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What is the entropy change of the universe for a rock if it falls from a height into a lake? The rock and the lake are at the same temperature According to my textbook, the entropy change of the universe is $+mgh/T$. I'm confused about why this happens.
after falling (without air resistance), wouldn't the rock possess $K_E = mgh$, which would then be transferred to the lake in the form of heat. Wouldn't this mean that the lake absorbs the same heat energy ($mgh$) from the rock to bring it to a standstill.
Would this not result in change in entropy of universe being $= 0$?
| The lake absorbs part of heat through the friction between the rock and the water's mulecules. At the same time, the internal energy of rock is changed accordingly. Therefore, this process (heating up the lake) is irreversible. So the total entropy increases.
| {
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Why does increasing tension in a string increase the speed of travelling waves? We know that $$ v = \sqrt\frac{T}{\mu} $$ meaning that increase in the tension of a string increases the velocity of the traveling wave. But how exactly does this happen? If we consider that the travelling wave is just a certain amount of energy and momentum ($C$) which is propagating then, I think that increasing tension (along one direction) stretches the string in that direction hence decreasing the density (along the other two directions) therefore for a fixed amount of momentum $C=mv$ to travel less $m$ means more $v$ for a fixed $C$. However, I am not sure of this interpretation.
| Increasing the string tension effectively reduces the remaining elastic capacity.
A "wave" or mechanical signal (such as a force or impulse) propagates through a perfectly rigid material at the speed of sound. If the material is not rigid but elastic, then for each particle along the string, that particle first must move a bit before the elastic force has been established to the next particle. This will take a longer time, and then you see a delayed propagation.
Elastic forces are delayed in their very nature - just try to hang a spring vertically and then let go of the top. The bottom will keep hanging stationary in its spot even while the top of the spring is rushing down towards it. The spring force in a properly "soft" of flexible/elastic spring takes a longer time to propagate than the speed that the top is falling with.
By adding tension to a string you are actually "pre-stretching" it. Try to pre-stretch a spring and then you'll feel that it is much harder to stretch it further - you have used some of its elastic capacity. Each particle along the string is now "less loose" so we have effectively reduced the elasticity and thus reduced the elastic behaviour.
Your own interpretation as a density change along the string is also correct, as far as I can see. I think you can use that as well.
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Differences between observations of light versus gravity for a satellite traveling by the sun A satellite travels in a geodesic by the sun with sufficient velocity to escape the sun's orbit. The distance of closest approach is 100 light seconds when the satellite's velocity is perpendicular to the sun as observed by light. At this point in time, t = 0, the satellite will compare 2 observations: the direction from the satellite to the center of the sun's light with the direction from the satellite to the perceived force of gravity from the sun. Are they perfectly aligned or is there any difference, even if very small? Want to isolate how the sunlight might be affected by the sun's gravitational field versus how propagation of the sun's gravitational field might be affected by the field itself.
| Answer from Safesphere's link: https://arxiv.org/abs/gr-qc/9909087v2, "Does eqn. (2.4) imply that gravity propagates instantaneously? ...it clearly does not...Indeed, the vector (2.5) does not point toward the “instantaneous”
position of the source, but only toward its position extrapolated from this retarded data...Indeed, it can be rigorously proven that no gravitational influence in general relativity can travel faster than the speed of light"
This resolves how celestial objects are attracted to the instantaneous position of other celestial objects by gravity (i.e. not delayed by propagation at c), but also how massive objects outside our light event horizon are also outside our gravitational event horizon. (See correctly answered Can gravity from a massive object outside our cosmological event horizon have any effect on Earth?) Thank you, Safesphere, for the excellent resource.
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Elastic collision with one moving object hitting a stationary object In an elastic collision, I understand that momentum is conserved and kinetic energy is conserved. If billiard ball of silver (with velocity $v_{(Ag)}$ impacts a stationary billiard ball of aluminum, I am trying to calculate the velocity of the aluminum ball after the collision, $v_{(Al)}$. After an elastic collision, the impactor is at rest and the impactee has the motion.
Using momentum, $= m \cdot v$
$$m_{(Ag)} \cdot v_{(Ag)} = m_{(Al)} \cdot v_{(Al)}$$
Assuming silver is 4x denser than aluminium, then using momentum, the aluminium ball should have velocity
$$v_{(Al)} = 4\cdot v_{(Ag)}$$
But if we use kinetic energy, $1/2 m \cdot v^2$
$$\frac12m_{(Ag)}\cdot v_{(ag)}^2=\frac12m_{(Al)}\cdot v_{(Al)}^2$$
$$v_{(Al)}^2=\frac{m_{(Ag)}}{m_{(Al)}}\cdot v_{(Ag)}^2$$
$$v_{(Al)}=\left(\frac{m_{(Ag)}}{m_{(Al)}}\right)^{\frac12}\cdot v_{(Ag)}$$
$$v_{(Al)}=2\cdot v_{(Ag)}$$
Somewhere I have lost some neuron connections in my brain because I cannot resolve this conflict. This is a perfectly elastic collision so both momentum and kinetic energy should be conserved.
I have read multiple threads including:
When is energy conserved in a collision and not momentum?
How to calculate velocities after collision?
How can I calculate the final velocities of two spheres after an elastic collision?
Calculating new velocities of $n$-dimensional particles after collision
Velocities in an elastic collision
Summation of the velocities before and after an elastic collision
| You say...
After an elastic collision, the impactor is at rest and the impactee has the motion.
That is only true if the masses of the two balls are equal. Clearly they will not be equal if one is made of silver and the other aluminium.
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Why does phase conjugation not result in superposition? I'm reading about phase conjugations in optics here while this is more extensive but it's in German. Nevertheless, the image might visualize it quite well:
I wonder why such a reflected phase conjugated wave, which travels back to its source (?!) doesn't add to the impinging wave up to a superposition wave?
Another question though it is quite fundamental: In which way does the phase describe the direction, respectively angle of reflection, here?
PS: Top left reads as "Original and result image", top right is a "phase conjugating mirror", bottom right is "regular mirror" and bottom left is "double distorted image".
"Verzerrt" means distorted.
| The outgoing (phase conjugated) wave will interact with the incoming wave (from the bottle; the interaction isn't visible in the picture though, just for clarity I guess). Just as two opposite traveling pulses on a rope will interact. When the pulses on the rope have traveled "through" one another, they will continue their (undisturbed) motion. The waves in your example are not pulses but long structures. Only the outcoming wave will reach your eye though (undisturbed like each of the two pulses on the rope). After it has interacted with the incoming wave (which goes into the mirror).
In the first image, you see only the on-the-mirror-impinging-bottle-distorted wave. in the second picture, you see only the wave that emerges from the mirror. These two do superimpose. That is, in the classical picture of light. This doesn't mean that there is no energy traveling in between though (even though the two waves might superimpose in such a way that the result is zero).
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Operator Product Expansions using the stress energy tensor I'm reading "String Theory Demystified" by David McMahon. On p.111 of the book, it is asserted that:
One operator product expansion of particular interest involves the energy-momentum tensor. In the complex plane: $$T_{zz}(z)= \ :\eta_{\mu \nu} \partial_z X^\mu \partial_z X^\nu:$$
where the $:$ indicates time ordering.
Then using this result, follows the computation of the OPE of the radially ordered product $T_{zz}(z) \partial_w X^\rho (w)$:
$$\langle R(T_{zz}(z)\partial_wX^\phi(w))\rangle = R(:\eta_{\mu\nu} \partial_z X^\mu(z) \partial_z X^\nu(z): \partial_wX^\rho (w) )\\
=\eta_{\mu\nu} \langle \partial_zX^\mu (z) \partial_wX^\rho(w)\rangle\partial_zX^\nu (z) \\+
\eta_{\mu\nu}\langle\partial_zX^\nu(z)\partial_wX^\rho(w)\rangle \partial_z X^\mu(z)$$
However, I don't understand the second equality in this equation. What are the rules to deduce this? I see obviously why the metric just comes out as it is treated as a constant but I don't quite get what is going with the operators.
| Ok, so it seems that this just Wick contracting the normally ordered fields with those that are not. So we obtain two possible contractions, which give the expressions in my question.
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Since the speed of light is constant and also the speed limit; would you, in your reference frame, have no upper bound on your speed? Let us imagine you are in a vacuum and after having maintained a speed of 0 km/s (standing still) you accelerate to 297,000 km/s (99%). You know this is now your speed because you have a speedometer telling you so. You then decide to maintain that speed for a while.
With the speed of light is always ~300,000 km/s faster than you, what is preventing you from (again in your reference frame) increasing your speed, as shown by a speedometer, an arbitrary amount faster than ~300,000 km/s? After all, the speed of light will always be always faster.
I feel like length contraction even backs this since it will make your space wheels tinier. You're essentially scaled down and your tiny wheels would have to rotate many more times to go the distance just 1 rotation would have taken you with your non-contracted length. This then would cause the speedometer to relay speeds faster than the speed of light.
| You do not need a speedometer; as long as you have fuel to "burn" you can add another $\delta v$, and so continue to accelerate. Eventually you will run out of fuel.
| {
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What is the distinction between work done by torque and force about an axis other than centre of mass? Suppose a point particle is undergoing circular motion about a fixed point such that a tangential force is accelerating its speed continuously. Now I believe that I can calculate the work done on this particle either by Force or by Torque.
However, for a rigid body, when I calculate work, I calculate it as:
*
*Work done by Torque in the frame of centre of mass to cause angular displacement added to the work done by the force in displacing the centre of mass.
or
*By work done by force in displacing its point of application
Now I understand that both of my methods 1 and 2 are equally valid for the circular motion case
But what I do not understand is that why isn't the work done by both force and torque (about point of rotation) added in the circular motion case? Why are these two works synonymous here, but not in method 1?
Is method 1 (as described above) applicable only only when torque is calculated about Centre of Mass and no other point?
| In the circular motion case we only have a particle, so the centre of mass itself is not defined. Since force is tangential, $dW=F ds$, so that $W=F s$ considering constant force. If you apply apply torque method you get $W=\tau \theta =F r\theta =F s$. Thus they are the same result. You can apply method (1) about any point, but be careful to consider the work of a pseudo torque due to the pseudo force that may arise if your chosen point is accelerated. You do not have to concern yourself with that if you choose COM as point of analysis because any pseudo force would pass through COM itself, hence creating no torque.
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Why is it said that antiparticles are a result of combining SR with Quantum theory? I did understand the historical reasons for the discovery of antiparticles in this context. But are antiparticles really a 'consequence' of combining special relativity and quantum theory? Why isn't it better to say that the existence of antiparticles are consistent with QM and SR?
| This is a very interesting question. Galileo once stated that "our Universe is a “grand book” written in the language of mathematics".
But are antiparticles really a 'consequence' of combining special relativity and quantum theory
Yes, they are, in that the relativistic mathematical formulation of quantum theory suggests that antiparticles emerge from this mathematics.
In going from standard quantum mechanics to quantum field theory, we introduce the Klein-Gordon equation $$\pm \sqrt{m^2c^4+{\vec{p}}^2c^2}\; \psi = i \hbar \frac{\partial}{\partial t} \psi$$ and the Dirac equation $$(\beta m c^2 + c {\vec{\alpha}}\cdot {\vec{p}})\Psi = i \hbar \frac{\partial}{\partial t} \Psi$$
These equations have solutions that represent both positive and negative energies. We are familiar with positive energies, but this appearance of negative energy solutions represented a possible conundrum, and so a closer look at was happening was required.
Negative energies were nonsensical, and to make a long story short, these negative energy solutions are explained by the existence of antiparticles.
Why isn't it better to say that the existence of antiparticles are consistent with QM and SR?
It's OK either way. The existence of antiparticles is consistent with the combination of QM and SR, and the mathematical formulation itself also predicts the existence of antiparticles.
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What would you see dropping a sufficiently strong chain with substantial length into a black hole? Here's a visual representation of the scenario
Here you can see we have a black hole on the left. The event horizon is the edge of the black hole. You are far away from the event horizon, and a chain is passing you by fast heading toward the black hole. Due to the size of the black hole and the makeup of the chain, the chain will not break before reaching the event horizon.
After a while, the following scene happens:
As the end of the chain approaches the event horizon, the chain slows down due to the immense gravity as it approaches a frozen state.
As I show in the diagram, you can observe this phenomenon as well as observe the chain moving fast by you toward the black hole.
As for my question..
How can the part of the chain near you appear to be moving quickly toward the black hole, while the end near the black hole is frozen (or close to it)? Where does all that chain go?
Let me ask the same question in another way..
If the distance between you and the black hole is 1000 units, and the chain appears to be almost frozen 1000 units away from you, how could you reconcile watching 10000 units of chain speed past you? How does that 10000 units appear to fit within a distance of 1000 units from your perspective?
| because at the even horizon light cant escape we will see that end disappear as there is no more light for us to see so it is still going into the black hole we just can't see it, as for it looking like it slows down that is just the was space-time has distorted the light, not from the physics being applied to the chain so it is still going at speed.
| {
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Equipotential as a circle
I just dont understand how with this configuration there could exist a equipotential as a circle. For the assumption that $R>>$ dipole size I think it is there for the approximation of potential due to a dipole $p$ at a distance $r$ $$V=\frac{pcos{\theta}}{4\pi \epsilon_{o} r^{2}}$$
This is all I could make it out.
I'm not asking to solve this question but please explain what I need to look out for and how should be the uniform field be in order for the above to work out?
| for the circle to be equipotential the electric field at every point should be perpendicular to it.
i've turned the diagram 45 degrees just so it is easy to understand.See the diagram carefully while reading so that you can understand the solution clearly.(dia 1)
as the net electric field should be perpendicular to the circle the component of electric field tangenial to the circle should be kpsin(theta)/r^3 in magnitude opposite to the arrow in diagram to cancel it.
and also the electric field should not depend on theta so that at every point electric field is same. it can be in two possible ways.(dia 2 and dia 3) (its just that E should be 90-theta degree to the tangent so that sin(theta) cancels so that it wont depend on theta.)
in first diagram the electric field is horizontal and does not depend on the theta whereas in second case the electric field is (180-2theta) to the horizontal whose direction depends on theta.
therefore the first diagram is the correct diagram and magnitude of E acting should be kp/r^3.
by tilting the diagram to the original position you can get diagram 4. hope you can solve it using this.
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Will a massive particle be attracted to an infinitely long beam of light? I was attended to Bonnor beams. A Bonnor beam is the general relativistic exact solution for the spacetime surrounding an infinitely long straight beam of light (it includes also the spacetime including the interior of the beam, but I'm mainly interested in the outside spacetime). It is said that two parallel beams, traveling in the same direction, will never converge. They will always stay parallel. Two anti-parallel traveling beams will converge though.
Does this mean that massive particles (with initial velocity zero wrt the beam) will only be dragged along the beam? That is will they acquire only a velocity parallel to the beam and no velocity perpendicular to the beam?
Do Bonnor beams, by the way, show that photons can exchange gravitons? If two opposite traveling beams converge they must.
| Looking at the original paper by Bonnor, he finds that geodesics of particles are affected by the beams, not just through gravitational attraction but a Coriolis-like force along them:
Note that this is velocity-dependent: a particle at rest would not start drifting along the beam. However, it is hard to avoid this since the beam is heavy (after all, it is an infinite mass distribution, it is a wonder it doesn't implode already) and will tend to attract a particle starting from rest: there will be radial velocity too.
The metric does not tell you anything about gravitons, since they are not part of the entirely classical theory of general relativity. But light certainly can bend spacetime.
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What is the role of the dilaton in Jackiw-Teitelboim 2D gravity? I read that the Einstein Hilbert action is topological in 2 dimensions. (What does that mean?). To write down a non-trivial action one introduces the dilaton field in JT gravity. Does this field have a physical interpretation?
I read some references but did not really understand them.
| For the following, I refer to https://arxiv.org/abs/1711.08482
Dilatonic gravity models often originate from some higher dimensional parent theory. For example, the following action
\begin{equation}
I = \frac{1}{16 \pi G_N} \int d^2x \sqrt{-h} \left[\Phi^2 R_h + \lambda (\partial \Phi)^2 - U \left( \frac{\Phi^2}{L^2} \right) \right]
\end{equation}
is a result of dimensionally reducing the Einstein-Maxwell action for a magnetically charged BH in 4d. The induced metric in this case is $h$ and $U$ is just some potential.
When counting dimensions, the dilaton $\Phi^2$ has a dimension of length$^2$ and is often seen as a measure for the area of the transverse 2-sphere, i.e. the 2 spatial directions over which we integrated.
When specified for JT gravity
\begin{equation}
I_G[g, \phi] = \frac{1}{16 \pi G_N} \left[\int_\mathcal{M} \sqrt{-g} \phi (R + 2) + 2 \int_{\partial \mathcal{M}} \sqrt{- \gamma} \phi_b (K - 1) \right] \\
\end{equation}
The action is topological in the sense that when evaluating the path integral, the dilaton $\phi$ acts as a Lagrange multiplier, setting $R=-2$. So we need to sum over all possible geometries with that curvature. However, the second term contains the dilaton on the boundary $\phi_b$ and when we avert our attention to the boundary, the dilaton becomes dynamical.
In a sense, this is the result of introducing a spacetime cutoff which moves the boundary inwards a bit. Because only at the boundary at spatial infinity gravity can be neglected, this cutoff boundary feels the effect of gravity and the dilaton becomes dynamical. This is often called the boundary particle.
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Tensor product of wavefunctions In a system of two non-interacting particles in a one-dimensional infinite square well, we represent the eigenstate of the whole system as the tensor product of the eigenstates of the individual particles: $|n_1,n_2\rangle=|n_1\rangle \otimes |n_2\rangle$, since both kets belong to different Hilbert spaces.
However, if we solve the problem by separation of variables in the time-independent Schrödinger equation
$$
\widehat{H}_{0}\left(x_{1}, x_{2}\right)=\widehat{H}_{0}\left(x_{1}\right)+\widehat{H}_{0}\left(x_{2}\right)
$$
$$\psi_{n_{1} n_{2}}\left(x_{1}, x_{2}\right)=\psi_{n_{1}}\left(x_{1}\right) \psi_{n_{2}}\left(x_{2}\right)=\frac{2}{L}\sin{\left(\frac{n_1\pi x_1}{L}\right)}\sin{\left(\frac{n_2\pi x_2}{L}\right)}$$
Why is not the tensor product used in case the approach is made through wavefunctions?
| I'd say that it is merely for historical and cultural reasons. Mathematically there is an isomorphism
$$
L^2[{\mathbb R}^3]\simeq L^2[{\mathbb R}]\otimes L^2[{\mathbb R}]\otimes L^2[{\mathbb R}],
$$
but we customarily use the first form for a single particle moving in 3d.
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What is the $y$-axis in an electromagnetic wave? Apologies if my question is unclear, any help to clarify it along the way is most welcome.
I'm confused about what we mean when we say electromagnetic 'waves' (say visible light). In the usual mental picture we have of a simple sine wave, what does the y-axis correspond to? In mechanical waves like water or sound, we can plot the vertical displacement of each particle along the x-axis as a value on the y-axis. Alternatively we can fix a specific particle in the water/air and take the x-axis to be time and the y-axis to be its physical displacement.
But for light traveling in a vacuum, there would be no such displacement of particles. So what does the y-axis correspond to? What do we mean by light being/behaving like a wave?
| Maybe this will help
Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This 3D animation shows a plane linearly polarized wave propagating from left to right. The electric and magnetic fields in such a wave are in-phase with each other, reaching minima and maxima together.
Here y is the axis of propagation of the wave. Because of polarization x,z are the axis where the electric and magnetic field amplitude changing with time is expressed for each point on the y axis. In diagrams the y axis could be considered an optical ray.
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Wave packet in quantum mechanics? When we talk about light waves or EM waves, we simply say that the wave packet is the superposition of other waves of different wavelengths. In quantum mechanics, we say the same thing; the superposition of many waves associated with electron form a wave packet. I don't understand this, because one wave is associated with one electron. It is only superimposed with other electron waves. Does an electron make superpositions with itself having different wavelengths? How is this possible?
| The state of a particle is given by the wave function (state vector $\psi$), which we get by solving the Schrödinger's equation. This wave function can be written as the superposition of many other simpler wave functions analogous to Fourier decomposition. Hence, the superposition of those simpler wave functions gives the overall wave function of a single electron.
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Why don't electrons accelerate in a circuit? In a circuit, electric field exerts force on electrons, so they must accelerate. Every text book I have read, points that electrons move with a constant drift velocity. How can this happen? Does Newton's law not apply there?
| Newton's law does apply but the electrons in the circuit aren't moving in a straight line like a car on a straight road for example. The motion of the electron is rather chaotic, they bounce around in the crystal and their average velocity is considered to be proportional to the Electric field.
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Multiple spin measurements in the same direction What will be the outcome of the experiment during which charged particles will go through the set of Stern Gerlach apparatus aligned in the same direction (say Up)?
[SG Up] < ?1
[ Source ] -> [SG Up] <
[SG Up] < ?2
Has been this experiment actually conducted? What was its outcome?
*
*for 1 "all up"
*for 2 "all down"
or something else?
| The Stern-Gerlach apparatus creates what is known as spin-path entanglement – ie. the particle's spin and its position become entangled. More specifically, if a particle flies from a SG apparatus up, then we are sure that its spin is also up, and vice versa.
If we measure spin along an axis and the incoming particle is already “polarized” along the same axis, we will get just one result with 100 % probability. The reason for this is simple: assume a particle's spin points along an axis $\vec s$ and the SG apparatus has an axis $\vec a$, then the probability of measuring “up” is given by $\cos^2 \frac{\alpha}{2}$, where $\alpha$ is the angle between $\vec s$ and $\vec a$. For $\alpha = 0$ the probability is $1$ and for $\alpha = 180^°$ the probability is $0$.
Therefore, your intuition is right. Since all the particles ariving at detector 1 are spin up, they will all bend upwards. Likewise, all the particles ariving at 2 are all spin down, therefore they'd be deflected down. If your source is giving off “unpolarized” particles with spin directed in random directions, then you'd get a 50-50 chance of measuring “1 up” and “2 down”.
But as @Charlie pointed out, this applies to “fast” consecutive measurements, ie. those where you don't apply any additional tricks to the flying particles. For example, if there was an additional homogeneous magnetic field between the SG apparatuses, then the Larmor precession would steer the spins away from the perfect zero angle and you could measure “1 down” and “2 up” with non-zero probability.
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Weightlessness and free fall on oil drop experiment Suppose I have an oil drop which is negatively charged placed in a box with the bottom of the box negatively charged. When this box and oil drop is placed on a table where the box is stationary, the oil drop is stationary in the middle of the box. Hence, we can conclude that the force of repulsion between the negatively charged oil drop and bottom surface of the box is EQUAL to the the weight.
So now, I drop the whole system from the top of a building. Assume there is no air resistance. Since the whole system is in free fall, is it correct to say that for the same oil drop and box, the oil drop will now move upwards towards the top of the box?
Since free fall is the same as being in outer space with zero gravity.
| Simple answer is, the oil droplet will slowly move upwards. Because, the oil droplet is floating in the middle of the box for electrostatic repulsion. 'The droplet can't start it's free fall until the box's surface or the box move downwards'— What does it mean? It means that that box starts it's free fall before the oil droplet;and so the box gains a higher velocity than the droplet due to gravitational acceleration. And if the box gains a higher velocity than the droplet, the droplet will start going upwards in the box.
| {
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Is Landau theory for phase transitions valid only for "order to disorder" phase transition? In the Landau theory we assume order parameter that is equal to zero at $T>T_c$ and none zero at $T<T_c$ which is valid only for order to disorder phase transition according to my understanding.
So that is mean that I can't use Landau theory on Liquid - Gas phase transition?
| Away from the critical point, the liquid-gas phase transition is a first order phase transition, so you must use a theoretical description that treats it as one. The standard formulation of the Landau theory does not treat first order phase transitions, but it can be modified so that it does.
The liquid-gas phase transition through the critical point is second order, and the Landau theory can be applied to it.
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Why is power transmission carried out at low current high voltage? My textbook states that power is transmitted at high voltage and low current since $P=I^2R$ and as the current has a small magnitude, the heat dissipated across the transmission lines is less than when we carry it out at high current and low voltage. But $P=I^2R$ can also be written as $P=V^2/R$ and hence a discrepancy would occur. Where am I going wrong?
| When we talk about high voltages we do not mean a high potential difference between the two ends of the wire. In any case we want to minimise the potential difference across the wire as this obviously means a loss of potential in the wire( the thing that is needed to power things)
in the equation
P=v^2/r
v is the potential difference across the wire, and R is the resistance along that stretch of wire. so by using this equation we can minimise the power losses by lowering the potential difference across the wires, Well how do we do this?
in P=IV the V is the potential at a point(measured against 0 potential)
so if the initial potential is increased at the start of the cable, for the same power, the current must be lower. Which, reduces losses in the wire as per I^2*r. Which means the potential difference ACROSS the cable is lower as less is being wasted as heat
You're confusing PD across the wire, vs the initial potential.
Which is exactly why "voltage" is such a bad word as it means different things
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Solving the Schroedinger equation with the initial condition as an energy eigenstate I was studying quantum mechanics by watching a video lecture series. In the lecture https://youtu.be/TWpyhsPAK14?list=PLUl4u3cNGP61-9PEhRognw5vryrSEVLPr&t=2784 , the professor tries to solve the Schroedinger equation with the initial condition of being in an energy eigen state.
$$\hat{E} \psi(x,0) = E \psi(x,0) \ \ \ \ \ \ \ eqn (1)$$
$$i\hbar\partial_t\psi(x,0) = \hat{E} \psi(x,0) = E \psi(x,0)\ \ \ \ \ \ \ eqn (2)$$
$$\partial_t\psi = \frac{-iE}{\hbar}\psi\ \ \ \ \ \ \ eqn (3)$$
$$\psi(x,t) = e^{-i\frac{Et}{\hbar}}\psi(x,0)\ \ \ \ \ \ \ eqn (4)$$
The above were the series of equations the professor wrote on the blackboard. I don't understand how you get from equation (2) to equation (3). Eqn (2) is valid only at $t=0$. How do you then get equation (3) which looks like it is valid for all $t$?
| By equation 2, $\psi$ at some infinitesimal time $\epsilon$ will be in the same state, multiplied by a phase $e^{-iE\epsilon/\hbar}$. Since it's the same state, you can use equation 1 again. And you can keep doing this forever.
But I feel like there should be a cleaner answer than this.
| {
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Friction between two horizontal rotating discs Let's say we have two rotating disks and one disk is given some angular velocity. As one disk rotates, due to friction the other disk should rotate as well. The two disks are completely identical (radii, mass etc). The two disks are lined up next to each other horizontally, and only the rims of the two disks are in contact.
In this setup, the normal force and the frictional force are the most important forces. But what is the exact relation between the forces and how can we calculate them?
| Initially, there is a relative motion between the two discs. So, the frictional force is given by $ F_{fr} = \mu_k N $, where $N$ is the normal force, and $\mu_k$ is the coefficient of kinetic friction. At one point, the relative velocity between the two discs becomes 0. At this stage, the frictional force also be 0.
Hope this answers your question.
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Why is 4-velocity not defined as the covariant derivative of position instead of the regular time derivative? The geodesic equation is usually written as
\begin{equation}
D_\tau u^\mu = 0
\end{equation}
where $D_\tau= u^\mu \nabla_\mu$ is the covariant proper time derivative and $u^\mu=\frac{dx^\mu}{d\tau}$ is the 4-velocity. I understand why the equations of motion for a free particle in curved spacetime have the covariant derivative $D_\tau$ and not the usual $\frac{d}{d\tau}$: As we move in curved spacetime we need to account for the effect that curvature has on vectors. Usually this term is called the acceleration
\begin{equation}
a^\mu=D_\tau u^\mu
\end{equation}
My question is: If velocity is defined in this way... Why is 4-velocity not defined in this way too? In other words, why is the formula for 4-velocity this
\begin{equation}
u^\mu =\frac{dx^\mu}{d\tau}
\end{equation}
and not this
\begin{equation}
u^\mu =D_\tau x^\mu
\end{equation}
| Because position is not a vector.
| {
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Why doesn't current flow through an open branch? I know that current doesn't flow through open branch because current can't flow through air due to its high resistance .But i was thinking , what's the problem if current flows through an open wire (assumed 0 resistance for matter of circuit solving). I mean, isn't there a possibility that the charge flowing through the wire keeps accumulating at its end as it can't flow through air as battery is incapable of flowing it through the air. Why do we have to entirely abandon that wire?
| Voltage is the driving force .
If you have a real ( non ideal) wire and connect it across a battery, due to its non zero resistance, there will be a continuous drop in voltage. If the battery is 5 V rated, the potential at one end of the wire is 5, dropping down to 2.5 V at its mid length and all the way to zero at the other end.
Now if you take another wire and just touch one end at the mid way point of the above wire( and leave the other end hanging, not connected to anything. By this way, you get an open branch) and wait for a second to let it achieve steady state, what happens?
The potential across the entire second wire is 2.5 V. Since current requires a potential difference , no current flows.
It is useful to think of charge as water flowing from the battery but it leads to confusions like the one you have ( i.e. does it get accumulated etc.)
However, your theory is not completely wrong since such a charge accumulation is exactly the reason a voltage is caused along a rod that is moving perpendicular to a magnetic field.
| {
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EM wave power dependency on frequency Does a (classical) radio wave with a given amplitude carry more power if that wave is at a higher frequency than at a lower frequency?
| Energy density in EM field is given by
$$
u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2 \mu_0} B^2
$$
and energy flux (energy crossing unit area per unit time is)
$$
{\bf S} = {\bf E} \times {\bf H}
= \frac{1}{\mu_0} {\bf E} \times {\bf B}
$$
where the second version applies in vacuum.
It follows that the answer to your question is no: there is no dependence on frequency if the amplitude is given.
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A clarification on acceleration and velocity This is one of those questions which require an answer that does not take practical limitations into account. It is a theoretical physics question, perhaps. If there are any loopholes used, please explicitly state them.
If the position is known as $x(t)$ from t=0 to t=1 second, how do I get the velocity at the initial and end points, since velocity at the end point will require $x(1-(\Delta t)/2)$ and $x(1+
(\Delta t)/2)$, which are added and divided by $\Delta t$ ?
It gets worse if I want to know the acceleration at the end point, which requires the $v(1+(\Delta t)/2)$ which in turn requires $x(1+(\Delta t))$, which is simply not available.
Is this an order thing or is it just neglected in calculus?
| Strictly speaking, the velocity at the end points is not defined, since you cannot determine either the left-hand or the right-hand limits to the change in position at those times as the time interval gets arbitrarily smaller.
Since velocity is the time-derivative of the position, $$v(t) = \frac{dx(t)}{dt}.$$ For this derivative to be defined at $t$, we must accordingly have
$$v(t_+) = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t} = \lim_{\Delta t \to 0} \frac{x(t - \Delta t) - x(t)}{-\Delta t} = v(t_-).$$ That is, both the right-hand as well as the left-hand derivatives must be defined, and they must be equal. In the interval $t \in (t_0, t_1)$, the right-hand derivative is not defined at $t = t_1$, whereas the left-hand derivative is not defined at $t = t_0$. Therefore, mathematically, the function $x(t)$ is not differentiable at the end-points: the velocity is not defined at those points.
The velocity, however, still exists on the interval $t \in (t_0, t_1)$. That is because, by definition of differentiability on an interval, the function $x(t)$ just needs to have a right-hand derivative at $t= t_0$ and a left-hand derivative at $t = t_1$ to be considered differentiable. This is assuming, of course, $x(t)$ is sufficiently nice and smooth everywhere in between. See this for more information on differentiability.
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What does an $x$ Watt bulb actually means? Let's say I have a 11W bulb in my home, connected to a 220V power supply. What exactly does it mean that this bulb is 11W? As far as I know, the Wattage is determined by the formula $W = V * I$, so it really depends on the voltage and the current. The current is determined by the resistance: $V = R * I$. So I don't understand what is the meaning of a 11W bulb?
| $11 W$ means the bulb uses 11 watts of power when it's operating at the rated voltage. If you run it for an hour for example, it will cost $0.011 kWh$, and your power company will bill you for ___ (check your local electricity prices).
You can also calculate how much current the bulb will draw since you know the voltage, as well as the resistance of the bulb (from $P = V^2/R$).
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If lower gear generates more torque at the wheel, why does it generate less acceleration/speed of the car compared to higher gear? If lower gear generates more torque (compared to higher gear) at the wheel, that would presumably mean the wheels can exert more force on the floor to propel the car forward. Thus, why does lower gear generate less acceleration/speed of the car compared to higher gear? I'm asking this to understand how changing to a lower gear helps with braking when a car is driving downhill.
Notes:
*
*Diameter of wheels of the car is constant
*Assume weight/resistive forces are constant for a car driving in lower gear and the same car driving in higher gear
|
Thus, why does lower gear generate less acceleration/speed of the car
compared to higher gear?
The wheels RPM depends on the motor RPM and gear ratio. $N_w = \frac{N_m}{i}$. The acceleration is the derivative with respect to time: $\frac{dN_w}{dt} = \frac{\frac{dN_m}{dt}}{i}$.
So, for smaller $i$ (higher gear), the (potential) acceleration is greater. The problem is that $\frac{dN_m}{dt}$ is not an independent variable. If we select a higher gear too soon, even pressing the accelerating pedal until the end, the motor RPM will accelerate slowly, canceling the advantage of the higher gear.
On the other hand, when driving downhill, depending on the slope, it is possible that the wheels accelerate the motor instead of the opposite. In this case, if we don't press the accelerating pedal, the amount of mixture in the cylinders is below the required to keep that RPM, and the play of opening and closing valves acts as a air cushion that helps to brake the car (instead of a sucession of explosions that deliver power). That effect increases for higher motors RPM, which can be reached using low gears.
| {
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Does a lens always act as a Fourier transform? I understand that putting a lens behind an aperture at the distance 1f, it will "get" the diffraction pattern to appear in the back focal plane. In this case the FT of the aperture plane happens within the realm of using Fraunhofer diffraction.
Does a lens generally Fourier transform its front focal plane into its back focal plane or is that just the special case with Fraunhofer diffraction and the idea of the lens as a FT is kind of misleading?
| An ideal lens takes a point source that is located on its focal plane, say $\mathcal {F}$ emitting monochromatic homocentric rays/spherical waves and transforms them into parallel rays/plane waves whose direction (propagation phase) depends on the location of the point source in the focal plane relative to the symmetry axis.
By reciprocity it also goes the other way around, that is an ideal lens converts incident parallel rays/plane waves into a point located on the focal plane on the other side of the lens. All this means that as far as the focal planes are concerned, say $\mathcal F$ on one side or $\mathcal F'$ on the other side, homocentric (congruent) rays on $\mathcal F$ are converted into plane waves and vice versa, and thus is a 2D Fourier transform. Point sources are just "Dirac delta", and their FT is constant, hence, plane waves, only their phases vary according to the location of the source. To show this you still need to consider that the system itself is linear shift invariant and linear superposition holds for rays as well as for wavefronts.
Note though that the result of the transformation is not a pure Fourier Transform for two reasons:
*
*there is no ideal lens whose focus is uniformly point-like across
its "focal plane"
*even if it had ideal foci the resulting transformed parallel
rays/planar wavefronts have an excess position dependent quadratic
phase shift (spatial phase modulation). The FT magnitude is correct
but its phase is off. Also, the result is not the FT of the aperture
field (ie., the field at the exit aperture of the lens) itself but
rather the aperture field also carries a spatial quadratic phase
modulation and then we take its FT after the modulation.
| {
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How to generate electric current without a permanent magnet? The question is pretty simple:
Can we build a device that coverts mechanical work in electric current1 without employing a permanent magnet and without access to any external source of current?
The restrictions in place seem to rule out the possibility of current generation via induction; and I cannot think of another practical method. I have heard that industrial alternators sometimes work with electromagnets, but we don't have access to any external source of current, so this path doesn't seem viable.
Do we really need stupid magnetic rocks to produce current? Unacceptable.
To be more specific and minimize to risk of misunderstandings: my question is more or less equivalent to the following one
Can we build a device, powered by hand via some sort of rotating lever, that produces electric current, crucially without employing any external current and without any permanent magnet?
[1]: Usable electric current, let's say sufficient to properly power up a lamp; doesn't matter if AC or DC.
| You have a lot of ways to convert other forms of energy to electricity without permanent magnets, you might be using one right now, batteries, wich uses chemical energy (Unless it is recharged)
Here is a list of what i can think of:
*
*Solar power
*Chemical reactions (A good example for that is batteries)
*Static electricity (Would be funny to see hydropower with a turbine spinning socks on carpets)
*Lighting strikes
*Atmospheric elctricity
| {
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Direction of propagation of electromagnetic waves I have a differential equation for an electromagentic wave propogating in the z-direction and oriented alone the x-axis:
$$\frac{d^2 E_x}{d z^2}+\omega^2 \mu \epsilon E_x=0$$
and if I say $k^2= \omega^2 \mu \epsilon$
the solution to the differential equation yield the following solution:
$$E_x=E_{x}^+ e^{-jkz} + E_{x}^- e^{jkz}$$
where $E_{x}^+$ and $E_{x}^-$ are arbitrary constants.
$E_{x}^+ e^{-jkz}$ is called the forward travelling wave and $E_{x}^+ e^{+jkz}$ is called the backward travelling wave.
Why is the exponent with the negative sign considered as the forward travelling wave while the one with the positive sign considered the backward travelling wave?
Does this have anything to do with stability issues?
| Your equation is incomplete and is not an electromagnetic wave equation. Any wave function $f$ has to depend on both position and time:
$$f = f(x,t)$$
as well as satisfy the wave equation
$$\frac{\partial^2 f}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 f}{\partial t^2}$$
Sinusoidal solutions to the wave equation will, in general, have $(kx \pm \omega t)$ as the argument. It should not be difficult to see that $(kx - \omega t)$ corresponds to a wave traveling in the $+x$ direction, and $(kx + \omega t)$ corresponds to a wave traveling in the $-x$ direction.
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How to identify series and parallel connections of capacitors in complex circuits? Like for this question (picture attached) how to identify if capacitors are in series or in parallel?
| Those batteries and capacitor are all in series. The current will follow through them one after each other.
If they were in parallel, the current could take two different paths (hence the parallel part).
Keep in mind that if you were to extend the circuit, this would change. But as it is, from the viewpoint of each battery, the two capacitors (and other battery) are in series.
This circuit simplifies to a 16 V source applied to a 3F and 5F capacitor in serie. This splits the voltage 10 V and 6 V (5 eighth and 3 eighth) between them, which is how they pick two of the multiple choice answers.
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Is pseudo force just an ad hoc number to explain motion in non-inertial frames? Consider an observer in a non-inertial frame $S$ who observes a particle's motion with a relative acceleration $\vec a_s$ and further calculates (or was told by his fellow observer in an inertial frame) the net real forces acting on it as $\vec F$.
Now the observer adds a pseudo force $\vec f_s$ to the net real forces acting on the particle to explain relative acceleration $\vec a_s$.
Question:
Is what I described above the "proper" method to calculate pseudo forces?
In other words, is a pseudo force just an arbitrary constant that arose out of our desperation to explain non-inertial motion?
|
In other words, is pseudo force just an arbitrary constant that arose out of our desperation to explain non-inertial motion?
Suppose we have a motion happening somewhere in space, now we can either observe it with a frame attached to the person in motion or one which is outside and is inertial. According to newton's laws, the acceleration measured by the person outside is given as:
$$ F_{net} = ma \tag{1}$$
Now, let's see the frame for the guy with accelerating frame, of course, it is naive to apply newton's laws here as they are only stated for inertial frames. However, there is a 'fix' by introducing a new function $G$ such that sum of forces minus the 'perceived' acceleration by a particle in the frame.
$$ F_{net} -ma'=G \tag{2}$$
Now, this is cool and all but how would we calculate G? Well that's easy, you see equation (1) and plug that information into (2):
$$ m(a-a') = G$$
Now what is $a-a'$ physically? This means the difference between acceleration as measured from an inertial frame and the acceleration measured in the accelerating frame of reference.
Here is a famous example, suppose you are in a car accelerating to the right and there is a pendulum attached to the roof, let's say there was someone standing on the ground outside and looking at the car. The discussion on this example is excellently done by Bob D in this post and linked.
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How do I normalise the wavefunction of a hydrogen 1s orbital to obtain the normalisation constant? The wavefunction I've been given for a 1s hydrogen orbital is:
$$ \Psi = A e^{-r} $$
And I need to normalize this to find the value of A. I understand to normalise this I would inset this wave function into:
$$\int_{-\infty} ^\infty \Psi^*\Psi dr = 1$$
When I do this, I get:
$$A^2 \int_{-\infty} ^\infty e^{-2r} dr = 1$$
From the literature, I believe that the normalisation constant of this wavefunction should be equal to $\frac{1}{\sqrt{\pi}}$ but I can't work out where the pi comes from? Or how to evaluate this integral since it is unlike any I have seen. Are my limits wrong? Is there a standard integral I don't know about?
Any help would be much appreciated!
| In 3d spherical coordinates, the integration is
$$
\int_0^\infty dr r^2 \int_0^{\pi} \sin\theta d\theta \int_0^{2\pi} d\phi \vert\psi(r,\theta,\phi)\vert^2
$$
since
\begin{align}
\int_{-\infty}^\infty dx \int_{-\infty}^\infty dy\int_{-\infty}^\infty dz
\to \int_0^\infty dr r^2 \int_0^{\pi} \sin\theta d\theta \int_0^{2\pi} d\phi
\end{align}
in spherical, i.e. the volume element
$$
dV=dx\,dy\,dz\to r^2\sin\theta dr\,d\theta\,d\phi
$$
in spherical. The range of the integration is the standard one for the spherical coordinates, where $0\le r< \infty$, $\theta\le 0 < \pi$, and $0\le \phi < 2\pi$ in the definition of the coordinates.
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I am moving right and rain is falling vertically down.Why should I hold umbrella at an angle? This is related to relative velocity.I get that, from my moving frame of reference rain is making an angle.But still... it doesn't make sense to hold umbrella at an angle when rain is falling vertically down.
| I guess you're asking for an intuitive explanation, because it seems to be mathematically clear.
Well, imagine a set of three columns of raindrops, each row containing three drops. Let's call columns A B C and rows 1 2 3, because they fall orderedly. You first encounter drop 1, then the second, then the third.
For a standing pedestrian, he will get hit by three drops: A1, A2 and A3, that way.
However, if you walk rightards, you first get drop A1, but then, while you walk forwards, row 2 has fallen down, so you will encounter drop B2, and then, for the same reason, you'll find C3. Those drops have been circled in the picture.
So, what you see is "like" if the rain were falling diagonally. The only way for those drops to find you is that the drops "tilt" towards you, but that's just your perception. A standing man will see vertical rain. It all depends on the reference frame.
| {
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Does bitcoin mining take work? I'm neither a professional in cryptocurrency nor physics, but an interesting idea occurred to me. Bitcoin involves mining, which generates a lot of heat as waste.
Is the amount of heat produced by a computer mining bitcoin the same as a traditional heater, if they have the same input power?
I know of course about the law of conservation of energy, but it feels that some "work" has been done on the information in the system, and thus the waste heat energy produced would be less.
I know this mining is not work in the sense $W=Fd$, but it does seem to be decreasing the entropy in the universe by organising information. It feels intuitive that some energy should be used up in the very computation itself. If not, then wouldn't computation be free and hence it could be used to decrease entropy (by organizing information) without using energy?
| I wouldn't say it is organising information
Bitcoin uses one thing called Proof of Work (PoW), which consists in varying some numbers until the hash function matches a target number of zeros.
You're not generating information differently than solving a numerical equation. The energy used in the PoW comes from the power supply. One part is inverted in the computer performance itself, and the rest is heat... but this is not too different than solving a numerical problem, or drawing in a really high resolution image, or any other computer action.
PoW consumes tons of energy, because it is made with that purpose: have a big obstacle so that you cannot get the block for free
There are many people searching for more sustainable alternatives, such as Proof of Useful Work (POUW)
| {
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Indistinguishability of Quantum States and its Consequences In the book Quantum Computation and Quantum Information, there is a discussion about how if states are not orthonormal then there is no quantum measurement capable of distinguishing the states.
I am interested in the consequences of this. What does this mean physically? How is this applied when manipulating quantum states?
I am sure there are other questions that stem from this that I am not even thinking of, but that would be interesting to explore. So any help in understanding the consequences of this would be greatly appreciated.
| I'm not sure if this will help but a very sharp former colleague once posed this question.
You have two boxes. The first contains 1000 horizontally polarized photons and 1000 vertically polarized photons while the second contains 1000 left circularly polarized photons and 1000 right circularly polarized photons. How do you tell which is which?
Since this is sort of a riddle, I'll make the answer below not visible by default.
Use a vertical polarization filter. If it lets exactly 1000 photons through then you opened the first box. If there's a slight imbalance then you opened the second box. Note than even though 1000 is the most likely value for the second box, binomial distributions are such that it is more likely for you to not see the most likely value :).
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When I walk down the stairs where does my potential energy go? When I leave my room I walk down three flights of stairs releasing about 7kJ of potential energy. Where does it go? Is it all getting dispersed into heat and sound? Is that heat being generated at the point of impact between my feet and the ground, or is it within my muscles?
Related question, how much energy do I consume by walking? Obviously there's the work I'm doing against air resistance, but I feel like that doesn't account for all the energy I use when walking.
| This is easily tested experimentally. If you walked down a longer staircase, such as subway escalator, as fast as possible, you would feel that most of the energy have been be dissipated as heat in shin muscles and tendons. Then it rapidly moves with blood flow into the rest of your body.
Energy consumption for 5km/h walking is 4 calories per kilometer per kilogram of weight.
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Mechanical energy in a body moving upwards Why is it that mechanical energy is always conserved, I mean when an object is thrown in air, why does the kinetic energy convert to potential energy and not any other form of energy?
| Mechanical energy is not always conserved. If you throw the object up hard enough to hit the ceiling then suddenly its remaining kinetic energy is converted into heat, sound waves and deformation energy, and it is left with just potential energy. When it falls back down and hits you on the head, same thing happens again.
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Anomalous Dimension and Infinitesimal Transformation In the theory of Renormalization Group (RG) transformations I ended up with the following equation
$$\left( \frac{Z(\mu)}{Z(\mu / s)} s^{d-2}\right)^{1/2} = 1+\left(\frac{1}{2}(d-2)-\gamma_{\phi}\right)\delta s$$
where $Z(\mu)$ is the wave function renormalization, $s$ is a parameter
, $d$ is the dimension of the theory and
$$\gamma_{\phi} = -\frac{\mu}{2Z}\frac{dZ}{d\mu}$$
is the anomalous dimension. How do I get this equation? I know I am supposed to consider an infinitesimal transformation $s = 1+\delta s$, but I do not see how to expand the LHS. Any ideas?
| Expanding the LHS=$\left(\displaystyle \frac{Z(\mu)}{Z(\mu/s)s^{d-2}} \right)^{1/2}$ in $s=1+\delta s$ to first order in $\delta s$ goes as follows:
*
*$\displaystyle \frac{\mu}{s}=\frac{\mu}{1+\delta s} \sim \mu (1-\delta s)$
*$Z(\mu/s) \sim Z(\mu (1-\delta s))\sim Z(\mu) - \displaystyle \mu \frac{dZ(\mu)}{d\mu} \delta s\quad$ Taylor expanded around $\delta s$ hence the factor $\mu$
*$\displaystyle \frac{Z(\mu)}{Z(\mu/s)} \sim 1 + \frac{\mu}{Z(\mu)} \displaystyle \frac{dZ(\mu)}{d\mu} \delta s$
*$\left[\displaystyle \frac{Z(\mu)}{Z(\mu/s)} s^{(d-2)}\right]^{1/2} \sim \left( 1 + \displaystyle \frac{1}{2}\frac{\mu}{Z(\mu)} \displaystyle \frac{dZ(\mu)}{d\mu} \delta s\right) \left(1+\displaystyle \frac{(d-2)}{2} \delta s \right)\qquad$ because $(1+\delta s)^{(d-2)/2}=e^{(d-2)/2\ln (1+\delta s)} \sim 1 + \displaystyle \frac{(d-2)}{2} \delta s$
Expanding the last product again to $\cal O (\delta s^2)$ the fourth equation yields the RHS:
RHS $= 1 + \left( \displaystyle \frac{(d-2)}{2} +\frac{1}{2}\frac{\mu}{Z(\mu)} \displaystyle \frac{dZ(\mu)}{d\mu} \right) \delta s = 1 + \left( \displaystyle \frac{1}{2}(d-2) -\gamma_\phi \right) \delta s$
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What is the definition of a magnet or a magnetic field? Electric forces are the forces which come about between two types of charges, positive and negative. Gravitational forces are the forces between matter. Nuclear forces are the forces which act on the atomic scale and are quantum mechanical forces, they act between nucleons.
And, magnetic forces are the forces between magnets? I felt this definition wasn't specific enough so I searched up the definition of a magnet.
"A magnet is a material or object that produces a magnetic field" Well, what is a magnetic field?
"A magnetic field is a vector field that describes the magnetic influence on magnetic materials"
I feel like this is a loop.
I am sorry, but I am not satisfied and feel like there is a more fundamental definition of magnetic field or magnet that I am not aware of. So, what is a magnet?
|
Electric forces are the forces which come about between two types of charges, positive and negative.
Gravitational forces are the forces between matter.
Nuclear forces are the forces which act on the atomic scale and are quantum mechanical forces, they act between nucleons.
To finish your list, magnetic forces are the forces between aligned magnetic dipoles of subatomic particles.
"A magnet is a material or object that produces a magnetic field" Well, what is a magnetic field? "A magnetic field is a vector field that describes the magnetic influence on magnetic materials" I feel like this is a loop.
Magnetic fields are obtained in two ways:
*
*in a permanent magnet, a part of its particles (electrons, protons, neutrons) is aligned and holds itself up to the specific Curie temperature of this material
*accelerated (circularly moving or linearly accelerated) electrons also align themselves with their magnetic dipoles and realise a common magnetic field.
By the way, the magnetic moments of the electron and the other atomic particles are intrinsic (constants independent of external circumstances). And you get an insightful idea of magnetism and electromagnetic induction if you put all these phenomena in relation to the fact that electrons are not only a charge but also magnetic dipoles.
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If a compass is placed near the middle of the magnet, where will the compass needle point? If we're thinking about a case where bar magnet is placed on a flat surface, with its N-pole on the right and a compass is placed just above the middle of the magnet, in which direction will the compass needle point? Will it go in circles? Here's a diagram:
| Mark H's answer is good, but I just want to add some clarity to what a compass does and what Earth's north and south mean, magnetically, and it's easier to do that with a picture, hence using an answer rather than a comment.
A compass in Earth's magnetic field points North, and a compass points in the direction of the magnetic field, which goes in arcs from the magnetic north pole to the magnetic south pole.
Presumably somewhere early in the history of physics before field lines were invented somebody got confused. Earth's magnetic field is such that the magnetic field lines, which by convention follow arcs from the north pole to the south pole, point north. That is to say, if Earth was a bar magnet, the south pole would be under the geogrpahic north pole, and the north pole would be under the geographic south pole.
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What are the physical principles at play when a glass is stuck to a wet table? I've decided to write a relatively detailed paper on the following situation, but I'm finding the topic quite hard-to-google. Imagine a glass table with water spilled onto it. Once a drinking glass is placed on top, it becomes quite difficult to remove. It is more than the weight of the drinking glass that you have to overcome in order to lift it from the table.
Some more specifics and details:
*
*The contact surface of the drinking glass is a circle.
*Both the table and the glass surfaces are completely flat.
*There is no air trapped between the table and the drinking glass when the latter is placed onto the puddle of water resting on the former.
So, main question: how can we model this situation, taking all relevant things into consideration, in terms of the force required to lift the drinking glass?
Secondary question bombardment:
*
*Does the thickness of the layer of water matter? (If so, in what way?)
*How thick is a layer of water anyway?
*Is atmospheric pressure at play here?
*I assume the surface tension/viscosity of the water too?
| Many glasses have a concave bottom, and there is likely to be some air trapped underneath. In any case, when you try to lift the glass, the pressure under the glass drops and you are working against the air pressure from above. Adhesion and surface tension may also play a roll. As you lift, water flows in from the edges to equalize the pressures, so there may be a lag which depends on the rate of flow (which will depend on the pressure difference).
Any tilt of the glass may also make a difference. For experimentation, you will want to try glasses with different shapes on the bottom. You will need a gripping device that lets you put the center of support above the center of gravity (or not), and a method of measuring the force as a function of time (and or motion). To look at the effect of surface tension, use a surface with a hole under the glass.
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Is it possible to convert a normal force into a vertical force of a moving object into the direction of travel? I have read some questions here about normal forces applied to moving objects but I still can't find an answer to the following scenario:
So there is an object with a constant velocity and on its way the object is impacted by varying normal forces . Inside the object is a damper which brings the object back on its original path.
Can I extract energy from the damper without slowing down the object?
If not, is it even possible to convert the normal force into a force into the direction of travel to increase its velocity?
If yes, what would happen in the "black box" to make it possible in a real technical system?
My approach so far:
On one hand this system seems to be like an object with friction and it's not possible to increase speed with friction. On the other hand kinetic energy is dependent from velocity so if the impact is faster than the damper, there would be a energy difference which can be extracted.
Thanks in advance!
| *
*You can imagine some kind of "regenerative damping" whereby the damping mechanism would store the energy of the shocks rather than dissipate it (actuating a magnet along a coil, etc.)
*However, since the energy of the shock is by definition a consequence of (and hence borrowed from) the kinetic energy of the body or vehicle considered, then in the best case scenario, feeding said energy back to said body would merely help to maintain its initial speed and not actually speed up the moving object.
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Unexpected different results from Newton's second law Please don't ban me. I read through Homework-like questions and I know they should ask about a specific physics concept and show some effort to work through the problem. I hope the question is ok.
I recently came across a mechanic basic problem which I wanted to solve following two different approaches. The odd thing is that the approaches bring different results, so probably there might be some wrong reasoning which I cannot explain. Can anyone explain why results are different?
Here's the figure of the system:
The question is about getting the friction static coefficient between the plane and the bar.
We know the bar weight $P=980N$, and the angle $\theta=60°$, the system is in equilibrium, and the static friction is at its maximum value.
Solution 1
Definition of static friction:
$f_{max} =\mu _{s} N\ \rightarrow \ \mu _{s} =\frac{f}{N}$
Newton's second law, forces perpendicular to the plane
$y:\ N=P\cdot \cos 60°$
Newton's second law, torques relative to the right edge of the bar:
$f\cdot L\sin 60°=P\frac{L}{2} \ \rightarrow \ f=\frac{P}{2\sin 60°}$
Static friction coeff:
$\boxed{\mu _{s} =\frac{1}{2\sin 60° \cos 60°}=1.2}$
Solution 2
Definition of static friction:
$f_{max} =\mu _{s} N\ \rightarrow \ \mu _{s} =\frac{f}{N}$
Newton's second law, forces perpendicular to the plane
$y:\ N=P\cdot \cos 60°$
Newton's second law, torques relative to the left edge of the bar:
$T\cdot L\sin 60°=P\cdot \frac{L}{2}\rightarrow T=\frac{P}{2\sin 60°}$
Newton's second law, forces parallel to the plane:
$x:\ f+T=P\sin 60°\rightarrow f=P\sin 60°-T=P\sin60°-\frac{P}{2\sin 60°}$
Static friction coeff:
$\boxed{\mu _{s} =\frac{\sin 60° -\frac{1}{2\sin 60°}}{\cos 60°}=0.6}$
The solutions for the static friction coeff. are different, in particular solution two is $\frac{1}{2}$ solution 1!!
| In the section "torques relative to the right edge of the bar:"
The torque from the $N$ should be included...best of luck getting the two ways to match.
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Is it necessary to use a collimated light beam when using a polarizer? I am using two polarizers (one for an halogen light source and one for a camera) to measure light reflected in parallel and perpendicular polarization setups. Should the light source used along with the polarizer be a collimated light beam or can it be a diffuse illumination ? (and, if it should, why?)
| Yes, you should collimate your beam. The reason for this is that polarizers work well at the angles of incidence they were designed for, and increasingly worse at other angles. The reason for this is that they rely on optical reflectivity (and interference for thin film polarizers) of the two orthogonal polarizations, which obviously depends on the angle of incidence. Often, there is an acceptance angle indicated on the polarizer's datasheet. For example, for Glan-Taylor polarizers this angle is usually about 1-2 degrees.
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Why doesn't the double-slit experiment violate the uncertainty principle? In the double-slit experiment, when an electron reaches the detector after passing through the holes, it has a certain momentum which we can measure with arbitrary accuracy.
From this data, we can calculate what momentum the electron had when it was passing through the hole.
So we can know the electron's position (as it was inside the hole) with arbitrary accuracy and also simultaneously know its momentum.
Doesn't this violate the uncertainty principle? Where have I gone wrong?
| The narrower the slits ($dx$), the broader the expected (measured and/or calculated) distribution of momentum ($dp$) of the photons passing said slits, so the product ($dx.dp$) cannot get arbitrarily small, therefore the Heisenberg principle ($dx dp \geq \hslash$) is respected in the double slit experiment.
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Why is Hartree-Fock considered a mean-field approach? In studying the Hartree-Fock method for solving systems of interacting particles, I have often found that the method is referred to as a mean-field approach. Wikipedia's page for instance says that the mean-field approximation is implied. I don't see why this is the case though.
In the HF method, one considers a single Slater determinant and minimizes the energy with respect to the orbitals that compose the determinant. The equations that follow this procedure are the HF equations and while it's true that they resemble the Schrodinger equations of independent electrons in a mean-field, they also have an exchange term, which makes those equations different from the Hartree equations, which actually do look like a mean-field approximation. In any case, I don't see where a mean-field approximation is implied.
| In HF we assume that we can separate the wavefunction into a product ansatz
$$
\Psi(r_1, r_2, ...)\Psi^*(r_1, r_2, ...)=|\Psi(r_1, r_2, ...)|^2=\rho(r_1, r_2, ...)
$$
$$
\rho(r_1, r_2, ...)\approx\rho_1(r_1)\rho_2(r_2)\cdots
$$
$$
\rho_1(r_1)=\psi_1(r_1)\psi^*_1(r_1)
$$
This allows us to define single particle potentials that are summed to form the mean field potential $V_H$,
$$
\begin{aligned}
V_H(r_i) &=\frac{-e}{4\pi\varepsilon_0}\sum_{j=1(\neq i)}^N \int dr_j \frac{\rho_j(r_j) }{|r_j-r_i|}\\[1.5em]
V_H(r_i) &=\frac{-e}{4\pi\varepsilon_0}\sum_{j=1(\neq i)}^N \int dr_j \frac{\psi_j(r_j)\psi^*_j(r_j) }{|r_j-r_i|}
\end{aligned}
$$
The potential energy of an particle is then calculated by interaction with this mean field potential of the remaining particles,
$$
\begin{aligned}
E_{pot,i} &= q_e\int_V dr_i \psi_i(r_i)\psi^*_i(r_i)\frac{e^2}{4\pi\varepsilon_0}\sum_{j=1(\neq i)}^N \int dr_j \frac{\psi_j(r_j)\psi^*_j(r_j) }{|r_j-r_i|}\\
&=q_e\int_V dr_i \psi_i(r_i)\psi^*_i(r_i)V_{H}(r_i)
\end{aligned}
$$
I would consider this the reason why HF is called a mean field method.
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Why does the electron not spin? The goto answer to that question is that the electron is a pointlike particle and cannot spin.
The electron is not pointlike though. It is described by a wavefunction. One can prepare the wavefunction to describe a very small electron, but not a point-like electron.
Is there a genuine answer to the problem?
| Particle physicists usually use the term "spin" to denote intrinsic angular momentum, which for a charged particle can give rise to a magnetic dipole moment. In this sense, the electron has spin, even though it is an elementary particle and, as far as we know, has no internal structure.
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Why aren't all quarks clumped together in one giant hadron? As far as I am aware, the strong interaction is attractive only, and its carrier, the gluon, is massless meaning it has unlimited range. If this is the case, how come we only observe quarks in pairs and triplets? What's preventing every quark in the universe from attracting every other quark and forming one ever-growing mass of colored particles?
| In quantum chromodynamics, the theory that describes quarks, there exists a quantum number called color charge or just "color", and all stable hadrons need to have neutral or "white" color.
All hadrons need to have this zero total color charge because of color confinement or quark confinement, such that the color force, or strong force, works only at short ranges, but dramatically increases with distance.
Neutral color particles that exist naturally, are hadrons with a quark of one color and an antiquark of the corresponding "anticolor" i.e., mesons, or three quarks of different colors i.e., baryons, so that its net color is white.
For example, in a proton there exists a quark that is red, one that is green, and another that is blue. Overall the proton color is neutral, or white.
In 2013, scientists at CERN discovered using highly energetic collisions, the existence of a quickly decaying tetraquark, which is a state containing four quarks. There is speculation on how the four quarks are bound in this four-quark state. A lot of physicists tend to agree that these are two mesons $\mid q_1\bar q_1
+ q_2\bar q_2\rangle$ loosely bound to each other. The overall color is still neutral. But again, these are not stable structures. And also pentaquarks, which were also made by high energy collisions, must have three quarks (eg., red+blue+green = white), and the two other quarks with color + anticolor (e.g., blue + antiblue = white).
So we see that quark forces are attractive only when in colorless states. We also see that quark states containing more than three quarks exist at high energies but nevertheless decay very quickly. There does not appear to be a reason why we could not synthesize particles containing multiple quarks (provided the overall state is colorless), though this would require high energy collisions. Given that the color force is very short range that increases in strength with distance, and since the average energy density of the universe is nowhere near the energy required for more than three-quark bound states, we do not see such states naturally.
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Antiunitary operators and compatibility with group structure (Wigner's theorem) From Wigner's theorem, we get that a physical symmetry can be described either by a unitary or antiunitary operator, eventually with a phase factor, as in here.
However we have to respect the group structure, so we need to have:
$$O(f)\circ O(g)=e^{i \cdot\phi(f,g)}\cdot O(f*g)\tag{1}$$
where $*$ is the group operation.
For unitary operators this is fine, but for antiunitary operators there is a problem: the left side of the equation is linear while the right side is antilinear.
Here my doubt comes. The argument of linearity is used to show that only unitary operators can describe continuous symmetries (e.g. spatial translations), but why does it not work for symmetries associated with a finite group (e.g. time reversal)?
I'm not asking why time reversal needs to be described by an antiunitary operator, I'm asking how come an antiunitary operator can satisfy $(1)$ if it's the representation of a finite group.
| I am not sure to understand the problem. What it is clear is that a (unitary or projective unitary) representation is not possible if it is made of antiunitary operators only. That is because, just in view of the representation rules, some operators should be antiunitary and unitary simultaneously as you point out. (In your example, if $f$ and $g$ are represented by anti unitary operators, their composition $f*g$ must be unitarily represented.) In fact, in the concrete cases where the time reversal operation is part of a larger symmetry group, not all operators representing the elements of the group are antiunitary according to the composition rule, even if the time reversal is represented anti unitarily.
Another issue is the possibility of existence of antiunitary elements in the image of the representation. Evidently an element $a$ cannot be represented by an antiunitary operator if, e.g., $a=b^2$, for some other element $b$. This is the case, in particular, when representing a Lie group and when we restrict ourselves to deal with the connected component of the group including the identity element: each element here can be constructed as a finite product of elements $b^2$ as above. That is the reason why, a posteriori, the time reversal symmetry does not belong to that component, as we know that it admits antiunitary representations.
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Why is everything not invisible if 99% space is empty? If every object is $99$% empty space, how is reflection possible? Why doesn't light just pass through?
Also light passes as a straight line, doesn't it? The wave nature doesn't say anything about its motion. Also, does light reflect after striking an electron or atom or what?
| Frame challenge: Yes, each atom is mostly empty space, but there are very many of them in the path of an idealized classical light ray, and even a small amount of obstruction in each adds up.
Even if electrons were little classical spheres with a radius equal to the "classical electron radius" (which is around 5 orders of magnitude smaller than an atom), and photons were point particles moving in straight lines, a photon would have scant chance of passing through a kilometer of air without colliding with an electron. [If we assume 2 grams of matter per mole of electrons, the numbers work out to about 100 kg/m² would block half of the light rays]. So we shouldn't be able to see the sun!
This doesn't mean that the other answers are wrong when they speak about how electrons are smeared out in space. But when you run the numbers, the real mystery needing to be explained by quantum effects is how anything manages to be transparent. For this, we need to take the wave nature of photons into account -- they cannot be correctly understood as simply "particles".
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If air is a bad conductor, how does fire heat up a room? If air is a bad heat conductor, how does fire heat up a room?
Could someone help me, as I really don't get this?
| Probably the most important heat transfer mechanism in the development and spread of a fire in room is radiation which can raise the temperature of the materials in the room to their ignition point. The main role of the air in the room is to provide oxygen to enable ignition and self-sustained combustion.
Hope this helps.
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Double Slit Interference pattern - horizontal or vertical? What determines whether in a double slit experiment the interference pattern will be horizontal or vertical for example? Is it mostly seen horizontal because the slits are small enough horizontally for interference pattern to be seen? Also, I sometimes see a "rectangular" interference pattern when using a lens to focus the light. How does that happen? Shouldn't we expect a circular pattern?
| If the slits are vertical, the light will diffract left and right, creating a horizontal pattern.
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Do $2s$ and $2p$ orbitals have same energy? While solving the Schrodinger equation for the H atom, we get $E_n$ depending exclusively on $n$ (actually on $\frac{1}{n^2}$). Then I thought 2s and 2p orbitals must have the same energy.
But while reading Molecular orbital theory in Atkin's Physical chemistry book, I found written that "$2s$ and $2p_z$ orbitals have distinctly different energies".
Can anyone please explain this?
| The hydrogen 2s and 2p orbitals do not have the same energy. They are separated by the tiny Lamb shift, which is only 1057 MHz or 4.3714 10$^{−12}$ eV. The origin of this energy difference lies in radiative (QED) corrections to the degenerate Dirac result.
Note that there is also hyperfine splitting due to interaction with the proton spin. This is different for 2s and 2p.
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Entanglement in 2D Harmonic Oscillator system Let's assume a 2 dimensional harmonic oscillator system with the Hamiltonian $\hat{H} = \frac{1}{2} p_x^2 + \frac{1}{2} p_y^2 + \frac{1}{2} \omega_x^2 x^2 + \frac{1}{2} \omega_y^2 y^2$ with the solution of the ground state being simply the product of the the ground state of each independent mode.
$\psi_{gs}(x,y) = \psi_{x}(x) \psi_{y}(y) \propto e^{-w_xx^2} e^{-w_yy^2} $ which is clearly a separable solution.
But when we rotate our coordinate system by $\pi/4$ we, get the new coordinates to be
$ x' = \frac{1}{\sqrt{2}}(x + y)$ and $y' = \frac{1}{\sqrt{2}}(x - y) $
and expressing the solution in those coordinates will lead to the solution
$\psi_{gs}(x',y') \neq \psi_{x'}(x') \psi_{y'}(y') $ and hence a signature of entanglement.
How can rotating a physical system (or just someone choosing a different coordinate system) lead to it being entangled?
| To add to @ZeroTheHero answer: one usually talks about entanglement in a narrower sense, as a state of two or more particles, since what we have here can be described as a superposition of one-particle states (whatever is the basis).
| {
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Experimental result of the atomic nucleus volume by scattering alpha particles from the atomic nucleus. Investigation of the electron volume by what? Rutherford's alpha particle experiments marked the beginning of the determination of the volume of the atomic nucleus.
How were the experiments conducted that led to the statement of the point-like electron?
| To enlarge slightly on Anna V's answer, here is what we mean by scattering experiments.
Let's imagine a tight beam of electrons zooming through a vacuum, aimed at a target consisting of a simple sort of atom, and separated into bunches of electrons. So... we are going to "machine-gun" the target atom with "bullets" consisting of electrons.
Then we surround the target with sensors that tell us at what angles the electron bullets bounce off the target atoms. This "ricochet" data tells us how big the target is and what shape it has, if any.
At low electron energies, the electron bullets bounce off the electrons that surround the target atoms, and we find out how big the atom is. As we increase the energy, the electron penetrates the electron cloud and bounces off the nucleus inside the cloud, and we find out how big the nucleus is. At higher energies, the electrons bounce off of individual particles (protons and neutrons) inside the nucleus, and now we know the diameter of a proton or neutron. More energy, and we start bouncing electrons off the quarks inside a single proton or neutron.
Although this is a simplified picture, in all cases the ricochet data tells us that the electron doesn't have anything "inside" it (unlike a proton, which has a well-defined diameter and internal structure in the form of three little quark guys running around in there) and that its own diameter is far smaller than anything we can measure with scattering experiments like this.
| {
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Uniqueness of the definition of Noether current On page 28 of Pierre Ramond Field theory - A modern primer the following is written:
"we remark that a conserved current does not have a unique definition since we can always add to it the four-divergence of an antisymmetric tensor [...] Also since $j$ [the Noether current] is conserved only after use of the equations of motion we have the freedom to add to it any quantity which vanishes by virtue of the equations of motion".
I do not understand what he means by saying, any quantity which vanishes by virtue of the equations of motion.
| I can give you an example:
$$
S=\int dt \left(\dot x^+\dot x^--x^+ x^-\right)
$$
has a conserved current associated with $x^{\pm}\rightarrow e^{\pm\rho} x^{\pm}$ given by
$$
j=x^+\dot x^--x^-\dot x^+
$$
This means that the current above will be conserved if the equations of motion are satisfied. Now if we add to this current a term of the form $\ddot x^++x^+$ the statement will stil be true, i.e. the current will stil be conserved. This is due to the fact that this term that I added is precisely the equations of motion.
| {
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Why does running spend more energy than walking? The study energy expenditure of walking and running concludes that running spends more energy than walking.
My understanding is that although running makes one feel more tired, that only indicates that the power was higher (since the time of displacement was shorter), but at the end of the day the total energy dispensed to move oneself forward by friction should be the same.
Given the study shows otherwise, what could be the flaw in my reasoning?
| A common stumbling block pointed out in the other answers and the comments is the inappropriateness of modeling a runner as a point-like object (a spherical runner in vacuum), as is done in simple classical mechanics. Indeed, kinematic or dynamic (in terms of Newton laws) description of a runner may require more complex models. On the other hand, energetic and thermodynamic arguments do not suffer from this limitation.
A rather general answer given in basic mechanics (and applicable to objects that are not point-like) is that power is proportional to force times velocity:
$$
P=\mathbf{F}\cdot\mathbf{v},
$$
that is, moving with the higher velocity, given the same force, results in more energy spent.
Whether the force is constant or increases with velocity depends on how we model running. It is our everyday experience that, if we do not apply any force (e.g., if we stop moving our legs), we stop. The likely reason is the air drag. We thus could write the Newton's equation as
$$
m\dot{v}=-F_{drag}(v) + F.
$$
The simplest choice for drag force is $F_{drag}(v)=m\nu v$, where $\nu$ is the drag coefficient. However, we may limit ourselves to a general statement that thsi force is directed against the direction of our motion and increases with velocity, i.e.,
$$
F_{drag}(v)>0,\\
\frac{d}{dv}F_{drag}(v) >0.
$$
The Newton equation gives
$$
F=F_{drag}(v) \Rightarrow P(v)=F_{drag}(v)v,\\
$$
and the power grows with velocity:
$$
\frac{dP}{dv} >0.
$$
Remark
*
*Note that this simple mechanical calculation ignores the energy expenditurs on moving the parts of our body. Since our movements are rather different when walking and running, the discussion above is actually more applicable to fast vs. slow walk than to running vs. walking.
| {
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Does real life have "update lag" for mirrors? This may sound like a ridiculous question, but it struck me as something that might be the case.
Suppose that you have a gigantic mirror mounted at a huge stadium. In front, there's a bunch of people facing the mirror, with a long distance between them and the mirror.
Behind them, there is a man making moves for them to follow by looking at him through the mirror.
Will they see his movements exactly when he makes them, just as if they had been simply facing him, or will there be some amount of "optical lag"?
| This is a simple distance problem the distance from your eye to the mirror and the distance of the man from the mirror added together. The size of the mirror makes NO difference at all in speed of light but some in how far the object is away and the prior answer was correct except he was using the speed of light in a vacuum which you are not in.
Now, the most important condition is that you are able see the man's gestures in the mirror. I would say that he would have to be no further than around 1 mile in total (his and your distance from mirror added together) for an average person to be able to see his gestures. So the formula would be 1 mile multiplied by the speed of light in atmosphere. Don't forget elevation; it affects air density.
| {
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Center-of-mass and relative coordinate integral transformation Assuming I have a function $f(r_s,r)$ where $\vec{r_s}$ is the center-of-mass- and $\vec{r}$ the relative-coordinate of a two particle system and I perform the following integral in spherical coordinates (assuming the function is only dependent on the absolute value of the coordinates, so that angle dependencies get trivially integrated over)
$$16\pi^2\int_0^\infty\int_0^\infty f(r_s,r)r_s^2r^2\,dr_s\,dr$$
I now want to rewrite the integral but in terms of the individual particle coordinates $r_1$ and $r_2$. The transformation rules are given by
$$\vec{r_s}=\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1m_2}\hspace{1em}\text{and}\hspace{1em}\vec{r}=\vec{r_2}-\vec{r_1}$$
I am unsure as on how to write the proper measure of the integral so that the result of both ends up being the same.
| It looks like you have a pair of 3D integration variables in spherical coordinates, albeit ones in which you've integrated out the angular coordinates. You might have to undo that and put those back in to the integral. Once you do, you can use a Jacobian (or a pair of Jacobians, depending on how you look at it) to convert to integrals in terms of the Cartesian components of $\vec{r}_s, \vec{r}$. From there, you can again use a Jacobian transformation using the equations you give above to convert to an integral in terms of $\vec{r}_1, \vec{r}_2$.
The Wikipedia article on the "Jacobian matrix and determinant" has details of the procedure.
| {
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Can we cool Earth by shooting powerful lasers into space? In a sense, the climate change discussion revolves around the unwanted warming of the earth's atmosphere as a whole.
It seems a bit too obvious to be true, but could we cool the atmosphere by simply shooting that unwanted energy somewhere else?
Energy might be collected from remote expanses where it would otherwise be somewhat pointless to harvest it due to lack of habitability and resulting anticipated losses due to transmission (ocean surface, ???)
If so, what would be a good place to shoot it?
| Your idea to cool the Earth by shooting photons off into space is actually what already happens now! However, instead of a laser, the Earth cools itself by blackbody radiation. The Earth radiates a $\mathrm{\approx 300~^\circ K}$ (room temperature) black body spectrum with a peak at $\mathrm{kT=\frac{1}{40}eV}$ which is just a little redder than an infrared laser. The Earth also receives radiation from the Sun and a $\mathrm{3 ~^\circ K}$ blackbody spectrum from the cosmic microwave background (CMBR). Overall the Earth is in equilibrium (its temperature is constant) so it radiates as much power as it receives.
The total power radiated by the Earth is:
$$\mathrm{Power=4\pi(R_{earth})^2\sigma T^4=2.3\times 10^5\quad Terawatts}$$
It would take some really big CW lasers to compete with this!!
| {
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Does an electron have multiple positions BEFORE being observed? I realize that an electron can only ever be detected at a single location in space when it is observed. That is, post-detection, an electron can only ever be at a single position in space. But prior to the detection, when the probability distribution of finding this electron is spread over space, can we interpret this as the electron existing at multiple locations in space? Or is it that the electron exists at one definite position even prior to measuring it but it's just that we don't know where it is and hence have to represent it as a probability distribution?
| An electron existing at different positions in space would raise more interpretation questions than those it is intended to reduce.
Here a partial list of difficulties.
If a full electron coexisted with its clones at different
positions, we would go into troubles with conservation laws:
*
*charge and spin conservation;
*conservation of energy: the coexistence of many copies of the original electron should imply their mutual (repulsive) interaction. Suddenly, after a position measurement, all the clones would disappear, and also such interaction energy would disappear. Notice that all known calculations of energy based on the presence of one electron at the time agree very well with the measurements of energy. Moreover, the number of clones would not be finite, thus implying infinite repulsive energy between the clones.
It would be possible to imagine the coexistence of partial clones of the original electron. That is the presence at different points of the space of a fraction of the electron. Also, in such a case, there would be the presence of a self-energy modifying the energy between two measurements of position. Such a self-energy would have observable effects in atomic spectroscopy, but they have never been observed.
Therefore, we are left with only one electron. The only allowed statement about its position in each state can only be a probability distribution of measuring one observable of the one electron.
| {
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About virtual displacement
Thornton Marion
The varied path represented by $\delta y$ can be thought of physically as a virtual displacement from the actual path consistent with all the forces and constraints (see Figure above).
The varied path $\delta y$, in fact, need not even correspond to a possible path of motion
Doesn't the second quote contradict the first. The first says the virtual path is a possible path, the second says it need not be?
| First of all, the first statement is wrong, the condition of being consistent with forces should be eliminated, retaining the consistency with constraints. Because if it is not eliminated, then the first statement is basically saying that the varied path is kinematically admissible, or in your phrase, possible path, which is wrong. So yes, the two statements contradict each other, but that is due to an incorrect definition of virtual displacement in the first statement. In the second statement, the reason why virtual displacements are not possible displacements, in general, is that virtual displacement is considered at a certain moment of time for a system with $dt=0$ while kinematically admissble/possible displacements have to take the change of the constraints with time into consideration. For instance, let suppose there is a particle being able to slide along a straight rigid rod rotating about a fixed point at either its ends. In this case, the virtual displacement vectors can only point along the rod, since by definition, we neglect the rotation of the rod, hence no angular velocity imposed on the particle by the rod. For possible displacements, we need to consider the rotation of the rod and add up the imposed angular velocity and any velocity of the particle on the road. Hence the possible displacement vectors will not point along the rod at the very least since rotation of the constraining rod is considered.
| {
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Proton Electron Merger Can somebody explain what would happen if an electron & a proton, very close to each other are left to "fall" to each other in a straight line?
|
Can somebody explain what would happen if an electron & a proton, very close to each other are left to "fall" to each other in a straight line?
One of the three solid evidences that classical electrodynamics and mechanics could not describe electrons, protons and atoms was exactly the fact that in classical electrodynamics the electron attracted by the proton charge would by acceleration fall on the proton neutralizing it, with a continuous electromagnetic radiation.
Instead there existed discrete frequencies , the atomic spectra. Quantum mechanics was invented, leading to fitting the hydrogen spectra with quantized energy solutions.
The other two experimental no-goes of classical physics that quantum mechanics explained mathematically at the time were the photoelectric effect and black body radiation.
| {
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Addition of velocities vs. Addition of forces Imagine two strings tied to a box.
Case 1: Two strings are pulled with the same $u$ velocity. The box will also move with velocity $u$.
Case 2 : Tension along $\text{a}$ string is $T$. Therefore total force acting on the box is $T+T=2T$. (Box is accelerating)
I think my problem is obvious. Both velocity and force are vectors. But why we can not get the velocity of the box in the first case as $u+u=2u$? (This is obviously wrong, but why?)
| Sorry for my poor english. French is my native language.
To define a vector, it is necessary to specify the vector space on which it is defined. In general, for a manifold, we have a tangent vector space at each point. In classical physics, space has an affine flat space structure and we define at each point a tangent vector space which, for an affine flat space, can all be identified with each other.
It is in this tangeant vector space that the addition of two vectors is defined. One thus defines at each point the vector space of the displacements. The vector sum of two displacements from a point is a displacement. By dividing by time, we go to velocity vectors.
On the other hand, there are difficulties in defining the derivative of the velocity : we compare vectors at different points. To be able to do this, we have to define a parallel transport and connection that allows a vector to be transported from one point to another. It is very easy in the case of the affine space of classical physics. More complicated in a variety: it is necessary to introduce a covariant derivative.
So even the addition of two forces at different points is not a simple consequence of the structure of vector space. We have to transport the vectors. And this is only simple for an affine flat space.
| {
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How to know the direction of the unit normal vector to an open surface? For an open surface shape (assume surface $x=4$) there is 2 possibilities for the unit normal vector direction either $(+î)$ or $(-î)$. so how can we know what direction it is supposed to be (ie $.. - î$ or $+î$)?
| The unit normal vector is part of the definition of a surface. To completely define a surface you must give (choice) its unit normal vector. Consequently you must be careful using it for calculations. For example the flux of a vector function for one choice would be the opposite of that of the other choice etc.
In Figure-01 above we see a surface in $\:\mathbb R^3\:$ represented by the parametric equation $\:\mathbf x\left(u,v\right)$. Note that $\:\mathbf x_{u}\boldsymbol{=}\partial \mathbf x/\partial u\:$ and $\:\mathbf x_{v}\boldsymbol{=}\partial \mathbf x/\partial v\:$ are tangent vectors to the parametric curves, $\mathbf{x}_{u}\boldsymbol{\times}\mathbf{x}_{v}\boldsymbol{\ne 0}$ and the unit normal vector shown is
\begin{equation}
\mathbf N \boldsymbol{=} \dfrac{ \mathbf{x}_{u}\boldsymbol{\times}\mathbf{x}_{v}}{\Vert \mathbf{x}_{u}\boldsymbol{\times}\mathbf{x}_{v}\Vert}
\tag{01}\label{01}
\end{equation}
But you could equally well choose as unit normal vector $\mathbf N' \boldsymbol{=-}\mathbf N$.
| {
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Eigenvalues of antiunitary operators I have sometimes come across the statement that antiunitary operators have no eigenvalues. For example, on page 34 in the book "Topological Insulators and Topological Superconductors" by Bernevig and Hughes, it is stated that
The preceding ( $T i T^{-1} = -i$ ) makes it clear that the time-reversal operator $T$ must be proportional to the operator of complex conjugation. Such operators are called antiunitary and, unlike unitary (sic.) operators, do not have eigenvalues.
I do not understand this statement. For example, consider the antiunitary operator $\sigma_x K$ where $K$ corresponds to complex conjugation and $\sigma_x$ is a Pauli matrix, then
\begin{equation}
\sigma_x K \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix} = \pm \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}
\end{equation}
Naively, I would therefore conclude that $\left( 1, \pm 1 \right)^T$ is an "eigenstate" of $\sigma_x K$ with "eigenvalue" $\pm 1$. If we multiply this eigenstate by a phase $e^{i\phi}$, it remains an eigenstate but its "eigenvalue" changes by $e^{-2i\phi}$. Hence, it seems that one can have eigenstates of an antiunitary operator but their eigenvalue is not a single scalar. Instead the eigenvalue corresponds to a circle.
Edit
I have found this paper which deals with the subject, but seems to contradict the original statement: https://arxiv.org/abs/1507.06545
If we consider the time-reversal operator again, since for spinless particles $T^2=1$, there exist eigenstates of $T$ without unique eigenvalues. However, for spin 1/2 particles, $T^2 = -1$ and there exist no eigenstates (see the answer of CosmasZachos).
| Your fine link has the answer for you in its section 2.2, illustrating that some antiunitary operators, like Fermi's spin flip, lack eigenvectors, as you may easily check.
But the counterexample you chose is of the $\vartheta ^2={\mathbb I}$ variety, and so $\vartheta$ does have the obvious eigenvectors: that's the point of Proposition 2.3 , corollary 2.4 !
Check your
$$
\sigma_x K \sigma_x K ={\mathbb I},
$$
in sharp contrast to
$$
i\sigma_y K i\sigma_y K =-{\mathbb I}.
$$
The first has eigenvectors with $\vartheta^2$ having a positive semidefinite spectrum, but the second doesn't.
| {
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Why is the gravitational potential energy lost not subtracted from the required work done in the given problem?
An elastic string of natural length $l \;\text{m}$ is suspended from a fixed point $O$. When a mass of $M \;\text{kg}$ is attached to the other end of the string, its extension is $\frac {l}{10} \;\text{m}$. Some work is done to produce an additional extension of $\frac{l}{10} \;\text{m}$. Show that the work done in producing this additional extension is $\frac{3Mgl}{20} \;\text{J}$.
My Attempt. I tried to apply the work-energy principle which says that the change in total energy of an object equals the work done on it. So, the required work done should be the elastic potential energy (EPE) gained minus the gravitational potential energy (GPE) lost, which gives unmatched $\frac{Mgl}{20} \;\text{J}$. Later, I found out that if I simply ignore the GPE I will get the desired answer. But why the GPE can be ignored? Isn't the additional GPE loss got stored in the EPE?
Comment. It is a high school mechanics problem, so please do not over-complicate things. Thank you in advance.
| The issue here is that energy of the spring-mass system is not conserved. The initial state of the system is the mass at rest and attached to the unstretched spring. If we release it, the mass will begin to gain kinetic energy and oscillate. It will never reach the final state of equilibrium where $mg=kx$. Therefore you cannot say that all of the gravitational potential energy goes into elastic potential energy. Work must be done by an external agent to bring the mass back to rest to reach the final state.
| {
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Why the phase shift of an electromagnetic wave or light after reflection is smaller than 0, being in the range $[-\pi, \pi]$? In some reports, the phase shift becomes in the range $[-\pi, \pi]$, for instance on this website. However, Wikipedia gives the range as $[0, \pi]$ instead. Which one is correct?
| The first link you quote is about Fresnel equations in general, which may describe both transmission or reflection off a boundary. In this case, the deflection angle is $\theta \in [-\pi, \pi]$ with respect to the boundary, with the negative angles being the transmitted (refracted) part, that is above the boundary. The positive angles correspond to the reflected part, that is below the boundary. For reflection, then, you have $\theta \in [0, \pi]$ $-$ your second link is specifically to Total Internal Reflection and hence the reduced range.
| {
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Why doesn't charge escape from capacitor? A charged spherical capacitor kept in air do not loose charge because air is a bad conductor and increase in charge results in Corona Discharge. Is it because the nucleus of air molecule repels the charge in sphere and after a limit the repulsion is less than the attraction by sphere leading to Corona Discharge. But if the same charged spherical capacitor is kept in vacuum, what would happen. Shouldn't charge repel each other and escape the spherical surface because now there is no nucleus of air molecule to repel those charges. Will the capacitance decrease or not.
| If the charge of an object does not change, it means the net flux of charges to and from it is zero. It all depends on the environment and the object itself. It is normal that an object has some charge even in the near vacuum of space. Sunlit surfaces for instance become charged positively due to ultraviolet radiation knocking electrons out of the material. Surfaces in shadow become charged negatively due to the fact that the flux of plasma electrons in space is higher than that of protons (due their smaller mass and thus higher velocities). The material charges up to a value so that the net flux of charges is zero and thus a steady state is reached. See http://holbert.faculty.asu.edu/eee560/spc-chrg.html for more information.
| {
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Does gravity get stronger the higher up you are on a mountain? So I saw this article stating that gravity is stronger on the top on the mountain due to there being more mass under you however I have read some questions other people have asked and most of the responses state that the mass is concentrated at the middle of the earth meaning gravity doesn't get stronger the higher up you go. I would like to know which one of these it is as the article is a pretty reliable source. Here is the link to the article https://nasaviz.gsfc.nasa.gov/11234
| I looked at the link you gave, I think it may not mean to say the higher up you go on a mountain, the stronger the gravitational field.
The link's meaning, I suppose (because they mentioned it is measured by satellites, I suppose it was measuring gravity fields at the altitude of the satellites, which supposedly stayed the same for the whole measurement) is something like this: they measured gravity at the same altitude around the globe, and found that at this altitude and at spots with Mountain ranges right below, such as the Himalayas, due to the high concentration of mass, the gravity field is stronger. While at the same altitude and at spots with Ocean trenches right below, such as the Mariana Trench, due to the low concentration of mass, they measure a weak gravity field, as expected.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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In 2D space, how to calculate the direction to hit a moving projectile from a position? Imagine a 2D space. A is where our missile is, and B is where our target is currently moving with a velocity of $v_{2}$. B will come close to A in a certain time and then move away like a comet to earth. Our missile can travel at a speed of $s_{1}$, where the target speed is $|v_{2}|$. The magnitude and direction of $v_{2}$ won't change. Similarly, for our missile, whose speed is constant, we need to calculate $v_{1}$ such that once the missile is launched, it will hit the moving target.
So we have $A$, $B$, $v_{2}$ and $s_{1}$. How can one find $v_{1}$ to hit the target?
The missile and target move with the equation
$$A = A + v_{1}\cdot\mathrm dt$$
$$B = B + v_{2}\cdot\mathrm dt$$
How do I calculate $v_{1}$?
| So we have $\vec A, \vec B, \vec v_2$ and $|\vec{v_1}| = s_1$. We can represent the vector $\vec{v_1}$ as follows:
$$
\vec d = \vec B - \vec A \\
\vec B + \vec v _2 t = \vec A + \vec v _1 t \\
\vec v_1 = \frac{\vec d + \vec v_2t}{t} \\
\vec v_1 ^2 = |\vec v_1|^2 = v_1^2= s_1^2\\
$$
Where t is the time after which the collision will take place. We can calculate time by squaring this expression:
$$
\vec d + \vec v _2 t = \vec v _1 t \\
d ^{2} + 2 (\vec v _2 \cdot \vec d)t + v_2^2t^2 = s_1^2t^2 \\
(v_2^2 - s_1^2)t^2 + 2(\vec v_2 \cdot \vec d)t + d^2 = 0
$$
Solve the quadratic equation for t > 0.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why does water contract on melting whereas gold, lead, etc. expand on melting? My book mentions that water contracts on melting, but the book doesn't give any reason why it does so. It is mentioned that:
$1\,\mathrm g$ of ice of volume $1.091\,\mathrm{cm}^3$ at $0^\circ\mathrm C$ contracts on melting to become $1\,\mathrm g$ of water of volume $1\,\mathrm{cm}^3$ at $0^\circ\mathrm C$.
I searched on the internet but I failed to find any useful insight. Could someone please explain why water contracts on melting?
| It's because of the crystal structure of the solids. When water freezes, the molecules form various structures of crystals which have empty gaps that cause the solid to be about 9% larger in volume than the liquid was. Metals usually form crystals when they freeze too, but they're often simpler crystals, if you will, and often don't have as much empty space in them as ice/snow does.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/653220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Why does twisting a cork make it easier to remove from a bottle? When we want to remove a cork from a bottle first we turn the cork. Turning in one direction makes it easier to remove in the axial direction.
Does anyone know something more about this?
| The initial twisting step is just ergonomically easier than the initial pull it would be require to get the cork moving. For this you have to overcome the static friction between cork and glass. Also there may be van der Waals forces that need to be broken before the cork starts moving. Once the cork moves the friction is lower and it can be pulled. Note that mechanical devices don not bother to twist, they just pull.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why can't the speed of gravitational waves be greater than the speed of light if the universe can expand faster than the speed of light? Since the expansion speed of the universe can be greater than the speed of light, why can't gravitational waves, which also uses space as the medium, travel faster than the speed of light?
| Gravitational waves propagate at the same speed as light and in a similar manner. The point here is that the speed $c$ is only measured locally. The speed of objects with mass is also only limited to be $\lt c$ locally. The relative speed of two objects separated by great distance is not well-defined. See this post for a detailed explanation.
The expansion of the universe does not give objects any "speed" because the universe does not lie in a higher-dimensional space. For the same reason, there is no "central point" in the expansion. However, that being said, gravitational waves do experience cosmological redshift just like light.
| {
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Is time taken by light to travel any distance 0 or finite? According to relativity, Light does not experience any time. So it must travel any distance in no time.
But, we know that light has finite speed $c$. So it should take finite time.
| Speed is measured with respect to some reference frame. Relativity does not allow for a reference frame that is comoving with a photon, so there's no sense within the theory talking about the time or speed "experienced" by the photon in "its own frame". From any valid frame, the speed of light is $c$.
The statement that light does not experience time is not really physically precise. It captures something about limiting behavior of certain formulas, but it is a limit that isn't realizable in the way that the simplified statement suggests.
| {
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Why don't we use absolute error while calculating the product of two uncertain quantities? I've found a rule that says, "When two quantities are multiplied, the error in the result is the sum of the relative error in the multipliers."
Here, why can't we use absolute error? And why do we've to add the relative errors? Why not multiply them?
Please give me an intuition to understand the multiplication of two uncertain quantities.
| You get the best intuition if you just take two easy numbers with a possible error and multiply it. Choose 100±1 and 200±4 the relative errors are 1/100 or 1% and 4/200=2%.
Now multiply and you get for the positive error 101*204=20604=20000+604 or an error of about 3%. Multiplying the absolute error would give you 1*4 instead of 604, multiplying the relative errors would give you 2/10000 or 0.02%.
Try it with any two other numbers.
| {
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Why do little chips break off so easily from strong neodymium magnets? I have some strong toy neodymium magnets. Typically after a while little chips start breaking off, unlike from most other small metal objects, like in this image.
It could of course be that neodymium is more brittle than metals used for other objects, or that they often hit each other much harder than in a fall due to their magnetism, or that they are just low-quality, but I was wondering if it could have to do with internal tensions that are not present in non-magnetic objects, maybe due to adjacent domains of different magnetization?
Does anyone know what could cause this?
| The Neodymium magnets that you use are not actually made from solid chunks of metal, but are rather made by compressing a large amount of powder into blocks through a process called sintering. This is the reason they are so brittle, which is not helped by the fact that they are also so strong!
Furthermore, such magnets are very vulnerable to corrosion. In order to deal with this, most of them have a protective coating of nickel (which gives them their familiar colour) to prevent exposure to the atmosphere. However, if they happen to chip, then those regions lose this coating and become exposed, which in turn leads to further corrosion and spalling.
| {
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The faster you move, does it take more and more energy to increase your speed at the same rate? I'd like to confirm this somewhat counterintuitive result. Starting with the definition of kinetic energy:
$$E = \frac{1}{2} mv^2$$
Assume a vacuum, no external forces, and starting from rest. Adding some energy $E$ to the system by burning some fuel (by firing a rocket, etc.), the following should be true.
$$\frac{1}{2} m{v_1}^2 = \frac{1}2 m{v_0}^2 + E$$
Solving for $v_1$,
$$v_1 = \sqrt{{v_0}^2+\frac{2E}m}$$
If $E$ remains constant, by burning fuel at a constant rate, $\Delta v$ decreases as $v$ increases.
| Yes, that's right, you can also differentiate both sides of
$$\frac{1}{2} mv_1^2 = \frac{1}2 mv_0^2 + E$$
with respect to time and putting $\frac{dE}{dt} = P$, the power supplied, you get to
$$mv\frac{dv}{dt} = P$$
so for a given power $\frac{dv}{dt}$ is inversely proportional to $v$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/656658",
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Proving that flux contribution due to rotation is zero for a triangular loop welded to infinite wire set up
Consider a triangular loop attached at the vertex to an infinitely long wire which has time varying current flowing in the +x direction
Adapted from JEE advanced paper-1 2016
One may find that the contribution due to the current varying is given as:
$$ V= \frac{\mu_o d}{\pi} \frac{di}{dt}$$
But now, suppose the set up is rotated about the axis of wire, what extra EMF would be generated? Apparently the answer is zero but I find it a bit tricky to understand. So far, I understand that this is due to $\vec{B}$ by the wire being a function of $r$ and hence all point at same distance of axis is equivalent (at least to the magnetic field).
I need to show some how prove that, if we were to rotate the set up about the vertex, then the $\vec{B}$ field at all points in the interior of the triangle would be the same as the original unrotated configuration for all possible rotation angles.
Preferably, I wish for a mathematical explanation but a physical one is fine as well if it detailed.
| Since then magnetic field produced by a current in the long wire forms circular loops around the wire, a rotation of the triangle around the wire would not result in a change in the flux.
| {
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How to understand the ambiguity of vector resolvation? When we solve problems where there is a pendulum suspended using a tight, inextensible string and the question asks about the tension developed in the string at the highest point of the bob's swing. The following is the conventional approach to solve the problem.
As you can see, even I resolved the tension and $mg$ into their respective components. The confusion I had was - here $T=mg\cos\theta$ and $mg=T\cos\theta$. How do I know which one to consider? Because they both make equal sense (to me at least) - their directions match perfectly.
| Newton's Second Law is $\vec{F} = m \vec{a}$. In terms of vector components, this becomes
$$
F_x = m a_x \qquad F_y = m a_y.
$$
Depending on how we set up our coordinates, we may have one or both of $a_x$ or $a_y$ non-vanishing. But we can use our freedom to pick our coordinate axes however we want to simplify our life. In particular, if we pick our coordinate axes so that $\vec{a}$ points along one of them, then the other component of $\vec{a}$ is zero. This makes one of the terms vanish, simplifying the algebra.
In the present case, we can pick our axes so the $x$-axis points along the arc that the bob will describe, and the $y$-axis points along the string. Since we are at the highest point of the swing, there is no centripetal acceleration; so the acceleration $\vec{a}$ will be in the $x$-direction only, with $a_y = 0$ and $a_x = a$. The equations then become
$$
mg \sin \theta = m a \qquad T - mg \cos \theta = 0.
$$
This system of equations is particularly easy to solve for the unknowns $T$ and $a$: we will have $T = mg \cos \theta$ and $a = g \sin \theta$.
But it's important to note that choosing different axes is not wrong, per se; it just makes the algebra harder. For example, suppose you chose horizontal and vertical axes instead. In this case, you would have $a_x = a \cos \theta$ & $a_y = - a \sin \theta$, and Newton's Second Law would be (in these components)
$$
T \sin \theta = m a \cos \theta \qquad T \cos \theta - mg = -m a \sin \theta.
$$
This system of equations is harder to solve for $a$ and $T$, but you can show that the solution for $a$ and $T$ is exactly the same as we got above. (Try it!)
| {
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Kinematics of Scattering Amplitudes in $\left(2, 2\right)$ Signature within the Amplituhedron I am just working my way through the concepts of Amplituhedron and often stumble across the phrase
[...] in $\left(2,2\right)$ signature $\lambda$, $\tilde{\lambda}$ are real and independent [...]
in various references (Jaroslav Trnka, 2014; page 7, Nima Arkani-Hamed, 2014; page 24 or Livia Ferro, 2020; page 3). What I don't quite understand is the following: If we take the following momentum four-vector:
\begin{equation}
p_\mu =
\begin{pmatrix}
-1, 0, 1, 0 \\
\end{pmatrix}
,
\end{equation}
and contract this with the Pauli Matrices in $\left(2, 2\right)$ signature (I am very sorry that I may not use the right terminology here), we obtain the following matrix:
\begin{equation}
P_{\alpha\dot{\alpha}} =\frac{1}{2}
\begin{pmatrix}
p_0 + p_3, p_1 - p_2 \\
p_1 + p_2, p_0 - p_3 \\
\end{pmatrix}
=
\begin{pmatrix}
-\frac{1}{2}, -\frac{1}{2} \\
\frac{1}{2}, -\frac{1}{2} \\
\end{pmatrix}
.
\end{equation}
However, unlike in $\left(3, 1\right)$ signature, the determinant of the matrix is not $0$, but $\frac{1}{2}$.
In all the YouTube videos and references I found, this $\left(2, 2\right)$ signature is never properly explained or referenced. Do you have any sources about the signature that can help me identify my mistakes or that can help me understand this concept better?
| The main points are:
*
*There is bijective isometry from the split-signature space $(\mathbb{R}^{2,2},||⋅||^2)$ to the space of $2\times 2$ real matrices $({\rm Mat}_{2\times 2}(\mathbb{R}),\det(⋅))$,
where
$$\begin{align} ||p||^2~=~&(p^0)^2-(p^1)^2+(p^2)^2-(p^3)^2~=~\det(P), \cr
p~=~&(p^0,p^1,p^2,p^3)~\in~\mathbb{R}^{2,2},\cr
P~=~&\sum_{\mu=0}^3p^{\mu}\sigma_{\mu}~\in~
{\rm span}_{\mathbb{R}}\{ \sigma_0,\sigma_1,i\sigma_2,\sigma_3\} ~=~{\rm Mat}_{2\times 2}(\mathbb{R}).\end{align}$$
*There is a bilinear map
$$ \mathbb{R}^2\times \mathbb{R}^2~\ni~(\lambda,\tilde{\lambda})\quad\mapsto\quad P~:=~\lambda\tilde{\lambda}^T~\in~ {\rm Mat}_{2\times 2} (\mathbb{R}),$$
from 2 Weyl spinors $\lambda,\tilde{\lambda}$ to a rank-1 operator $P$, which necessarily has vanishing determinant, i.e. the corresponding momentum is light-like/null.
| {
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What are clover fermions? I've seen the term been used quite a lot when reading about lattice gauge theory calculations. So far what I've gathered is the following, from this source [1].
Lorentz invariance of the action is broken on a discretized lattice. When calculating any quantity on the lattice, the continuum limit $a\rightarrow 0$ must be extracted, where $a$ is the lattice spacing. However this convergence is often slow, with $O(a)$ error terms, so one way to speed it up is to introduce an $O(a)$ irrelevant operator to the Lagrangian density, and tune its coefficient so that the lattice artifacts disappear in certain physical quantities (masses, cross sections, etc.).
$$\mathcal{L}\rightarrow \mathcal{L}+\int c_{\textrm{sw}}\mathcal{L}_c\, $$
$$\mathcal{L}_c=-\frac{a}{4} \bar \psi \sigma_{\mu\nu}\psi F^{\mu\nu}$$
The tuneable coefficient $c_{\textrm{sw}}$ is called the Sheikholeslami-Wohlert coefficient, and is a function of the lattice spacing through the gauge coupling $g$. The correction term $\mathcal{L}_c$ is called the "clover term".
So this would tell me what the clover-term is, the clover-improved action, but what is a clover fermion?
[1] "Clover fermions in the adjoint representation and simulations of supersymmetric Yang-Mills theory" arXiv:1311.6312 [hep-lat]
| A clover fermion is a fermion described by the Wilson fermion action plus the clover-term. Clover fermion is just a short-hand version of Wilson clover fermion. Other terms which describe the same fermion discretization are clover-improved Wilson fermion or $\mathrm{O}(a)$ improved Wilson fermion.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does dispersion occur? In a vacuum, electromagnetic waves of all frequencies travel with the same phase speed, so they propagate with a fixed shape once determined. In a dispersive medium, waves of different frequencies travel with different phase speeds and this causes the wave packet to change shape when propagating. Certainly, the dispersion phenomenon is due to the medium, but what is its property responsible for dispersion?
| Electromagnetic field induces polarisation and magnétisation in the media, which are not an instantaneous response. This results in k-vector being frequency-dependent, hence the group velocity,
$$
v_g=\frac{d\omega}{dk}=\left(\frac{dk}{d\omega}\right)^{-1}
$$
is different from the phase velocity
$$
v_{ph}=\frac{\omega}{k},
$$
which is what we call dispersion.
Update
Dispersion and causality section of the Wikipedia article on permittivity gives a rather good review of the relevant .EM equations
| {
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Amount of force required it to tip over a cone Say I have a cone of height $h$, radius $r$, and mass $m$.
How can I determine the amount of force required to tip it over (to have it fall completely to the other side), say exerted (horizontally) at the top of the cone? And in addition, how does the position at which I exert the force affect the amount of force I will need to tip it over?
Any advice on how I would aproach this problem is appreciated.
| For rotational static equilibrium, all torques about the CG must balance. Whatever amount of force is required to make the normal vector be applied at the rim of the cone is the load required for tipping.
| {
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Why would partial derivatives not commute inside an axionic cosmic string? In this 1985 paper by Callan and Harvey, Eq. $11$ seems to claim that in the presence of an infinitely extended string-like topological defect, partial derivatives do not commute on the string:
$$
[\partial_x, \partial_y]\, \theta = 2\pi \delta(x)\delta(y)\,. \tag{11}
$$
This is apparently "because of the topology of the axion string." I do not quite follow. Can someone please explain the reasoning to me in a bit more detail? In other words, I would like a more explicit derivation of the above equation. Thank you.
| The integral
$$
I_\gamma=\oint_\gamma d\theta= \oint (\partial_x\theta dx +\partial_y \theta dy)= \oint\nabla \theta \cdot d{\bf r}
$$
is $2\pi$ if the loop $\gamma$ encloses the origin and zero if it does not. Now use Stokes' theorem
$$
I_\gamma = \int_\Omega \nabla\times (\nabla\theta) d^2r , \quad \partial \Omega=\gamma.
$$
Interpreting the fact that $I_\gamma$ is $2\pi$ or zero depending on whether the origin lies within $\Omega$ as a statement about distributions, we read off that $\nabla\times (\nabla \theta)= 2\pi \delta^2(x,y)$.
In components this is $\partial_x \partial_y \theta - \partial_y\partial_x \theta = 2\pi \delta^2(x,y)$.
| {
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Can a metal sheet roof be vibrated at audio frequencies? So I'm renting a cheap place for now due to certain circumstances. Problem is outside noise is excessive because there is no ceiling and the roof is only a metal sheet... Lately I play white noise to get some peace but it's not as effective as I want it to be... So I was wondering if I built my own speaker (come) to attach to the metal roof an play white noise, would the whole roof become some giant speaker emitting white noise in turn blocking outside noise since the roof is the main culprit for noise coming in, the walls are fairly thin.
I know this is a weird physics question but I need my peace to work on my projects, so I'd appreciate any information
| If you plan to stay there for long it's probably best to add a ceiling inside the metal sheet. In-between put a thick layer (20cm or more) of sound absorbing material such as Rockwool.
If you use the right material it'll act as a heat insulator in winter. Living spaces lose most heat through the ceiling.
| {
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Why does the normal force do no work in the falling stick problem? I'm studying the next problem what I found in a book.
A stick of length $l$ and mass $M$, initially upright on a frictionless table,
starts falling. The problem is to find the speed of the center of mass as
a function of the angle $\theta$ from the vertical.
I have analyzed the solution using energy methods, and I understand the problem almost completely, but I still do not understand why the normal force does not exert work.
Part of the solution that I found is below:
The key lies in realizing that because there are no horizontal forces, the center of mass must fall straight down. Since we must find velocity as a function of position, it is natural to apply energy methods.
The sketch shows the stick after it has rotated through angle $\theta$ and the center of mass has fallen distance $y$.
The initial energy is
$$\begin{aligned}E&=K_0+U_0\\&=\frac{Mgl}2\end{aligned}$$
The kinetic energy at a later time is
$$K=\frac12l_0\dot\theta^2+\frac12M\dot y^2$$
and the corresponding potential energy is
$$U=Mg\left(\frac l2-y\right)$$
Because there are no dissipative forces, mechanical energy is conserved and $K+U=K_0+U_0=Mgl/2$. Hence
$$\frac12M\dot y^2+\frac12l_0\dot\theta^2+Mg\left(\frac12-y\right)=Mg\frac l2$$
We can eliminate $\dot\theta$ by using the constraint equation. The sketch shows that
$$y=\frac l2(1-\cos\theta)$$
Hence
$$\dot y=\frac l2\sin\theta\;\dot\theta$$
Question: Why does the normal force not do any work?
| The bottom of the stick, the part in contact with the table, is moving only horizontally. The normal force is applied vertically to the bottom of the stick. Therefore, the dot product of the velocity of the bottom of the stick and the normal force is 0, so no work is done.
| {
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Torques acting on a unicyclist When a unicyclist rounds a bend, he or she has to tilt in order to generate a frictional force, the friction acting as the centripetal force constraining the unicyclist to a circular path. However, this tilting causes the combined weight of the unicyclist and the unicycle to have a torque about the point of contact of the tyre of the unicycle and the ground. What balances this torque? It cannot be the frictional force nor can it be the normal reaction on the tyre since both forces act on the point of contact and thus have no torque about it. On the other hand, if the torque remains unbalanced, the unicyclist will eventually crash to the floor yet this doesn't happen in many cases.
I am assuming the motion is observed by a non-accelerating observer, so there is no centrifugal force or any other fictitious force.
|
On the other hand, if the torque remains unbalanced, the unicyclist will eventually crash to the floor
Not quite. An unbalanced torque must mean that angular momentum is changing. But angular momentum can be expressed in many different ways, not just rotation.
In the case of the unicyclist if we look at the point on the ground that the cycle passes over in a turn, then we demand that the gravity modify the angular momentum about that point. This happens by the cyclist accelerating away. Linear velocity (if not in line) can contribute to angular momentum.
Another way of looking at this is in the frame of the cyclist. Here the cyclist appears to maintain the turn with no counter-torques. But since the cyclist is circling, this is not an inertial frame. There will be a fictional force appear in this frame. The force will exactly counter the torque from gravity.
| {
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Electric field above a ring of charge To find the electric field at a point $p$ which is at a distance $h$ above the center of a ring of total charge $q$ with radius $r$, one can integrate the charge density over the circumference of the ring and get:
$$E = \frac{qh}{4\pi\epsilon_o(r^2+h^2)^{\frac{3}{2}}}$$
Another approach is to sum up the total charge on the circumference and multiply it by the distance between each point on the circumference and point $p$. The distance from $p$ to any point on the circumference is constant and is equal to:
$$\sqrt{r^2+h^2}$$
Since the horizontal components of the field cancel out, the field can be calculated as:
$$E = \frac{q}{4\pi\epsilon_o d^2} = \frac{q}{4\pi\epsilon_o\,(r^2+h^2)}$$
The two approaches yield different results, so the second must be wrong. But where?
| What you miss in the second method is that the vertical component of the field is not equal to the total magnitude of the field. As you said, the horizontal components cancel out so you have to sum the vertical components only. You can tell that the second formula is wrong with no calculation. The field in the center of the ring should be zero (the field of each piece is all horizontal) and the formula does not produce this result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Conservation of Angular Momentum when potential energy is function of position I have read that the angular momentum $(L)$ of a body is conserved if its potential energy is a function of solely its position vector. For example, the motion of planets on their orbits. I have two questions regarding this concept:
*
*How do we prove this?
*Is the converse also true?
| The conservation of angular momentum is due to the symmetry of some physical systems under rotation, and is a special case of Noether's theorem.
To understand this, consider the Euler-Lagrange equation for a system:
$$\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} = 0$$
Where $q_i$ and $\dot{q}_i$ are the generalized positions and velocities of particles in the system, and $L \left( q_i, \dot{q}_i, t \right)$ is the Lagrangian of the system, which equals the difference of total kinetic and potential energies in Newtonian mechanics:
$$\displaystyle{ L \left( q_i, \dot{q}_i, t \right) = \sum_i T \left( q_i \right) - V \left( \dot{q}_i \right) }$$
Now, the quantity $\frac{\partial L}{\partial \dot{q}_i}$ can be thought of as the conjugate momentum, $p_i$ , of position $q_i$, so we may rewrite the Euler-Lagrange equation as,
$$\frac{\partial L}{\partial q_i} = \frac{dp_i}{dt}$$
Obviously, if the Lagrangian $L$ is not a function of some coordinate $q_i$, the corresponding conjugate momentum is conserved,
$\frac{dp_i}{dt} = \frac{\partial L}{\partial q_i} = 0$
Such coordinates are also known as 'cyclic' coordinates. Thus, if a system is symmetric under rotations, all angle parameters become cyclic coordinates, and the corresponding conjugate momenta are conserved. These momenta happen to be the components of the angular momentum vector.
For example, consider the motion of one particle, characterized by two coordinates which we shall take as polar coordinates $\left( r, \theta \right)$. A system with rotational symmetry then has a Lagrangian of the form,
$L = \frac{1}{2} mv^2 - V \left( r \right) \\ L = \frac{1}{2} m r^2 \dot{\theta}^2 - V \left( r \right)$
Thus, the conjugate momentum of $\theta$ is,
$p_\theta = \frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta} = mvr$
And $\frac{d p_\theta}{dt} = \frac{\partial L}{\partial \theta} = 0$
Voila, this conserved quantity is nothing but angular momentum. Similarly, symmetry under translation in space gives rise to conservation of linear momentum; translation in time implies conservation of energy; and so on.
In Newtonian mechanics, the kinetic energy term in the Lagrangian only depends on generalized velocities, so only the potential energy term needs to be independent of some coordinate for conservation of its conjugate momentum.
However, in general, it is not always possible to separate the Lagrangian into kinetic and potential energies, so we must sure that each term is not a function of generalized position.
| {
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"url": "https://physics.stackexchange.com/questions/659322",
"timestamp": "2023-03-29T00:00:00",
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Local decrease of entropy, does it require life? Universal entropy can decrease only locally at the expense of bigger increase elsewhere.
Can this occur in a lifeless environment or does it necessarily require living organisms to do it?
Can this occur spontaneously or does it have to be an intentionally arranged process, like building a refrigerator?
My assumption is that you need to spend purposeful effort to decrease entropy locally. You need to spend energy to create differences in energy density and you need to have a reason why you do it. Living organisms use energy to create and maintain their internal order for the reason of survival. Inanimate matter has no reason to do anything. Causal uncontrolled processes always go towards higher entropy.
This question seems to enter the grey zone between physics and philosophy. Does a local decrease of entropy require intentional control over the course of events?
| It happens about 50 times in a second in any internal combustion engine, including cars. Without life. Life has no specific meaning in thermodynamical sense.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is space really expanding? In a book called "Einstein, Relativity and Absolute Simultaneity" there was this sentence by Smith:
There is no observational evidence for a space expansion hypothesis. What is observed are superclusters of clusters of galaxies receding from each other with a velocity that is proportional to its distance.
He goes on to say space is Euclidean and infinite. Wouldn't this mean Big Bang was a explosion in spacetime rather than a expansion of spacetime as it is often told?
Is Smith just wrong or don't we know yet?
| According to its Introduction, Einstein, Relativity and Absolute Simultaneity is a volume of essays “devoted, for the most part, to arguing that simultaneity is absolute” (as the title suggests). This is not mainstream physics. Since the book’s editors (William Lane Craig and Quentin Smith) are/were philosophers rather than physicists, its value as a cosmology textbook is doubtful.
So, yes, according to the weight of the available evidence and the consensus of mainstream physics, Smith is wrong.
| {
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"url": "https://physics.stackexchange.com/questions/659661",
"timestamp": "2023-03-29T00:00:00",
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How can spacetime be expanding faster than the speed of light? How can spacetime be expanding faster than the speed of light when the speed of light is the speed limit of the universe?
| The idea that space expands faster than c is a fundamentally flawed concept. The expansion of space is measured in units of m/s/Mpc. A quantity measured in those units cannot be compared to a quantity measured in m/s. There is no sense in which a greater than or less than comparison is even valid between such incompatible quantities.
| {
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"url": "https://physics.stackexchange.com/questions/659753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does electron physically move in an interband transition? How does electron move say from ground state energy level to first excited state? Is there any actual displacement in terms of motion?Is there a way to logically think about this by the help of creation and annihilation operators?
I know this is a quantum mechanical point of view but is there some way we can think in a classical sense?
| Energy levels are eigenstates of the Hamiltonian, so their probability density remains constant in time. But a superposition of two (non-degenerate) Hamiltonian eigenstates has time-dependent relative phase, which makes the density change with period $\propto E_1-E_2$. Speaking classically, this means something like an oscillatory motion.
A resonant transition between two energy levels happens through gradual increase of the amplitude of the target level and corresponding decrease of the amplitude of initial level. After the target level has amplitude $1$ and the initial level's amplitude goes to $0$, if no decoherence happens, this system will transition back to the initial state and then repeat the process again in Rabi cycles.
So, the transition from the ground state will begin by a small oscillation of the peaks of the probability density. Then this oscillation becomes larger, and after some time the probability density peaks from the target state appear. These peaks grow, while those of the ground state diminish. After some time the oscillation stops, so that we get the probability density of the target state. Then the reverse transition happens, unless something inhibits it.
Classically this looks a bit counterintuitive: the most active "motion" (change of probability density in time) happens during the transition, not when the system completely gets to the target, more energetic, state. And the initial and final states look "motionless", because they are stationary.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/660105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Conceptual: Pressure change in closed very long pipe Let's say I have a very long pipe with a very small diameter with one closed end and a tight fitting piston.
If I pull the piston to increase the volume and/or push the piston to reduce the volume, will the pressure change be felt throughout the volume or only in the vicinity of the piston after a very long time? Assume that the system is kept at a constant temperature.
As per ideal gas law, it seems to me that the pressure change should be felt throughout the volume. However, a colleague suggested that if the loss through the pipe is large then the change is only felt in the vicinity of the piston. Which of these explanations are correct if we observe the pressure after a very long time?
| As you are pushing the piston, there will be a greater pressure (or lower pressure, if you pull it backward) in the region in front of the piston, and the pressure in the whole pipe will equalize after a time depending on how long the pipe is once you stop pushing.
The pressure will be the same throughout the volume after this time. If there is no other force that would push on the gas particles to one end (for example, gravity if the pipe is oriented vertically), then there is no reason why the pressure would vary along the pipe after a long time and after the pushing/pulling stops.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/660224",
"timestamp": "2023-03-29T00:00:00",
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Issue about rotational and translational kinetic energy of a pendulum Let’s say we have a pendulum that consist of a light string hanging a disk-like object. It is allowed to undergo simple harmonic motion with small oscillations.
My question: Is the energy of the disk pendulum at anytime written as
*
*(a) $$E_\text{total}= \frac12mv^2 + mgh + \frac12Iω^2,$$ where $v$ is tangential velocity of the center of mass of the pendulum and $I$ is the moment of inertia of disk $+ m(\text{length of string})^2,$ or
*(b) $$E_\text{total}= mgh + \frac12Iω^2$$ and $I$ is the moment of inertia of disk $+ m(\text{length of string})^2$?
| The kinetic energy of a rigid body is invariant (does not change) with the location where it is measured if the parallel axis theorem is used to transfer mass moment of inertia from point to point.
For your example, consider the following two locations
*
*About the center of mass the object has mass moment of inertia of $I$ and kinetic energy $$ K = \tfrac{1}{2} m v^2 + \tfrac{1}{2} I \omega^2 $$
*About the pivot point the system has mass moment of inertia of $I+m \ell^2$ and kinetic energy $$\require{cancel} K = \cancel{ \tfrac{1}{2}m 0^2} + \tfrac{1}{2} (I + m \ell^2) \omega^2$$
Both of the above calculations yield the same result, as you can prove yourself by using the kinematic relationship $v = \ell \omega$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/660429",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Where does pseudo force act at? It is known that, to apply Newton's laws in a non-inertial frame, we use the concept of pseudo force. We also know that force is a bound vector. Hence, is there a general way to determine where the pseudo force vector would be located at?
| In a linearly accelerated reference frame the same pseudo force acts uniformly (at any given instant) on all particles in an extended body. This set of pseudo forces can be replaced by a single force acting at the body’s centre of mass.
In a rotating reference frame the pseudo forces will not be uniform, so you have to determine the pseudo force acting on each particle individually and then integrate across the body as a whole.
It is no coincidence that this parallels the analysis of the motion of an extended body in a uniform or non-uniform gravitational field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/660611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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