub.html )
Is it necessary or not that there exists an amount of water around the ship that weights at least the same as the weight of the ship?
| Some of the comments on the answers here show that the Archimedes Principle is being thought of by many people as a force that somehow arises literally as a direct result of the fluid's being displaced by a boat. In fact, Archimedes Principle is only a mnemonic for the results of the full calculation of what is actually happenning, which is the transmission of normal force across fluid - floated object interface.
The whole story here is the sum of the normal pressure forces on the floated / steeped body over its submerged boundary, and this is quite independent of whether the body of fluid weights more or less than the imaginarily displaced fluid, exactly as described in Sammy Gerbil's Answer. In my answer here I derive the general expression:
$$\mathbf{F} = \int_V\boldsymbol{\nabla}(p(\mathbf{r}))\,\mathrm{d}V\tag{1}$$
for the sum of these normal pressure forces where we are to imagine the pressure field $p(\mathbf{r})$ that would be present in the fluid within the surface if the fluid weren't being displaced by the body taking up the volume $V$. This imaginary displaced fluid device comes from the application of the (Gauss) divergence theorem to the first principles expression for the sum of the pressure normal forces on the body:
$$\mathbf{F} = \int_{\partial V}\,p(\mathbf{r})\, \mathbf{\hat{n}}(\mathbf{r})\,\mathrm{d} S\tag{2}$$
an expression which is patently free of imaginary displaced fluids. (2) and (1) are readily shown to give the Archimedes Principle if we put the equilibrium condition for a fluid in a gravitational field $\nabla p(\mathbf{r}) = \rho\,\mathbf{g}(\mathbf{r})$ into (1).
| {
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Are the diffusion terms conservative? Generally the diffusion terms are of the form $$D = \dfrac{\partial}{\partial x} \left(\mu \dfrac{\partial u}{\partial x} \right) .$$
Is this this term conservative or nonconservative?
| In terms of fluid dynamics, a conservation law is one in which the net flux in is equal to the net flux out. This is typically represented as the PDE,1
$$
\frac{\partial u}{\partial t}+\nabla\cdot\mathbf F=S\tag{1}
$$
where $u$ is the conserved quantity, $\mathbf F$ the flux and $S$ the source term.2 In your case, $\mathbf F=-\mu\nabla u$, so it is a conservative term because it satisfies (1).
Note, though, that the domain of dependence for a diffusion equation at a point $\left(\mathbf x,\,t\right)$ is the entire domain at all previous times. This differs from the convective equation where the domain of dependence is along characteristics (lines that satisfy $du/dt=0$).
1. This can be equivalently written as an integral equation.
2. Often times $S=0$.
| {
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4-Vector Potential Notation How am I supposed to interpret this notation:
$$F^{uv} = \partial^uA^v-\partial^vA^u$$
I know that $\partial^u = (\frac{1}{c}\frac{\partial}{\partial t},-
\vec\nabla)$
So for example for the potential $$A=\left(\begin{matrix}
0 & 0 & 0& E_z\\ 0 & 0 & B_y & 0\\ 0 & -B_x & 0 & 0\\ E_z & 0&0&0\end{matrix}\right)$$
So to compute $F^{23} = \partial^u\left(\begin{matrix}0\\B_y\\0\\0\end{matrix}\right) -\partial^v \left(\begin{matrix}0&0&B_y&0\end{matrix}\right) = 0 - - B = B$.
Is this the correct way to do it? I'm just getting confused by the notation.
| The indices come with ordering $(0,1,2,3)$ so that $\partial^0=\frac{1}{c}\frac{\partial}{\partial t}$, $\partial^1=-\frac{\partial}{\partial x}$ etc.
$A^\mu$ is a 4-vector with components $(A^0,A^1,A^2,A^3)$,
not a matrix as your notation suggest. Thus, in your specific example,
$$
F^{23}=\partial^{2}A^3-\partial^3 A^2=
-\frac{\partial}{\partial y}A_z+\frac{\partial}{\partial z}A_y\, .
$$
| {
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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?
|
I know that this might make sense from a mathematical point of view, but what would it actually look like?
Why not start with the data, which tells us what it actually looks like?
The observations tell us that all galactic clusters are receding from each other. In an explosion in three dimensions, one could track the paths and find the center from which the explosion started. The observations tell us that they are all * receding from the earth* There is no reason or evidence why the earth would be the center of a cosmic explosion!
That is what gave rise to the "big bang" model,formulated in the four dimensions of time and space in General Relativity.
Since Georges Lemaître first noted in 1927 that an expanding universe could be traced back in time to an originating single point, scientists have built on his idea of cosmic expansion. While the scientific community was once divided between supporters of two different expanding universe theories, the Big Bang and the Steady State theory, empirical evidence provides strong support for the former.
The center is at time=0, where the three space dimensions are also at (0,0,0) and a singularity that gave rise to the explosion observed . This singularity has been modified by introducing an effective quantum mechanical theory for gravity near this origin.
| {
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Sound wave coherent time Put an ordinary light bulb in front of a double slit and you will not observe interference due to the short coherent time compared with the observation time.
What about sound waves? What is the typical coherent time of sound waves? Is it usually much longer than the observation time?
| Sound waves are macroscopic phenomena, and not affected by the quantum effects in the same way as the double slit experiment with light.
Sound waves do "degrade" in real life because when they travel through non-uniform media (e.g. in air, scattering may be caused by temperature gradients, convection currents, variations in humidity, etc) but at scales of a several wavelengths and cycles of vibration (which typically means distances of the order of meters and times of the order of seconds) the interference patterns are stable for as long as the sounds are being generated.
| {
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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?
| A human parachuting from $h=4000\,\mathrm m$ (cf. http://adventure.howstuffworks.com/skydiving1.htm) needs to get rid of their potential energy $mgh$.
If we assume that all this energy is used to evenly heat up the skydiver, who essentially consists of water with its well-known specific heat capacity $c_{H_2O}=4182\frac{\mathrm J}{\mathrm{kg}\cdot \mathrm{K}}$, the temperature rises by
$$ \frac{gh}{c_{H_2O}}= \frac{9.81\cdot 4000}{4182}\,\mathrm K<10\,\mathrm K,$$
not enough to incinerate.
One might object that the heating would occur mainly on the front instead of evenly, but the skydive is long enough for much heat to be transported across the body or even convected away form it by the surrounding air.
Note that the mass of the skydiver did not enter into the above simple estimate, only the staring altitude and the specific heat.
A re-entering spacecraft, on the other hand, not only starts form a higher altitude (more than 100 kilometers), but additionally has to get rid of its substantial kinetic energy. If we don't want to look up the numbers for orbital speed, let's try from memory. One always hears that one orbit takes about 90 minutes, hence the velocity must be at least (using the slightly smaller Earth circumference)
$v\ge \frac{40000\,\mathrm{km}}{5400\,\mathrm s}\approx 7400\,\frac{\mathrm m}{\mathrm s} $, so the energy per kilogramm of mass is $\frac12v^2\approx 55\,\frac{\mathrm{MJ}}{\mathrm{kg}}$.
We see that this is a lot, lot more than the mere $40\,\frac{\mathrm{kJ}}{\mathrm{kg}}$ of our skydiver.
The spaceship might get rid of much of the kinetic energy by using its rocket engine, but that would be unwise: It took a whole lot of fuel to bring the rocket into orbit in the first place (or rather: into orbital speed, after all the potential energy is small compared to the kinetic energy in low earth orbit); hence it takes (almost) as much fuel to achieve the same $\Delta v$ upon re-entry. But in order to have so much fuel available in orbit, it must have been transported there in the first place. This is infeasible because of the fuel-to-payload ratio.
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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.
| The heat gets "smeared out" everywhere. It dissipates all around the Earth. This raises the Earth's temperature by an absolutely miniscule amount, because the amount of heat involved is tiny compared to the incoming energy from the sun. And because the Earth is a little bit warmer, it radiates heat into space a little bit faster, so the Earth reaches equilibrium temperature again, a fraction of a Kelvin higher than it would without all that human activity.
In very large cities, this manifests as an urban heat island effect, where the temperature in the city can be one or two Kelvin higher than the temperature outside the city, when windspeeds are low. However, when windspeeds pick up, the warm air gets blown out of the city faster, and so the heat dissipates around the countryside over such a wide area that it becomes unnoticeable.
See also this related question and answers on the Earth Science StackExchange: How does anthropogenic heating affect global warming?
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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.
| What you say is correct. But, it's not what Hook's law tells you.
Hook's law says, the force exerted on a mass, permanently or temporarily attached to the spring, is proportional to the difference between the instantaneous length of the spring and the equilibrium length. In the direction which points toward the equilibrium.
When the external agent is removed, the displacement of the spring from it's equilibrium hasn't changed. The force is still pointing to the equilibrium state. And, everything is fine.
| {
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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?
| After searching more for an answer to this, I found the following:
It appears that folding does allow the paper towel to absorb more water, due to the fact that the water is stored between the folds of the paper towel (what the speaker refers to as "interstitial suspension", I mistakenly called this "interstitial forces"). Because of this, the amount of folds does appear to play a role (i.e. more folds is better, to a point). There also appears to be an increase to the rate at which the towel absorbs water, but this is not related to the aforementioned suspension.
Other factors are also at play as mentioned above, such as that more liquid is absorbed at each pass, reduction in wasted edges, etc.
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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
| Setting $\hbar=1$ for simplicity, the matrices you need are the $S=1$ matrices. One easily obtains
$$
S_z=\left(\begin{array}{ccc}
1&0&0\\
0&0&0\\
0&0&-1\end{array}\right)\, ,\quad
S_+=\sqrt{2}\left(\begin{array}{ccc}
0&1&0\\
0&0&1\\
0&0&0\end{array}\right)\, ,\quad S_+=S_-^\dagger,
$$
from which one recovers $S_x$ and $S_y$ by inverting $S_\pm=S_x\pm i S_y$.
The matrix for $S^2$ will be $2\times I$ where $I$ is the $3\times 3$ unit matrix.
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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?
| This dispersion curve from Wikipedia Commons shows how the index of refraction varies with wavelength. It is this dispersion property which results in the splitting of light into a variety of colors, as observed in prism demonstrations and rainbows.
We see that if Dense Flint is used for a lens, there will be quite a bit of color splitting. With the crown glasses, there is less, and in this graph, the dispersion (as well as the visible-range indices of refraction) are smallest for fluorite.
High quality fluorite lenses are prized by photographers for their very low chromatic aberration. They are also very expensive, with a 70-200 mm zoom going for US$1200.
Regarding your 2nd question, the focal point is defined to be the location at which paraxial rays converge.
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Guided waves equations In Griffiths's Introduction to Electrodynamics, monochromatic guided waves are proposed to have the form
$$\mathbf{\tilde{E}}(x,y,z,t)=\mathbf{\tilde{E}}_0(x,y)e^{i(kz-\omega t)}$$
$$\mathbf{\tilde{B}}(x,y,z,t)=\mathbf{\tilde{B}}_0(x,y)e^{i(kz-\omega t)}$$
where
$$\mathbf{\tilde{E}}_0=E_x\mathbf{\hat{x}}+E_y\mathbf{\hat{y}}+E_z\mathbf{\hat{z}}$$
$$\mathbf{\tilde{B}}_0=B_x\mathbf{\hat{x}}+B_y\mathbf{\hat{y}}+B_z\mathbf{\hat{z}}$$
Then the following is stated:
In every denominator the expression $(\omega /c)^2-k^2$ appears. But, as far as I know, $\omega /c=k$, so
$$(\omega /c)^2-k^2=k^2-k^2=0$$
What am I missing here?
| $\omega=kc$ for plane waves which have spatial variation only along $k$. In a waveguide, the field varies in other directions as well and the relation is incorrect.
You can think about a waveguide mode as being composed of a standing waves pattern in the plane of the waveguide combined with the propagation along the normal direction. If you found the total wavenumber $k_0$ including these standing wave components as well, you would get the $\omega=k_0c$ relation. Since $k$ here only corresponds to the propagating part of the wave vector and not the total wavenumber the relation $\omega=kc$ does not hold.
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How molecules radiate heat as electromagnetic wave? an object of higher temperature radiate infrared rays as a way to decrease the temperature. how a molecule produce a electromagnetic wave? in atoms electromagnetic radiation is caused by electrons. what is responsible in molecules?
|
in atoms electromagnetic radiation is caused by electrons. what is responsible in molecules?
There exist atomic and molecular orbitals of the electrons composing atoms and molecules.
This means that the charge distribution around an atom or a molecule in space is uniform only for some quantum numbers. Otherwise there is a shape to the charge distribution of the electrons, allowing positive charge regions from the positive charge of the nucleus. The positive and negative charge regions cause the attraction that pairs (or clusters) of atoms have and build into a molecule. Electrons in molecules also have spatial charge distributions and this is what creates the attraction for bonding.
This bonding has extra degrees of freedom of rotation and vibration, and the electrons are in energy levels that are almost continuous, thus can have small transitions of energy leading to infrared radiation.
The positive and negative field regions will also give rise to radiation in a gas, when molecules bounce off each other.
So different models are used to predict the radiation, which ultimately is the black body radiation, depending on the type of matter under study, gas, or fluid or solid.
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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?
| An inertial frame is equivalent to a frame's velocity at any given time. An accelerating frame still has intertial frames for the same reason that we can calculate instanteous slopes of a function. An accelerating frame is changing inertial frames constantly but that doesnt mean it isn't an inertial frame at a given point it time.
| {
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Is the magnetic field of a moving electron caused by length contraction in the direction of motion? Consider an electron moving relative to us. Because the space in the electron's rest frame is contracted relative to us in the direction of the electron's velocity, the electric field lines are squeezed in the same direction, so the electric field "density" is bigger perpendicular to the electron's motion (but smaller (zero?) in the direction parallel to its motion). Is this the qualitative source of the magnetic field?
| The electric field for a charged particle moving in the $x$ direction of motion is Lorentz transformed as
$$
E'_x~=~\gamma E_x,~E'_y~=~\gamma(E_y~-~vB_z/c^2),~E'_z~=~\gamma(E_Z~+~vB_y/c^2).
$$
The motion of charge transforms the electric field components into magnetic field components.
| {
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Fluid velocity in a vertical pipe Consider a pipe with length $L$ and uniform radius $A$ is held vertically. According to the continuity equation, the velocity of water going into the pipe seems to be the same as the velocity of water coming out. But according to Bernoulli's equation:
$$P_{atm}+\frac{1}{2}\rho v_1^2+\rho gL=P_{atm}+\frac{1}{2}\rho v_2^2$$
$$v_2=\sqrt{v_1^2+2gL}$$
Which means that the e water would come out faster, which makes much sence. What is wrong with my equations?
| Consider a vertical pipe connected to a shallow reservoir:
There are mainly 3 points of interest:
*
*The outlet of the vertical pipe:
The heads are:
$P_{atm}+\frac{1}{2}ρv^2$
*The inlet of the vertical pipe:
By continuity, the stream has same velocity at inlet and outlet
$P_{in}+\frac{1}{2}ρv^2 + ρgL = P_{atm}+\frac{1}{2}ρv^2$
$P_{in} = P_{atm} - ρgL$
*At the surface of the reservoir:
$P_{atm} + ρgL = P_{in} + \frac{1}{2}ρv^2 + ρgL $
$v = \sqrt{2gL}$
which is the text book answer of Torricelli's law
https://en.wikipedia.org/wiki/Torricelli%27s_law
The principle behind Bernoulli's equation is conservation of energy. The problem can be understood this way:
On water surface of the reservoir, the velocity is zero, but when water is entering the pipe, it gains velocity. Since there is no change in elevation in a shallow reservoir, energy must be converted from pressure head to kinetic head.
As water travels along the pipe, the elevation head decreases but the kinetic head has to remain unchanged due to continuity, so energy is converted from elevation head to pressure head.
| {
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Can a line of finite point charge configuration be in equilibrium? Can we arrange a finite number of point charges in a line so that they are all in equilibrium? Earnshaw's theorem proves that there cannot be a stable equilibrium point in an electric field, but what about unstable equilibrium? If not, how to prove it?
| If I put a charge of $+4q$ at $x=+x_0$ and $x=-x_0$ and have charge of $-q$ at the origin then the force on the charge at the origin is 0 by symmetry and the force on the outer charges is $$\frac{1}{4\pi\epsilon_0}\left(\frac{-4q^2}{x_0^2} + \frac{16q^2}{(2x_0)^2}\right) = 0$$
so the system is (unstable) equilibrium.
| {
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Relation between perturbation theory and Taylor expansion in QM So I am looking at non-degenerate perturbation theory. The idea is that the perturbing term in the Hamiltonian is small so you somehow expand the energies and wave functions in this small term and collect orders.
Now I did an exercise in which you apply perturbation theory to a system, which is solvable. You then show by Taylor expanding the analytical result of the energies that the first order perturbation term is equal to the first order term in the Taylor expansion. Should this be obvious? I know that the first order perturbation theory was derived based on expanding the energies in the small perturbing term but somehow I cannot see that it is exactly equivalent to simply calculating the first order term in the energy.
| Yes.
Suppose
$$
H=H_0+\epsilon H_1=\left(\begin{array}{cc}A & 0 \\ 0 & B\end{array}\right)+
\epsilon\left(\begin{array}{cc}a & b \\ b & c\end{array}\right)
$$
where $a, b$ and $c$ are real for simplicity. You can easily work out that
the exact eigenvalues are
$$
\lambda_\pm =
\frac{1}{2}\left(A + B +\epsilon (a+ c) \pm
\sqrt{(A-B)^2+\epsilon(A-B)(a-c)+\epsilon^2((a-c)^2+4b^2)}\right)\, .
$$
Expanding - say - the first eigenvalue in powers of small $\epsilon$
and assuming that A-B is "large enough" to factor it from the square root gives
$$
A+ a \epsilon +\epsilon ^2 \frac{b^2 }{A-B}
$$
which is just $E_1^{0}+\epsilon \langle 1\vert H_1\vert 1\rangle +
\epsilon^2 \frac{\vert \langle 1\vert H_1\vert 2\rangle\vert^2}{E^{0}_1-E^{0}_2}
$
as "predicted" by perturbation theory.
I presume what you did must be basically equivalent to the above. This is a good exercise because you can also see why the unperturbed energies must be such that $A\ne B$:
if $A=B$ then $H_0$ is basically the unit matrix and the eigenvectors will be those of $H_1$ since any similarity transformation will not affect a multiple of the unit matrix. In addition, the case $A=B$ results in a simplification in the discriminant, which no longer contains $A$ or $B$, making the assumption on the difference $A-B$ invalid.
This can be generalized to any hermitian matrix. The difficulty is that it's not so easy (and in fact in general impossible) to write down a closed form for the eigenvalues, but the $2\times 2$ example shows that, if you could, you must recover term by term the results of perturbation theory.
After all, the eigenvalues cannot depend on how you calculate them, i.e. exactly first and then expand, expand first and then add the terms.
| {
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time dilation in a gear chain system Thought experiment...
suppose we have a mechanism connecting two chain gears as follows:
gear A rotates at 2 rpm. gear B is near a massive star. because of time dilation, time is slower at B, lets say half the time at A. therefore, as seen from observer at A, the gear at B will rotate only 1 rpm.
therefore there is mismatch of speed on each gear, how to explain this?
disregard elasticity/ mass of chain/ or friction... thank you
| 1) Special Relativistic Effects (for relative velocity) for observer at A
1a) Belt (and wheel) travelling towards them appears time dilated and length contracted.
1b) Belt (and wheel) travelling away from them appears time dilated and length contracted.
2) General Relativistic, (gravitational effects) for observer A
2a) B appears time dilated due to acceleration against gravity
2b) Belt at B and gear at B appears stretched, due to the warping of space.
you can find more about effect 2b) here length contraction in a gravitational field
| {
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Is hermiticity a basis-dependent concept? I have looked in wikipedia:
Hermitian matrix
and
Self-adjoint operator,
but I still am confused about this.
Is the equation:
$$ \langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x \in \text{Domain of } A.$$
independent of basis?
| The definition that you have cited is indeed basis-independent as it only makes reference to the inner product $\langle\cdot,\cdot\rangle$ and the domain of $A$, neither of which is basis-dependent.
Note that "symmetric" in your above sense and "self-adjoint" in the broader sense are connected by the Hellinger-Toeplitz theorem which says that if the domain is the full Hilbert space, then the operator is self-adjoint: and this in turn means that what physicists mean by "self-adjoint" or "Hermitian" is in fact your notion of "symmetric;" operators like the Hamiltonian usually are not defined over the whole Hilbert space since they're symmetric but not bounded.
This has been argued to lead to a sort of mathematical incompleteness of quantum mechanics (Warning: PDF, warning: philosophy), however it doesn't impact most day-to-day applications in physics and isn't even present in most references about it. Here's a set of lecture notes which mentions it.
| {
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Pauli- Villars regularization in the Electron Vertex Function: Evaluation I'm studying one loop contribution for electron vertex function form Peskin and Schroeder's book " An introduction to quantum field theory " Section: 6.3. I have some troubles with Pauli- Villars regularization and getting the final results, so any help will be appreciated ..
Starting from:
$I = \delta\Gamma^\mu(p',p)= 2 i e^2 \int \frac{d^4l}{(2\pi)^4} \int^1_0 dx dy dz \delta (x+y+z-1) \frac{2}{D^3} ~ \times \bar{u}(p') \Big[\gamma^\mu . \Big(-\frac{1}{2} l^2 +(1-x)(1-y)q^2+(1-4z+z^2)m^2\Big) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} (2 m^2 z (1-z)) \Big]u(p)~~~~~~~~~(6.47)$
where $D= l^2-\Delta+i\epsilon, ~~~~~~~~\Delta=-xyq^2+(1-z)^2m^2$, while :
$ \Delta_\Lambda = -xyq^2+(1-z)^2m^2+ z \Lambda^2 $.
After momentum integration and Pauli- Villars regularization, this equals:
$I= \frac{\alpha}{2\pi} \int^1_0 dx dy dz \delta (x+y+z-1) ~ \times \bar{u}(p') \Big(\gamma^\mu . \Big[ \log \frac{z\Lambda^2}{\Delta} + \frac{1}{\Delta}\Big((1-x)(1-y)q^2+(1-4z+z^2)m^2\Big)\Big] + i \frac{\sigma^{\mu\nu}q_\nu}{2m\Delta} \Big[2 m^2 z (1-z)\Big] \Big)u(p)~~~~~~~~~(6.54)$
*
*Here it's suppose it substitute by $\log\frac{\Delta_\Lambda}{\Delta} = \log\frac{-xy q^2+(1-z)^2m^2+z\Lambda^2}{\Delta}$ , but why instead it substitute only by $\log\frac{z\Lambda^2}{\Delta}$ ?
*Then how can we reach for :
$ F_1(q^2) = 1 + \frac{\alpha}{2\pi} \int^1_0 dx dy dz \delta (x+y+z-1) ~ \times
\Big[ \log \Big( \frac{m^2(1-z)^2}{m^2(1-z)^2-q^2xy}\Big) + \frac{m^2(1-4z+z^2)+q^2(1-x)(1-y)}{m^2(1-z)^2-q^2xy+\mu^2z} - \frac{m^2(1-4z+z^2)}{m^2(1-z)^2+\mu^2 z}\Big] ~~~~~~~ (6.56)$
In deed I'm little bit confused, how did we get $1$ term ?, where $\log z\Lambda^2$ had gone ? now $\log m^2(1-z)^2$ in the nominator, the part of $\Delta_\Lambda$ which didn't written in the previous equation , also in this $\log$ part of the nominator, where's the $q^2$ term ?
I tried to read the book explanation, but I can not understand too much so have any one made this exercise before ?
| All they've done is taken the limit of large $\Lambda$. Then
\begin{align}
\Delta_\Lambda=-xyq^2+(1-z)^2m^2+z\Lambda^2\rightarrow z\Lambda^2
\end{align}
and
\begin{align}
\log\left(\frac{\Delta_\Lambda}{\Delta}\right)\rightarrow \log\left(\frac{z\Lambda^2}{\Delta} \right)+\mathcal{O}(\Lambda^{-2}).
\end{align}
This is what they mean when they say just after equation (6.53) that "the convergent terms are modified by terms of $\mathcal{O}(\Lambda^{-2})$, which we ignore."
| {
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2D elliptic gaussian beam on tilted wall I can calculate laser's beam cross section at every point along z-axis. But how to do a "projection" of this beam on tilted wall? I see option to use ray tracing, but there is possibly a better way.
Ordinary projection using Camera Matrix does not takes into account beam propagation till the wall (from first point of contact of the cross section, see figure). Are there some other types of projections?
| You can do a coordinate transformation $x \rightarrow x', y \rightarrow y', z \rightarrow z'$ for a coordinate system in which the $x'y'$ plane is aligned with the wall.
| {
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How can a generic potential transform under Lorentz transformations? The standard relativistic particle Lagrangian is
$$L = - \sum m_i \sqrt{1-v_i^2} - V(x_i).$$
The first term contributes a scalar to the action, as it should, but the second term is not clearly a Lorentz scalar; it contributes $\int V(x) \, dt$, so to make it Lorentz invariant we need to know how $V(x)$ transforms.
For example, in the case of electromagnetism, we have $V = q \phi$. The resolution to this problem is that the full coupling is actually $q (\phi - \mathbf{A} \cdot \mathbf{v})$. Then the potential contributes $\int A_\mu dx^\mu$ to the action, which is manifestly Lorentz invariant.
I'd like to know how this works for a general potential $V$, besides just the electric potential. (Examples might include $V = kx^2/2$ or $V = -Gm_1 m_2/r$.) Does it make sense in general to make $V$ part of a four-potential, or perhaps more general? If so, what do the other components physically represent?
| Your question has no answer since there is no sensible classical relativistic multi-particle picture. A corresponding no-go theorem was proved by Currie, Jordan and Sudarshan, Reviews of Modern Physics 35 (1963), 350.
See also https://physics.stackexchange.com/a/32401/7924
| {
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What is an induced electric field? I have read in many books about induced current in a coil (Faraday's law), and also the motional emf across a moving conductor in a magnetic field. But somewhere I read about induced electric field due to a time varying magnetic field.
And I think that Induction of electric field is the fundamental phenomenon, and induced emf and current are the results of it
I am just a novice in physics.
Could someone explain me how these phenomenon (Induction of emf and Induction of electric field) are related to each other?
| You are right that a changing magnetic field creates (induces) an electric field, this is an actual law of nature.
Now if you put a conductor where the magnetic field is changing, you will get a current due to the produced electric field.
But in the case of the moving conductor moving through a magnetic field the reason is different. The reason for the produced current is Lorentz force, the electrons in the conductor are pushed by Lorentz force and hence you get the current.
Notice in this case, even if the conductor is moving through the magnetic field the magnetic field is NOT changing so electric field is not produced, the reason must therefore be the Lorentz force.
Whenever you get confused just check whether the magnetic field is changing or not. If the magnetic field is changing then the reason for the current must be an induced electric field, if it is not changing the reason must be Lorentz force.
| {
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Is short circuit technically the same as overloading? Taking the simplest circuit: battery and resistors.
If I connect lots of resistors in parallel, wouldn't that increase the current to an extent that it would be technically be very similar to shorting the circuit?
| If you have $N$ resistors in parallel, all of which have a resistance of $R$, the total equivalent resistance will be
$$ \left( \frac 1 R + \cdots + \frac 1 R \right)^{-1} = \left( \frac N R \right)^{-1} = \frac R N \;. $$
So yes, if you take a sufficiently large amount of (identical) resistors in parallel, it's the same as not having any resistors at all.
| {
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Why is perturbation theory always implemented around $\alpha=0$? In the perturbative approach to field theory we expand whatever we are computing on a power expansion in some coupling
$$
\sum^nd_n\alpha^n
$$
then in principle we can compute all the $d_n$. This series is in general expected not to be convergent, but it is hoped that it at least is an asymptotic expansion of the true thing when $\alpha\to0$.
My question is, why is the perturbative approach only implemented around $\alpha=0$? I mean, we could make expansions around any given $\alpha_0$ and obtain expansions that (would be hoped) to be asymptotic to the real thing when $\alpha\to\alpha_0$?
$$
\sum^nb_n(\alpha-\alpha_0)^n
$$
why is perturbation theory always implemented around $\alpha=0$?
| It is because one wants to do a perturbation theory around the known solution. Of course, you can denote by $\alpha$ anything you want, but usually it is some coupling constant. The limit $\alpha \rightarrow 0$ corresponds to a free theory, which can be easily solved. After it is done, one computes the corrections to the known solution to the free theory in the form of the perturbation series in $\alpha$.
| {
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Why does a wet towel become more hot than a dry one? I remember a long time ago my mum had told me not to use a wet tea towel when taking food out of the oven because you can burn your hands - lo and behold, it came time to make dinner and I did not head my mother's warning from all them years ago and my hand got burned. Why does a wet tea towel heat up more quickly when removing items from the oven?
Is it because the water molecules are more excited when heated?
| Water has a high heat capacity and is also a pretty good conductor of heat.
The dry towel has a lot of air pockets and doesn't have a lot of heat capacity in general.
Essentially you're bridging the air gap that was acting as an insulator. The water also provides a sink for the energy instead of just going into the air. That sink then feeds into your hand. The liquid will also increase the surface area of contact with your hand and the hot object.
Just to clarify a bit on the title question: the term "more hot" may be confusing. The wet towel doesn't have to be a higher temperature than the dry towel (it probably still will be). It just has to transfer the heat into your hand faster (which is still a definition of "more hot" some people just may take that as temperature).
| {
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relation between mass of black holes and tempereture? As we know, by increasing the mass or energy of a black hole, its event horizon radius will increase, but why its temperature should increase too?
Really I want the relation between mass and temperature of black hole!
| Higher temperature means it's hotter for a lower mass, i.e., it emits more and higher freq radiation. Micro black holes which evaporate away quickly do so at the end in a burst as they are the hottest. You have to get the details from the paper.
The temperature is the one that would define that of a black body radiator, with the same probability density function for the radiation radiated. For black holes Hawking found that it is inversely proportional to the mass. He also determined that it is proportional to the surface gravity - defined as the acceleration that you'd have to give a body at the horizon (or right outside it) to keep it there without going in. Small black holes have a very high curvature of spacetime near them, thus a very high surface gravity and temperature, and emit at higher frequencies as they get smaller, till they dissappear in a burst.
As they get smaller they will send out more gamma ray radiation.
See about black holes temperature and thermodynamics at https://en.m.wikipedia.org/wiki/Black_hole_thermodynamics
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Synchronized Clocks in Inertial frame Will the synchronized clocks placed in an inertial frame remain synchronized forever?
| Yes, they will.
"Synchronization" in a strict sense means that both clocks are said to define one and the same inertial frame. At the moment of synchronization they are said to share same velocity and direction. They will carry the inertia defined at the moment of sychonization with them, all the time. Any acceleration one of the synchronized clocks will experience will be the same acceleration as the other one experiences under the same circumstances within the defined frame.
To test this statement imagine one clock in a sports car, the other one as a stationary clock. The fact that a pendumlum measuring time will be accelerated in the sports-car thus differing from a pendumlum in the stationary clock does not spoil the synchronization as any change due to acceleration can be dealt with calculating (Lorentz transformation).
If it's not mechanical pendumlum and movement that is being used to measure time but the speed of light there are not even divergent accelerations of clocks in respect of velocity to be calculated upon as light cannot be accelerated beyond the speed of light. However changes in respect of the once defined direction of the light beam have to be considered ("time delation").
This shows that inertial frames are being defined by synchronization (in a very narrow sense). A defined inertial frame will remain defined, as you can always calculate on any acceleration any participating particle experiences, be it a clock.
| {
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Doppler effect and apparent frequency What is meant by "apparent frequency"?
I mean the answer we get by applying the formula; what does it signify?
If it is the frequency received by the observer, does it mean that the observer receives the same frequency no matter what the distance of the source?
Shouldn't distance of play a role?
| Actually, in an ideal situation distance of the source will not matter. In real life, the farther the source, the greater the damping effect on the wave. The sound loses energy and hence the amplitude and intensity fall.
However, this does not affect the frequency. Frequency is given by the formula fλ=v where symbols have usual meaning. When observer moves towards the sound source, he receives more waves per second, irrespective of the distance. When sound source moves towards observer distance between consecutive waves i.e wavelength decreases and hence frequency increases, depending only on speed of source and not distance. Everything related to frequency depends upon the speeds of the waves, source and observer.( I chose not to go in detail about the workings of the Doppler effect as that is not actually the focus of your question. I hope that you have a basic idea of this phenomenon).
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How to calculate rate of water heating with pulsed near infrared laser? I'm designing something that requires me to heat a small volume $( <10 ml)$ of water with a $2100 nm$ pulsed near-infrared laser with settings between $1 J$ and $1.5 J$ and $5-15 Hz$. I'm heating the water in the range of $20-70$ degrees C. I thought that it would be fairly easy to calculate the rate of heating as below:
$$Q = mc\Delta T$$
$$\Delta T = \frac{Q}{mc}$$
$$Q = Eft$$
Therefore,
$$\Delta T = \frac{Eft}{mc}$$
$$or$$
$$dT = \left(\frac{Ef}{mc}\right)dt$$
Experimentally, all of my heating curves are linear but all of my experimental heating rates are $0.6\left(\frac{Ef}{mc}\right)dt$ almost perfectly. Am I stupid and missing something simple up? Is there a coefficient specific to pulsed laser heating that would give me $0.6$ somewhere? I'm assuming that all of the energy from the laser is absorbed by the water because of the low penetration of infrared through water and my distance of $\approx 2 cm$.
| No, there is no specific effect related to heating with pulses compared with any other heating method. Your calculation should give the correct result. You should treat this as a systematic error and look for a physical explanation.
Perhaps your assumption of perfect absorption is incorrect? Absorption is not affected by pulse repetition rate, and probably is only very weakly dependent on temperature because decrease in density - ie the "concentration" of absorbers - is small.
The 1st graph in the wikipedia article on Absorption of EMR by water gives an absorption coefficient of about $\alpha \approx 2000 m^{-1}$ at $2.1 \mu m$. Your absorption length is $x=0.02m$ so over this length you should expect transmission of about $e^{-\alpha x} = e^{-40} \approx 4\times 10^{-18}$. This confirms your assumption that almost all the energy from the beam should be absorbed, so there must be some other explanation(s) for the energy loss.
Reflection at the air/glass front face of the heating cell would only account for about 3.5% of the loss. I am assuming the cell is made of glass.
Have you included in $mc$ the heat capacity of the cell which holds the water? (I presume not, since you take c=4.2.) And also of the thermometer?
If the time taken to heat the water is "long", and the apparatus is not thermally insulated, losses due to conduction/convection/radiation could be significant.
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Maxwell energy Distribution in One dimension Maybe my question is too specific but I could not find the answer.
Abstract from:
Yield Optimization and Time Structure of Femtosecond Laser Plasma
$\kappa \alpha$ Sources
The generation of femtosecond $\kappa \alpha$ x-rays from laser-irradiated plasmas is studied with a view to optimizing photon number and pulse duration. Using analytical and numerical models of hot electron generation and subsequent transport in a range of materials, it is shown that an optimum laser intensity $I_{opt}=7×10^9Z^{4.4}$ exists for maximum Kα yield. Furthermore, it is demonstrated that bulk targets are unsuitable for generating sub-ps x-ray pulses: instead, design criteria are proposed for achieving $\kappa \alpha$ pulse durations ≤ 100 fs using foils of $≈2μm$ thickness.
I have found in Yield Optimization and Time Structure of Femtosecond Laser Plasma
$\kappa \alpha$ Sources: PRL one dimensional Maxwell energy distribution of electrons as
$f(E) dE =\frac{1}{\sqrt{EkT}}\exp\left(-\frac{E}{kT}\right) $
This expression diverges for $E\rightarrow 0$.
How one reaches to this expression I could not understand. Usually in 3D case the exponent is multiplied by $\sqrt{E}$.
I don't think that it is a typographical error
They have also written the total energy content as
$En=\int n f\left(E\right)E dE$
From dimensions of this expression it looks like the particle distribution per unit energy.
Is there a specific physical significance of $f\left(E\right)$
Please help.
| Yes, there is. You can look here. It is the Maxwell-Boltzmann distribution of the energy per degree of freedom. This is a chi-square distribution and note that
$$\int_0^\infty \frac{1}{\sqrt{EkT}}e^{-\frac{E}{kT}}dE=
\frac{2}{\sqrt{kT}}\int_0^\infty e^{-\frac{x^2}{kT}}dx$$
that is finite as it should. The authors of the paper are somewhat cavaliers about normalization factors. Note also, that any power of $E$ grants a finite integral reducing to known Gaussian integrals.
| {
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How do we know which lighter elements fuse to form heavier elements? Formation of lighter elements is fairly straightforward: $^4$He + $^4$He + $^4$He = $^{12}$C. However for each heavier element (heavier than Fe in particular), is there exactly one combination of (two lighter) elements that can be fused? Is it possible that any combination of atomic weights can be combined to form a heavier element; i.e., can any combination of sufficient atomic masses combine to form a given heavy element?
| You are, in fact, correct : there's more than one way to combine elements to form other heavy elements.
What we can predict to some extent is the relative probability of different combination (different fusion reactions). We also have knowledge of how relatively stable the different nuclei are. Some reactions are much less likely (by orders of magnitude) than others, and some of the fusion products are themselves very unstable.
The subject area, in case you need to know, is called Stellar Nucleosynthesis. There is a link on that page to an old paper by Burbridge, Fowler and Hoyle which gives a good flavor of how this is worked out. Things have advanced somewhat since that paper, but it's an interesting read (it's a long paper, almost a book - 100 pages).
It's not really practical to describe all the possible reactions in a star so we tend to just concentrate on the main ones.
We can check our predictions using evidence from various sources on the cosmic and stellar abundance of various elements. So we get to know that our theoretical calculations, while we can't be necessarily exact, are in the right ballpark.
| {
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Why does a wave not interfere with its secondary wavelets? Huygens principal says that every point wavefront is the source of a secondary wavelet. If this is true, why do those wavelets not interfere with the main wave? Shouldn't waves looks like a circle of interfering circles?
| I suggest you find a good source, and sit down and draw the wavelets out to understand it visually. As user45664 pointed out, thinking about the "main wave" is a bit misleading. Because, in this picture that you posted, you need to account for Huygen's Principle:
"...Every point on a propagating wavefront serves as the source ... of [another] wavefront..."
Namely, at the right of the circles, you gotta draw more circles for each point on each circle And then more circles on top of the new circles!! Huygen's says that all these circles will constructively and deconstructively combine to form a radiating plane-wave "trough" corresponding to the oscillatory nature of $\vec{E}$ and $\vec{B}$.
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$
As an aside:
Huygen's Principle "... naturally has several shortcomings, one of which is that it doesn't overtly incorporate the concept of interference and perforce cannot deal with lateral scattering. Moreover, the idea that the secondary wavelets propagate at a speed determined by the medium ... is a happy guess." - Optics 4th edition, Eugene Hecht
| {
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Proof of the generalized equipartition theorem The generalized equipartition theorem (where variables need not be quadratic) states that if $x_i$ is a canonical variable (position or momentum variable), then
$$\left\langle x_i \frac{\partial \mathcal{H}}{\partial x_j}\right\rangle = \delta_{ij}\ k T$$
where the average $\langle \cdot \rangle$ is taken over an equilibrium probability density $\rho(p,q)$:
$$\langle f(p,q) \rangle = \int dp dq \ \rho(p,q) \ f(p,q)$$
In the most general case this probability density is the canonical ensemble's. For the theorem to hold ergodicity is also required. However, I'm having trouble finding a rigorous prove where the assumptions are explicitly used in the derivation.
Could you provide such a prove, or a reference to a paper/book where it can be found?
| J. A. S. Lima and A. R. Plastino published the article On the classical energy equipartition theorem back in 1999. In their article they derive a generalized equipartition theorem. Their generalized approach is valid for systems with arbitrary distribution functions and for systems with non-quadratic terms in the Hamiltionian. A link to their article is https://doi.org/10.1590/S0103-97332000000100019. This article should answer most of your questions.
Their article can be summarized as follows:
Suppose there is a system with $f$ degrees of freedom and the Hamiltonian is
\begin{equation}
\mathcal{H} = g(x_1,...,x_L) + h.
\end{equation}
$(x_1,...,x_L)$ is a subset of the phase-space coordinates for positions and momentas. $g$ is homogeneous such that
\begin{equation}
g\left(\lambda x_{1}, \ldots, \lambda x_{L}\right)=\lambda^{r} g\left(x_{1}, \ldots, x_{L}\right).
\end{equation}
For systems that are distributed according to a Boltzmann distribution you can derive the expression
\begin{equation}
\langle g\rangle=\frac{L}{r} k_B T.
\end{equation}
Here $k_B$ is the Boltzmann constant and $T$ is the temperature. This is the generalized form of the equipartition theorem that you are looking for. For example, for a monovalent ideal gas you have $r=2$ and $L=3N$ such that you recover the famous result $U = \frac{3}{2}N k_B T$ where U is the internal energy of the gas.
| {
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What constitute the secondary particles beam when a high energy proton beam hits a target material? Basically I want to know what particles emerge along with high energy photons(not sure about it) as the second beam when a specific high energy proton beam is incidented upon a target material such as iridium or gold. Can the composition of secondary beam be calculated theoretically?
Does the ratio of photon energy and particles energy of the secondary beam hold any specific value for a given energy of primary beam?
| News to me about the photon. I thought that was still theoretical.
Since we are talking unproven things, the rest of your question is hard to definitively answer. I'm going with pions.
| {
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Why does a force not do any work if it's perpendicular to the motion? I have a book that says the Moon's orbit is [in this context assumed to be] circular. The Earth does no work on the moon. The gravitational force is perpendicular to the motion. Why is there no work done if support force is perpendicular to the motion?
| As others have already explained. Well its because there is no displacement in the direction of the gravitational force. It is assumed that the orbit stays the same during our observation. The Earth is pulling the moon towards the centre but the moon is moving in a circular orbit, with no displacement towards the centre.
A simple analogy is a block released from a particular height. Now as it falls down gravitational force does work on it, by pulling it down through a distance and raising its kinetic energy. Now suppose as it moves down you apply a constant horizontal force to the right, now the block is moving slanted path, because of the resultant force due to the two forces. Now if we consider the horizontal motion, gravitational force is not the cause of that, it is simply your hand. So your hand does work for the horizontal displacement, not the vertical one which is done by gravity.
So, in the case of the moon, Earth's gravitational force is not the cause of why it moves in a circle, it only provides the centripetal force necessary and centripetal force causes no whatsoever displacement to the moon.
| {
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Speed of block sliding on frictionless ramps Here's the question:
My book says the answer is C. How is it not A?
I know that all the potential energy is transferred to kinetic energy. With algebra, knowing Kinetic energy is (1/2) * m * v^2 and gravitational potential is mgh, I solve for h which results in (v^2)/2g
Ok so since this is a proportional reasoning problem my focus is that h = v^2 meaning height is directly proportional to the square of velocity. That being said, if we half the velocity, that means that some value (let's use p for the variable) multiplied by height gets me to (1/2v)^2.
That being said, if we refer back to the equation h = (v)^2 and v is being halved, it should look like this:
h * p (<- growth factor) = (0.5v)^2. Now it looks like the right side of the equation grew by a factor of (1/4). Think about it. If your velocity is 10, the right side of the equation becomes 100. If you half that velocity, the right side of that equation becomes 25. You can see that the right side of the equation grew by a factor of 1/4. That means that $p$ (my growth factor variable) should also be (1/4). $h * 1/4 = h/4$. NOT $3h/4$. Where did I go wrong?
| The potential energy at the start is $mgh$, which is converted into kinetic energy of $\frac 12mv^2$ at the bottom. When the velocity is $\frac v2$ the kinetic energy is $\frac 12m(\frac v2)^2=\frac 18mv^2$. As $\frac 14$ of the energy is now kinetic, the rest must be potential, so the potential energy is $\frac 34mgh$ the the height is $\frac 34h$
| {
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What is the significance of the phase constant in the Simple Harmonic Motion equation? The displacement of a particle performing simple harmonic motion is given by $x = A \sin(\omega t + \phi)$ , where $A$ is the amplitude, $\omega$ is the frequency, $t$ is the time, and $\phi$ is the phase constant. What is the significance of $\phi$. How is it used? Please explain the meaning of the phase constant
| The equation of motion for a simple harmonic oscillator is
$$
\ddot x+\omega^2 x=0
$$
and the most general solution to this is
$$
x(t) = A_1 \cos \omega t + A_2 \sin \omega t
$$
Note there are two constants of integration that correspond to the equation being a second order differential equation. More physically, the velocity is given by
$$
v(t) = - \omega A_1 \sin \omega t + \omega A_2 \cos \omega t
$$
and the two constants of integration are fixed by the configuration of the system at any given time. Said differently, if you know the position and velocity at time $t_0$ you can solve for $A_1$ and $A_2$. You should do that as an exercise.
Now note that the expression you have can be written as
$$
x(t) = A \sin \phi \cos \omega t + A \cos \phi \sin \omega t
$$
and you can therefore relate $A$ and $\phi$ to $A_1$ and $A_2$ and from there to the position and velocity at time $t_0$.
| {
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Uncertainty principle and the speed of light If an EM wave only gives us a probability of where a photon may be at a given moment, and the HUP tells us that we can't know the exact location of the photon. Then would it be correct to say that a photon does not travel in a straight line?
If this is true, wouldn't the photon's crooked path mean that it must travel faster than $c$ for us to measure its straight line speed at $c$?
Here's a comparison to explain my question further: If two sprinters run a 100 yard dash in 10 seconds, but one sprinter is required to run in a zigzag manner wouldn't that runner need to run faster than the straight line sprinter for both runners to run the race in 10 seconds.
Does the crooked traveling photon need to exceed $c$ for us to measure its straight line speed at $c$?
| Your crooked path argument holds for a classical electromagnetic wave going through a medium. That is why the speed of light in a medium can be less than c, the speed of light in vacuum. At the photon level it is that photons effectively go a larger distance at velocity c, as they interact individually with the lattice of the material, and a lower group velocity is measured.
In vacuum there are no interactions for the photon, which is a point particle, and its velocity is c and the same is true for the light beam. The Heisenberg uncertainty does not enter.
The validation of this is the validation of the special theory of relativity which has been tested by the myriads of experiments in physics.
| {
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Acceleration of car. One dimensional motion easy problem A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
My attempt at solving the problem:
$$a(x) = \frac{v - u}{t}$$
where
$v =$ final velocity
$u =$ initial velocity
$$$$
I get the answer as $4.05 \space ms^{-2}$
But the correct answer given to the problem is $8.10 \space ms^{-2}$.
They used a different equation to reach that answer.
Did I use the wrong equation? I have the average velocity and not the instantaneous veolcity?
| That equation is not the most straight foreward for this situation but it can be used, here's how. If you divide the displacement, 110 m, by the time of 5.21 s you will get 21.1 m/s. But that is the average velocity over the entire displacement. The average veloctiy, when v is changing uniformly, is found by adding the initial and final velocities and dividing by two. Since the initial velocity is 0 m/s, you will need to double the 21.1 m/s for the final velocity such that the average will be the calculated 21.1 m/s. So the final velocity, and therefore the change in velocity over that displacement, is 42.2 m/s. When you divide that delta v by the time over which it occurs, you will have how much delta v per second, acceleration.
While it's true there is another equation that does this all in one step, it would be good for you to understand the process I've outlined above.
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Ampere's Law Confusion I had this question recently in a test. Different methods are yielding different answers. Can someone point out the mistake?
We are given 4 infinite wires carrying current out of the plane as shown. Find $$\int_{-\infty}^{+\infty} \vec{B}\cdot\,\mathrm d\vec{x} ,$$ (along x axis)
My logic for line integral along the infinity part being zero is that by using Biot-Savart law, the field produced by the current carrying wires would definitely tend to 0 at infinity.
The answer given is $$ \bar u (-3) $$ , which seems like average of both the values. Can someone point out my mistake?
| When you do the loop integral about one set of wires you are ignoring the other set of wires. Going from $-\infty$ to $+\infty$ around your first loop, you "collect" half of the B field due to one set of currents (the other half comes when you go back in the other direction - your assumption that it's zero "because you are far away" is wrong. You know it is, because a complete loop integral "at infinity" must give you the same value as if you were close).
The actual field is of course the sum of the fields due to the four wires. So you add the two loop integrals, and divide by two (because you only go halfway around the loops).
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A question on Collison of macroscopic particles
Hello,
In the above question I could solve for average elastic force by taking velocity with respect to wall and finding change in momentum of the ball after that divided change jn momentum by time interval. Answer comes out to be option b.
But as it is written in the question that collision is elastic, therefore, Kinetic energy before the collision should be equal to kinetic energy after the collision and option d should also be correct. But the correct answers according to the book are b and c.
Please Explain why d is incorrect and c is correct.
| I think d is incorrect because the kinetic energy increases due to the massive moving wall hitting against the ball.
Kinetic energy before the collision should be equal to kinetic energy
after the collision
will only be true if the wall is static, not moving.
Just like hitting a ball with a bat.
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Time reversal and Antilinear operators I'm struggeling to solve this question. Can anyone help me please?
Let us consider a generic quantum mechanical system governed by
the Hamiltonian $\ H(t) $.
In what follows we denote the evolution operator by
$\ u(t, t_0) $. Hence,
$\ |Ψ(t)> = u(t, t_0)|Ψ_0> $
satisfies the time-dependent Schrodinger equation, where $\ |Ψ_0> $ represents the
wave function at $\ t = t_0$.
(a) Prove the unitarity condition
$\ u^† (t, t_0)u(t, t_0) = u(t, t_0)u^†(t, t0) = I $ .
(b) Now let us assume that the system under study exhibits a symmetry
represented by an antilinear (and antiunitary) operator $U$ which has nothing
to do with the time reversal. Show that in this case $\ [u(t, t0), U] = 0 $, and
thus the system is unstable since its spectrum is unbounded from below.
(c) Show that instability disappears if antilinear $U$ includes time reversal.
The conclusion from (b) and (c) is that symmetries represented by antilinear
operators are possible, but they necessarily involve time reversal.
| Let's make $u(t)=e^{it}$ ($H$ is hermitian)
Then $u^\dagger u=e^{-it}e^{it}=e^{it}e^{-it}=u u^{\dagger}=1$
Let $e^{it}$ and $U$ act on $\phi$:
$$(e^{it} U-U e^{it})\phi=(-e^{it}U+ e^{-it}U)\phi,\quad \text{or}\quad (-e^{it}+e^{-it})U\phi.$$
Now $U\phi=-U\phi$, so $(-e^{it}+ e^{-it})=-(-e^{it}+e^{-it})$ and therefore $0$, so (b) holds
When we include time reversal in $U$, let $U e^{it}$ act on $\phi(0)$, which becomes $-\phi(t)$.
Try it from there yourself.
| {
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Equation of motion from Polyakov action If I write (my understanding of) how to derive the equations of motion from the Polyakov action, I come up missing a term.
Beginning with the basic Polyakov action
\begin{equation}
S = -\frac{T}{2}\int\sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu\,d^2\xi
\end{equation}
the Lagrangian is
\begin{equation}
\mathcal{L}=-\frac{T}{2}\sqrt{-h}h^{ab}\partial_aX^\mu\partial_bX^\nu
\end{equation}
which after plugging into the Euler-Lagrange equation is
\begin{equation}
0-\frac{\partial}{\partial \xi^b}\left[-\frac{T}{2} \sqrt{-h}h^{ab}\partial_aX^\nu\right]=0
\end{equation}
which leads to (my result of)
\begin{equation}
\partial_b\left[ \sqrt{-h}h^{ab}\partial_a X^\nu\right] =0
\end{equation}
The equation of motion should be
\begin{equation}
\Box X^\nu = \frac{1}{\sqrt{-h}}\partial_b\left[ \sqrt{-h}h^{ab}\partial_a X^\nu\right] =0
\end{equation}
My question is that I do not see where the $1/\sqrt{-h}$ comes from. (This result is ok for the Polyakov action on its own, but I want to be able to add a mass term to the action which will make the $1/\sqrt{-h}$ matter then.)
I think that there is something fundamental that I am missing about deriving the equation of motion. I understand that the operator $\nabla_\mu X^\nu=\partial_\mu X^\nu +\Gamma^\nu_{\mu\lambda}X^\lambda$, but I am bothered by deriving the equations of motion when the Lagrangian density is $L(X^\nu,\partial_aX^\nu)$ instead of $L(X^\nu,\nabla_aX^\nu)$. If I derive the Euler-Lagrange equation of motion from varying an action containing $L(X^\nu,\partial_aX^\nu)$:
\begin{eqnarray}
\delta S &=& \int \delta L (X^\nu,\partial_a X^\nu)\sqrt{-g}d^nX \\
&=& \int \left[ \frac{\delta L}{\delta(\partial_aX^\nu)}\delta(\partial_aX^\nu) + \delta X^\nu\nabla_a\frac{\delta L}{\delta(\partial_aX^\nu)}\right]\sqrt{-g}d^nX + \int \left[ \frac{\delta L}{\delta X^\nu}-\nabla_a\frac{\delta L}{\delta (\partial_a X^\nu)}\right]\delta X^\nu\sqrt{-g}d^nX
\end{eqnarray}
can I still send the surface term (the first term above) to zero? Thanks for any insight.
| It's easier to compute the equations of motion by using variations of the inverse metric.
$$ \frac{\delta S}{\delta h^{ab}}=0 $$
You can use the distributive property of the variational operator and make use of the identity
$$ \frac{1}{\sqrt{-h}}\frac{\delta\sqrt{-h}}{\delta h^{ab}}=-\frac{1}{2}h_{ab} $$
| {
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Short question about moment force I have a question about moment Forces. Let $\mathbf{e_1}$, $\mathbf{e_2}$ be the unit vectors defining a Cartesion coordinate system $Oxy$.
Let $\mathbf{F}$ be the force applied at point $A$.We have:
$$\mathbf{F} = F_x \ \mathbf{e_1} + F_y \ \mathbf{e_2}$$
where
$$\begin{cases} F_x &= a\\F_y &=-b \end{cases}$$
The moment of force $\mathbf{F}$ about point $O$ at point $A$ is, by definition:
$$\mathbf{\mathcal{M}_{/O}}\left(\mathbf{F}(A)\right) = \mathbf{OA} \times \mathbf{F}$$
Therefore the magnitude $M$ of this moment force is
$$M = \left(\begin{array}{c} x \\y\end{array}\right) \times
\left(\begin{array}{c} F_x \\F_y\end{array}\right)=
\left(\begin{array}{c} x \\y\end{array}\right) \times
\left(\begin{array}{c} a \\-b\end{array}\right) =
-(xb+ya)$$
However, we can also have
$$M = \left(\begin{array}{c} y \\x\end{array}\right) \times
\left(\begin{array}{c} F_y \\F_x\end{array}\right)=
\left(\begin{array}{c} y \\x\end{array}\right) \times
\left(\begin{array}{c} -b \\a\end{array}\right) =
xb+ya$$
How to define which one to use? What is the convention used for the 1st and 2nd equations? It should depend on the convention used for a positive moment but I can't figure out how it's done.
Edit: Added my intuitive answer
I'll post my intuition just below but ... this is not really solid as it is only intuitive. I'd still like a solid proof.
*
*Convention used: Moments are positive when rotation is clockwise (opposite of the geometrical convention)
*For a positive moment, as rotation is clockwise, The vector along the rotation axis must be pointing outward (away) (defined by $\mathbf{e_3}$)
*Therefore, $\mathbf{u_r} = \mathbf{e_2} \times \mathbf{e_1} = - (\mathbf{e_1} \times \mathbf{e_2})$ . Then, coordinates of $\mathbf{F}$ and $\mathbf{OA}$ are defined by $(\mathbf{e_2},\mathbf{e_1})$ and not $(\mathbf{e_1},\mathbf{e_2})$
*Equation 2 for $M=xb+ya$ is correct for this convention (opposite to the geometrical/mathematical one)
Is this correct? How to demonstrate it?
Thanks!
| For 2D problems with cross products always make it a 3D problem with the z-coordinate 0. The cross product is uniquely defined for 3D problems.
This results in the following 2D cross products
$$ \begin{align}
\omega \times (x,y) = \begin{pmatrix} 0\\0\\ \omega \end{pmatrix} \times \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} & = (-\omega y, \omega x) \\
(v_x,v_y) \times z = \begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} & = (v_y z, -v_x z) \\
(x,y) \times (F_x,F_y) = \begin{pmatrix} x\\ y \\ 0 \end{pmatrix} \times \begin{pmatrix} F_x \\ F_y \\ 0 \end{pmatrix} &= F_y x - F_x y
\end{align} $$
| {
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Glasses underwater So I figured the refraction of the cornea is based on the index of air and the vitrous humor to make a perfect image. Underwater this is messed with because water has the same index as the eye. Hence you can't see clearly underwater without making an air pocket with scuba goggles for example.
Would it be possible to compensate this by wearing regular glasses with a diopter that is equivalent to the difference in index between air and the eye, normally?
| Trivially no. To see this, construct the glasses out of a material with the same refractive index as water.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why doesn't current decrease in series combination? I know that the question is quite stupid but I want to get an insight of this case. consider 3 resistors connected in series with a battery, after the current passes through resistor 1 it loses some of its energy, the kinetic energy of the charge carriers will definitely also decrease and so does the drift velocity then why doesn't the current decrease? Its quite confusing.
| Current is due to the drift velocity of electrons . In the Transient State when the current sets up there is an accumulation of electrons at places like bends in the wire. There is an electric field ( small) in the wire that drives the current. At any place where there is an accumulation of charges the outflowing current will be less than the inflowing current and the field will act in such a way so as to equalize the current. It so happens at the ends of the resistor. But all this happens in such a quick time that for all practical purposes the current is same in every brach of the circuit.
It is just like the electrons which have reached the end of the first resistor communicate the presence of the resistor to the electrons coming behind them and convey them that they have to come slowly because there is a resistor ahead.
Griffith explains this very efficiently. Also there a certain beautiful answers here concerning these questionS !
| {
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Do physicists use particle "energy" to refer to kinetic energy? In 1963, this paper was written about the effects of radiation on solar panels. The paper states that:
When electrons at energies greater than 145 KeV and protons at energies
greater than 98eV bombard a silicon crystal, they can displace an atom from the
crystal lattice, producing a lattice vacancy and a recoil atom which comes to rest
as an interstitial atom.
However, the resting energies of electrons and protons are far greater than this, at roughly 511 KeV and 938 MeV respectively. I concluded that the paper was referring to kinetic energy rather than total energy, and adjusted my calculations based on this conjecture.
So: Was I correct to assume that the paper referred to kinetic energy, or was it instead some other measure of the particles' energy?
More generally, is there a standard meaning for a particle's "energy" when referring to such particles moving at relativistic speeds?
| Special relativity theory gives you $(mc^2)^2 = E^2 - (pc)^2$, where $m$ is the rest mass of particle (511 keV for electrons), $p$ is the momentum and $E$ the total energy. When you have the natural unit system where $c = 1$, the equation becomes
$m^2 = E^2 - p^2$,
which might be strange looking for you.
When an electron has an energy of 145 keV, it must be the kinetic energy only, there is just no way that you can choose $p$ such that $E$ could be smaller than $mc^2$. For higher energies, it could become ambiguous. If you have $E \gg mc^2$, then the actual difference will be small because the particles are ultra-relativistic anyway.
In a particle physics context, where particles can annihilate each other, energy is usually meant as the total energy including rest mass, just like the above equation says.
| {
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Translation of Vectors I am a bit confused about translation of vectors. In the first class in physics itself we are told that we can translate vectors as we like to the desired position to do whatever that we are trying to do. For example, if someone draws two random vectors then to get the sum, we translate them, make a parallelogram and draw the diagonal as the resultant.
However I have some doubts on this. In the following example, clearly we cannot translate the vectors.
Consider this rigid body. We want to calculate the torque about origin of a force.
Now if we translate the force vector, then we would obtain the following.
Obviously the situation are very different and its not equivalent.
So are we really allowed to translate vectors?
| Yes, we are allowed to translate vectors. To put it simply, think of coordinate geometry. When you shift the origin, it doesn't effect the orientation and length of a line segment. Similarly, in vectors, translation doesn't effect the vector as a vector is defined by its magnitude and direction. You can move it to anywhere in space and it will remain the same, provided these 2 properties don't change.
In the example you gave in the problem, although force vector doesn't change and is translated, the position vector of the point of application changes . b and a are completely different vectors in terms of magnitude and direction and hence their respective vector products with force vector are different and hence, the torque is different.
To summarize, translation of vectors means that the vector doesn't change if you move it to different points in space. However, the torque on a body also depends upon the position vector from centre of mass, which changes in this case. This doesn't contradict translation of vectors, since vector product depends upon both the vectors, not only force.
Update-Point of application matters only because position vector depends upon it and hence torque.
| {
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Dirac delta function property in a scattering proof I'm studying the proof for the decoherence of the off diagonal elements of a density matrix through scattering with the environment and I'm stuck at a certain point:
My problem is A1.14 relation. (A1.13 as well to be honest, but I guess that the $(2\pi/L)^3$ is just sort of a normalization where $2\pi$ is, maybe $2\pi\hbar$ from the uncertainty relation and $L^3$ is the "volume of the scattering")
Sadly I have no idea why $\delta^2(q-q')=\delta(q-q')L$. Oh, $q$ should be a generic module of a momentum (maybe before the scattering) and $q'$ should be the module associated to the momentum of the particle after the scattering.
Since both these relations are said to be "usual replacements" I hope somebody recognise them and can help me out.
| In one dimension we have
$$(2 \pi)\delta(p) = \int_{-\infty}^\infty \mathrm{d}x~e^{ipx}$$
And so, formally,
$$ (2\pi)\delta(0) = \int_{-\infty}^\infty \mathrm{d}x$$
If we imagine that we are actually working in a large but finite space with length $L$, then we can intepret $(2 \pi) \delta(0)$ as this length $L$. The relevance of this is that when we have the square of a delta function, the first will enforce the condition in the argument of the delta function, and this will leave the second delta function as $\delta(0)$. This is technically undefined, so we need to write it as something sensible – in this case, as the length of the space we're considering.
| {
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Infinite array of capacitors and inductors You may be familiar with the surprising result one gets when calculating the equivalent of an infinite array of resistors. What if we change this circuit and replace the resistors with capacitors and inductors?
Following the notation given in the link I've provided above, let's replace $R_1$ with $C$ and $R_2$ with $L$ so that
$$R_1 \rightarrow Z_1=\frac{-j}{\omega C}$$
$$R_2 \rightarrow Z_2=j\omega L$$
where $j$ is the imaginary unit. Again using the result written in that link, we get the following equation:
$$Z_{eq}^2-\frac{j}{\omega C}Z_{eq}-\frac{L}{C}=0$$
Solving this quadratic equation we get that:
$$Z_{eq}=\frac{\frac{j}{\omega C}\pm\sqrt{\frac{-1}{\omega^2C^2}+4\frac{L}{C}}}{2}=\pm\sqrt{\frac{L}{C}-\frac{1}{(2\omega C)^2}}+j\frac{1}{2\omega C}$$
So an array of ideal capacitors and inductors lead to a complex (not imaginary) equivalent impedance if $L>\frac{1}{4C\omega^2}$. This means that, if the circuit was fed with a source, actual power would be dissipated, even though each of the individual impedances are purely reactive. How does this make sense?
| Initially I thought you had just rediscovered the Telegrapher's Equations - but then I realized you had your capacitors and inductors "the other way around" from that more usual scenario (described here)
Even though your situation is unusual, there is a way to understand what is happening. The capacitors in your network are charging up - and while some of that charging is transient, some of it "goes on forever" because of the infinite extent of the network. This charging of the capacitors means there is a mechanism for storing energy - and I think that's what your equations are telling you.
This is much easier to understand when you switch your capacitors (to be $R_2$ in your diagram) and inductors (to be $R_1$). When you do that, you end up with an expression for the impedance that will tend to $Z=\sqrt{\frac{L}{C}}$ when you make $L$ and $C$ infinitesimal while preserving their ratio (which is what happens when you consider a transmission line as being made up of many small inductors and capacitors).
When you have an ordinary transmission line, a pulse will propagate, and energy will be stored per unit length. The storing of energy is indistinguishable from dissipation of energy (until you get a reflection or some other mechanism to extract the energy again).
| {
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If magnetic fields are created by moving charges, how do magnets work? This is probably is stupid question but I think it must clear up some misconception I have. Magnets, presumably have magnetic fields. But where are the moving charges? Don't we need a current?
|
How do magnets work?
The subatomic particles electron, proton and neutron have magnetic dipole moments. This magnetic fields are intrinsic properties, means this fields exist under all circumstances. In nonmagnetic materials the sum of all magnetic dipole moments is zero.
In permanent magnets the sum over all this magnetic dipole moments is not zero. To produce strong magnets the material will be milled to powder and than pressed under the influence of a strong external magnetic field. During this production process some electrons are "frozen" with their magnetic fields, all in the same direction along the external magnetic field. Behind a certain temperature the and by the help of the thermic oscillations of the atoms the alignment of the magnetic dipole moments gets lost.
Now your question
Magnets, presumably have magnetic fields. But where are the moving charges
is answered in a way that moving charges are not involved.
| {
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Is the Moon in a "Freefall" Around the Earth?
The force of gravity keeps our Moon in orbit around Earth. Is it correct to say that the Moon is in “free fall” around Earth? Why or why not?
I think the answer is yes. The moon is falling towards the Earth due to gravity; but, it's also orbiting the Earth as fast as it's falling towards it. This balance between the 2 forces means the moon is essentially "freefalling" towards the Earth. Is my thinking correct? Thanks.
| Yes, it is in free fall towards the Earth and that is what allowed Newton to unify the terrestrial and the celestial mechanics, one of the greatest advances on Science ever. He did that by assuming that both the Moon and a falling apple (allegedly) falls toward Earths due to a gravitational force. Using his newly discovered law of universal gravitation he calculated the forces between Earth and Moon and Earth and the apple. The ratio between the accelerations obtained is
$$\frac{a_m}{a_a}=\frac{a_m}{g}=\left(\frac{R}{d}\right)^2,$$
where $a_m$ and $a_a$ are the accelerations of the Moon and apple, $R$ is the Earth's radius and $d$ is the Earth-Moon distance.
The acceleration $a_m$ was obtained by the centripetal acceleration of a circular orbit,
$$a_m=\omega^2d=\frac{4\pi^2}{T}d,$$
whre $T$ is the period of the orbit. The Earth-Moon distance as well as the Earth's radius was known at Newton's time so he was able to verify that the two expressions above agree and that was a confirmation of his theory.
But why does the Moon not strike the Earth? Instead of a kinematic approach one can use a dynamic approach to answer this. The Moon's angular momentum (with respect to the Earth) is non-vanishing and due to the fact that the force is central, it has to be constant in time. In order to hit the Earth, the Moon's angular momentum has to go to zero, but this is not possible unless there are another force (besides the Earth's pulling) causing a non zero torque on the satellite.
| {
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Sea quark parton annihilation? Consider the figure below1:
This can be read as follows (please correct me if I am wrong): two particles come in and 'fragment', a parton from each particle $C$ and $D$ annihilate to form the particle $X$. An intuitive guess is that the partons $C$ and $D$ must correspond to valence quarks whist the 'jets' of $A$ and $B$ must contain the remaining valence quarks and the usual gluon and sea quarks. My question is this: Is such a reaction possible but in which $C$ and $D$ represent sea quarks (or even gluons)? Please can you explain either way.
1Image adapted from that given on page 20 of http://www.hep.man.ac.uk/u/hanl/lecture/Lecture1_LHC+TeVatron.PDF
| Assume proton proton scattering: something has to be exchanged in order to interact. Because they are strong interacting particles they exchange at least a gluon with the out going protons absorbing the energy and breaking out into jets.
A gluon leaves from a quarq of one of the protons and gives energy to one of the quarks of the other resulting into three jets.
For your diagram, a triple gluon vertex is allowed, as for example:
where your C and D are the two exchange gluons going into a third one. The event will have at least three jets, the two from the "spectator" protons and the third from the gluon process I (the diagram is written for hypothetical new particles searched for).
Your C and D cannot be a quark and antiquark unless you draw the corresponding anti to them and do something with them. There are quantum numbers that have to be kept track of, particularly baryon number . Two quarks cannot annihilate, they can interact.
| {
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Experimentally measure velocity/momentum of a particle in quantum mechanics In the context of quantum mechanics one cannot measure the velocity of a particle by measuring its position at two quick instants of time and dividing by the time interval. That is,
$$ v = \frac{x_2 - x_1}{t_2 - t_1} $$ does not hold as just after the first measurement the wavefunction of the particle "collapses".
So, experimentally how exactly do we measure the veolcity (or say momentum) of a particle?
One way that occurs to me is to measure the particle's de Broglie wavelength $\lambda$ and use $$p = \frac{h}{\lambda}$$ and $$v = \frac{p}{m}$$
to determine the particle's velocity. Is this the way it is done? Is there any other way?
| A particle's velocity can be measured just as you've described. If you shoot the particles through apertures that are much larger than their wavelength, the wavelike effects are minimal and they continue with basically the same momentum. If you squeeze the target aperture, though, particles that pass through continue with the same speed, but in different directions. This is what the position/momentum uncertainty means - as the aperture narrows, we know more and more about where that particle is when it goes through. Therefore, we will know less and less about which way it goes afterwards.
There are many ways to measure the energy of a particle. You're right to recognize the relationship between the wavelength and the momentum, but these values are related algebraically. The energy and momentum are related (for massive and mass-less particles) by $E^2 = (p c)^2 + (m c^2)^2$. While we can measure the speed of photons in a vacuum, that is a defined unit, so we're really measuring the length of a meter when we perform that experiment.
| {
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How does the temperature of an ideal gas exhausting into vacuum vary? Since a gas at a certain pressure exhausting into vacuum has no atmospheric pressure to push against, there shouldn't be any adiabatic cooling taking place.
But looking at the energy conservation:
$TdS=dU+VdP+PdV$
$TdS = 0$ {Adiabatic process}
$PdV = 0$ {No change in volume of container}
Hence: $dU=-VdP$, leading to a lower temperature?
| TdS is not zero for all adiabatic processes. It is only zero for all adiabatic reversible processes, and the process you describe is not reversible. If you have an adiabatic chamber with gas on one side of a barrier and vacuum on the other side and you suddenly remove the barrier, the gas does not do any work on the walls of the chamber and no heat comes through the walls, so the change in internal energy of the chamber contents (the gas) is zero, and its temperature remains constant. But the pressure of the gas has decreased, so its entropy has increased.
The process described by @Jan Lalinsky would occur if, rather than removing the barrier suddenly, a small pinhole were made in the barrier so that the gas escaped into the vacuum compartment very slowly. In this case, the gas in the original pressurized chamber will experience an adiabatic reversible expansion, and its temperature will drop. However, if the barrier between the chambers were adiabatic, the gas that escaped into the chamber that was originally under vacuum would actually have a temperature higher than the original gas temperature when the pressures equilibrate, since the overall change in internal energy for the combination of the two chambers would be zero. On the other hand, if the barrier were not adiabatic, and the system were allowed to re-equilibrate (both thermally and mechanically), the final state would be the same as if the barrier had been removed suddenly, and the final temperature and internal energy would match the initial temperature and internal energy of the gas before the barrier was removed or the pinhole was applied.
| {
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Why a 3D Universe? The universe as it stands is perceived by us to be 3 dimensional. Why is it that the universe formed into 3D space and not 4D or 5D space for example?
| Assuming you're asking for a "first principles" explanation, there currently is none. That the world around us has three spatial dimensions is an obvious experimental fact, a raw input to our theories, not a derived statement.
Neither general relativity nor quantum field theory, which are our currently most fundamental theories, nor speculative theroies like string theory make four dimensions/three spatial dimensions really special or unique - the first two can be thought of in any dimension, and string theory naturally lives in 10 or 26 dimensions and must be "trimmed down" to the four dimensions we live in, which is possible, but not somehow implied as necessary by the theory itself.
| {
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Bell's theorem and how it solves the EPR paradox Could someone explain to me how Bell's theorem solves the EPR paradox and 'spooky action at a distance'?
From what I understand, when measuring a state, say spin up in the x direction, the wave function collapses and the other particle must be spin down in the x direction instantaneously no matter how far it is from its entangled partner.
I can't seem to put all the pieces together with local realism and local hidden variables.
| Bells inequality is derived by assuming 'local reality'. This inequality has been shown to be violated by experiment and by theory (quantum mechanics). So the conclusion is: nature doesnt always posess local reality. Assuming local reality made EPR a paradox, so without this it is resolved.
| {
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Current as the time derivative of the charge I have been told that the current $i$ can be defined as
$ i = \displaystyle\frac{dq}{dt} $,
where $q$ is the charge and $t$ is the time.
I do not understand this definition because, if the charges are moving so that the net charge remains constant in an infinitesimally thin cross-section of a wire, $q$ is constant with time and hence $dq/dt = 0$. That result would mean that no constant current can exist unless the charge change has a linear dependence with time (i.e. $q = q(t) \propto t$).
As I assume my reasoning is wrong, where is my mistake?
Thank you.
| Electric current can be understood through the analogy of water flow.
Just as the 'water current' in a river is the amount of water which passes a point in a given time interval, similarly, the electric current can be understood as the amount of electric charge passing through an area in a given time interval.
Of course, the analogy is not entirely correct, since charges can be either positive or negative, but the analogy is useful if we 'define' the electric current in terms of the 'flow' of positive charges
So, for example, considering the electric current through a wire of cross section $A$, the amount of charge ${\Delta q}$ passing through this surface in a time interval ${\Delta t}$ allows us to 'define' the average electric current as:
$I_{avg}=\frac{\Delta q}{\Delta t}$
If amount of charge passing through the area every second remains constant, then $\Delta q$ is constant so the current is constant.
If the charge flow varies with time, then we can define the instantaneous current by taking the limit as $\Delta t\rightarrow0$ to get:
$i=\frac{dq}{dt}$
The concept of the 'instantaneous' current through a circuit is further complicated when we consider the current through a capacitor element, which requires the introduction of a displacement current to account for changes in electric displacement, the full description of which is given by Maxwell's equations.
| {
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Condensation of Water. Classroom Controversy In our test there was a question that went like so:
Question 4 You have a glass of iced water on an unshaded picnic table and went for a walk for 30 minutes. When you return you noticed
the glass has water on the outside of it.
a. In terms of heat transfer explain what has happened to the glass of water.
The majority of the class understood this question and answered it correctly.
b. Would there have been more or less water on the outside of the glass if the picnic table was in the shade? Explain.
This question caused a lot of controversy with the majority of the students (including myself) believing that the shade would have caused more water on the outside. Whereas the teacher and a few students thought that the sun would have caused more water on the outside.
The reasoning that the teacher provided was not very convincing and so we have come to this forum to ask what is the correct answer to part b and most of all WHY?
We are 16 -17 years of age if you need to know the level for the explanation.
| Consider the possibility that there is no "correct" answer.
Maybe the purpose of the question was just to get you to think about and discuss a physically interesting situation.
Condensation will depend on at least the following factors
*
*Air humidity
*Air temperature
*Air circulation
*Water temperature
It is possible for direct sun light to affect all of these factors and the effects on condensation will work in different directions. Which effect dominates may depend on several factors not specified in the question. Even if there is a "correct" answer in most situations, it may not be possible to confidently conclude this based on a simple physical argument.
| {
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