Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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What happens if we rotate one gear and this rotates another and this to another and so on. Will the last gear rotate faster than light? what happens if we rotate a small gear attached to a larger one and the large one rotates another small one attached to a large one and the large one rotates another small attached to a large one and so on
Will the last gear rotate faster than light?
| No. Rigidity in the sense that is frequently assumed in Newtonian mechanics breaks down in special relativity. A torque applied to a gear at one point doesn't immediately start turning each point in the gear at the same angular velocity. The "information" propagates in the form of small deformations within the gear, roughly at the speed of sound, which is less than the speed of light.
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What is the difference between a force and a net force? I read in Newton's first law, it states that an object will continue to have a constant velocity unless acted upon by a force whilst for other articles, it states "unless acted upon by a net force." Which one is correct? Are they both interchangeable? Is there any difference between these two concepts?
|
[...] whilst for other articles, it states "unless acted upon by a net
force."
The latter is really the correct expression, the former really being a 'lazyism'.
Consider the following body, subject to three forces as an example:
Newton's First Law refers to the net force $\mathbf{F}$, calculated as the vector sum:
$$\mathbf{F}=\displaystyle\sum_{i=1}^n\mathbf{F_i}=\mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3}$$
The individual forces, in that sense, don't matter, only the so-called resultant force matters.
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How does the Brownian motion of air molecules compare to the threshold of human hearing as a function of frequency? This fantastic question essentially asks what is the noise floor of air? Both the answer given on that thread and the value stated by Microsoft are around -23 or -24 dBSPL.
However, overall loudness is only one metric. What does the amplitude of the noise in dBSPL look like when graphed out as a function of frequency in the audible range? How does the shape and level of the curve defined by that graph compare to the threshold of human hearing as described by the equal loudness contour?
| My understanding is that we are talking here about the molecules tapping on the eardrum, i.e. the shot noise - similar to the noise of the rain drops. It is a white noise, i.e. its spectrum has the same amplitude at all frequencies.
On a deeper level however there are at least two characteristic timescales that would limit the width of the spectrum:
*
*the collision time between the molecule and the wall
*the intermolecular scattering time, which characterizes the density fluctuations of the air
it is also important to note that the air consists of several types of molecules (nitrogen, oxygen, carbon dioxide and some others), each of which is characterized by its own scattering times.
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Difference between vibration and oscillation considering a point in space Most answers and articles I've read so far try to give real world examples such as a spring or a pendulum. However, I'm trying to understand the core difference between the two terms in the most fundamental context.
Do we need the term vibration if we can treat a point in space as if it oscillates but in much shorter distances? (Here the phrase "much shorter" sounds too vague which should be unacceptable in physics.) Or does the term vibration imply that the concept of time and a relative range restriction (to a particular movement we're dealing with) together is involved with the movement?
It definitely seems possible to pass my undergraduate exams without going into this much detail, but I feel like there's a gap in my understanding and I would be very glad if you could give me a more complete picture.
Thank you.
| Lets look at the words:
Vibration is a mechanical phenomenon whereby oscillations occur about an equilibrium point. The word comes from Latin vibrationem ("shaking, brandishing"). The oscillations may be periodic, such as the motion of a pendulum—or random, such as the movement of a tire on a gravel road.
and
Oscillation is the repetitive variation, typically in time, of some measure about a central value (often a point of equilibrium) or between two or more different states. The term vibration is precisely used to describe mechanical oscillation. Familiar examples of oscillation include a swinging pendulum and alternating current.
So vibration is an oscilation in a specific medium, a mechanical phenomenon, an example of an oscillation.
| {
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Have we probed beyond Extremely Low Frequency (ELF) light to see is there is anything out in the universe emitting it? I could only find some articles talking about radios used to communicate with submarines and a few now-defunct transmission towers for ELF. I would think this would be something that was studied to see if anything produced such frequencies in the universe but I couldn't find anything.
Do we see emissions in the electromagnetic spectrum just drop off at a certain point and that is why the research isn't going on?
| Extremely low frequencies (anything below kHz) cannot travel through interplanetary space, because the
plasma frequency is too high.
The plasma frequency is given by
$$\nu_p = \left( \frac{e^2 n_e}{4\pi^2 \epsilon_0 m_e}\right)^{1/2} = 9000 \left(\frac{n_e}{{\rm cm}^{-3}}\right)^{1/2}\ {\rm Hz},$$
where $n_e$ is the electron number density and $m_e$ is the electron mass. If waves have a frequency below the plasma frequency then they will be reflected.
The value of $n_e$ varies from place to place. In the Earth's ionosphere, typical values might be $n_e \sim 10^{5}$ to $10^{6}$ cm$^{-3}$. This means that waves with $\nu <$ a few MHz from outer space will not penetrate the Earth's ionosphere.
Even if we could put a radio telescope in space (or on the Moon?), there are sufficient electrons in the interplanetary medium near the Earth ($n_e \sim 10$ cm$^{-3}$) that waves with $\nu < 30$ kHz will not reach it from outer space.
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Is a capacitor a dipole? A few more questions about understanding dipoles I recently learned about dipoles, according to its definition I was wondering if a capacitor can be considered also as a dipole?
Also I was wondering what is the physical meaning of the dipole moment $\vec{p}=qd$?
And my last question is what is the motivation of studying dipoles? What is so special about it? From what I have learned it is just 2 equal and opposite charges that are a distance $d$ apart and they create an electric field according to what expected.
| I will answer your first question. A capacitor is not necessarily a dipole. For example, a conducting wire surrounded by a conductor separated by an insulator is a capacitor but not a dipole.
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Does the Ryu-Takayanagi conjecture only apply to vacua? Does the Ryu-Takayanagi conjecture only apply to vacua, or does it also apply to arbitrary excited states?
For excited CFT states, the entanglement entropy can be proportional to the volume, not surface area, or it can even be arbitrarily large.
| The Ryu-Takayanagi (RT) formula certainly applies to more general backgrounds, but it doesn't hold for arbitrary excited states. Remember that generic high energy excited states in the CFT will correspond to a bulk which isn’t at all semiclassical and will look very quantum, i.e. no nice bulk geometry. But of the CFT states that correspond to semiclassical bulk geometries (i.e. quantum fields with classical gravity), the RT formula will apply.
Remember that RT says that the entanglement entropy of a boundary region is equal to the area of a minimal surface in the bulk. But if the bulk is an AdS black hole, that minimal surface can pick up an extensive contribution from the black hole horizon. The larger the black hole in the bulk, the larger this extensive contribution will be. You are correct that entanglement entropy can be proportional to the volume (e.g. take a random state in your Hilbert space), but these will be the states that are very quantum in the bulk.
Regarding RT in specific bulk geometries: The original paper by Ryu and Takayanagi discusses some more general backgrounds other than the AdS vacuum. Also see this review. More general (time-dependent) backgrounds are discussed in the covariant generalization of RT. Also, different types of corrections to the RT formula are known (both quantum corrections and corrections from higher derivative terms), which I think are explained in Sec 5 of this review.
Also, the RT formula isn’t really a conjecture (at least no further than holography is a conjecture) as it has been derived from the bulk gravitational path integral in this paper.
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${\rm 2D}$ isotropic oscillator: Is ${\rm SO(4)}$ a subgroup of ${\rm Sp}(4,{\rm R})$? Consider the ${\rm 2D}$ isotropic oscillator. The hamiltonian is $$H=\frac{1}{2}(p_x^2+p_y^2+x^2+y^2)$$ and the phase space is $4$ dimensional. In this case, the set of all linear canonical transformations that preserve the form of Hamilton's equations form a group ${\rm Sp}(4,{\rm R})$. On the other hand, the group of orthogonal transformations ${\rm SO(4)}$, in the phase space, leaves the hamiltonian $H$ invariant. Now, there can be a subset of canonical transformations that leaves the Hmiltnian invariant in addition to preserving the form of Hamilton's equations.
*
*Will it be correct to assert that ${\rm SO(4)}$ is a subgroup of ${\rm Sp}(4,{\rm R})$?
| No, $SO(2n,\mathbb{F})\subseteq Sp(2n,\mathbb{F})$ [understood via their standard embedding into $GL(2n,\mathbb{F})$] is only true for $n=1$. This fact can be proved by considering the corresponding Lie algebras.
Specifically, $Sp(4,\mathbb{R})$ is (the double cover of) the restricted anti de Sitter group $SO^+(3,2;\mathbb{R})$, cf. e.g. my Math.SE answer here.
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Is there a smooth transition from inverse quadratic gravity to linear gravity? I can't remember exactly what it was, but I remember going through a problem in physics related to gravity on and inside a sphere, and found that inside, gravity acts linearly as a result of some triple integral cancellation with an assumption on uniform density.
Suppose Earth itself was a perfect sphere and you could pass through it. Does gravity actually spontaneously transition from inverse quadratically to linearly? Would it really be some magic spontaneous switch in forces? Or is there something in the math that can explain a gradual transition from a quadratic factor to a linear factor?
| For a spherical body of mass $M$ and radius $R$, assuming the density is constant you can calculate the gravitational field strength at any distance $r$ from the centre of the body using Gauss's law. If you work through the algebra (which is a good exercise) you'll eventually find
$$g = \begin{cases}
-\frac{MGr}{R^3}, &\qquad r \le R, \\
-\frac{GM}{r^2}, &\qquad r > R.
\end{cases}$$
If you look at both expressions for $r = R$ you'll see that they agree. This means that the gravitational potential is continuous at $r = R$. If you take the first derivative with respect to $r$ of both expressions, however, you'll find that they are not the same. This means that the potential is not smooth in the mathematical sense of a smooth function.
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Finding direction of field when using Gauss law I have seen this stack question Applying Gauss' Law to find Electric Field but I got confused when I saw the comment and the answer because, clearly the flux is a scalar quantity (due to dot product). So after using Gauss law to find magnitude, how do I find the direction in which the field vector points?
| If you apply Gauss' law to obtain the electric field, the direction of the electric field should be clear before applying Gauss' law, otherwise you will not be able to do the product $\vec{E}\cdot \vec{dS}$ and get the magnitude out of the flux integral. To determine the direction use symmetry arguments; try placing the vector in an arbitrary direction and think: is this direction privileged in any way? Does it make any sense at all that the electric field has a component in this direction given the symmetry of the charge distribution?
| {
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Time evolution of the standard deviation of an operator How would I find the time evolution of the standard deviation of an operator? For example, how might I find the time evolution $\sigma_x (t)$ of the standard deviation
$\sigma_x = \sqrt{ \langle \hat{x}^2 \rangle - \langle \hat{x} \rangle^2}$
of the position operator $\hat{x}$ given a state $| 0 \rangle$ representing a particle in the ground state of a harmonic oscillator?
Can I multiply the initial value
$\sigma_x (0) = \sqrt{ \dfrac{\hbar}{2 m \omega} }$
by the generator of time translation
$\hat{U} = \large e^{\frac{-i \hat{H} t}{\hbar}}$? I've also tried switching to the Heisenberg picture and applying the Heisenberg equations of motion but have been unable to reach a conclusion. Any help would be greatly appreciated, thanks.
| Adding to @Vadim answer's: note that when a system is in an eigenstate of the Hamiltonian, it is stationary. If an operator does not have an explicit time dependence, its expectation value will be constant. This is true regardless of the details of the Hamiltonian or the operator.
To see this, you can note that if the state is an eigenstate $H|\psi\rangle = E |\psi\rangle$ then $|\psi(t)\rangle = \exp(-iEt/\hbar) |\psi\rangle$, meaning that for any observable
$$ \langle \hat{O}(t) \rangle = \langle \psi(t) | \hat{O} | \psi(t) \rangle = \langle \psi | \hat{O} | \psi \rangle$$
independent of time.
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Pseudo Force and Inertial and Non-Inertial frames In the figure given below is block placed on an incline $\theta$. Now the lift is accelerating upwards with an acceleration $a_0$. Now if we make our measurements from the lift frame we will have to apply a pseudo force $-ma_0$. Which will have two components one in the direction of $Mg\cos\theta$. And other in the direction of $Mg\sin\theta$. Now $Mg\sin\theta+Ma_0\sin\theta=Ma_\text{net}$. Where $a_\text{net}$ is the net acceleration in that direction.
Now let's observe it from the ground or an inertial frame here the object has a net upward acceleration which has a component opposite to $Mg\sin\theta$. Therefore $Mg\sin\theta=- Ma_0\sin\theta$. Now what I thought was that this is not possible and hence there is another force acting opposite to $Mg\sin\theta$, $Ma_\text{net}$. Now this doesn't make any sense to me if there is a force acting opposite to $Mg\sin\theta$, and the net is also in that direction, then won't the object move upwards on the incline. Now that doesn't make any sense. Can someone tell me what Is happening and from where is this $a_\text{net}$ coming from when observing in the inertial frame?
| To avoid confusion caused by gravity, we will assume the lab frame is in an inertial frame, floating in space far from Earth. $F = ma$ works in this frame. In these coordinates, there is no net force on an object that stays at $x = 0$.
Working in the lab frame, you see the lift accelerated upward. Given no friction, the inclined plane exerts a normal forces on the block. This has an upward component and a leftward component. The leftward component makes the block slide down the plane as the upward component lifts it. The block accelerates upward as it slides along the plane, but not as fast as the lift.
To repeat the exercise in the lift frame, you have to pretend the lift is not accelerating. You choose a frame of reference where $x^{'} = 0$ is attached to the lift. It stays still, while the point $x = 0$ accelerates downward.
But now you are working in a frame where $F^{'} = ma^{'}$ gives the wrong answer. When $F^{'} = 0$, you see the block accelerating downward to keep up with the lab frame's $x = 0$. To make $F^{'} = ma^{'}$ work, you need to pretend that there is a downward force to explain the pretended downward acceleration.
In the lift frame, the downward force presses the block into the frictionless inclined plane. The plane presses back with a normal reaction force that has an upward and leftward component. The leftward component makes the block slide down the plane as the sum of the upward component and pretend forces accelerates it downward. The block accelerates downward as it slides along the plane.
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Has the ballistic motion of an electron in gravitational field ever been measured? Reading this question I thought of an argument that an electron's trajectory would bend in the gravitational field despite the electron's being incapable of strong interaction; this would then disprove the conjecture stated in that question. But then I couldn't find any references to actual experiments that have been done to measure this ballistic motion of an electron in gravitational field.
Has such motion actually been measured?
| After searching online for a while, I found this paper, entitled "EXPERIMENTAL COMPARISON OF THE GRAVITATIONAL FORCE ON FREELY FALLING ELECTRONS AND METALLIC ELECTRONS". The paper refers to two earlier papers:
1.F.C.Witteborn, L.V.Knight,and W.M. Fairbank, in Proceedings of the Ninth International Conference on Low Temperature Physics, edited by J.G.Daunt, D.0. Edwards, F.J. Milford, and M. Yaqub {Plenum Press, New York, 1965), p.1248.
*F.C.Witteborn, thesis, Stanford University, 1965 (unpublished).
There are a bunch of related papers here.
Bottom line: it appears that measurements have been done.
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The Von Neumann interpretation of the Double Slit Experiment First, I am a philosopher not a physicist, so apologies for limits in my understanding of quantum physics. My interest is in the philosophical implications of physics. I've just joined looking for physicists I can ask questions to, but my first question hasn't worked ... so I am going to rework it and ask it in an extremely simple way.
In the double slit experiment, if the experiment is repeated with the device measuring ‘which-way’ information switched on or off, but without the experimenter knowing when it it on or off, can we tell from the pattern on the screen when it was on and when it was off?
Thank you.
| "Consciousness causes Collapse" is based on the idea that the measuring device should be described by a waveform as well as the experiment. Before measurement is done the system under examination is in a mixed state, say
|up> + |down>
Once the measurement is done, the combined system + measuring device is in the state
|up>|measuring device says 'up> + |down>|measuring device says 'down'>
I.e. the final waveform is also in a mixed state, and the waveform never collapses. Von Neumann conjectured that, since we only see one outcome, it therefore does collapse at some point and perhaps it is the interaction with consciousness that causes the collapse.
CCC is only one of many possible solutions to the so-called measurement problem - it cannot be proved or disproved - it is matter of interpretation.
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Why don't the nuclear fusion processes inside the sun produce electron antineutrinos $(\bar\nu_e)$? Why don't the nuclear fusion processes inside the sun produce $\bar\nu_e$ despite having the same mass as $\nu_e$? Is the reason as simple as "there is no production channel for $\bar\nu_e$s." ?
| You must convert proton to neutron somehow (pp chain or CNO chain) and this means by charge conservation that a positron must be emitted, and by lepton number conservation to balance lepton number that anti-electron is accompanied by neutrino and not anti-nu.
| {
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Why does water keep dripping out of the bucket even after the faucet is turned off? So I just noticed that when I filled my bucket with water until it overflowed and then I turned the faucet off, the water kept dripping for like 20 seconds. Why does this happens? Shouldn't it have stopped dripping sooner after I turned the faucet off?
| If you are talking about the water overflowing from the bucket, it is because water has a surface tension. This lets the surface, or skin of the water rise slightly above the rim of the bucket before the gravity pulls the water enough to break the surface tension and start it flowing over the edge. Once water stops flowing into the bucket it will still flow out for a short while until it's surface tension equals the gravity pulling it over the edge.
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Why is internal resistance of battery considered outside the terminals although it is present between the terminals inside the battery In ideal battery the internal resistance is zero whereas in non-ideal battery there is some internal resistance now this internal resistance is due to the battery material (electrolyte) and is present inside the battery between the terminals then why do we represent and eventually do calculations by considering that internal resistance to be connected with battery terminals externally. I’m totally unable to get the point. Please help
| It's just 'empirical' (a rule which is good enough to predict observations, but is not necessarily related to the actual cause). Observations of the current driven by a battery and its voltage show a roughly linear relationship which can't be explained by the external resistance in the circuit alone, but are consistent with an 'ideal' battery plus an additional resistance. Although the chemical processes which cause this (now labelled) 'internal resistance' are entirely different from those causing resistance in the external circuit, a linear approximation works well well enough and explains what is seen. Once we decide to use this as an assumption, then we can use 'ohms law' and draw the circuits as an 'ideal battery' plus an 'internal resistance' and get useable results. But it is simply a convention which gets good enough answers, and shouldn't be confused with 'where the resistance sits' in reality.
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Deflection of an electron from a charged wire and from an interference grid: In both cases the electron radiates? Near a charged wire a moving electron is deflected. Since the deflection is an acceleration, the electrons emit photons.
What about an electron's radiation, which is affected by an interference grating?
|
Since the deflection is an acceleration, the electrons emit photons.
Not in quantum mechanics/quantum field teory, emission is not obligatory, as it is in classical electrodynamics, when there is a dp/dt. It is an interaction between a particle and another particle, as is the case in diffraction, the other particle being the atoms at the grating, and this exchanges a virtual photon and the momentum is transferred to the solid lattice. In the case of the electron with the wire current the dp/dt will be taken by the wire lattice.
The case of bremsstrahlung has extra electromagnetic vertices. The electron experiments with gratings are at very low energies, and any such radiation is negligible.
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Commuting the time evolution operator Given the time evolution operator $U(t, t_0)$, I don't understand why it is true that for a time-independent operator Q,
$$[Q, U(t, t_{0})] = 0 \Leftrightarrow [Q, H(t)] = 0 $$
where H is the Hamiltonian.
| You can derive the Heisenberg equation from the Schrödinger equation.
$$\frac{dQ_H}{dt}= \frac{d(U^\dagger Q_S U)}{dt}$$
Then after you write it complete and define $H_H=U^\dagger H_S U$ yo see the following:
$$\frac{dQ_H}{dt}=-i[Q_H,H_H]$$
And finally you can say $U$ and $H$ always commute for time dependent $H$ and at the end you can write your result.
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Sanity check: can we define "arbitrary" hamiltonians? I just want to do a sanity check on my understanding of Hamiltonian mechanics:
My understanding is: For any number $n$, take the phase space $\mathbb R^{2n}$, and take any arbitrary differentiable function $H:\mathbb R^{2n}\to \mathbb R$ to be the Hamiltonian. Then all of the standard results about Hamiltonian mechanics will apply to the system generated by $\dot q=H_p, \dot p=-H_q$ (In particular Liouville's theorem applies, and Noether's theorem applies to the Lagrangian obtained from the Legendre transform). There are no further regularity conditions needed to do Hamiltonian mechanics on this system.
Is this correct?
| *
*We only need differentiability of the Hamiltonian $H(q,p,t)$ as long as we stay in the Hamiltonian formulation. In particular, Noether's theorem works with the Hamiltonian action, cf. this Phys.SE post.
*However to guarantee the existence of a regular Lagrangian formulation in $n$ variables $(q^1,\ldots,q^n)$ (via a Legendre transformation) we need to impose that the Hessian $\frac{\partial^2 H}{\partial p_i\partial p_j}$ has maximal rank, i.e. is invertible.
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Energy of a rotating wheel Why does a rotating wheel have kinetic energy $K=\frac{1}{2}mv^2$ associated with its movement?
I will clarify.
Let's say I have a rotating wheel in an empty void which has an angular velocity $\omega$, then its total energy will be $E=\frac{1}{2} I \omega^2$, as the wheel is not moving. Now suppose I put this wheel on a surface. The wheel starts moving and, from what I understand, the total energy associated to the wheel is $E=\frac{1}{2} m v^2+ \frac{1}{2} I\omega^2$. How is that possible since energy cannot be created nor destroyed? The wheel is the same, so how can it have two different amounts of energy?
| You are correct that before contact with the surface the wheel has $KE=\frac{1}{2}I\omega^2$ and after contact with the surface the wheel has $KE=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2$. Since energy is conserved and since $m$ and $I$ are unchanged that implies that $\omega$ must decrease in order to compensate for the linear KE.
As the spinning wheel contacts the stationary surface there will be a frictional force between the surface and the wheel which will act to oppose the slipping. In this case the direction opposing slipping is forward. The forward force will provide both a linear force to increase linear momentum and a rotational torque to decrease the rotational motion. This leads to a decrease in $\omega$ and an increase in $v$
Ideally the decrease in rotational KE will be exactly equal to the increase in linear KE. However, in practice there will be some skidding at the beginning and therefore some dissipation to heat.
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Why is $\Delta x$ or $\Delta p$ constant for a particular $\psi_n$? We were asked to calculate $\Delta x \Delta p$ for the $\psi_0,\psi_1$ of the harmonic oscillator.And so we calculated the answers and verified that $$\langle T \rangle +\langle V\rangle = (n+1/2)\hbar\omega\tag{1}$$ and that indeed they do follow the uncertainty limit.
But why can't we chose what to measure more precisely, position or momentum? Why is it that if we ever try to measure a particle resembling a harmonic oscillator at very low temperatures (at ground state) $\Delta x$ has to be $\sqrt{\hbar/2m\omega}$ and $\Delta p$ has to be $\sqrt{\hbar m \omega/2}$?.
Shouldn't it depend on the observer's wish, what he/she chooses to measure more precisely keeping in mind that the uncertainties follow HU?
Harmonic Oscillator is just an example.
| Quantum mechanical measurements are carried not on a single object, but an ensemble of objects prepared in the same state. Measurement on a single object produces a specific value, e.g., of the position - $x_i$. Measurements on the ensemble of $N$ objects produce
$$\{x_i | i=1...N\},$$
which allow calculating sample mean and variance:
$$\bar{x} = \frac{1}{N}\sum_{i=1}^Nx_i,\\
(\Delta x)^2 = \frac{1}{N-1}\sum_{i=1}^N(x_i - \bar{x})^2.
$$
For very big $N$ these should ultimately converge to their values estimated from the probability distribution $w(x) = |\psi(x)|^2$:
$$\langle x\rangle = \int dx xw(x),\\
\sigma_x^2 = \langle (x-\langle x\rangle)^2\rangle - \int dx (x-\langle x\rangle)^2w(x).$$
This is really about probability, statistics, and measurement theory, not necessarily in the context of quantum mechanics.
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What is the name of the shape of the iron core in a transformer? I'm researching on transformers and curious to know about what this shape is called (the actual core block, ignore the wires around it).
| That can be called a core-type transformer (with a single window) in contrast to the shell-type transformer. Note also that those types of cores are also constructed with laminated materials to reduce eddy current losses.
The linked definitions from the Electrotechnical Vocabulary of the International Electrotechnical Commission (IEC) give the above terms as deprecated, but they are well established nonetheless.
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In the generation of X-Rays, why the incoming electron generated from anode knocks out the K shell electron rather than outer shell electrons? When the high energy beam of particles or photon hits the cathode, electrons from $K$ shell are knocked in the generation of characteristic x-rays. Why do inner electrons get knocked out?
| There is a probability that any of the electrons are knocked out.
However only the most energetic photons resulting from electron transitions would be classed as being in the X-ray part of the electromagnetic spectrum.
If the charge on the nucleus is low, eg for a hydrogen atom, the transitions to the inner most energy levels result in the emission of UV photons.
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Why does the string of a simple pendulum need to be perfectly flexible? The book says:
"The string should be perfectly flexible, if we like to neglect the effects of different velocities of the different parts of the string during the oscillation."
Can anyone explain what actually is going on over here? I mean how does the different velocities of different parts of the string affect the system?
| Imagine that the string was not perfectly flexible but instead was a length of steel wire, rigidly fixed at its upper end. Then when the weight on its other end tries to swing back and forth, the springiness of the wire applies an extra restoring force in addition to that generated by gravity and the resonant frequency of the pendulum will shift to a higher value.
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Does "adding a constant to the potential energy" imply "adding a total time derivative to the Lagrangian." I am trying to understand why Taylor says in his Classical Mechanics text, "we can always subtract a constant from the potential without effecting any physics."
I assume "doesn't effect any physics" means the equations of motion are unaltered — as is the result of adding a total time derivative to a Lagrangian.
How are the two statements related? Explicitly please.
| You can add $\dot q=\frac{dq}{dt}$ to the Lagrangian without harm, and this has nothing to do with a potential.
Since the EOM are obtained from derivatives of the Lagrangian, adding any constant (to the potential energy or otherwise) will have no effect.
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To what precision is the heat capacity an extensive quantity We know that heat capacity is an extensive quantity, basically meaning for double the amount of substance you need double the energy to increase temperature. To what extend is this actually true, like:
*
*Are there e.g. (measurable) surface effects?
*Have (precision) experiments been done on this?
| Experimentally, I think it is easier to do the opposite: cooling bits of heated material in water inside a insulated container and measure the temperature change for example.
When trying to increase the mass of material heated in a furnace to the same temperature, and measure the associated increment of energy (say gas or electric), the losses through the furnace walls may change, complicating the computation.
Surface effects act on the kinetics of heating / cooling, not in the temperature change, and plays no role in the heat capacity, as I understand.
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Conserved quantity for any motion in 1D Newton's second law for a nonrelativistic particle of mass $m$ in 1D, reads, $$F \bigg(x, \frac{\mathrm{d}x}{\mathrm{d} t}, t \bigg)=m \frac{\mathrm{d}^2 x}{\mathrm{d} t^2}$$, where $F$ is the net force function. Now, if we assume that $F$ is time-independent and that $v=\frac{\mathrm{d} x}{\mathrm{d} t}$, can be written as a function of position $x$, $v=v(x)$, and using the chain rule we find the equation $$mv\frac{\mathrm{d} v}{\mathrm{d} x}=F(x,v).$$
After some manipulation, we arrive at the differential form $mv\mathrm{d}v-F(x,v)\mathrm{d}x=0$. In order for this differential form to be exact, it needs to satisfy the equation $$\frac{\partial}{\partial x} (mv)=-\frac{\partial F}{\partial v}=0$$, which shows that this differential can only be exact iff $F$ does not depend on velocity. In order to include forces that depend on velocity, one needs to multiply the differential from by an integrating factor $\lambda=\lambda(x,v)$, to obtain $\lambda mv\mathrm{d}v-\lambda F(x,v)\mathrm{d}x=0$. Let $A$ be the general conserved quantity for 1D motion. Then $A$ must satisfy $$\frac{\partial A}{\partial v}=\lambda mv, \frac{\partial A}{\partial x}=-\lambda F(x,v).$$
Now, here comes my question. How can one obtain an explicit formula for $A$ in terms of integrals of $F$ and $mv$. What will be in general the physical interpretation of $A$ and the integrating factor $\lambda$?
| It is all essentially about the energy conservation. When forces do not depend on velocity, they are conservative, and characterized by the potential energy, so the total energy is conserved. When the forces do depend on the velocity, these may be either forces like Lorentz force that do not change the total energy or forces like viscous friction, when the energy is not conserved and conserved quantity does not exist (there is no proof in the question of the existence of a non-trivial integrating factor).
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Locally flatness in general relativity My professor made following statement:
The spacetime of GR is curved in the presence of strong gravitational fields. The effects of curvature
manifest themselves at large distances. Locally, one can choose a flat Minkowskian metric.
I dont get it:
I thought, gravitation is expressed by curvature. If I sit on my chair, this will be due to gravitation. But this is not an effect shown on large distances (as e.g earth and sun).
| Sitting in your chair, you are not in an inertial reference frame. The equivalence principle states that a local accelerating frame is not distinguishable from the effect of gravity. This has no bearing on curvature. If you are sitting in your chair, you have chosen an accelerating frame. You have not chosen a Minkowski metric (even though your accelerating frame is flat). If you want to chose a Minkowski metric, you must chose a frame in free fall, not in a chair.
Your professor is not talking about local frames. Curvature is not seen in local frames (just as the curvature of the surface of the Earth is not seen in a town map). Curvature is seen in the gravity of the Earth because an inertial frame (one in free fall, not sitting in a chair) on one side of the Earth, does not remain in uniform motion or at rest with respect to an inertial frame on the other side of the Earth.
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Two-particle system wave function I have a similar question as Two particles system
That is:
why for two-particle without interaction will have wave function $\psi(x_1,x_2)=\psi_a(x_1)\psi_b(x_2)$
And when we exchange it will have the form
$\psi(x_2,x_1)=\pm\psi(x_1,x_2)$.
and expression
$\psi(x_1,x_2)=A[\psi_a(x_1)\psi_b(x_2)\pm\psi_a(x_2)\psi_b(x_1)]$
I was a bit confused by the first answer in the post above,why the last phase is irrelevant, so you get just the product of individual wavefunctions in $\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$ . Since the point here is $\phi$ is not a constant,it depend on position $(x_1,x_2)$,even if it's constant,why we can ignore it?
And I get lost by the solution to the second question provided on the post,that is why $\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)$ implies $\Psi(x_2,x_1)=e^{i\phi}\Psi(x_1,x_2)$ ,since $e^{i\phi(x_1,x_2)}$ is function of ordered pair $(x_1,x_2)$,when we exchange $(x_1,x_2) \to (x_2,x_1)$why it has the same form?
I found another post it seems more reasonable solution
| This is done for identical particles (really in QM we cannot distinguish between the two particles for ex. electrons or bosons)
consider some operator $\hat{\rho}$ which swaps two particles A and B.
$\hat{\rho} \psi(A,B)= e^{\iota\theta} \psi(A,B) $,
where $\psi(A,B)$ is the amplitude of wavefunction, which under swapping operation picks up a phase.
Now if we operate it twice we must get the same wavefunction,
$\hat{\rho}\hat{\rho} \psi(A,B)=\psi(B,A) = (e^{\iota\theta})^2 \psi(A,B) $
so $(e^{\iota\theta})^2 = 1$
so $e^{\iota\theta} = \pm 1$
Hence we get $\psi(B,A) = \pm \psi(A,B)$
Now suppose our particles are in states $\psi(A), \phi(B)$, to make then indistinguishable under swapping of A and B, we write them as superposition,
$\psi(A,B) = C[\psi(A)\phi(B)\pm\psi(B)\psi(A)]$,
now you try swapping them, you will get
$\psi(B,A) = \pm \psi(A,B)$
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Is it correct to say that; an object moving at a constant velocity is moving in a straight line? I think it is correct because:
v = displacement/time, if the direction changes, the displacement changes so the velocity is not constant anymore.
Please correct me if I'm wrong
| Yes. Velocity has both magnitude as well as direction, so a constant velocity would imply constant direction as well. Constant speed, however doesn't necessarily mean straight-line motion (e.g. uniform motion along a circular path).
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Angular momentum and Potential field In addition to total energy, which components of the angular momentum $(\mathbf{L}_x,\mathbf{L}_y,\mathbf{L}_z)$ are conserved in the presence of an external potential field $V(x,y,z)$? I've learnt that it is usually the $\mathbf{L}z$, but how should I prove it? There seems no way to find a relation between $V$ and $\mathbf{L}$. Please help !!
| The general statement is that if $V$ is rotationally symmetric about a given axis, then the component of $\vec{L}$ parallel to that axis is conserved. This is easiest to demonstrate by rewriting the potential in terms of polar coordinates and showing that the torque is always perpendicular to the axis of symmetry.
Alternately, this can be viewed as a consequence of Noether's theorem, which I am honor-bound to mention in this context because it's a fascinating theorem.
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Failure of wave theory in explaining photoelectric effect What property of waves makes it unable to explain to the photoelectric effect, and how does that property make the wave theory fail in explaining the photoelectric effect?
| If light was only a wave increasing the intensity of light would have increase the kinetic energy of photo electrons as intensity of the wave represents it's energy but it was found out that kinetic energy is unaffected by the change in intensity , it simply increases the number of electrons (all with the same energy) hitting the metal.
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Why is the field quantization applied to the free particle? This post claims that there is no real photon (particle) with a plane wave solution well-defined momentum state).
It makes sense somehow to me. I can think of several arguments:
*
*The plane wave solution doesn't have well-defined probability as it exists in infinite space with equal probability. We will never be able to detect those particles.
*Our probability interpretation breaks down with the plane wave solution. It's not square-integrable.
Even though it seems non-sense, we apply quantization rule to the free particle which makes quantization as the imaginary process which doesn't exist in reality.
Besides 'it works' argument, do we have any explanation why this quantization process makes sense?
|
Besides 'it works' argument, do we have any explanation why this quantization process makes sense?
Quantum field theory (QFT) is used for interacting elementary particles, and also as a tool/model for other quantum mechanical states (condensed matter for example).
Your premise:
we apply quantization rule to the free particle
Is wrong:
QFT was not developed to study free particles, because as you say the plane wave solutions cannot describe the probability of seeing a particle at ( x,y,x,t). To describe with fields a single particle one has to use the wave packet solution,.
Besides 'it works' argument, do we have any explanation why this quantization process makes sense?
The "it works" is reserved for the Feynman diagram use of the QFT tool/model, .
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What is the point of a voltage divider if you can't drive anything with it? The voltage divider formula is only valid if there is no current drawn across the output voltage, so how could they be used practically? Since using the voltage for anything would require drawing current, that would invalidate the formula. So what's the point; how can they be used?
| This is perhaps a niche concern, but one advantage of voltage dividers is that they also cleanly divide down the noise.(*)
In very low temperature electronics experiments, you might want to drive your very sensitive device that can't handle more than 10 microvolts of applied voltage, but your voltage source might be way more noisy than this. So, you instead start out with 1 volt at your voltage source, and divide it down 100000x.
As you say, the resistance of the divider does end up adding to the final measured resistance of your device. But, this can just be subtracted off after the measurement is done.
(*) - they can also add back in a lot of noise if one is not careful of ground loops: the voltage divider adds its local ground voltage noise fully into the output.
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Specific note that is not clear for me in and the derivation of maxwell equation $\oint \vec{B} \cdot d \vec{r}=\mu_{0} I_{e n c}$ I know this is not the full equation but right now in this path of the course that what we learned so far.
We studied that a wire along the $z$ axis produces magnetic field $\vec{B}=\frac{\mu_{0} I}{2 \pi \rho} \hat{\varphi}$
then for every closed loop that goes around the wire we can write $$\oint \vec{B} \cdot d \vec{r}=\int \frac{\mu_{0} I}{2 \pi \rho} \rho d \varphi=\mu_{0} I$$
He mentioned that, the critical point in the proof is that, we can say the magnetic field behaves like $\rho^{-1}$ which is not clear for me why is it so critical. What would happen if it did not behave like that?
In addition he wrote that if the loop that we take does not go along the wire, the circulation of $\vec{B}$ will be equal to zero. This is not clear for me physically or mathematically.
for example loop like that :
while the black dot is the wire in the origin.
| Given,
$$\oint \vec{B} \cdot d \vec{r}=\int \frac{\mu_{0} I}{2 \pi \rho} \rho d \varphi=\mu_{0} I$$
There can be 2 reasons your teacher said that the $\rho^{-1}$ is crucial for the proof.
*
*Consider the equation, $\int \frac{\mu_{0} I}{2 \pi \rho^{n}} \rho d \varphi=\mu_{0} I$
where n is a real number. This clearly upon integration yields that current I is directly proportional to $\rho^{1-n}$ which is exactly not the case. It can be relevant if a non uniform current density $J$ is given.
*It is just a take away point, i.e. as one moves far away from the wire the field strength decreases.
For the second part of your question (the diagram).
Use this $$\oint \vec{B} \cdot d \vec{r}=\int \frac{\mu_{0} I}{2 \pi \rho} \rho d \varphi=\mu_{0} I_{enclosed}$$
Here enclosed means the net current flowing in or out of the given loop. Say you have 5 wires and a circular loop around it, if two carry current $i$ along +z axis and 3 carry $2i$ current along - z axis then the net $I_{enclosed}= 2*i - 3*2i= -4i$
Analogously, in this case the loop encloses no current, and hence there will be null value for the circulation of B.
Mathematically, if you consider the previous proof that, if a wire produces the magnetic field $$\frac{\mu_{0} I}{2 \pi \rho}$$ you can take the circulation manually as well. That is,
$$\oint \vec{B} \cdot d \vec{r}=\int\frac{\mu_{0} I}{2 \pi \rho_{1}}*\rho_{1}d\theta+0-\int\frac{\mu_{0} I}{2 \pi \rho_{2}}*\rho_{2}d\theta+0=0$$
Here $\rho_{1},\rho_{2}$ are the radial distances of each part of the circular portion of the wire in the diagram.
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Why symmetric potential has even bound ground state and does odd ground state exist? I have read that symmetric potential has even bound ground state, but I don't know how to derive it? The only conclusion I can derive is for even potential I can take real wavefunction.
I also want to ask, if odd bound ground state ever exist? I have never seen any.
| One particle wave function for a bound ground state does not have nodes. This is obviously not the case for the wave functions of several fermions, which necessarily have zeros to satisfy the Pauli principle. Note also that the wave functions of the eigenstates of a scalar potential can also always be chosen real.
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Photon description of quantum-optical interference experiments I am currently studying the textbook The Quantum Theory of Light, third edition, by R. Loudon. In the introduction, the author says the following:
In the customary photon description of quantum-optical interference experiments, it is never the photons themselves that interfere, one with another, but rather the probability amplitudes that describe their propagation from the input to the output. The two paths of the standard interference experiments provide a sample illustration, but more sophisticated examples occur in higher-order measurements covered in the main text.
The first sentence is a bit unclear. Is the author saying that it is never the photons themselves that interfere with one another, but rather the probability amplitudes (of the photons) that interfere with each other (which sounds weird, since the photons themselves are probability amplitudes, right?)? Or is the author saying that the photons (in the form of probability amplitudes) never interfere with each other at all, and that the photon propagation from input to output is fully described by the probability amplitude (that is, photons do not affect each other at all)? Or is it saying both?
I would greatly appreciate it if people would please take the time to clarify this.
| The term photon applies to an electromagnetic wave packet of finite size and a total energy determined by the frequency of the wave. The strength of the two fields at any point determines the probability that all of the energy and momentum of the packet will be absorbed by some other entity (often an electron) at that point. Since this “collapse of the wave” is difficult to conceive, the common assumption is that, instead of being distributed as energy density in the fields, the energy (and momentum) of the packet is carried by a “point-like particle” which apparently wanders at random throughout the packet. In the quote you referred to, the author was using the term photon to denote the point-like particle while leaving any interference effects to the wave.
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Where is energy dissipated on charging a capacitor? The energy lost on charging a capacitor can be easily found from the change in energy of the components of the circuit and the energy supplied by the battery.
On charging a capacitor I know that the energy loss appears as heat in the internal resistance of the battery and the wires.
But what if I take (Purely theoretically) a battery with 0 internal resistance and wires with 0 resistance?
I can't see why the capacitor won't be charged, so from the calculation there must be energy dissipated in the circuit.
So in that case, where is the heat dissipated?
Or is this ideal case faulty?
| When charging a capacitor the energy from the battery is transferred to the capacitor. If the wires have resistance, some of this energy is lost, i.e. dissipated. If the resistance is zero, there are no losses - but there is still the energy transfer from the battery to the capacitor.
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Why didn't heavier elements settle at the core of the solar system? As the solar system formed, why didn't all of the heavier elements such as iron, collect where the sun is leaving the lighter elements in the outer solar system?
| Most of the iron and heavy elements in the solar system are in the Sun.
They aren't all in the Sun, because the material from which the solar system formed was well-mixed and turbulent.
The reason that the concentration of heavier elements is higher in the small, rocky, inner planets (the gas giants have an abundance mixture similar to the Sun), is that firstly, only heavier, "refractory" elements could condense into solids in the hotter inner parts of the protosolar nebula; and secondly, once these had clumped to form protoplanets, they were unable to capture and hold onto the vast amounts of hydrogen and helium in the early protoplanetary nebula, because of their low gravities and high temperatures.
The sinking/diffusion you refer to requires, stratified, stable fluids. This condition was not satisfied in the proto-solar-system, but was satisfied in the early history of the planets; hence the ample evidence for chemical differentiation in most of them, with heavier elements sinking towards the middle.
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Does seeing an object modify it state? I took a dive into the concept of information in physics and from what I have read, I which someone to confirm if looking at something really modify the state of that thing.(modify state here refers to any small change in the object "information arrangement")
| Having read the comments to the post I think you're talking about collapses. In quantum mechanics, a physical state is, in general, in a superposition of states with each state corresponding to a specific possible value you could measure. When you do measure however you only get one result and an "explanation" is that the unobserved wavefunction, which is in a superposition, collapses to a specific state, the one that corresponds to the value you are measuring. This is what people usually mean when saying that looking at a state changes it. Effectively this is what happens.
You should note however that this explanation doesn't convince a lot of people. Why should the state change at all? This problem is known as the measurement problem and is one of the biggest one in physics.
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Can circularly polarized light resonate in a Fabry-Perot cavity? I am curious as to how a standing wave of circularly polarized light may be forced to resonate in a Fabry Perot cavity as in this paper - https://doi.org/10.1103/PhysRevD.98.035021
I know that in a standard Fabry Perot cavity with incident linearly polarized light, the standing wave produced has the magnetic and electric field components 90 degrees out of phase (and of course perpendicular).
Is it also true that a circularly polarized beam resonant in a Fabry Perot cavity capped with quarter wave plates has a structure where the magnetic field component is 90 degrees out of phase with the electric field component?
| Circularly polarised light is two linearly polarised electromagnetic waves with their electric fields at right angles to one another and $\pi/2$ out of phase.
So if it works out for each component of the electric/magnetic field what does superposition imply?
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What is that which is magnetic force? I believe a magnetic field is caused by movement of charged particles.
But, what is it that is actually making the force between two magnets repel or attract to another? For repelling, I'm mentally visualizing a water balloon between two magnets. What is the water balloon composed of? Is it purely the "magnetic field"? If so, seems like more work needs to be done.
| The magnetic force can be understood as a relativistic effect on the electric force. It results from Lorentz transformation of the electrostatic force for a stationary charged particle.
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In what sense is a quantum damped harmonic oscillator dissipative? The classical Hamiltonian of a damped harmonic oscillator $$H=\frac{p^2}{2m}e^{-\gamma t}+\frac{1}{2}m\omega^2e^{\gamma t}x^2,~(\gamma>0)\tag{1}$$ when promoted to quantum version, remains hermitian. Therefore, the time evolution of the system is unitary and probability conserving. The Heisenberg equation of motion, for the operators $x$ and $p$ derived from this Hamiltonian matches perfectly with the classical Hamilton's EoM, in appearance: $$\dot{x}=\frac{p}{m}e^{-\gamma t}, ~\dot{p}=m\omega^2xe^{\gamma t}.\tag{2}$$
Question Quantum mechanically, how to show that this is a dissipative system? Note that this system has no stationary states.
| I'll use the convention of writing the exponent as $\gamma t / m$ rather than $\gamma t$.
The actual energy of the HO is
$$ E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega^2x^2 = \frac{1}{2m}p^2\mathrm{e}^{-2\gamma t/m} + \frac{1}{2}m\omega^2x^2 = \mathrm{e}^{-\gamma t/m} H$$
since $p = \partial_{\dot{x}}L = \mathrm{e}^{\gamma t/m}m\dot{x}$.
Ehrenfest's theorem means that
$$ \frac{\mathrm{d}}{\mathrm{d}t}\langle E\rangle = -\mathrm{i}\langle [E,H]\rangle + \langle \partial_t E\rangle = -\frac{\gamma}{m}\mathrm{e}^{-\gamma t/m}\langle H\rangle + \mathrm{e}^{-\gamma t / m}\langle \partial_t H\rangle= -\frac{\gamma}{m}\langle E\rangle + \mathrm{e}^{-\gamma t / m}\langle \partial_t H\rangle,$$
so as $t\to \infty$ (meaning we can ignore the second term, not a literal limit) we have that $\langle E\rangle(t) \to \mathrm{e}^{-\gamma t / m }\langle E\rangle (0)$, same as in the classical case.
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Where can i get the step by step expression for CMB anisotropy? I never able to wrap my head around alm term used in expression
If you know the expression, kindly explain but also give textbook references. It would be helpful.
| Any function on a sphere can be expanded as a linear combination of spherical harmonics $Y_{\ell m}$. The numbers $a_{\ell m}$ tell you “how much” of each spherical harmonic is needed. This is similar to a Fourier series for decomposing any periodic function $f(x)$ into its sinusoidal “harmonics”.
Conceptually, the spherical harmonics are a complete orthonormal basis for the infinite-dimensional vector space of functions on a sphere, in the same way that $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are a complete orthonormal basis for three-dimensional Euclidean vector space (and in the same way that $\cos{nx}$ and $\sin{nx}$ are a complete orthonormal basis for periodic functions in one dimension). The $a_{\ell m}$ are the components of $T$ in this basis.
One classic reference book for such things is Methods of Mathematical Physics by Courant and Hilbert.
The book or paper you referenced has a typo. They meant that the $a_{00}$ term measures the mean temperature. This is because the first spherical harmonic $Y_{00}$ is a constant over the sphere. The other terms represent angular fluctuations in temperature around the mean, on smaller and smaller angular scales as $\ell$ increases.
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Can Maxwell's equations be generalized to all fields? For having studied both classical and quantum optics, I regard Maxwell's equations as the grand "cheat sheet" from which (almost) all optical/photonic phenomena can be derived. Yet, I also know that wave-particle duality extends to all other fields and particles in the standard model. I'm therefore left with a nagging sense that Maxwell's equations should---up to differing units---be universal (cf. e.g., Gauss' law for gravitation).
I expect that some behaviors (say, the Aharonov-Bohm effect) won't be observable since some particles in the standard model are charged while others aren't, or that monopoles may exist while others don't, etc. That said, don't we have any evidence that the overall template of Maxwell's equations is universal?
| This is because of special relativity. The fact that the speed of light is finite can be packaged nicely im saying that we live in a space where the notion of proper distance makes sense:
\begin{equation}
ds^2=-c^2dt^2+ dx^2+dy^2+dz^2
\end{equation}
In particular, the only things we can talk about are things which are Lorentz/Poincare invariant/covariant. In other words, the symmetries of Minkowski space fix the form of Maxwell’s equations and their analog for other matter. This is summarized by the so-called representation theory of the Poincare group. These are described by quantum numbers like spin and momentum. Maxwell theory is a spin-1 massless representation for example, and the Maxwell equations are the simplest equations of motion consistent with special relativity. The Dirac equation comes from the spin-1/2 representations, and the Klein-Gordon equation comes from the spin-0 representation. If you loosen up your assumptions and impose the equivalence principle instead, you’ll end up getting Einstein gravity. So your suspicion has a natural explanation, the possibilities consistent with the symmetries we see in nature are quite constrained.
| {
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How does light get energy when moving from one medium to another? It is said that the speed of light increases when moving from optically denser medium to rarer medium but
as light can be considered as a particle how does it get the energy to increase it velocity
| I would just add to Emilio Pisanty's comment by saying that the frequency remaining a constant as light travels across a boundary separating two media is required because it is required that the tangential component of E be continuous across the boundary.
In turn, that the tangential component of E must be continuous across the boundary separating two media can be found by computing the line integral along any path through the two media and using Stoke's theorem.
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Are there any physical processes of which we have a full understanding? Are there any physical processes of which we have a full understanding? For instance, we know that each orbit has a different energy, and electrons can move to a higher orbit by absorbing energy and drop to a lower orbit by emitting energy. However, do we understand everything related to atomic orbits? For example, what determines the shape of the orbits, do orbits sometimes change shape, is it fixed, do they cross each other, do they change depending on the proximity of other atoms or electrons? How much is there to learn about simple physical concepts and processes related to atomic orbits?
Is there a concept related to the lack of knowledge we have of just about anything related to physics outside of maybe classical physics (I am assuming we understand 99% of everything related to processes in classical physics)?
| No for philosophical reasons.
The closest science can come to knowing fundamentally reality would be a Theory of Everything. But TOE's are a dime a dozen. It-from-bit, constructor theory, mathematical universe, geometric unity. None of them make unique predictions, or ones we could test.
Science can just make more accurate and accurate models. Of course having a TOE that explains everything makes it tempting to say you know what is real, but even then as finite beings you could argue we can only test a finite number of times! Let alone the uniqueness problem above.
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Lagrangian potential for Newtonian gravity In the Wikipedia site for Lagrangian (field theory) the Lagrangian density for Newtonian gravity is given by
$${\cal L}(\mathbf{x},t) = \frac{1}{2}\rho(\mathbf{x},t)\mathbf{v}^2 -\rho(\mathbf{x},t) \Phi(\mathbf{x},t) – \frac{1}{8\pi G}(\nabla\Phi(\mathbf{x},t))^2$$
I understand how variation of this Lagrangian leads to the correct Poisson equation $\nabla^2\Phi=4πG\rho$.
However, if ${\cal L}$ is simply the differential form of $T-V$, I would have thought that just the first two terms would be sufficient. Where does the third term in ${\cal L}$ come from?
| The last term represents the energy that is carried in the gravitational field itself. If you are familiar with electrostatics, this would be the equivalent of the statement that the energy stored in an electric field is $\frac{1}{2}\epsilon|\mathbf{E}|^2=\frac{1}{2}\epsilon(-\nabla V)^2$, where $V$ is the electric potential (the electric equivalent to $\phi$).
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Why does mass-spring system tend or prefer to oscillate with natural frequency? Can you explain it theoretically rather the derived version of it? Why the mass spring system prefers to oscillate at natural frequency, why does it rather oscillate at other frequencies? Every object in the universe is always vibrating, is it true? if it is the mass spring system at equilibrium, the atoms/molecules inside the systems are still vibrating with a natural frequency.
| The frequency depends on the spring stiffness, k, and the mass, m, as the math shows. For a more intuitive explanation, suppose we watch as the mass moves through the middle (equilibrium) position and the spring starts to stretch. There are two competing factors: the stiffness of the spring, which wants to pull the mass back to the equilibrium position, and the inertia of the mass, which wants to have the mass continue to stretch the spring. For a given system (k and m), these competing factors will determine how often the mass goes back and forth i.e. the frequency. As the equation for the period of oscillation, T, shows, the larger the mass the larger is T; the larger the stiffness, the smaller is T. To get the actual dependence on m and k does require the math.
| {
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Relativity without constancy of light speed Using homogeneity of space, isotropy of space and the principle of relativity (without the constancy of light speed), one can derive:
$$x' = \frac{x-vt}{\sqrt{1+\kappa v^2}}$$
$$t' = \frac{t+\kappa vx}{\sqrt{1+\kappa v^2}}$$
$\kappa = 0$ denotes Galilean and $\kappa < 0$ denotes Lorentz Transformation.
What does $\kappa > 0$ denote? Is it physically possible? I was told that it is self-inconsistent. Can somebody help me with a proof of this?
| There has been much discussion on one-way speeds of light and simultaneity etc. in the philosophy of physics literature. Most famously, Reichenbach introduced a parameter $\epsilon$, which gives the (one-way) light speed in opposite directions as $c/2\epsilon$ and $c/2(1-\epsilon)$. Here $c$ is the "two-way" speed of light, which can actually be experimentally measured.
One way of interpreting this discussion is as ordinary relativity but described on a "tilted" hyperplane: one which is not orthogonal to the observer's 4-velocity. This is the approach of papers like Ungar 1991, see equation 9 for the one-way Lorentz transformations. I haven't analysed your $\kappa$ parameter specifically. But it is certainly consistent to describe relativity using coordinates that are tilted relative to given observer(s).
| {
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Polarization of sunlight by the earth's atmosphere I have just learned that sunlight traveling through the earth's atmosphere picks up a net polarization by collisions with molecules ($O_2$, $N_2$, etc.) that the photons encounter. One would think that those molecules would have to possess some degree of common alignment in order to produce light that possesses nonrandom polarization. Since those molecules are randomly positioned in the atmosphere, how can a net polarization of the transmitted light be produced?
|
One would think that those molecules would have to possess some degree of common alignment in order to produce light that possesses nonrandom polarization.
Molecules are quantum mechanical entities and light interacting with individual molecules should be thought of in photons. Nevertheless, the classical elecromagnetic light with its description by the maxwell equations emerges from the quantum framework correctly, so it is better for collective detection, as is the bulk polarization, to think in terms of classical light as the answer by David explains.
So the differences in polarization come from the fact that the rays from the sun are unidirectional, the perpendicular to the ray defines a plane for the E field , and depending on the scattering angle different components will scatter differently.
Look at the classical formulation of Raleigh scattering
Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle, therefore, becomes a small radiating dipole whose radiation we see as scattered light. The particles may be individual atoms or molecules; it can occur when light travels through transparent solids and liquids, but is most prominently seen in gases.
The same results will come out by thinking photons and molecules, in a much more complicated mathematically way.
The "common alignment" you seek to produce polarization comes from the directionality of the light, and the stratification of levels in the atmosphere.
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Origin of $\ell \leq n-1$ orbital rule I am wondering about the origin of the $\ell \leq n-1$ orbital filling rule. For the hydrogen atom, I believe the reason is because in the spatial wave function there is the term
$$\psi \propto \sqrt{(n-\ell-1)!}$$
so if $\ell > n-1$, by the definition of the factorial, $\psi$ goes to $0$. However, what about atoms with more protons and electrons? Will we always have this sort of term in our wavefunction or is there are more general argument for why this inequality has to be true?
| TL;DR: The quantization condition
$$n_r ~:=~ n-\ell -1 ~\in\mathbb{N}_0 \tag{A}$$
follows by looking for normalizable wavefunction solutions $R(r)$ to the radial TISE.
In more detail: Schematically, one first solves $R(r)$ in the regions for small & large radial coordinate $r$. After factoring out the newly found asymptotic behaviours, one obtains a function $v(r)$, where (due to the TISE) the coefficients of its power series satisfy a recursion relation. It turns out that the series $v(r)$ must truncate in order for the solution $R(r)$ not to alter its asymptotic behaviour. This leads to the quantization condition (A), cf. e.g. Ref. 1.
References:
*
*D. Griffiths, Intro to QM, 1995; subsection 4.2.1.
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Does dark matter follow all principles of regular physics? Is dark matter bound by all the laws of regular physics?
i.e. laws of thermodynamics, speed of light, length contraction, mass-energy relation.
What about Newton's laws of motion (since all of Newton's laws assume an interaction between particles)?
| If dark matter consists of particles it would follow all the rules of regular physics.
If dark matter is caused by entropic gravity though (also called emergent gravity; see for example this discussion about Erik Verlinde's theory) dark matter doesn't consist out of dark particles. I think there is proof that contradicts it, but suppose it's true.
In this case, it might be obvious to you that dark matter will not conform to regular physics because of the simple fact it isn't a regular theory from which it originates.
For example, the dark matter in Verlinde's theory will stay inside the blob of matter following upon two blobs of matter that collided (e.g. the bullet cluster) while in the case of dark matter being particles, two blobs of dark matter would appear on both sides of the normal matter blob.
| {
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Why are aerodynamic / streamlined shapes always stumpy at the front? I'm building an autonomous boat, to which I now add a keel below it with a weight at the bottom. I was wondering about the shape that weight should get. Most of the time aerodynamic shapes take some shape like this:
The usual explanation is that the long pointy tail prevents turbulence. I understand that, but I haven't found a reason why the front of the shape is so stumpy. I would expect a shape such as this to be way more aerodynamic:
Why then, are shapes that have good reason to be aero-/hydrodynamic/streamlined (wings/submarines/etc) always more or less shaped like a drop with a stumpy front?
| You are correct if your boat will only travel in a straight line.
In real life the motion of the boat will often have a yaw angle, so that it is moving slightly "sideways" relative to the water. For example it is impossible to make a turn and avoid this situation.
If the front is too sharp, the result will be that the flow can not "get round the sharp corner" to flow along both sides of the boat, without creating a lot of turbulence and waves which increase the drag on the boat.
| {
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Difference between Mixed and Pure states Suppose that there is a system of two photons 1 and 2, each of which is in a mixed state $1/2|R\rangle\langle R| + 1/2 |L \rangle\langle L|$, where $|R \rangle$ and $\langle L|$ are two orthonormal pure polarization states. (So the composite state would be a product of two mixed states.)
How is this case different from the case where each of 1 and 2 is in a pure state $1/\sqrt{2}(|R\rangle+|L\rangle)$? Can you distinguish these two cases experimentally? Because it looks like the measurement outcomes should be the same in both cases: either both are in $|R\rangle$, both are in $|L\rangle$, or one is in $|R\rangle$ and the other in $|L\rangle$. Please educate me.
| The thing is whether those two photons are independent or not.
A single photon can be $|L\rangle$ or $|R\rangle$, which are pure states, or any linear combination of them (normalized).
However, whtn you have two photons as a compound system, you can have these 4 pure states: $$|LL\rangle, |LR\rangle, |RL\rangle, |RR\rangle$$
This notation is obviously $|LR\rangle = |L\rangle_1 \otimes |R\rangle_2 $
And your final state can be any combination of those 4 primitive states.
If you are working with the two photons together, you should write any state as a lienar combination of your basis.
So if you want to say that each photon is in $\frac{1}{\sqrt2} \left(|L\rangle + |R\rangle \right)$, it's okay, but you should write it as a linear combination of the 4 basis vectors, because it will be much easier to work with them.
| {
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Why is maximal kinetic energy lost in a perfectly inelastic collision? A perfectly inelastic collision is one where both of the colliding objects stick together and move as one.
My question is, why, of all possible combinations of final velocities that conserve momentum, does this one lead to the greatest loss in kinetic energy?
One reasoning I found was that this is the only combination in which the total kinetic energy of the system becomes 0 in some frame of reference (com frame). But just because the KE is 0 in some frame doesn't mean that it is the least possible in every other frame, does it?
| For a perfectly inelastic collision, there is a reference frame (the center of momentum frame) in which the final state has zero kinetic energy, and zero is the absolute minimum.
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Is it possible the many-worlds hypothesis explains dark matter? To provide context for the title, is it possible that dark matter is just gravity from other 'branches' of the universal wave function that have split from ours, that are weakly interacting with the gravity in our branch? Is this an idea anyone has explored before?
| Interesting idea, but ... no. It does not work like that. The branching idea is a bit misleading. Instead, the Many Worlds interpretations refers to the terms in the superposition of the wave function of the universe. Each terms in this superposition is viewed as a "world," which is perhaps a bit misleading. So all possible "worlds" are present in the wave function all the time, but with different coefficients.
This wave function undergoes unitary evolution. What it means is that whenever an interaction takes place the coefficients in the superposition are changed through a unitary process. Such a unitary change maintains the overall normalization of the total wave function. So it is not really the case that the universe branches off into many universes.
How gravity enters the picture is not fully understood yet. For that, one needs a theory of quantum gravity, which is still a work in progress. However, it is reasonable to argue that it would involve the entire wave function and not just the individual terms. So the "worlds" should not give different gravity effects. They should all act together to produce a coherent gravity effect. However, this understanding is still rather speculative.
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Force on charged particle when moving along magnetic field lines I was thinking, what happens to the force on a charged particle when it is moving along the magnetic field lines?
I am familiar with the right-hand-rule and it seems to me that RHR does not apply in this situation. Since the RHR, the field lines, direction of particle moving and the force on the particle are all perpendicular to each other.
But the question I am asking is, what happens to the force when the direction of a particle moving is parallel to the field lines or along the field lines?
Here is a picture of what I mean:
The electron, moving down or along the field lines with a velocity of v.
| Because $\mathbf F=q\mathbf v\times\mathbf B$, if $\mathbf v$ is parallel to $\mathbf B$ then $\mathbf F=0$ and the particle experiences no magnetic force.
This (which uses a RHR to get the directions right) is always valid.
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What is the direction of the frictional force acting on a rolling wheel? Let's keep things simple and assume that the wheel does not slide. Assume you are in a car that moves with constant speed. Obviously, the wheels exert a tangential force to the surface on the road in the points of contact, P. According to Newton's third law action equals minus reaction so the road exerts a force to the wheels in the direction of motion (please, see fig. 1). So the friction acts in the direction of motion of the car. It seems reasonable since if there were no friction then the wheels would be turning but the car would not move. However, there is one thing I can not understand. Let's now imagine a wheel rolling (without slipping) on a smooth surface with initial velocity V0 (here I mean that t). The same argument should hold (the wheel exert a tangential force to the surface on the road in the point of contact, P. According to Newton's third law action equals minus reaction so the road exerts a force to the wheel in the direction of motion). So the frictional force should again act in the direction of motion. Nevertheless, the wheel slows down with time which means that frictional force should point in the opposite direction to the direction of motion. Could someone give a QUALITATIVE explanation of the seeming paradox? I would also ask to use JUST the "language" of forces (not to apply the concept of the torque since it makes the explanation less vivid).
| If a car is gaining speed, the force from the road is forward. If it is breaking, the force is backward. If the speed is constant, the force is forward and large enough to overcome various sources of drag.
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Why is the electromotive force (EMF) highest in the loop when $\theta$ = $90$? My question is in regard of the following snippet provided by my textbook.
So why is the electromotive force (EMF) highest in the loop when $\theta$ = $90$ or $270$?
So the magnitude of the induced EMF will be determined by the rate at which the loop is rotating, according to Faraday's Law. EMF will be maximum when the rate of change of flux is at maximum.
But why does this means that the loop has moved to a position parallel to the magnetic field and the flux through the loop is zero? Since there is no magnetic field penetrates the area at that instant shouldn't there be no current? In turns, shouldn't there be no magnetic field?
| "Since there is no magnetic field penetrates the area at that instant shouldn't there be no current?"
I'm afraid that this is a bit like saying about a ball thrown upwards that has reached its highest point: "Since there is no velocity there can be no acceleration". In other words, you do need to consider carefully that it is rate of change of flux that matters, irrespective of the actual value of the flux. This rate of change at any instant is the slope of the tangent to the top graph at that instant. So just consider how that slope changes with time!
[There is another (more fundamental) way of understanding why the emf is a maximum when the coil has this orientation. Two sides of the coil (AB and CD) are parallel to the field lines; the other two, BC and DA are at right angles to the field. Because the coil is rotating, BC and DA are moving in opposite directions, at right angles to the field. Their velocity component at right angles to the field is greater than for any other orientation of the coil. Therefore the Lorentz force urging the free electrons in these sides of the coil is a maximum, so the e.m.f. is a maximum.]
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What principles of physics restrict a disintegration ray? If a vehicle were to move horizontally through solid stone by separating it into pieces, moving the pieces behind it, and reassembling them precisely as they were, the net change in energy ought to be zero. The actual cost of boring a hole appears to depend on the immense inefficiency of grinding up material through friction or, more exotically, converting it to vapor with a laser drill.
Suppose you have some wild notion such as using coherent phonons produced by a computer-adjusted array to constructively interfere into sudden shear waves in front of the vehicle every millimeter or so. Is there any principle of physics you could use to say that this needs to cost no less than some specified amount of energy per gram of rock moved, or can't be done at all?
| A basic principle behind the minimum energy requirement is that of a potential barrier.
Consider a very simplified model: the object to disintegrate is a pair of oppositely charged particles which are at a small distance apart. There is an electrostatic attraction between them, so they have less energy being close than they would have if they were far away.
Specifically, the energy $E$ of the system of two stationary particles is in the form
$$
E = - \frac{k}{r}
$$
where $k$ is some constant, while $r$ is the separation between the particles.
Disintegrating the object in this context would entail moving these far apart (say, with very large $r$, so that $E \approx 0$).
Now you can see that, even if you later go back to the situation with small $r$ (and thus negative energy), if you want to pass through the condition of the two particles being far apart you must give them enough energy to reach it.
This is all for a perfectly efficient process: the thing is, molecules and atoms in materials are bound together so that the energy they have when they are close is less than that they have when they are far away.
For a real-world process one would surely have to account for all kinds of inefficiencies, but the number of bonds to break and re-form will give a hard lower limit on the energy needed.
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Heat energy change in adiabatic process We know that when a system gains heat energy from the surroundings through conducting walls of the container , then the temperature of the system also increase till it reaches an equilibrium value .
Whereas in an adiabatic compression of a gas where temperature increases but the change in heat energy of the system is zero .
It must have some value as the state variable of gas i.e. temp. Changed and also we know that ∆Q = s.m.∆T
| The definition of adiabatic expansion is that the heat exchanged by the surroundings and the system is zero.
Also,
∆Q = ∆U + work done
The work done in an adiabatic process is compensated by the change in internal energy, making sure that no other external factors supply energy to the system.
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Lorentz boost expressed as Hyperbolic versors At this link https://en.wikipedia.org/wiki/Versor#Hyperbolic_versor it is claimed that an hyperbolic versor, defined as:
$$
\exp(a \mathbf{r})=\cosh a+\mathbf{r}\sinh a
$$
where $||\mathbf{r}||=1$ correspond to a Lorentz boost. But I cannot work out a proof. Can anyone help?
I assume one starts by applying the exponential to a 4-vector $\mathbf{s}$ as follows:
$$
\mathbf{s}'=\exp(\frac{a}{2} \mathbf{r})\mathbf{s}\exp(-\frac{a}{2} \mathbf{r})
$$
Then I get
$$
\begin{align}
\mathbf{s}'&= (\cosh a/2+\mathbf{r}\sinh a/2 )\mathbf{s} (\cosh a/2 -\mathbf{r}\sinh a/2)\\
&= (\cosh a/2+\mathbf{r}\sinh a/2 ) (\mathbf{s}\cosh a/2 -\mathbf{s}\mathbf{r}\sinh a/2)\\
&=\mathbf{s} \cosh^2a/2-\mathbf{s}\mathbf{r}\cosh a/2\sinh a/2+ \mathbf{r}\mathbf{s}\sinh a/2\cosh a/2 - \mathbf{r}\mathbf{s}\mathbf{r}\sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+\cosh a/2\sinh a/2(-\mathbf{s}\mathbf{r}+ \mathbf{r}\mathbf{s}) - \mathbf{r}\mathbf{s}\mathbf{r}\sinh^2a/2
\end{align}
$$
edit (based on answer):
$$
\begin{align}
\mathbf{s}'&=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp) \sinh^2a/2
\end{align}
$$
edit2:
$$
\begin{align}
\mathbf{s}'&=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp+2s_\perp-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel+s_\perp-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (\mathbf{s}-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2-\mathbf{s}\sinh^2a/2+ \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\
&=\mathbf{s} + \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\
&=\mathbf{s} + \mathbf{r}s_\perp \sinh a +s_\perp (\cosh a -1)\\
&=s_\parallel + ( \cosh a + \mathbf{r} \sinh a )s_\perp
\end{align}
$$
Is this a Lorentz boost? How do I show that it is?
| It isn't a Lorentz boost by itself, but it can be used to easily derive one (at least in 1+1 dimensions). It's similar to the expression of spatial rotations in term of Euler's formula:
$$ x'+iy'=e^{i\phi}(x+iy),\ \left[e^{i\phi}=\cos\phi+i\sin\phi,\ i^{2}=-1\right]$$
which gives
$$ x'+iy'=x\cos\phi-y\sin\phi+iy\cos\phi+ix\sin\phi$$
separating real and imaganiry parts gives:
$$\begin{aligned}x'=x\cos\phi-y\sin\phi\\
y'=y\cos\phi+x\sin\phi
\end{aligned}$$
Instead of complex numbers we now use the hyperbolic versor together with split complex numbers, and replace y by t:
$$ x'+rt'=e^{r\eta}(x+rt),\ \left[e^{r\eta}=\cosh\eta+r\sinh\eta,r^{2}=+1\right] $$
which gives
$$ x'+rt'=x\cosh\eta+t\sinh\eta+rt\cosh\eta+rx\sinh\eta $$
separating real and split-complex parts gives:
$$\begin{aligned}x'=x\cosh\eta+t\sinh\eta\\
t'=t\cosh\eta+x\sinh\eta
\end{aligned}$$
In order to use hyperbolic versors in more dimensions, see split-quaternions (2+1 dimensions) and biquaternions (3+1 dimensions).
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Difference between Doppler effect and relativistic Doppler effect? When the source of a sound wave is moving in the direction of the sound wave it is a different scenario than if the receiver of the sound wave is moving in the direction where the sound waves are coming from. Why is this not the case in the relativistic Doppler effect, why is it unimportant whether the source or the receiver is moving?
| For sound, when we speak of the observer moving or the source moving we mean moving with respect to the medium through which, and by means of which, the sound travels, so the two cases can be distinguished. The relativistic Doppler effect is for electromagnetic waves, which don't use a medium. So the only source or observer velocity that can be relevant is the relative velocity between source and observer.
[The speed of light is the same in the source's frame of reference (that is the frame in which the source is stationary) as in the observer's frame.]
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Does position-momentum uncertainty principle apply to a two-body wavefunction when we frame boost to its center of mass reference frame? The heart of my question is below in bold. The rest is clarifying information or additional points of discussion - in case my assumptions are the heart of my misunderstanding.
In a two body attractive quantum mechanical system (namely the simplified Hydrogen atom - ignoring spin etc.), it can be shown that this reduces into two Hamiltonians. We find one Hamiltonian for the center of mass, and one for the behavior of the individual particles.
Assuming the individual particles are bound by their mutually attractive potential, the Hamiltonian has discrete Energy Eigenstates. The center of mass Hamiltonian, $H_{com}$, is a free particle solution. Does position-momentum uncertainty principle still apply to the wavefunction describing solutions to $H_{com}$ ?
In my undergrad QM class, the professor said that when we're in the center of mass reference frame, we ignore $H_{com}$. When I pressed him further on this, he said it wouldn't make sense for the atom's wavefunction to "spread out" in the center of mass frame (due to a $k$-space distribution of $\psi$). This didn't sit well with me, because being in the center of mass frame of the atom implies we know the center of mass momentum exactly (or to extreme accuracy), and therefore can't know the exact position (or only have an extremely inaccurate knowledge).
| Yes it holds because the corresponding operators satisfy the standard commutation relations. It can be proved in various ways. The most elementary way is to describe the system using the so-called Jacobi coordinates (in the phase space) which include the coordinates of CM and the total momentum. Assuming that the position and momentum operstors of every single particle satisfy the standard commutation rules, it arises that the position of CM and the total momentum satisfy the same relations.
However the typical time a wavefunction spends in spreading out in space, when it freely evolves, is related to the value of the mass of the system: large masses imply large times. This implies that the wavefunction of the center of mass (assuming the total wavefunction factorised) needs a typical time larger than the times of the single particles, if these are not subjected to mutual interactions.
| {
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Why rotating objects stops? Sorry for asking this simple question, but really I couldn't find a good document discuss what I need exactly.
I am implementing a flight simulation, but my question is related to physics rather than aerodynamics so I find to ask the question to physics experts.
Suppose that I am having a cuboid (Simple form of the plane) with the following dimension:
Length: 14.8m
Height: 4.8m
Depth: 10.0m
The coordinate system is X is right, y is Up and z depth (inside the paper).
I applied a torque on the Y axis, the rectangle begins to gain angular velocity and it rotates in the XZ plane around its center of gravity.
Every thing works fine for now, but after I remove the torque the cuboid should stops i.e. the angular velocity should be decreased till reaches zero. How this happens?
I think this should be due to the moment of inertia, as I am using 3D coordinate system the inertia should be inertia tensor and the right way to calculate the inertia tensor from these dimensions.
If what I thought is right so I need the equation for how the inertia tensor is affecting the angular velocity till the angular velocity reaches zero.
If I am not right, what is the force that affect the cuboid to stop rotating?
| for your flight simulator you can applied braking torque and then stop the simulation when the angular velocity is zero.
your equation
$$I_y\ddot\varphi(t)=\tau_m(t)+\tau_b(t)$$
where $I_y$ is the inertia about the y axes and $\tau_m$ is the applied torque to accelerate the cuboid and $\tau_b$ to decelerate the cuboid
Simulation
$\tau(t)=\tau_m(t)+\tau_b(t)$
Angular velocity $\dot\varphi$
| {
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What do maufacturers do to ensure that a valve is capable of containing hydrogen gas? I am a researcher working to develop a hydrogen fuel system for use in a specialized internal combustion engine. The biggest challenge from an engineering standpoint with a project such as this is the inherently volatile nature and small size of the hydrogen molecule. My research has led me in a variety of different directions and provided a great deal of information about hydrogen's properties as an element and as an energy source, however after over two years of research, I have thus far found no information whatsoever on what specific manufacturing techniques are employed to ensure minimal leakage, if any.
Any constructive input regarding the formatting or content of this question would be greatly appreciated, as this is my first post on this site.
| I am also in the hydrogen business and yes this is a big problem. We create gas by electrolysis on 16 stainless plates, each about 20cm x 20cm. We use neoprene gaskets about 1cm wide and 2.2mm thick, in between each of the plates to maintain proper separation. All tightened with 8 bolts. For fittings, we coat the threads with Sikaflex adhesive. Even the quality of the neoprene is important; if the neoprene is not perfectly uniform, it reduces our gas production and can lead to leaks. In cold weather the neoprene will shrink and cause major leaks, so it's important that you put the unit in the fridge before final tightening of the bolts. We have tried many other gasket materials and many other adhesives, but have settled on this.
| {
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Why do motorcycles front wheels lift of the ground when accelerating quickly When motorcycles accelerate quickly they do a wheelie. Where is the torque provided to lift the motorcycle?
| It is due to the conservation of angular momentum. As soon as the motorcycle is accelerated, there's a rotation of the rear wheels, associated with it will be an angular momentum. So, one is introducing an angular momentum into the system. As there was no angular momentum before accelerating, the net angular momentum should also be zero after accelerating.
Therefore, to conserve the angular momentum, there will be rotation in the reverse direction about the axis of the rear wheel. The rotation of wheels is in the anti-clockwise direction as the bike moves ahead. So, there will be a rotation in a clockwise direction i.e. front wheel lifting up.
Hope this helps!
| {
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In Rayleigh scattering do the frequencies scattered correspond to the discrete energy level transitions available to atmospheric particles? I have read that for Rayleigh scattering the photons are absorbed by the atmospheric particles then re-emitted (albeit very quickly) and the shorter wavelengths scatter better which is why the sky is blue. Am I right in saying that the sky is blue because the energy level transitions available in oxygen allow it to absorb and re-emit (scatter) the blue light from the sun? And that the reason why we don't get a violet sky, from the more abundant nitrogen in the atmosphere (nitrogen emission spectrum when mixed appears violet), is that there is simply less violet light in the sun's spectrum?
| Rayleigh scattering is scattering. The photons are not absorbed and the scattering can take place at all wavelengths. The scattering is caused by both oxygen and nitrogen in the atmosphere and they have very similar Rayleigh scattering cross-sections.
The sky is not violet because there is little violet in the solar spectrum, which is what is being scattered, and the eye is not very sensitive to that violet light. See Why is the sky not purple?
| {
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Probability in classical physics I have read lots of thing on probability in QM and the different ways of intending it. Now, I am wondering how physicists intend probability in classical physics. To be more specific, I have read some articles about the fact that probability in classical physics is seen by physicists as bayesian probability. I am not sure if the majority of physicists agree with this idea. How probability is generally intended in classical physics? And are there other consistent interpretation of probability in classical physics than subjective degrees of belief? Please let me know about them. Moreover, please suggest me if there are any articles that talk about this theme. Thanks.
| For better understanding of yours, the implementation of Probability, both in QM and CM can be presented as following:
In QM, the probability of any attribute can be used for two different aspects:
*
*For determining the distribution of constituents, within a system, &
*For determining individual attributes of the constituents within a system (like, position, momentum, energy etc. )
(Note that: In QM, where position, momentum, and so other attributes follow Principle of Uncertainty, the show their Eigenvalues in a given instance, and the implementation of probability here is pretty much different, than that in representing any distribution)
However, in Classical Mechanics, the probability is mostly implemented in determining a given distribution of any material, especially in the context of Statistical Mechanics, and that is due to the random motion, like Brownian Motion of atoms and molecules within a matter, and this is same in QM.
But, a classical particle isn't uncertain for any of the attributes, as it provides a deterministic knowledge about the system. (Which does not provide the perfect information about a system and for that we need QM.)
Hope it helps!
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How do I categorize material as having uniform or non-uniform tension? From what I've learned tension uniform/constant throughout the body only in massless objects eg: massless ropes, but when is it non-uniform?
EDIT: By uniform/constant tension I mean the value of tension in the object is the same at any point in the object. Non-uniform tension just means the opposite i.e tension at one point isn't equal to tension at another point in the object.
| Massless ropes will normally have uniform tension, since the weight of the rope itself doesn't factor into any tension calculation. Ropes with mass tend to have nonuniform tension, since their own weight needs to be accounted for - the top of the rope has to support everything below it, while the bottom of the rope doesn't have to support anything at all.
You can also have constant tension in a rope with mass, so long as that mass isn't hanging under the force of gravity, which would result in different masses above and below different points and unequal tension. To do this, simply lay the rope horizontally on the floor, and pull both ends - now you have a rope with mass and uniform tension.
| {
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Lorentz Force on a Current Carrying Wire Does the Lorentz Force on a Current Carrying wire given by the equation
$$\mathbf{F} = I \int \text{d}\ell \times \mathbf{B}$$
constitute an action reaction pair? That is, if i have two arbitrarily shaped current carrying wires, is it true that force on any one of them due to the magnetic field of the other is equal to the force on the other due to the magnetic field of the first?
| edited
No, it seems that the law of action is true for the cherry-picked case of parallel wires as shown by @Sagigever's response. For instance, this is certainly not true for the case of magnetic Lorentz force exerted by two charges (yellow in figure) moving perpendicular to each other (along the blue and red lines). In this case, the $\vec{F_{12}}$ is perpendicular to $\vec{F_{21}}$ (the dotted lines represent the magnetic field due to the charges).
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Do photons lose energy after radiation pressure is applied to a perfect reflector? I was readng Wikipedia article (English one) about radiation pressure because there is something I still cannot figure out. As I understand it, radiation pressure emerges from conservation of momentum. Photons or electromagnetic waves possess momentum and when they are absorbed, reflected or even emitted, the aborber/reflector/emitter experiences a pressure that is proportional to the irradiance (in watts per square meter).
OK. In the case of a perfect reflector that does nothing but reflecting incoming radiation, it would seem the reflector is pushed and the radiation is reflected in opposite directions. But the radiation has not lost anything, it has only changed direction. And I do not understand how something can be set in motion (the reflector), which amounts to do work, yet the source of this work does not lose energy. I mean, if the radiation, after a U turn, meets a second reflector, it would have pushed two reflectors a way, yet it would continue in its original direction as if nothing happened...? What am I missing here? Should not photons lose something?
| Your intuition is correct: each photon loses a very small amount of energy when it reflects from a perfect teflector that can move. The reflected photons will have slightly longer wavelengths than their incident counterparts.
| {
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Why do fluids not accelerate? A fluid flowing in a horizontal pipe must be flowing at a constant velocity because of the conservation of mass.
However, considering how there would be a pressure and hence force acting behind the fluid, for it to have a constant velocity, there must be an equal force slowing it down (depicted as $F?$).
I can't see a force that would be as big as the driving force. Can someone explain to me what this force is and how it's created?
| The fluid is accelerating. The continuity equation simply states that at any instant $A_1v_1=A_2v_2$ where $A_1$ and $A_2$ are the cross-sectional areas of the upper pipe and lower pipe respectively, with $v_1$ $v_2$ being the fluid velocities in the same. The potential energy of the fluid stored in the upper pipe is being converted to kinetic energy of the fluid flow in the bottom pipe. When the fluid in the upper pipe is at it's highest point, $v_2$ will be the greatest, and gradually this velocity decreases as the fluid height in the upper pipe decreases.
| {
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Accelerometer and Static Force I'm studying the principles of accelerometer and given below is what is stated in LiveScience
An accelerometer is an electromechanical device used to measure acceleration forces. Such forces may be static, like the continuous force of gravity or, as is the case with many mobile devices, dynamic to sense movement or vibrations. Acceleration is the measurement of the change in velocity, or speed divided by time.
So it says that it measures acceleration forces due to a static force, but how can a static force induce an acceleration on an object because a static force virtually keeps an object at rest.
An object at rest does not have an acceleration does it?
| Acceleration and Force are two different concepts. The first has S.I. Units m/s$^2$ and the second Newtons.
Accelerometers measure acceleration. Force sensors measure forces.
A static force (such as the force of gravity) is a force with constant magnitude. therefore a constant force induces a constant acceleration.
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Can rolling without slipping occur without friction? If a body is rolling without slipping is it necessary that there is friction acting on it ? I encountered a question in which there is a spherical body and a force is being applied on its top point ...so if there is only force then it should do translation motion only .. If there is friction also then it then only it can rotate with translation am I right?
| Rolling without slipping occurs when for the revolutionary body (RB, i.e. sphere, cylinder, disc, ring etc) the following relation holds:
$$v=R\omega$$
where:
*
*$v$ is the translational velocity,
*$\omega$ is the angular velocity,
*$R$ is the object's radius.
Now imagine we make a sphere rotate at $\omega$ and we then lower it carefully on a frictionless surface, so that the translational velocity vector is parallel to the surface and perpendicular to $\vec{\omega}$ and translational velocity scalar is $v=R\omega$.
With no forces or torques acting on the sphere because the surface is infinitely smooth, the relationship $v=R\omega$ holds forever!
Of course one may question whether such a motion really constitutes rolling without slipping. It looks more like slipping without rolling.
But if $v<R\omega$ or $v>R\omega$ then only friction can 'correct' this until $v=R\omega$.
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Wave motion in a transverse wave I was learning about wave motion and how in transverse waves, each particle executes SHM up and down. If that is the case, how is it so that energy is still transferred onto the next particle? The logical answer should be that it disturbs the other particle, but if it moves up and down how does that happen?
| It depends on the type of wave. In the derivation of waves in a string, we come to the equations:
$$\frac{\partial ^2 y}{\partial t^2} = \frac{|\mathbf T|}{\rho} \frac{\partial ^2 y}{\partial x^2}$$
$$\frac{\partial ^2 x}{\partial t^2} = \frac{1}{\rho} \frac{\partial T}{\partial x}$$
The first one is the transverse wave as such, propagating in the $x$ direction.
But the second equation shows some movement also in the $x$ direction. As the tension is almost constant ($\frac{\partial T}{\partial x}$ is small), that oscillatory movement is very small compared to the transverse one.
Also for water waves, besides the transverse (up and down) movement there is also some horizontal one, as shown here.
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Do elementary particles last forever? I have heard that not even black holes last forever, because of Hawking radiation. But what about elementary particles? Will an electron, for example, exist for all time?
| Even the so-called stable elementary particles might not last forever,
when there are other particles around.
Take for example electrons:
*
*Electron-positron annihilation:
The electron may be hit by a positron, both disappear, and 2 photons will appear.
$$e^- + e^+ \to \gamma + \gamma$$
*Electron capture:
The electron may be captured by a proton (from a proton-rich radioactive nucleus),
thus giving a neutron and a neutrino.
$$p + e^- \to n + \nu_e$$
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Can rockets fly without burning any fuel with the help of gases under extreme pressure only? Why is it necessary to burn the hydrogen fuel coming out of the engine for the lift of rockets?
If it is done to create a greater reaction force on the rocket then why can't we get the same lift with just adjusting the speed of the hydrogen gas going out of the engine like we can release them at a great pressure (and also by adjusting the size of the nozzle opening) and thus at a greater speed?
Is it possible for rockets to fly without burning the fuel and just releasing the fuel with a great force? (I know the rockets are too massive).
How does the ISP of the ordinary rocket engines compare with the one in my question ? Most of the answers have done the comparison (and a great thanks for that), but help me with the numerical difference in the ISP's.
(Compare it using any desired values of the amount of fuel and other required things for taking off.)
| Cold gas thrusters, generally using compressed nitrogen, are sometimes used for control, such as adjusting orientation in orbit.
In that case the low thrust is, if anything, an advantage, as it makes precise control easier.
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How to derive this relation? According to A relativist's toolkit by Poisson, the expansion of null radial geodetic in the Schwarzschild spacetime is $$\theta=\dfrac{2}{r}$$
How to derive this expression?
The expansion is defined as $$\theta=k^\alpha_{\phantom{\alpha};\alpha} = \dfrac{1}{\sqrt{-g}} \left( \sqrt{-g}k^\alpha \right)_{,\alpha}$$
and $k_\alpha=-\partial_\alpha \left[t-\int dr \left(1-\dfrac{2m}{r}\right)^{-1}\right]$
If I substitute all the values of the Schwarzschild metric, I get:
$$\theta = -\dfrac{2m}{(r-2m)^2}\neq\dfrac{2}{r}$$
Where do I go wrong?
| You probably made some trivial error. For example, did you forget to raise the index on $k$?
$$k_\alpha=\left(-1,\frac{1}{1-2m/r},0,0\right)\tag1$$
$$k^\alpha=\left(\frac{1}{1-2m/r},1,0,0\right)\tag2$$
$$\sqrt{-g}=r^2\sin\theta\tag3$$
In the calculation of $\Theta$, only the $\alpha=r$ term contributes:
$$\Theta=\frac{1}{\sqrt{-g}}\left( \sqrt{-g}\,k^\alpha \right)_{,\alpha}=\frac{1}{r^2\sin\theta}\left(r^2\sin\theta\right)_{,r}=\frac2r\tag4.$$
| {
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Rutherford scattering experiment, part 1 Around 1906-1914 several classical experiments on scattering of $\alpha$-particles on gold and platinum foils have been performed by Rutherford, Geiger, and Marsden. In standard literature on the subject it is tacitly assumed that the atoms of gold and platinum remained at rest after the interaction with $\alpha$-particles.
My question is why it is the case, particularly what was known at the time of experiments? Was it known at the time that atoms of gold (or platinum) interact with each other and form a crystal lattice?
| First, while your textbook may assume that the target atoms remained at rest (it makes the math easier), Rutherford, Geiger, and Marsden knew how to do the full kinematics problem and did not assume that. For gold and platinum it makes almost no difference in the analysis since those elements are ~50x more massive than an $\alpha$ particle (He ion).
Second, crystallography had a long history before then, and x-ray diffraction was demonstrated in 1912 (see Wikipedia on that), so, yes, scientists such as Rutherford and his group would have been reasonably sure that their targets were polycrystalline at the least. Still, Rutherford scattering results do not depend on the target being crystalline - crystals, amorphous, liquid, or gas targets will show similar scattering of the $\alpha$ particles.
| {
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Infinite Coulomb force when charges contact The Coulomb force between charged particles is inversely proportional to the square of the distance. Yet, why don't we observe the infinite force when the distance approaches zero? Say we can bring two positively charged glass rods and make them touch each other. We don't observe a very large amount of repelling force.
| A similar statement can be made about the gravitational force. Thus when we take two point particles of mass $m$ and $m'$ arbitrarily close to each other the gravitational force between goes to infinity.
Thus either one of three things can happen:
*
*Either the gravitational inverse square law must be modified on very small scales
*Or a new force kicks in to stop particles of matter approaching arbitrarily closely.
*There is some new kind of physics
Roger Boscovitch, a Croatian physicist of the 18th C, took the second option deduced the existence of interatomic forces. This new force, although he didn't know it at the time, was an aspect of the electromagnetic field.
Similarly here, when we bring an electron arbitrarily close to a proton, quantum mechanics kicks in and the electron and proton form a bound system - a hydrogen atom. This is option 3, the new physics of quantum physics steps in.
| {
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How does the Sun generate its energy? We all know that the sun generates its energy from nuclear fusion in the core. The electromagnetic radiation produced slowly travels upwards, while constantly being absorbed and re-emitted by the charged ions, until it reaches the photosphere, where it can basically travel freely (because there are less charged ions), until it travels out into space and then into our eyes.
But I just realized that this doesn't seem to be consistent with the exercise where we calculate the surface temperature of the sun using the Stefan-Boltzmann law. This law is a consequence of blackbody radiation theory, and so by using this law, we're now assuming that the sun's energy comes from the thermal motion of the particles in the photosphere. But as explained in the first paragraph, the energy actually comes from nuclear fusion deep in the core. I am probably stupid for not figuring out how these two explanations of the energy are consistent, but apparently I can't and I need help. Is it because the radiation from the core is absorbed by the photosphere, and then being re-emitted as blackbody radiation? Or is it because of something else?
| The sun light received at earth, is approximately fitted with the black body curve:
Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power.
That the law used the surface area for the definition
Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time
does not mean that the radiated energy comes only from the surface of the body under consideration, whether at low temperatures or at sun temperature. The surface area is necessary to compute the power. Black body radiation comes from the whole body.
The fact that the radiation from the sun follows more or less black body radiation means that the radiation comes from the whole sun, the one coming from the center taking a large number of years to come out finally added to the one of the plasma of the surface give the observed curve. That it is not an exact fit to a black body is due to these various mechanisms by which the radiation is produced.
| {
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Are distributions of position and momentum assumed to be independent in quantum mechanics? Given a wave-function of a single particle we can calculate probability density for positions. We can also calculate probability density for momenta. Are these probability densities assumed to be always independent?
Or, in other words, if we measure position and momentum of a particle (for example electron in hydrogen being in the ground state), should we expect that these two random quantities independent?
| To say that position and momentum are not independent random variables in QM really undersells the point. It is not that we cannot measure them independently, the Heisenberg uncertainty principle states that we cannot measure both quantities simultaneously at all.
What this means is that the distribution of momentum depends not only on the outcome of a measurement of position, it depends on whether we measured position at all. We cannot measure momentum and then ask what we would have measured if we had also tried to measure position, because that would change the outcome of the momentum measurement.
| {
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Weird projectile motion question The question is as follows:
A ball is thrown from a point $O$ towards a vertical wall in such a way that, after rebounding from the wall, it returns to $O$ without striking the ground. The ball’s initial velocity has magnitude $U$ and is at an angle $θ$ above the horizontal. When the ball strikes the wall, the horizontal component of its velocity is reversed and halved, but the vertical component is unchanged.
(i) Show that $U^2\sin{2\theta}=3gb$, where $b$ is the horizontal distance of the wall from $O$.
(ii) The point $P$ at which the ball strikes the wall is at a height $\frac{2}{9}b$ above the level of $O$. Find $U$ in terms of $b$ and $g$.
(iii) The ball is thrown again from $O$ with the same speed $U$, strikes the wall at the point $Q$, different from $P$ and returns to $O$ without striking the ground. Find, in terms of $b$, the height of $Q$ above the ground.
I found parts (i) and (ii) relatively straight-forward to solve, and I happened to get $U=\sqrt{5gb}$ for part (ii),
My question is: How is at possible that a particle is projected with the same speed from the same point able to follow the same trajectory both ways but hit a different point on the wall? Or am I missing something here?
| I'm not sure I understand the problem exactly, so do let me know if this doesn't answer your question:
If you've solved the first part, you should be convinced that the particle returns to $O$ when it's projected at a value of $\theta$ that satisfies the following equation: $$\sin{2\theta} = \frac{3 g b}{U^2}.$$ It can be shown that this equation has two roots in the regime $0<\theta<\pi/2$. See, for example, this Math.SE answer to Two roots of arcsin(x)
in the range [0,2π]. Essentially, it all boils down to the fact that $\sin{\theta} = \sin{(\pi-\theta)},$ and therefore that $$\sin{2\theta} = \sin(\pi - 2\theta) = \sin\Big(2\left(\frac{\pi}{2} - \theta\right)\Big).$$
In other words, $\theta$ and $\pi/2 - \theta$ are both solutions to the equation, and therefore there are two values of $\theta$ that satisfy the specified relation and consequently two heights that do, too!
| {
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Time transfer from proper to coordinate: apparent Special / General Relativity mismatching in theory In SR we've learned that the time dilation for an observer moving clock w.r.t one fixed in a frame at rest is
$$\tau = \gamma \tau_0 = \frac{\tau_0}{\left(1-v^2/c^2\right)^{1/2}}$$
ref: "Special Relativity - A.P. French" and many others
In this case being gamma > 1, it implies delta t < delta tau
No moving to GR, the basic starting expression for calculating the elapsed coordinate time from proper time for a observer clock located in a mass gravitational field and moving with velocity v w.r.t a frame at rest in the body mass center is (approximating square root at first order for v << c)
$$\Delta t = \int_A^B \left(1+\frac 1 {c^2} U + \frac{1}{2c^2} v^2\right)d\tau$$
ref: "Relativistic time transfer - ITU-R TF.2118-0" and many other
To note that all terms in the integral are positive, also excluding the presence of gravity (U=0), meaning that it would always result delta t > delta tau
This is an opposite result w.r.t. SR expression!
Can anyone clarify this (apparent) contradiction? Thanks in advance.
|
Can anyone clarify this (apparent) contradiction?
It's just different nomenclature. There is no contradiction.
French's equation 4-5 is
$$\tau = \gamma \tau_0 = \frac {\tau_0} {\left(1-v^2/c^2\right)^{1/2}}$$
Note that French's equation 4-5 uses two taus, $\tau$ and $\tau_0$, to represent the time difference between two events as measured by two different observers. The latter ($\tau_0$) represent the time difference as measured by an observer at rest with respect to the two events. The former ($\tau$) represents the time difference as measured by an observer moving with respect to the stationary observer.
French's $\tau$ is coordinate time ($\Delta t$ in more modern nomenclature) while his $\tau_0$ is proper time ($\Delta \tau$ in more modern nomenclature). A more modern way to write French's equation 4-5 is thus
$$\Delta t = \gamma \Delta \tau = \frac {\Delta \tau} {\left(1-v^2/c^2\right)^{1/2}}$$
With this modernized rewrite it is obvious that there is no contradiction.
| {
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Intuition for the Ricci tensor of a Reissner-Nordström black hole The Reissner-Nordström metric is given by $$g = -h(r)\,{\rm d}t^2 + h(r)^{-1}\,{\rm d}r^2 + r^2\,{\rm d}\Omega^2,$$where ${\rm d}\Omega^2$ is the round metric on a unit sphere $\Bbb S^2$ and $h(r) = 1-2mr^{-1}+qr^{-2}$, where $m\geq 0$ is a mass and $q\in \Bbb R$ is an electric charge ($q=0$ gives a Schwarzschild black hole, $q=m=0$ gives empty Minkowski space). One can compute that $R_{ij} = -qr^{-4}g_{ij}$ for $i,j\in \{t,r\}$ and $R_{ij} = qr^{-4}g_{ij}$ for $i,j\in \{\theta,\varphi\}$.
Is there a physical interpretation for the fact that we get different signs for different sets of indices? Or in other words, why should we expect this to happen?
| Actually, the Reissner-Nordström metric has
$$h(r)=1-2m/r+q^2/r^2\tag1.$$
(The charge is squared.)
The Ricci tensor is diagonal with
$$R_{tt}=-\frac{q^2}{r^4}g_{tt}\tag{2a},$$
$$R_{rr}=-\frac{q^2}{r^4}g_{rr}\tag{2b},$$
$$R_{\theta\theta}=\frac{q^2}{r^4}g_{\theta\theta}\tag{2c},$$
$$R_{\phi\phi}=\frac{q^2}{r^4}g_{\phi\phi}\tag{2d}.$$
This can be understood based on the form of the energy-momentum tensor $T_{\mu\nu}$. It is purely electromagnetic, with
$$4\pi T_{\mu\nu}=F_{\mu\alpha}F_\nu{}^\alpha-\frac14g_{\mu\nu}F^2\tag{3}$$
where $F_{\mu\nu}$ is the electromagnetic field tensor due to the hole’s charge $q$.
The electromagnetic field of a Reissner-Nordström hole consists of a radial electric field and no magnetic field. Its nonzero tensor components are only $F_{tr}$ and $F_{rt}$. In this case one finds that the nonzero components of (3) are
$$4\pi T_{tt}=\frac14g_{tt}F^2\tag{4a},$$
$$4\pi T_{rr}=\frac14g_{rr}F^2\tag{4b},$$
$$4\pi T_{\theta\theta}=-\frac14g_{\theta\theta}F^2\tag{4c},$$
$$4\pi T_{\phi\phi}=-\frac14g_{\phi\phi}F^2\tag{4d}$$
where
$$F^2=F_{\mu\nu}F^{\mu\nu}=-2\frac{q^2}{r^4}\tag{5}.$$
Finally, the Ricci scalar $R$ vanishes (which happens because the electromagnetic energy-momentum tensor is traceless), so the Einstein field equations simplify to
$$R_{\mu\nu}=8\pi T_{\mu\nu}\tag{6}.$$
Combining (6), (4), and (5) gives (2).
So the short answer is that the relationship you found is because the electric field is radial and the magnetic field is zero. Such an electromagnetic field has a mixed energy-momentum tensor of the form $T^\mu{}_\nu\propto\text{diag}(-1,-1,1,1)$ in spherical coordinates $(t,r,\theta,\phi)$, and so the mixed Ricci tensor has the same form.
| {
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Why can we see light further than it shines? This is a question I’ve been thinking about for a while. If I’m standing in the middle of a straight road, during night, I can see a car coming towards me because of its lights even if it is kilometers away. Notwithstanding, the driver can not see me because the car will brighten the road only few hundred meters further. What physics properties of the light causes this phenomenon?
| I have never seen such a load of unnecessary verbage in answer to a question!
We see the light from stars millions of light years away from us but only a complete dum-dum would expect an observer on a planet orbiting a distant star to see that star's light reflected off say an astronaut doing an EVA in Earth orbit.
The answer is that there is simply not enough light reflecting back towards the source. It's as simple as that.
| {
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Electronic band structures I am looking for step-by-step, clear, detailed and rigorous explanations of methodologies to calculate electronic bands of semiconductor, GaAs as an example, by solving the Schrödinger equation with the k.p method. Everything I could find so far assumes some degree of prerequisite knowledge. Any advice on readings?
| People often get annoyed when somebody cites wikipedia, but I would like to point out its less known feature: articles have references, which are often worth exploring. Though here, predictably, one of the main references is the Kittel's book. I suppose that there are more in depth treatments devoted to this subject, but it is worth noting that band structure calculations is a topic that is a) old, b) highly complex, and c) very material-specific (so one might have turn to research literature rather than books).
| {
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In Charles' law how to change volume in order to observe the change in temperature? At constant pressure,
$$ V\propto T$$
Letting the piston freely moving fixes the pressure of the gas to the atmospheric;
then heating the gas causes the $T$ to increase,
this causes the $V$ to increase by the same factor.
However, in the other direction, how can one maintain the pressure constant while changing volume? Feels it is impossible to change volume and maintain the pressure constant at the same time...
(to observe how the temperature changes)
|
However, in the other direction, how can one maintain the pressure
constant while changing volume? Feels it is impossible to change
volume and maintain the pressure constant at the same time
Think about it this way. The pressure against the walls of the vessel is proportional to the rate at which the gas molecules collide with the walls. If you increase the temperature of the gas the average speed of the gas molecules increases. That would increase the pressure if the volume was fixed since it would increase the collision rate.
However, if you allow the volume to increase at the same time you increase the temperature, the distance the molecules have to travel between collisions with the walls increases. The increase in distance between collisions with the wall compensates for the increase in the speed of the molecules, thus, the pressure of the gas remains constant.
Hope this helps.
| {
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Will a plastic feel less heavy when I put it in a bucket of water and carry it? If I'm carrying a bucket of water in one hand and a piece of plastic in the other, and then I decide to keep the plastic in the bucket of water (it floats). Will it feel less heavy in the second case?
I think it will feel the same because it's mass adds up to the bucket's mass and will be pulled by gravity with the same extent. But somehow I can't get my mind off from the fact that it's weight is already balanced by the up-thrust.
Is there a simple way to explain how this works? It would be clearer if you helped me with some free body diagrams or an analogy or something simple.
| Assuming the densities of the water in the bucket and the density of the plastic are constant, and that no water spills out of the bucket when the plastic is placed in the bucket, then
$$W_{Tot}=ρ_{w}V_{w}+ρ_{p}V_{p}$$
Where $W_{Tot}$ = Total weight of the combination of the water and plastic in the bucket (ignoring the weight of the empty bucket, i.e., the container), in terms of the volumes and densities of the water and plastic, denoted by subscripts $w$ and $p$ respectively.
The first term on the right side is the weight of the plastic outside the bucket and the second term is the weight of the bucket of water without the plastic floating in it. Although the weight of the plastic decreases when placed in bucket due to the buoyant force acting upward, at the same time the weight of the bucket that now includes the plastic increases by an equal amount, so that the combined weight is still the same as the sum of the weights of each by themselves.
The above assumes that no water spills out of the bucket when the plastic is placed in the water. If before putting the plastic in the bucket the bucket was filled to the rim with water, then the water displaced by the plastic will spill out of the bucket, reducing the total weight. In this case, the reduction in total weight exactly equals the reduction in the weight of the plastic floating in the water.
Hope this helps.
| {
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How do Electrons in a Crystal Lattice form Pair bonds? How do electrons around adjacent atoms in a crystal lattice form pair bonds when they are both negative and should repel each other?
| Strictly speaking, the bonds are not formed between the electrons, but between atoms, which are electrically neutral, because they also contain positively charged nuclei. It is difficult to give a deeper answer, since there exist many types of chemical bonds and the answer is different for each of them. A good start could be reading about covalent bond in the hydrogen molecule, typically discussed in QM textbooks.
Update
The bond types differ from one type of crystal to another, depending on the type of atoms and the crystal structure. For example, in diamond, silicon, GaAs, all the bonds are covalent, like in organic chains. Although in GaAs they are polar covalent, since the nearest neighbors are different types of atoms (otherwise GaAs and diamond have the same crystal structure). In graphite, the bonds are covalent within each graphene layer, but the coupling between the layers is via van-der-Waals forces. Metals are usually held together by metallic bonds, where electrons are shared by all the ions - one can think of it as a generalization of a covalent bond to more than two atoms (i.e. N electrons are shared by N ions).
| {
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Why do aluminium sheets wrinkle easily but are hard to un-wrinkle? I notice when I buy aluminium sheets (the kind used for wrapping food) they come in straight, smooth rolls. Once I use them though, they become wrinkled, and they are impossible to un-wrinkle. Image of a wrinkled sheet:
Why is this the case? Only thing I can think of is the Second Law of Thermodynamics, but I can't see why that would be applicable. It's still the same sheet of aluminium, after all, and it's not like it's getting mixed with something else (like when making coffee). Furthermore, the manufacturers are able to make it unwrinkled, and it doesn't become wrinkled until it's perturbed. I'm tagging the question with thermodynamics anyway, because I don't know what else it could be.
| When aluminum is plastically deformed as during the wrinkling process, the aluminum right in the wrinkles gets stronger than the undeformed aluminum. This process is called strain hardening and is most easily demonstrated with a piece of aluminum wire. If you bend the wire into an angle and then try to unbend it again, the bent zone resists unbending more strongly than the wire near the bend, and so the originally bent portion stays bent and the adjacent portions of the wire bend instead. If you then try to unbend them, the process just propagates outwards along the wire and instead of getting a straight piece of wire, you get a wire with a bunch of wavy bends in it.
You can unbend the wire if you squeeze it between two flat surfaces really hard and roll the wire between them, and at least in principle it is also possible to "iron out" the wrinkles in a piece of aluminum sheet too.
| {
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Can non-harmful eddy currents be used to heat water? I think after 'googling' some web sources that eddy currents are strictly localized in a volume of iron surraunded by a loaded inductive coil so can not cause an electric shock to a person while the person touches a point of that iron that is farther from the coil if the iron piece is long enough.My oppinion is if they cannot be transferred to a person directly they also should not be transferred by means of water. So my question is as follows:can eddy currents be used to heat water passing through a hollow iron piece (pipe) and so be used as an instant induction water heater?(Ofcourse an iron surface should be treated with anti-corosive layer) .
| yes you can. eddy current induction in hollow steel pipes is used to heat-treat the pipe and any liquid flowing through the pipe at the same time is going to get heated up by contact with the pipe's walls.
| {
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Why the inner core of planet Earth is not composed of the densest materials? I read on various sources the inner core of our planet is said to be mostly composed of a iron-nickel alloy, plus some lighter elements like silicon. On the other hand, I know that in general lighter elements go to the upper layers and heavier sink, because of buoyancy. So I am wondering why the densest materials are not considered in the inner core composition, namely osmium 22.59 g/cm3, iridium 22.56 g/cm3, platinum, all the way down lead.
Of course some of those heavy materials are rare. But even if they are rare, I am not sure this is the reason why they are not mentioned at all when describing the inner core.
I found a related question but not equivalent in my view because it does not address the core composition per se:
Why heavy elements don't sink to the core?
| Two years on and Winston´s question is still not answered.
Osmium and iridium are distinctly heavier than all other elements and their natural alloy, osmiridium, should find its way (by hydrostatics over millions of years) to the centre of the Earth.
Cox estimates the abundance of Os at $10^{-4}$ ppm and Ir at $10^{-6}$ ppm. Supposing an osmiridium alloy with a density of $22.58\;\mathrm{g}/\mathrm{cc}$, reflecting relative abundance, that means most of the terrestrial osmiridium should be in a sphere about $2\;\mathrm{km}$ in radius at the centre of the earth . . .
I suspect it is.
| {
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Why do waves, specifically light, diffract through a slit? I've been wondering this for a while now, and have thus far only come across answers that seem to use an equation as an explanation. I've also looked at Huygens' principle (albeit not in-depth), but this doesn't make much logical sense to me.
I'd sincerely appreciate if anyone could shed any light on the topic for me (pardon the pun :P), or try to explain Huygens' principle to me.
Thanks in advance!
| A plane wave, by construction, has all frequency components in the direction perpendicular to propagation. Now you let that impinge on a slit (or multiple slits) - you are now altering those frequency components, and what comes through the slit(s) is no longer a plane wave but the Fourier transform of the real space slits.
Now, the link to Huygen is pretty straightforward - it is all related to the surviving frequency components between the plane wave (a solid line of the wave sources) and what comes through. The math is all the same.
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Why are the capacitors in this circuit in parallel but not in series?
In the circuit, the capacitors are said to be connected in parallel. Why is that so?
Edit: The switch will be closed and C2 is fully charged by C1 and no more current will flow between C1 and C2. The question asks for the voltages and charges hold by C1 and C2. In the solution, it is mentioned that C1 and C2 are connected in parallel (V1 = V2), which is the part I don't quite understand.
| This is a nice question, Consider this circuit first, that build from one resistor and battery,
We can apply kirchhoff's law and we can get that $$V_R=V$$ while $V$ is the battery voltage, thus we can say that the resistor is connected parallel to the battery.
Now check the following circuit,
We can see that applying kirchoff's law here will yield $$V= V_{R_1} + V_{R_2} +V_{R_3}$$
so now the resistors is connected in serieis.
According to your question, this case is similar to the first circuit I mentioned, because the capacitor $C_1$ is charged, it can used as a battery source for the circuit and $C_2$ will be in the role of the resistor (it's just for analogous for the first circuit, of course there is difference between capacitor and resistor) , thus we can get from kirchoff's law that $$V_{C_1}= V_{C_2}$$ wich means that the capacitors are connected in parallel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/579916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Electric Potential Energy - How a charge can be brought from infinity to a point without accelerating it? Question:
My Book says:
ELECTRIC POTENTIAL ENERGY: Electric potential energy of a charge($q_o$) at a point(A) in the electric field due to any charge is given by the work done by an external force to displace $q_o$ without acceleration from infinity to that point(A).
$\color{red}{\text{How it is possible to displace a charge from infinity to a point without accelerating it}?}$
Suppose a point test charge q is located at:
(1) Infinity
Then the moment an external force $F_{external}$, the charge gets accelerated.
(2)A point in Q's electric field
Then the charge Q would exert an electrostatic force($F_Q$) and point charge q would accelerate due to this force. My book says now an external force $F_{external}$ is exerted to move it without any acceleration. Book does not specifies which force is larger.
Now three cases arrive:
$$(1) F_{external}>F_Q$$$$(2)F_{external}=F_Q$$ $$(3) F_{external}<F_Q$$
In (1) and (2) cases there would be some net force, so there would be acceleration.
In (3) case the net force would be zero so the charge would be at rest.
| Only because $q_0$ is very far away, it doesn't mean that it must be at rest with respect to the point $A$. I can suppose that no matter how far I imagine $q_0$, it has a radial velocity $v$ in direction to $A$.
Suppose the charges have opposed signals. Some control device attached to $q_0$ makes sure that the increasing force due to decreasing distance is balanced by an equal force opposing to it. So, the net force on $q_0$ is always zero, and the work done can be thought be made by the eletrostatic force or the opposing force, because they are always equal in modulus.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/579974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why we cannot use Gauss's Law to find the Electric Field of a finite charged sheet? I could not understand why we can't use Gauss's Law to find the Electric Field of a finite charged sheet?
| In order to calculate the electric field using only the Gauss law (rather than by solving the Poisson equation) one usually exploits the symmetry of the problem. In textbook cases this usually means that one can guess such a Gaussian surface, that the electric field has the same magnitude and direction in respect to the surface everywhere on the surface. This is not the case for a finite charge sheet, where the electric field lines at the edges of the sheet are likely to have different shape then those leaving closer to the center of the sheet - this is known as fringe effects (I couldn't find an image for a charged sheet, but for parallel plate capacitors the effect has been well studied.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/580186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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