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Notation for the divergence of a rank 2 tensor I am studying advanced fluid mechanics and sometimes you see equations written in index notation like
$$ Dv_i= \partial_t v_i +v_j\partial_jv_i$$
but sometimes you find this arrow/vector notation (what is the name of this notation?)
$$ D \vec{v} = \partial_t \vec{v} + (\vec{v}\cdot \vec{\nabla})\vec{v} $$
which is what I used when I learn vector calculus.
My problem arises when you have tensors, like the stress tensor, let us call it $\bar{\sigma}$. For example, one of Navier-Stokes equations (stationary flow) reads
$$ (\vec{v}\cdot \vec{\nabla})\vec{v} = -\vec{\nabla}\cdot\bar{\sigma} = -\partial_j\sigma_{ij} $$
Is there a reasoning on why the divergence on 2-rank tensors acts on the right (as if $\nabla$ is a vector column) ? How does one apply this to higher order tensors?
| In this answer I use $x=x_1, y=x_2, z=x_3$ and Einstein notation. Lets take tensor A
$$A =
\begin{bmatrix}
a_{11} && a_{12} && a_{13} \\
a_{21} && a_{22} && a_{23} \\
a_{31} && a_{32} && a_{33} \\
\end{bmatrix}
$$
On wikipedia in this article I found following information (in article they use S instead A) for cartesian coordinate system:
$$
\nabla\cdot A = \cfrac{\partial A_{ki}}{\partial x_k}~\mathbf{e}_i = A_{ki,k}~\mathbf{e}_i =
\begin{bmatrix}
\frac{\partial a_{11}}{\partial x} + \frac{\partial a_{21}}{\partial y} + \frac{\partial a_{31}}{\partial z} \\
\frac{\partial a_{12}}{\partial x} + \frac{\partial a_{22}}{\partial y} + \frac{\partial a_{32}}{\partial z} \\
\frac{\partial a_{13}}{\partial x} + \frac{\partial a_{23}}{\partial y} + \frac{\partial a_{33}}{\partial z} \\
\end{bmatrix}
$$
The result is contravariant (column) vector. But in this article is mention that $\mathrm{div}(A) \neq \nabla\cdot A$ and
$$
\mathrm{div}(A) = \nabla\cdot A^T = \cfrac{\partial A_{ik}}{\partial x_k}~\mathbf{e}_i = A_{ik,k}~\mathbf{e}_i =
\begin{bmatrix}
\frac{\partial a_{11}}{\partial x} + \frac{\partial a_{12}}{\partial y} + \frac{\partial a_{13}}{\partial z} \\
\frac{\partial a_{21}}{\partial x} + \frac{\partial a_{22}}{\partial y} + \frac{\partial a_{23}}{\partial z} \\
\frac{\partial a_{31}}{\partial x} + \frac{\partial a_{32}}{\partial y} + \frac{\partial a_{33}}{\partial z} \\
\end{bmatrix}
$$
When A is symetric: $a_{ij}=a_{ji}$ then $\mathrm{div}(A) = \nabla\cdot A$
Wiki also mention that some authors use alternative definition: $\nabla\cdot A = \cfrac{\partial A_{ik}}{\partial x_k}~\mathbf{e}_i $ probably only for case when A is symmetric (for which that alternative definition is equal to original). However alternative definition is NOT compatible with general curvilinear definition which I found on wiki too:
$$
\nabla\cdot A
= \left(\cfrac{\partial A_{ki}}{\partial x_k}- A_{li}~\Gamma_{kk}^l - A_{kl}~\Gamma_{ki}^l\right)~\mathbf{g}^i
$$
Currently I don't know what is exact definition of: - probably it will give something like
| {
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Do black holes violate the conservation of mass? This question has confused me for quite some time now. I have searched it up online, and the basic answer is: 'Mass is a form of energy. When black holes die they release the amount of energy that they should. Mass is conserved.' But there's a problem with that answer. Energy is conserved, but mass isn't; it's turned into another energy store. From what I've learned in school, the amount of mass in the universe is always the same, and that's conservation of mass. But if mass is just another type of energy and can be transferred into other types of energy, mass is certainly not conserved, and the conservation of mass doesn't exist. It's like saying conservation of kinetic energy. Kinetic energy can be transferred into the electrical or thermal or other energy stores, and although the overall energy in the universe is the same, the amount of kinetic energy has changed? Am I being stupid or is my life a lie?
| Conservation of mass is not a fundamental law of nature so the violation of it is not a problem. Mass is just a form of energy, and energy is the quantity that is supposed to be conserved. So since black holes do not violate the conservation of energy the problem you are worried about does not exist.
| {
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Potential in open branch of a parallel circuit with grounding
If the switch is still open, what will the electric potential at Q be, i.e. negative, positive or zero?
Is there a potential difference across the grounded point and point Q, or R3, even though the branch is open? I suppose the current would be zero in that branch. Why would there be a voltage even though the branch is open?
| In the lower part of your circuit you are dealing with resistors (including the connecting wires).
If a current $I$ passes through a resistor $R$ then there is a potential difference across the resistor $V=IR$.
No current passes through the lowest branch of the circuit which includes the switch because there is no conducting path in that branch.
So there is no potential difference between the earthed node and the left hand side of the switch which is therefore at zero potential.
There is no potential difference between node $Q$ and the right hand side of the switch so the potential of the right hand side of the switch will be the same as that of node $Q$.
There is a current through resistor $R_2$ going from left to right due to the battery in the circuit.
When passing through a resistor current flows from a higher potential (earth $= 0$) to a lower potential (at node $Q$).
| {
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Initial values of creation/annihilation operators I have a question about creation/annihilation operators. For example, if I have an evolution equation for annihilation operator of photon
$$ \frac{da_k}{dt} = -i \omega_k a_k$$
I obviously obtain
$$a_k(t) = a_k(0) e^{-i \omega_kt} $$
I not fully understand how to find initial value of $a_k$. Should we just find it from expression of canonical variables $P_k$ and $Q_k$ or maybe I should go to Schrodinger representation since $a_k(0)$ does not depend on time?
Or there is another way?
| The problem is that you are not very specific here. Let me try to infer as much as I can, and then you need to correct me. You are working in the Heisenberg picture here. That means your state is independent of time but you time-evolve your observable instead. Your notation suggests that you are considering a periodic problem with a single band of frequency $\omega_k$ (as there is no band index). $a_k$ annihilates one oscillator mode and $a_k^{\dagger}$ creates one. $k$ could be an element of the Brillouin torus or of the continuum, it is not clear which.
You mention that this is the evolution equation of a photon. That means you already have a universally accepted definition for creation and annihilation operators (one for each polarization in fact), and these satisfy commutation relations with a transverse delta function. So you are already given $a_k(0)$, and this is the thing that you plug in. In the absence of interactions, the number of photons is conserved, so the evolution equation will look like the one you suggested. Once you add interaction, then this will evidently be no longer true.
You will be able to find more details in chapter 13 of Herbert Spohn's book Dynamics of Charged Particles and Their Radiation Field.
| {
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Physical example of line charge Electric field due to an infinite line charge, sheet of charge, point charge, etc are popular problems solved in most text on Gauss's law of electromagnetism.
My question: does an (exact or approximate) example of "infinite/finite line of charge" exist in the physical world?
While we find application of sheet of charge (though finite, not infinite) in case of capacitors, and i can imagine the physical presence of point charge and spherical charge, etc, but a line of charge with uniformly distributed charge density, which basically means a thin conductor with charge Q per unit length - can we have such a thing?
As i understand:
*
*If we connect a battery to a straight wire, with circuit closed, we get a current, but still, any section of the wire is charge-less. So, not an example of line charge.
*If we connect a battery to a wire, with circuit not closed, the charges inside the conductor will move within so as to cancel the applied electric field. So again the conductor won't have uniform charge, so not an example of line charge.
*If we connect ac voltage to a wire, we get sinusoidal charge variation along the wire, so, again not an example.
Can anyone please give a realistic example, which can come close to a line of uniform charge.
| The magnetic field due to a current in a wire can be derived by considered it to be the resultant of Coulombic fields arising from (a) a line of positive ion-cores in the wire and (b) a moving line of electrons. Bearing in mind length contraction, it can be shown (for example Resnick: Intro to Special Relativity) that the charge densities for (a) and (b) cannot be equal and opposite in any but one inertial frame of reference, hence 'a residual field' outside the wire, having (it turns out) the characteristics of a magnetic field.
I suspect that this case technically meets your requirement, but that you were looking for a straightforward case of a line of charge. At least my example – a current-carrying wire – is pretty everyday, though the analysis may not be!
| {
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Question regarding Lorentz Transformation formula So the Lorentz Transformation formation equation are
$$x′=\gamma(x−vt),$$
*
*Does $x′$ and t represent time and position at one event(one instance) or do they represent two events- meaning is $x'$ actually $x_2-x_1$
*I am rather confused on when to use the Lorentz equation for solving problems? Can't I just use the time dilation and length contraction formulas depending on the problem's constraints?
| The equation you've given is not the full Lorentz transformation. There is also a transformation giving t'.
Does x' and t represent time and position at one event(one instance) or do they represent two events- meaning is x′ actually x2−x1
It works for either purpose.
Can't I just use the time dilation and length contraction formulas depending on the problem's constraints?
No. If length contraction and time dilation were all there was, then it would be equivalent to simply changing the units used for time and distance. The Lorentz transformation also says that different observers don't agree on whether events happened in the same place (also true for Galilean relativity) and that they don't agree on whether events happened at the same time.
| {
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What is meant by "collective behavior" in the definition of plasma?
"Plasmas are many-body systems, with enough mobile charged particles to cause some collective behavior ." [M.S. Murillo and J.C.Weisheit Physics Reports 302, 1-65 (1998)].
In the above definition what is meant by "collective behavior" ?
| Plasmas are controlled by long-range interactions (i.e., Coulomb potentials and magnetic fields) and so the particles respond accordingly. Since electric fields do work to get rid of themselves, a plasma will reach what is called a quasi-neutral state, i.e., equal number densities of oppositely charged particles or:
$$
\sum_{s} \ Z_{s} \ n_{s} = 0
$$
where $Z_{s}$ is the charge state of species $s$ and $n_{s}$ is the number density of species $s$.
This can manifest in a fluid-like behavior and the amount which the fluid-like behavior dominates depends upon the plasma parameters like number density, magnetic field, or plasma $\beta$ given by:
$$
\beta = \tfrac{ 2 \ \mu_{o} \ n_{o} \ k_{B} \ T }{ B_{o}^{2} }
$$
where $n_{o}$ is the total number density, $T$ is the total temperature (see definition at https://physics.stackexchange.com/a/375611/59023), and $B_{o}$ is the magnetic field magnitude.
Highly collisional plasmas, e.g., solar photosphere, also behave like fluids as they are collisionally mediated. That is, the relevant frequencies like the cyclotron, $\Omega_{cs}$, and plasma, $\omega_{ps}$, frequencies are at or below particle-particle Coulomb collision frequencies (see definitions at https://physics.stackexchange.com/a/268594/59023).
| {
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Do scalar quantities have magnitude only? I've heard that vector quantities have both magnitude and direction but I've never heard that scalar quantities have magnitude only. Magnitude of vector quantities cannot be negative but what about scalar quantities, like temperature (-1°C)?
If scalar quantities don't have magnitude then what is their "magnitude" called?
Also does the magnitude of a vector quantity include units with the numerical value or only the numerical value?
| A scalar $x$ has magnitude $|x|$, also known as the absolute value. Celarly, $x \neq |x|$ for negative $x$, but instead of saying the scalar has a direction, we would say that it has a sign (+ or -), which is a much simpler concept. Maybe your confusion arises from the fact that a two-dimensional vector can be described through two scalars, one being magnitude and one being direction? Because this means that all magnitudes are scalars, but a negative scalar does not correspond to a magnitude.
As for units: A vector of two-dimensional direction could look like $\bar v = ( 1 \text{ km}, 2 \text{ km})$. Its magnitude is $$|\bar v| = \sqrt{(1 \text{ km})^2 + (2 \text{ km})^2} = \sqrt{5} \text{ km},$$
so yes, the magnitude includes units.
| {
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Is there a higher dimension analogue of Noether's theorem? So I have recently read the proof of Noether's theorem from the book variation calculus by Gelfand. Basically, what I have already seen is that for any single integral functional, if we have a transformation that keeps the functional invariant, we can derive a quantity that doesn't change along any solution of the Euler equations of the functional.
My question: Is there an analogue that work for multiple integral functional? That is, the corresponding system of Euler Lagrange equations are not ODEs, but rather PDEs. Can we define a quantity that is invariant on the whole solutions of these PDEs? The same argument used in Gelfand's proof for single integral functional clearly doesn't work. Can we have something that doesn't change not only with respect with one variable t, but is unchanged everywhere on the whole space like $R^n$ as long as we have a killing vector field for the functional?
| In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time.
For example, where as in the one dimensional case you would write
$$
S = \int dt L
$$
in field theory you would write
$$
S = \int d^4 x \mathcal{L}.
$$
The equation of motion will be a PDE.
Noether's theorem, instead of giving you a conserved quantity $Q$ which satisfies $\dot Q = 0$, would now give you a conserved current $J^\mu$ (where $\mu = 0, 1, 2, 3$ and $\mu = 0$ is the time component and $\mu = 1,2,3$ are the space components) which satisfies $\sum_\mu \frac{d}{d x^\mu} J^\mu = 0$. You can still also find the a conserved quantity $Q$, which satisfies $\dot Q = 0$, defined by
$$
Q = \int d^3 x J^0
$$
and integrating over any fixed time.
| {
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Experimental tests of Schrodinger evolution, position distribution, in square well and other simple systems? Have the energy eigenfunctions in position space ever been experimentally tested for the simplest system undergraduates encounter when learning quantum mechanics, the square well? If not, what is the best example of the experimental verification of position space energy eigenstates or Schrodinger evolution, for simple systems? Almost always the 2-slit experiment is mentioned, or tests of hydrogen such as this one (for which the precision of theory-experiment agreement is quite low), but are there any other tests that test the probability distributions that undergraduates typically derive when learning quantum mechanics?
| You may wish to consider the exeriment by Crommie, Lutz and Eigler (Nature 363, 524 (1993)), who look at standig waves between step edges in a surface two-dimensional electron gas using scanning tunneling microscopy. Other experiments by the same IBM group have shown the standing waves in quantum corrals. Generally, STM images of patterned surface 2D electron gases taken at cryogenic temperatures have shown beautiful images of the 'wave functions' (better, the probability distributions, or even better, the local density of states) of confined systems within the last 30 years.
Another technique uses electron tunneling in semiconductor heterostructures to image the wave functions of quantum dots, as shown in this paper, for example (there are more papers in this direction by the same and other groups). These states are the bound states of so-called 'artificial atoms' (another name for few-electron quantum dots).
I found additional experiments such as this one using photoelectron spectroscopy to image the wave functions of adsorbed molecules.
Although these experiments may not be exactly what you are looking for, because they are rather complicated to explain in detail to undergraduates, they may be used as a 'teaser' to make them curious about learning advanced methods to eventually understand them.
| {
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Why is there so much iron? We all know where iron comes from. However, as I am reading up on supernovas, I started to wonder why there is as much iron as there is in the universe.
*
*Neither brown dwarfs nor white dwarfs deposit iron.
*Type I supernovas leave no remnant so I can see where there would be iron released.
*Type II supernovas leave either a neutron star or a black hole. As I understand it, the iron ash core collapses and the shock wave blows the rest of the star apart. Therefore no iron is released. (I know some would be made in the explosion along with all of the elements up to uranium. But would that account for all of the iron in the universe?)
*Hypernovas will deposit iron, but they seem to be really rare.
Do Type I supernovas happen so frequently that iron is this common? Or am I missing something?
| The nucleosynthesis in the inner of the stars generates energy: The huge amounts of energy form Helium from hydrogen. The star then start generating carbon from helium and so an. This finishes with iron. To generate with larger atomic numbers the star needs more energy. Most of them are generated in supernovae, where there is a lot more energy.
| {
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Confused about forces acting in this simple statics problem I'm trying to solve this statics problem to find the internal tension that results, in each member:
It shows 3 idealised weightless rigid members hinged in a vertical plane, in an isoceles triangle. Two of the members carry identical loads at their midpoint (W = Mg). The bottom member is constrained in such a way that the internal tension exactly balances the "spreading" forces developed by the two members it supports.
But when I try and calculate it (which should be simple!) I get an indeterminate solution. it looks like there should be another constraint, beyond the usual ones, but I can't see any constraint that I'm missing.
As a thought experiment, it should have a single, static, unique, solution. If I built a rigid triangle like this, from 3 lengths of wood, dangled weights from 2 of them, and somehow balanced it vertically with one bottom corner on a brick and the other bottom corner level with it on a roller skate (or just stood it up in an ice-rink), that would match the setup. The resulting concoction would clearly have a static equilibrium (ignoring any bending or overbalancing!) and therefore a unique + static set of tensions, for any given reasonable lengths of wood and magnitudes of weight.
My attempt at solving
By symmetry, I should be able to analyse just one of the diagonal members, to solve for all tensions. So I just draw one diagonal strut, and the reactions/forces at its ends:
For simplicity I'm using resolved reaction forces F1 and F2 for the support at the apex from the other diagonal member, and F2 and F3 for the horizontal internal tension and roller support at the bottom right corner (the horizontal components must be equal and opposite):
*
*Vertically:2W = F1 + F3=> F3 = 2W - F1
*Horizontally: zero net force (only 2 forces, equal and opposite)
*Moments round top-left end:
F3.(2d) = (2W).d + F2.(2h)
=> F3d = Wd + F2h
Substituting from (1):( 2W - F1 ).d = Wd + F2h
=> Wd = F1d + F2h
*Moments round bottom-right end:
(2W).d = F1.(2d) + F2.(2h)
=> Wd = F1d + F2hSame equation as (3).
So I now have 2 equations in 3 unknowns. I'm guessing there should be an extra constraint, but I can't see it.
| $F_1=0$. Reasoning: forces on left diagonal must be symmetric to right diagonal. Hence $F'_1$ on the left diagonal equals $F_1$. But $F_1$ and $F'_1$ are action and reaction forces. Hence are opposites.
| {
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Counterterms in quantum brownian motion In the part "Quantum Brownian motion" of the book, The theroy of open quantum systems written by Breuer, the author investigates on the Caldeira-Leggett model:
The Hamiltonian of the particle is
$H_{S}=\frac{1}{2m}p^{2}+V(x)$
The particle is coupled to a bath consisting of a large number of harmonic oscillators with masses $m_{n}$ and frequencies $\omega_{n}$
$H_{B}=\sum_{n}\hbar\omega_{n}(b_{n}^{\dagger}b_{n}+\frac{1}{2})=\sum_{n}\frac{1}{2m_{n}}p_{n}^{2}+\frac{1}{2}m_{n}\omega_{n}^{2}x_{n}^{2}$
where $b_{n}$ and $b_{n}^{\dagger}$ are teh annihilation and creation operators of the bath modes, while $x_{n}$ and $p_{n}$ are corresponding coordinates and canonically momenta.
The interaction is assumed to be:
$H_{I}=-x\sum_{n}k_{n}x_{n}$
where $k_{n}$ are coupling constants.
The following is my question:
The author says this type of interaction will yield a renormalization of the potential $V(x)$ of the Brownian particle. Hence we introduce a counter-term
$H_{c}=x^{2}\sum_{n}\frac{k_{n}^{2}}{2m_{n}\omega_{n}^{2}}$
to compensate for the renormalization resulted from interaction.
So my question is:
In this case, there seems to be no infinities to be absorbed by counter-terms like that in QFT. And the renormalization caused by the interaction seems to have real physical meanning(in most case in QFT, they have not), like Lamb-shift. So how could we introduce counter-terms to cancel the renormalization in this case?
| There is nothing quantum or explicitly divergent here, but even at the classical level the extra "push" from the interaction with the bath oscillators changes the potential seen by the particle so that it is no longer $V(x)$. The counterterm is designed to cancell this effect so that the parameters in $V(x)$ coincide what would be measured experimentally.
See the discussion on page 155 and the footnote on page 156 of the follwoing lecture notes: https://courses.physics.illinois.edu/phys508/fa2018/amaster.pdf
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Why is $\chi_{[\mu}\nabla_\nu \chi_{\sigma ]} = 0$ at the Killing horizon? Let $\chi$ be a Killing vector field that is null along a Killing horizon $\Sigma$
Why is $\chi_{[\mu}\nabla_\nu \chi_{\sigma ]} = 0$ at $\Sigma$?
| This is a partial answer. It assumes that the Killing field $\chi$ is normal to the Killing horizon. This implies that $\chi$ is null along the horizon, so it is consistent with the condition given in the OP. However, a Killing field that is null along the horizon is not necessarily orthogonal to the horizon; that's why this is only a partial answer.
Given a function $\Phi$ on spacetime and a constant $c$, the equation $\Phi=c$ defines a hypersurface. The vector field
$$
n_a=\partial_a\Phi=\nabla_a\Phi
\tag{1}
$$
is orthogonal to that hypersurface. Now, if $\chi_a$ is any vector field orthogonal to this family of hypersurfaces, we must have
$$
\chi_a=mn_a
\tag{2}
$$
for some function $m$. If $\chi$ is normal to the Killing horizon, then it can be written in the form (2) along the horizon.
Equation (2) implies
$$
\nabla_b\chi_a=\nabla_b(m\nabla_a\Phi)
=
(\nabla_b m)(\nabla_a\Phi)
+
m\nabla_b\nabla_a\Phi.
\tag{3}
$$
Now consider $\chi_{[c}\nabla_b \chi_{a ]}$, where the square brackets denote complete antisymmetrization (which is what I'm assuming they mean in the OP). According to the preceding equations,
$$
\chi_{[c}\nabla_b \chi_{a ]}
=
m(\nabla_{[c}\Phi)(\nabla_b m)(\nabla_{a]}\Phi)
+
m^2\nabla_{[c}\Phi\nabla_b\nabla_{a]}\Phi.
\tag{4}
$$
The first term in (4) is zero because of the identity
$$
(\nabla_{[c}\Phi)(\nabla_{a]}\Phi) = 0,
\tag{5a}
$$
and the second term in (4) is zero because of the identity
$$
\nabla_{[b}\nabla_{a]}\Phi = 0
\hskip2cm
\text{(zero torsion)}.
\tag{5b}
$$
Altogether, this shows that
$$
\chi_{[c}\nabla_b \chi_{a ]} = 0
\tag{6}
$$
whenever the vector field $\chi$ is orthogonal to the given hypersurface. This is (part of) the Frobenius theorem.
This derivation of equation (6) is outlined in section 2.3.3 in Poisson (2002), "An advanced course in general relativity," https://www.physics.uoguelph.ca/poisson/research/agr.pdf.
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Particle accelerators and the amount of radiation they emit Do particle accelerators emit a lot of radiation? If they do, can someone give me a rough estimate of how much?
I'm also curious as to what kind of radiation they give off.
| The idea of a storage ring is to store the beam, so giving off radiation means either losing beam, or losing energy. The later is accomplished via synchrotron radiation (accelerating charged particles in a circle), and is easily computed from the current, the energy, and the bending radius.
In the usual LINAC (linear accelerator), the beam is radiation, period. It travels in a vacuum and hits a target, producing secondaries (some of which are the experiment at hand, most of which are radiation).
My recollection of SLAC was that we used 40 $\mu$A at 4 GeV, which means a power of:
$$ P = IV = (40\,\mu A)(4\,GV) = 160\,kW$$
Over the course of an experiment we wound up exposing a temporary building several hundred yards behind End Station A to near the 100 mrem limit of annual radiation.
LAMPF (now LANSCE) was 1 mA of protons at 1 GeV, or 1 MW. There was a target that was only accessible by robots: if you saw it line-of-sight, you died.
The beams generally activate much of the laboratory material, including air ($^{15}$O) and esp. copper in vacuum connections, but the aluminum beam pipe not so much.
For comparison, the 20,000 Ci (!) $^{60}$Co source JPL has for space-radiation testing is "only" making 300 W thermal.
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Lorentz contraction of the wavelength of light I couldn't find this question on the suggested "similar questions". If this has been asked before please direct me to answer. My question is "why isn't the wavelength of light,which is in the direction of motion, going at the speed of light Lorentz contracted to zero instead of its value?"
| The Lorentz contraction factor $\gamma$ describes the length of an object in relation to the object's length in its own rest frame. Light doesn't have a rest frame, so the same analysis doesn't apply.
| {
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Strong empirical falsification of quantum mechanics based on vacuum energy density It is well known that the observed energy density of the vacuum is many orders of magnitude less than the value calculated by quantum field theory. Published values range between 60 and 120 orders of magnitude, depending on which assumptions are made in the calculations. Why is this not universally acknowledged as a strong empirical falsification of quantum mechanics?
| This is a very good question. As noted above it is not a prediction of "ordinary" quantum mechanics but of QED/QFT. As it is probably the wrongest prediction ever, something fundamental is wrong. However since predictions of QED tend to be very accurately confirmed by experiment, it cannot be a complete falsification. Something in the theory is false, however. Unresolved so far.
| {
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Find the time when tangential acceleration is equal to radial acceleration
A particle begins to move along a circular path of radius R with a constant magnitude tangential acceleration of $a_t$. After time $t$ it's the centripetal acceleration is equal in magnitude to tangential acceleration. Find $t$?
My Attempt:
$$a_t = \frac{dv}{dt}$$
$$a_r = \frac{v^2}{R}$$
Where,
$R =$ Radius of the circle
$a_r = $ Radial acceleration of the circle
(As given in the question)
$$a_t = a_r$$
$$\frac{dv}{dt} = \frac{v^2}{R}$$
I think integrating will give us the $t$, but what should be the limits?
Integrating from $t=0$ and $v=0$ doesn't work well. It just gives infinity.
If my way doesn't work, then how do I approach this problem?
Any help would be appreciated.
| You don't want to treat your final equation as a differential equation you need to solve. That equation is only going to be valid at one instant in time, whereas if you integrate it you will be assuming it holds at all times, and hence you won't have a constant $a_t$. Really what you have for a single time $T$ is
$$a_t(T)=a_r(T)$$
$$\left. \frac{\text dv}{\text dt}\right|_{t=T}=\frac{\left(v(T)\right)^2}{R}$$
You need to determine what $v(t)$ will be based on $\dot v=a_t$ and then it should be fairly easy from there. I'll leave the rest for you to do.
| {
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Violation of Stefan's law when shining a light on a black body Suppose there is a black body in a dark room and the room temperature is constant. Now a ray light is shone upon it. Since a black body absorbs all radiation that falls upon it, it must absorb more radiation than it did when no light fell on it, but according to Stefan's law, the radiation absorbed depends only on the ambient temperature (which is constant).
I feel like I haven't completely understood the concept of black bodies since the two explanations are contradicting each other.
| Before the light was shone on the black body the temperature of the black body was constant because the rate at which energy was radiated out to the surroundings was equal to the rate at which it absorbed energy from the surroundings.
When the light was shone on the black body continuously and absorbed by the black body this would represent an increase in the energy content of the black body and that would manifest itself as an increase in the temperature of the black body above that of its surroundings.
This would mean that the black body would start radiating out more energy than it did before and so there would be a net outflow of radiation from the black body.
Eventually, an equilibrium state would be achieved when the rate at which the black body is gaining energy from the light is balanced by the net loss of radiant energy by the black body to its surroundings.
This would require the black body to be at a higher temperature than that of its surroundings.
| {
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How do I find the pulling force required to lift an object in a system of 3 pulleys?
In the image above, assume that the person is lifting the 12.5 kg weight at a constant velocity and the pulleys are frictionless. In this case, what would the pulling force be?
Here's what I've tried:
I assumed that the resultant force on the pulley A would be equal to the tension T. Since the velocity is constant, the acceleration is 0.
T - 125 = 12.5 * 0
=> T = 125 N
So the resultant force on the pulley A is 125 N.
Here's the part I'm unsure about:
If the resultant force on the pulley A is 125 N, then should the tension on both sides of pulley B also be 125 N? And if so, would the tension on both sides of pulley C also be 125 N (which would result in a pulling force of 125 N)? Or would the resultant force from pulley A be divided somehow?
| While analyzing tensions in the ropes is a valid way to solve this problem, one can also consider that:
$$ F = \frac{dE}{dx}$$
so lifting the weight a distance $x$ off the ground (which is plain old lifting, $F=mg$), you have to pull the rope a distance $x/2$, so the force should be $2mg$.
Note that the main purpose of pulleys B and C is obfuscation.
| {
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Buoyancy question Suppose we have a ball of volume V and a block of the same volume V whith same density. We submerge the ball into some kind of liquid so that $\frac{1}{2}V$ is submerged into the liquid. We do the same thing with the block. Now $\frac{1}{2}V$ of both block and ball is submerged into the liquid. At this moment we can say that the buoyant force on both objects is the same. If we now push both the ball and the block by little bit into the liquid by the same volume, the force acting on the bottom of the ball would increase (because it is now a little deeper in the liquid) and so will the force acting on the bottom of the block. If we now look at the geometry of the objects, wouldn't there be additional force on the ball pushing it downward because of the curvature of the ball which wouldn't appear on the block? Can we still say the buoyant force is the same on both objects? I added a picture and highlighted the area where I think the force would act in red.
| This diagram might help you understand what is going on?
The initial position of the sphere is shown in grey and the new position of the sphere is shown in orange.
The pressure on all parts of the sphere below $XY$ increases by $h\rho g$ when in the new position $X'Y'$ and those parts all contribute to an increase in the net force upwards.
The section of the sphere $X''Y''Y'X'$ which was originally in the air is now in the liquid with the change in pressure on that surface ranging from $0$ along $X''Y''$ to $h\rho g$ along $X'Y'$ and the surface area which has become immersed is much less than the surface area already in the liquid.
This newly submerged part of the sphere thus contributed a net downward force which is less than the increase in the upward force contributed by the permanently submerged part of the sphere.
| {
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Velocity of efflux when fluid is itself moving In efflux velocity given by Torricelli equation:
$$ v^2=2gh $$
what is reference frame for this $v$. Is it fluid or ground? I am just a high school level student, so please forgive my ignorance.
| The velocity $v$ is the velocity relative to the container and the fluid in it. That is, the fluid in the container is taken to be at rest, so $v$ is the velocity increase as the fluid is forced through the hole.
Torricelli's law is simpler than you (probably) think. It just uses conservation of energy to say the increase in kinetic energy must be equal to the decrease in gravitational potential energy. If you consider some small part of the fluid with a mass $m$ then as it falls a distance $h$ the decrease in gravitational potential energy is:
$$ \Delta U = mgh $$
And the increase in kinetic energy as the water accelerates to a speed $v$ through the hole is:
$$ \Delta T = \tfrac{1}{2} mv^2 $$
Since total energy has to be conserved these must be equal, and equating them gives:
$$ v^2 = 2gh $$
| {
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Effect of earth's rotation in ballistics For this purpose, let's consider earth's rotations constant. Do earth rotation momentum get transfered to any object (a missile for example) that get's lauched? If so, why do we have to consider earth rotation when lauching the missiles? Wouldn't just follow earth rotation? (Btw, sorry for any grammar mistakes, I'm from a non-english speaking country).
| Earth's rotation velocity does get transferred to any object (a missile for example) that gets launched. The reason why we have to consider Earth's rotation when launching the missiles is the target rotates together with the Earth. Let us consider the following example: a missile is launched from point A to deliver some payload to point B. The typical flight time of an intercontinental ballistic missile is about half an hour. Due to Earth's rotation, point B travels up to 800 km in half an hour (depending on its latitude). So if one does not take Earth's rotation into account, the payload will be delivered to a point far away from B.
One can take Earth's rotation into account, for example, using a frame of reference rotating with the Earth. In this case, one needs to introduce fictitious centripetal and Coriolis forces. Littlewood gave the following example: allegedly, during a battle at Falkland Islands in 1914 (during World War I), German ships were destroyed from a maximum distance, but it required a lot of time, as initially the projectiles missed the targets by 100 yards, as corrections of about 50 yards for the Coriolis force were calculated for the latitude of 50 deg. North, whereas the Falkland Islands are at a latitude of about 50 deg. South. So ignoring or miscalculating the effect of Earth's rotation can be pretty risky.
| {
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Feynman diagram for semileptonic decay of neutral kaon I am unsure how to draw a feynman diagram for a reaction that occurs as follows
$$ K^0 --> l^+\nu_l\pi^- $$
Any tips would be helpful.
| Because neutral currents have not been observed and flavor changing can only be done with the emission of W bosons the strange antiquark will decay to an up antiquark and W- boson . The W- boson will decay to an electron and an antielectron neutrino. The pion formed by the up antiquark and the down quark will decay to a neutral pion and a W- boson.The W- boson will decay to an electron and an electron antineutrino.The neutral pion will decay to 2 gamma rays and all is left is 2 electrons and 2 electron antineutrinos and some light.
| {
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Why does a single X-ray photon generate 1620 electrons when it hits CCD detector? Fe55 is one radioactive isotope.
It emits X ray photons : mainly k-alpha & k-beta lines.
Why does a single X-ray photon generate 1620 electrons when it hits CCD detector ?
While, in photo-electric effect, single photon generates single electron i.e. a photo-electron.
| The iron K-alpha edge is at around 7,000 eV and that bandgap of Silicon is about 1.1 eV.
Let's assume that all that happens when the x-ray hits the ccd is that the x-ray is completely absorbed to create excited electrons. Then we can expect 7000/1.1 or about 6300 electrons from a single x-ray photon. This is a rough estimate, but you can see that x-rays should be easily able to create many electrons at a detector.
Edit: Just to be clear to OP, as Jon Custer mentioned the initial process is a single excited electron (i.e. leading order absorption $H_{int} \sim \mathbf{p}\cdot\mathbf{A}$ term), but the lifetime of this electron is so short that it creates more excited electrons almost immediately (within attoseconds).
| {
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Why doesn't a table tennis ball float on a surface of steel balls? How do we calculate buoyancy here? Place the beaker full of steel balls and submerge the table tennis ball under the steel balls. The table tennis ball does not float up. Why does it not float up? Do table tennis balls float when the diameter of steel balls is reduced? How to calculate the buoyancy of steel balls?
Would it come up without friction?
| Well, what if the steel balls were extremely small, say molecular size. In that case, the constraining annulus would look like a polished steel collar, and would likely hold down the ball even if the glass was shattered, underwater, in a swimming pool. . (The van der Waals forces, and metallic bonds, would account for that.)
But this example given, shows discreet balls of intermediate size, and unless they are magnetized, their coupling with the container is what allows restraint of the tennis ball.
If THIS setup was in the bottom of a deeper pool, and the beaker was shattered,
the steel balls would run radially away, and the tennis ball would pop up.
(Note: I answered this as if there was water in the beaker along with the steel balls and tennis ball. But the answer is not changed by my error.)
The term "van der Waals force" is sometimes used loosely for all intermolecular forces.
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Clarification on isotope notation I am a bit confused about the idea of isotope notation. I know that the top number is the mass number, equivalent to the number of neutrons + protons, while the bottom is the atomic number, equivalent to the number of protons.
What I don't understand is why during beta decay, the electron emitted is represented as having an atomic number of -1. What exactly does it mean to have a negative atomic number? Can the atomic number be considered as just the charge?
Also, my understanding is that the -1 charge balances the gain in the charge due to the proton being formed from a neutron. However, where does this electron originate from if the nucleus originally only contains protons and neutrons?
Thanks for any help, I appreciate it.
|
Can the atomic number be considered as just the charge?
It's not standard to talk about an electron as having an atomic number. However, if you interpret these numbers as charges, then the system you've been taught makes sense.
However, where does this electron originate from if the nucleus originally only contains protons and electrons?
Processes like these can create and destroy particles.
| {
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Minimum Angular Velocity
A bead is free to slide on a vertical circular frame of radius $R$ comes to equilibrium when $\cosθ = g/Rω²$.
The minimum value of angular velocity comes out to be $\sqrt{g/R}$, which we can find out by balancing Gravitational and centripetal force with Normal reaction to bead from the frame.
Why can't the angular velocity have values between 0 and $\sqrt{g/R}$?
| If your formula was correct, then the maximum $\cosθ = 1 = g/(Rω^2)$. However, it is not correct. Assuming $θ$ is measured up from the (downward) vertical to the radius going to the bead, then: $N(\cosθ) -mg =0$ and $N(\sinθ) = m(Rω^2)$.
Solve the first equation for $N$ and put it into the second. Then: $(mg/\cosθ)\sinθ = m(Rω^2)$ and $(\cosθ)/( \sinθ) = g/(Rω^2)$. The angular velocity is not limited.
| {
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Unable to get interference from milk using Time domain OCT (Michelson Morley interferometer) approach I was able to construct time domain Optical coherence tomography based on Michelson Morley interferometer using a LED. While testing the interferometer, I used glass slide, layers of scotch tape. I was able to get fringes from. but when I try to use milk drop as a sample I don’t see any fringes. To be noted I use, 4x objective lens of NA 0.1 and I use 12bit camera. Please let me know what may be the potential reason for not not getting fringes from the Milk.
| Possibly the milk particles are too small, the light randomly scatters throughout it, the returned scattered light then has random phase variations, and the fringes are washed out.
Ideally, you’ll have optical interfaces from which the light can reflect with a particular phase delay relative to the reference beam.
| {
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How do I Fit a Resonance Curve with Respect to Known Data? In an experiment, I collected data points $ (ω,υ(ω))$ that are theoretically modelled by the equation:
$$ υ(ω)=\frac{\omega \, C}{\sqrt{(\omega^2-\omega_0^2)^2+γ^2 \omega^2}} \,.$$
How can I fit the data to the above correlation? And how can I extract $\gamma$ through this process?
| If we put:
$$Y = \frac{\omega^2}{u(\omega)^2}$$
and
$$X = \omega^2$$
the equation becomes:
$$Y =\frac{X^2}{C^2} +\frac{(\gamma^2 - 2 \omega_0^2)}{C^2} X + \frac{\omega_0^4}{C^2}$$
You can then extract the coefficients using polynomial fitting. To get the least-squares fit right, you have to compute the errors in $Y$ and $X$ for each data point from the measurement errors in $\omega$ and $u(\omega)$.
| {
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Is it possible to trigger a nuclear reaction with physical force? In Mission Impossible Fallout, they're dealing with 3 plutonium cores. If one of those cores was thrown against a wall by Ethan Hunt, could it start a chain reaction and explode? For that matter, could any blunt force cause the nuclear reaction to initiate or would it require a radio active trigger?
| As written in direct comments to the question: in the case of fission bomb designs the critical mass state is achieved by setting off a chemical explosion. .
As written in comments: the problem is to design the chemical explosion in such a way that you get a significant yield of nuclear chain reaction.
By contrast:
If there is some mishap in which two lumps of nuclear bomb grade material are brought together accidentally then the immediate heat from the first fission events will drive those lumps away from each other.
I think that means that accidentally bringing lumps of nuclear fission material together will result in the tiniest nuclear explosion, causing radiation poisoning of all the people in the same room, but not much else.
Bringing lumps together on purpose, but with only the force of a human throwing the stuff will have a similar result; radiation poisoning of the people in the room, not much else.
To get a significant yield you have to do something extremely violent and extremely precise. The lumps of fissile material must be brought in contact faster than the speed at which the nuclear chain reaction developes. That is very, very difficult.
| {
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Micro-world Perception: What does a microbe perceive as it's being, say, sloshed in a glass of water? I've wondered how micro-organisms perceive the larger world, it's forces and the
consequences the relatively massive forces have on such tiny objects.
Let's say E coli (0.5 micro meters width, 2 micro-meters length) in a pure stream of water, moving with an acceleration of 2 m/s^2.
I've used a micro-organism in this question solely to validate perception. From a biological viewpoint, the microbe may not percieve the forces, nevertheless, what are the effects on the body.
I'm interested in the effects on a small scale on any micro-object.
| As pointed out by PM 2Ring, as animals get smaller, the effects of the viscosity of the water become greater. A human pushes his or her way through the water; a goldfish wiggles their way through the water, and an animal the size of a water flea (1/10th of an inch, or less) crawls through the water as if it were pancake syrup (to us). A big paramecium (1/100th of an inch or less) engages the water with its cilia and drags itself through it.
To a bacterium, water becomes a matrix in which it is embedded and although there are bacteria with single whip-like cilia for propulsion, their mobility is limited. When the water/matrix is accelerated, they are carried along with it unless the force causing the acceleration is (by our standards) really big- hundreds to thousands of G's. In this case, they sink through the water in the opposite direction of the G force and can also get sheared open and destroyed when the force varies strongly with position on a scale length of order ~several bacterium lengths or less.
| {
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Observing the conserved canonical momenta Suppose I have a Lagrangian $\mathcal{L}[\phi]$ with $\phi$ a cyclic variable, which means that the Lagrangian is symmetric under shift of $\phi\rightarrow\phi+c\quad$.
The equation of motion will be simply the conservation of the canonical momentum:$$\partial_{\mu}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}[\phi]:=\partial_{\mu}J^{\mu}=0$$
My question is: can in general the value of $J^{\mu}$ enter any observable?
More precisely, will any operator containing $J^{\mu}$ be generated at loop level?
If so, how will such operator be made generally?
Or somewhat equivalently: is the scalar $J_{\mu}J^{\mu}$ relevant in any way?
Edit: I realised that since $\mathcal{L}\supset J^{\mu}\partial_{\mu}\phi\;,\;\;$ the combination $J^{\mu}\partial_{\mu}\phi$ will enter the stress energy tensor. Is this the only way $J^{\mu}$ enters any observable?
| Generally I can observe $J^{\mu}$ whenever I add to the Lagrangian a coupling of the field $\phi$ to some other field, say $\psi$, in such a way that there is a vertex containing one $(\partial^{\mu})\phi$ line and only a vector combination of $\psi$ lines, say $b_{\mu}(\psi)$.
This vertex generates an operator $\langle \,J^{\mu}\,\,b_{\mu}(\psi)\,\rangle$ which depends on $J^{\mu}$.
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Are microwaves sinusoidal or is that just a model? I know that projectiles are parabolas because I can derive that from constant acceleration. And the height of a Ferris wheel rider vs. time is demonstrably a sine wave. What is the underlying thing that tells us microwaves are sinusoidal vs. some other periodic shape? I'm teaching trig.
| The wave equation for light tells us that solutions are sinusoidal:
$$ \frac{\partial^2 E}{\partial t^2} = c^2\frac{\partial^2 E}{\partial x^2}$$
(here, in one dimensions with the electric field $E(x, t)$ a function of position $x$ and time $t$).
If you guess a sinusoidal solution:
$$ E(x, t) = e^{i(kx-\omega t+\phi)}$$
(note:
$$e^{i(kx-\omega t+\phi)} = \cos{(kx-\omega t+\phi)}+i\sin{(kx-\omega t+\phi)} $$
is the most general solution that is sinusoidal in space and time)
and plug it in, you get:
$$ -\omega^2E(x, t) = -c^2k^2E(x, t) $$
which means:
$$ \omega = ck =2\pi c/\lambda $$
is the relation between frequency and wavelength ($\lambda = 2\pi/k$).
Hence you can specify a solution by its frequency $\omega$.
This represents and infinitely long wave with infinite duration. In the real world, boundary conditions matter, so this is not the actual solution. Nevertheless, the machinery of Fourier analysis allows us to build any solution, periodic or not, as a superposition of different frequencies.
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Why we fill dU/dT value in Cv(specific heat at constant volume) only and why not in Cp? According to equipartition of energy, the energy ossociated with each degree of freedom is $\frac{K_{b}T}{2}$ for one molecule .
For 'x' molecule which has degree of freedom f it's energy is given by
$U= \frac{f k_{b}Tx}{2} = \frac{f k_{b}RTn}{2}$ , where n is no. of moles
for small change in temperature
$\frac{dU}{dT} = \frac{fnR}{2}$
$C_{v} =\frac{dU}{ndT} = \frac{fR}{2}$ this much given in text book but why we can not fill dU/dT in $C_{p}$ from which we get $C_{p} = \frac{fR}{2}$.
| Because specific heat at constant pressure is defined in terms of enthalpy $h$ and not internal energy $u$. It is defined as
$$C_{P}=\biggl(\frac {δh}{δT}\biggr)_P$$
PROOF:
Since
$$h=u+Pv$$
$$\biggl(\frac{δh}{δT}\biggr)_{P}=\biggl(\frac{δu}{δT}\biggr)_{P}+\biggl(\frac{δ(Pv)}{δT}\biggr)_{P}$$
$$\biggl(\frac{δh}{δT}\biggr)_{P}=\biggl(\frac{δQ}{δT}\biggr)_{P}-\biggl(\frac{Pδv}{δT}\biggr)_{P}+\biggl(\frac{Pδv}{δT}\biggr)_{P}+\biggl(\frac{vδP}{δT}\biggr)_{P}$$
The last term is zero, therefore
$$\biggl(\frac{δh}{δT}\biggr)_{P}=\biggl(\frac{δQ}{δT}\biggr)_{P}=C_{P}$$
Hope this helps.
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Rocket Leaving Earth's Gravitational Field I stumbled across this seemingly simple question that really stumped me on further thought:
A rocket is intended to leave the Earth's gravitational field. The fuel in its main engine is a little less than necessary, and an auxiliary engine, only capable of operating for a short amount of time, has to be used as well. When is it best to switch on the auxiliary engine: at take-off, or when the rocket has nearly stopped with respect to the Earth, or does it not matter?
My understanding is that to escape the earth's field, there must be sufficient kinetic energy so that the total energy is positive. My first instinct was that, assuming the auxiliary engine operates at a constant power, it will cause the same change in energy regardless of how far it is from the earth, therefore it doesn't matter. However, it seems that where this change in energy occurs is actually important. Can someone help me understand what the correct answer is?
| In principle, I'll say that it is the same. We can imagine the small auxiliary engine like a little bit of more fuel in the main tank, which is the same as all the other fuel.
In practice, it depends on how efficient the engine is:
Rocket engines generate thrust by pushing back the exhaust gases from fuel burn. More efficient engines throw the exhaust faster, thus generating more thrust per unit of fuel. Also the engine dead weight go in the equation.
Fuel is carried by the rocket and is heavy; then the optimal solution will be:
- Use the less efficient engine first, so you don't need to carry up all that ineffective fuel & the heavy engine
- Use the most weight-efficient engine after.
If you can fire them at the same time, the answer by Paulo Gil works.
| {
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How can the big bang be "observed"? The Big Bang Observer is a proposed spacecraft to study gravitational waves. Especially the one that are thought to originate from the Big Bang. Question. How can we observe waves that has an origin at the beginning of time? Doesn't it implies then that we were there when it happened so that the wave were created and then travelled to us?
| The aim of the Big Bang Observer would be to study gravitational waves generated during the inflationary period that took place some $10^{-32}$ seconds after the Big Bang.
One of the frequently misunderstood aspects of the Big Bang is that it didn't happen at some point, with the universe expanding outwards from that point, but instead happened everywhere in space. This is discussed in Did the Big Bang happen at a point? This means if you pick any random point in the universe the Big Bang happened at that point (as well as everywhere else) some 13.7 billion years ago. So if that point happens to be 13.7 billion light years away at present time the gravitational waves generated during inflation 13.7 billion years ago are reaching us right now.
This is what is meant by observing the Big Bang. We would be literally observing the gravitational waves that were generated $10^{-32}$ seconds after the Bang Bang and have spent the time since then travelling towards us.
| {
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Quantum energy levels of a point mass rotating about a fixed point The question is: A particle of mass m is attached to a fixed point in space by a massless rigid rod of length a and can freely rotate about this point. Find the quantum energy levels of the system. What is the degeneracy of each energy level?
I used rotational kinetic energy:
$E=\frac{1}{2}I\omega^2=\frac{L^2}{2I}$
and then substituted $I=ma^2$ and $L=\hbar \sqrt{l(l+1)}$ to get:
$E_l=\frac{\hbar^2l(l+1)}{2ma^2}$.
So the energies are quantized as expected. But what is the degeneracy of each level? Plugging in a bunch of values for $l$ doesn't show any $l$s with similar energy so far. Is it correct that the degeneracy of each level is $0$?
| As @probably_someone wrote in his comment:
It might be easier to think of the equivalent problem: a particle is constrained to move on a fixed sphere, with no other forces besides the constraint (which is what is meant by "free" in this context).
Now I guess that the non-degenerate energy levels you wrote down are correct if we confine the rotation to one plane. But there is an infinity of planes in which the rod can rotate, which means there is an infinite degeneracy for all energy levels.
| {
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Do electrons in a flash of lightning oscillate between Earth and the clouds (or between clouds)? In this video, around 2:24, one can see a flash of lightning that keeps the same form for a little time span. Does this mean that it takes a while before the discharge is complete (which I can't imagine) or that the electrons move to and fro between the Earth and cloud (or between the clouds themselves) before the discharge is complete? I can imagine that extra electrons move behind the discharge which on arrival on Earth flash back, taking extra electrons back, etc. But in this case, the flash should dim over time. So why the flash remains in a steady shape that long?
| Lightning is a rapid transition from a non-conducting to a conductive state.lightning creates an ionized, electrically conductive channel through air inside the cloud,between clouds or between cloud and earth.The high electric field accelerates the electrons between the channel.When they collide with air molecules, they create additional ions and newly freed electrons which are also accelerated.So it takes some time.Electrons will only flow against the electric field.they don't oscillate.
Once the gap breaks down, current flow is limited by the available charge (for an electrostatic discharge) or by the impedance of the external power supply. If the power supply continues to supply current, the spark will evolve into a continuous discharge called an electric arc
| {
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Potential by Assembling Charges For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$.
In order to find potential of this sphere at the surface, why is my approach giving different answers?
Approach 1:
$$\rho = \frac{3Q}{4 \pi R^{3}}$$
$$q = \frac{4}{3} \pi x^{3} \rho = Q \frac{x^{3}}{R^3}$$
Potential at the surface would be $$V = \frac{q}{4 \pi \epsilon_0 x} = \frac{Q x^{2}}{4 \pi \epsilon_0 R^{3}}$$
Approach 2:
$$\rho = \frac{3Q}{4 \pi R^{3}}$$
$$q = \frac{4}{3} \pi x^{3} \rho = Q \frac{x^{3}}{R^3}$$
$$E = \frac{Q x}{4 \pi \epsilon_0 R^{3}}$$ (From Gauss' Law)
Potential at the surface would be $$V = -\int{\vec{E} \cdot \vec{dx}} = -\frac{Q}{4 \pi \epsilon_0 R^{3}} \int_{0}^{x}{xdx} = -\frac{Q x^{2}}{8 \pi \epsilon_0 R^{3}}$$
Why is the answer different in both the cases?
| The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.
I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.
Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.
The area under the graph $\int E\,dx$ is related to the change in potential.
In essence what you have done is found that areas $A$ and $B$ are not the same.
PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?
| {
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Density of supermassive black holes So I know that supermassive black holes can have "densities" less than water because black hole density scales as $1/m^2$ since $R_s = 2GM/c^2$. I am trying to reconcile this with the fact that black holes are the most compact objects for a given mass.
For instance, if I had 6 billion solar masses of material and squeezed it to the Schwarzschild radius, I would have a black hole. However, by my statement above, if I had 6 billion solar masses of water, this would occupy a volume less than the Schwarzschild radius, but would not be a black hole. How does this work? I know black holes actually have singularities (etc), but where is the flaw in my classical model of just density = M/V?
| If I'm understanding the question correctly, then you're trying to reconcile two statements: (1) A Schwarzschild black hole consists of empty space. (2) A sufficiently compact distribution of matter is a black hole.
These statements are not in contradiction to each other, because a Schwarzschild black hole is just one model of a black hole. It's a model of an eternal black hole, which has always existed. It isn't a model of an astrophysical black hole that forms by gravitational collapse. But if you let the process of collapse run to completion, and you stop feeding the black hole, it will end up looking just like an eternal black hole.
| {
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Why couldn't they take pictures of a closer black hole? The latest photos of the M87 black hole capture light from around a black hole at the center of the Messier 87 galaxy, which is 16.4 Mpc ($5.06 \times 10^{20}$km) from our milky way.
Why couldn't / didn't the scientists involved take photos of black holes less distant, for example those at the center of our Milky Way or Andromeda (0.77 Mpc) or Triangulum Galaxy? Wouldn't these black holes appear larger and the photos have greater detail / resolution and be easier to capture?
My intuition would be that maybe black holes at the center of closer galaxies aren't as large, or maybe they have more matter in the way / aren't directly aligned with our view from earth making it harder to capture them, but I don't know for sure.
| Since this isn't covered by Rob Jeffries' answer, let me add that Sagittarius A* (the black hole in the centre of Milky Way) was considered, but as explained by Heino Falcke at press conference revealing the photo (quoted after Deccan Herald)
Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear,
On the other hand both M87 and Sgr A* were "photographed" (i.e. data required was captured) it was just "easier" task to process data for M87 so we may expect picture of Sgr A* as the next one in some time.
As stated on EHT webpage
We study supermassive black holes Sgr A* and M87 because their apparent sizes are much larger than those of stellar-mass black holes when viewed from the Earth, so they are easier to study.
| {
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Integrability of a non-integrable quantum spin model at critical point Is it right, that non-integrable quantum spin models in one dimension become integrable at their critical points? Or do they stay nonintegrable at the critical point also? Are there any examples known? In the field of 2d classical models, the three-state Potts model is not in general integrable, but this model is integrable at the critical point.
| If the non-integrable quantum spin chain at the critical points can be described as a conformal field theory (not always the case), we can say that the model is "integrable''. Because CFT can be seen as an "integrable'' theory since it can be solved exactly and Yang-Baxter relation is satisfied naturally. If the critical points cannot be described by CFT, there is no general guarantee whether the underlying field theory is integrable or not.
| {
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Do working physicists consider Newtonian mechanics to be "falsified"? In the comments for the question Falsification in Math vs Science, a dispute around the question of "Have Newtonian Mechanics been falsified?"
That's a bit of a vague question, so attempting to narrow it a bit:
*
*Are any of Newton's three laws considered to be 'falsified theories' by any 'working physicists'? If so, what evidence do they have that they believe falsifies those three theories?
*If the three laws are still unfalsified, are there any other concepts that form a part of "Newtonian Mechanics" that we consider to be falsified?
| "Falsified" is more philosophical than scientific distinction.
Newton laws have been falsified somehow, but we still use them, since usually they are a good approximation, and are easier to use than relativity or quantum mechanics.
The "action at distance" of Newton potentials has been falsified (finite speed of light...) but again, we use it every day.
So, in practical terms, no, Newton laws are still not falsified, in the sense that are not totally discredited in the scientific community. Classical mechanics is still in the curriculum of all universities, in a form more or less identical that 200 years ago (Before Relativity, quantum mechanics, field theory).
Most concept in physics fit more in the category of "methods" rather than "paradigms", so can be used over and over again. And all current methods and laws fails and give "false" results, when used outside their range of applicability.
The typical example of "falsified" theory is the Ptolemaic system of Sun & planets rotating around the Earth. However, philosopher usually omits the facts that:
*
*Ptolemaic system was experimentally pretty good at calculating planet motions
*Most mathematical and experimental methods of the new Heliocentric paradigm are the same of the old Ptolemaic
So the falsification was more on the point of view, rather than in the methods.
| {
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Expansion of an ideal gas at constant pressure I approach these expansion problems like so:
The gas and the surroundings(piston+outside) are at the same pressure at first. We heat the gas. The pressure rises inside the syringe a bit. The gas expands so the pressure remains constant. Then I use P(the constant pressure of the gas) *dV. What I want to confirm is my reasoning on using this equation. It was derived assuming P(internal) = constant. But it does change momentarily. Is the reason we ignore it in the "a bit" nature?
Also for compression, the force exerted on the gas by surroundings (piston+outside) is taken as the force the gas exerts on the piston. Is this Newton's third law?
| If the heat addition occurs very slowly such that the pressure and temperature gradients in the gas approach zero, the process can be considered quasi-static and the pressure of the gas will always be very close to the external pressure. If the process is also frictionless, then we can say it is a reversible isobaric expansion. At every point along the process the ideal gas equation applies
$$\frac{V}{T}=\frac{nR}{P}$$
since $P$ is constant
$$\frac{V}{T}=constant$$
Since the external force applied to the gas by the piston and atmosphere is always approximately equal to and opposite the force the piston and atmosphere apply to the gas, Newton;s third law applies.
Hope this helps.
| {
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Angular momentum of a body about a point rotating about its own axes I want to calculate angular momentum of a sphere about point O. The sphere is rotating about its two axes with angular velocities $w_1$ and $w_2$.
I know that angular momentum = $m\vec{r}\times\vec{v} + Iw$, where v is velocity of centre of mass. Here, v=0, therefore angular momentum of COM = 0. But, the body is itself rotating. Now, which angular momentum should I take?
$Iw_1$ ,
$Iw_2$ ,
$Iw_1 + Iw_2$ ,
Components of $Iw_1$ and $Iw_2$ along r. or What?
| A rigid body can only have one rotation axis. When the angular velocity vector has multiple non-zero components, like $$\vec{\omega} = \pmatrix{ \omega_1 & \omega_2 & 0}$$
then the magnitude of rotation is described by the length of the vector $$ \omega = \| \vec{\omega} \| = \sqrt{ \omega_1 ^2 + \omega_2 ^2 } $$
The rotation axis direction is the unit vector along $\vec{\omega}$
$$ \hat{\rm rot} = \frac{ \vec{\omega} }{ \| \vec{\omega} \|} = \pmatrix{ \frac{\omega_1}{\sqrt{ \omega_1 ^2 + \omega_2 ^2 }} \\ \frac{\omega_2}{\sqrt{ \omega_1 ^2 + \omega_2 ^2 }} \\ 0 } $$
In case of a sphere (where the mass moment of inertia is uniform with direction) the angular momentum magnitude is $$ L = {I}\, \omega ={I}\, \sqrt{ \omega_1 ^2 + \omega_2 ^2 } $$
or by component
$$ \begin{matrix} L_1 = I\, \omega_1 \\ L_2 = I\, \omega_2 \\ L_3 = 0 \end{matrix} $$
and $$ L = \sqrt{ L_1^2 + L_2^2 + L_3^2 } $$
In general, it is easier to consider the vector form of the above with a matrix/vector equation
$$ \vec{L} = \mathrm{I}\, \vec{\omega} $$
$$ \pmatrix{L_1 \\ L_2 \\ L_3 } = \begin{vmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{vmatrix} \pmatrix{\omega_1 \\ \omega_2 \\ \omega_3} $$
For a sphere $I_1 = I_2 = I_3 = I$.
Even more useless information below:
Rotation of a rigid body happens along a line in space (the rotation axis). The location of this line relative to the COM is given by $$ \vec{r}_{\rm rot} = \frac { \vec{\omega} \times \vec{v} }{ \| \vec{ \omega} \|^2 } $$ where $\vec{v}$ is the velocity vector of the COM.
Corollary to this is the fact the momentum happens along a line in space (the axis of percussion), in such a way that a single impact along this line can instantaneously immobilize a rotating rigid body. This axis has direction along the linear momentum $\vec{p} = m \vec{v}$ and is located relative to the COM at $$ \vec{r}_{\rm imp} = \frac{ \vec{p} \times \vec{L} }{ \| \vec{p} \|^2}$$ where $\vec{L} = \mathrm{I}\, \vec{\omega}$ is the angular momentum vector at the COM.
Welcome to the introduction of screw theory in mechanics.
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What is the electric potential inside a point charge? We know that electric potential of spherical charge is inverse proportion with $r$ from $V = \frac{kq}{r}$ , So if $r$ is getting less then electric potential will be higher.
But , What about point charge ? Is it infinity inside it ? My teacher told me that it's zero but I am not believing that.
| When dimensions become very small, we are no longer in the realm of classical physics where potentials have mathematical singularities as in the classical 1/r Coulomb potential. Point particles belong to the realm of quantum mechanics and there the laws and computational rules are different.
The electron is a charged point particle in the standard model of particle physics, but when interacting with a proton, the potential near the point location is not accessible. What happens the electron is bound to the proton forming the hydrogen atom, and it cannot fall on the proton as the classical 1/r potential indicates. There is a lowest energy state, and it stays there in orbitals, which are instead of a track, probable locations where it may be found.
So it has no meaning for quantum dimensions to be asking the value of the 1/r potential. Its functional form has been used to calculate the orbitals, and even when there is a probability of overlap, as seen in the S orbitals, there are quantum number conservation rules and energy conservation rules that keep the orbitals stable.
Actually, the stability of the atom is one of the reasons that quantum mechanics had to be invented. In classical physics because of the 1/r attraction the electrons would fall on the nucleus, neutralizing them, and no chemistry would exist.
The proton cannot become neutral when the electron overlaps with it in space, because quantum mechanically the neutral particle is the neutron, which is heavier , and also electron lepton number has to be conserved. ( there is electron capture in nuclei where there is extra energy to allow for the reaction , but that is another story)
One needs to study quantum mechanics and nuclear/particle physics at the realm of point particles.
| {
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How do you calculate the error propagation in a Helmholtz coil? I'm having some trouble with the error propagation expression in a Helmholtz-coil.
Specifically $\frac{\frac{∂B}{∂R}}{B}$.
If anyone could help, I'd be really grateful.
The formula used is
$$B(x) = \frac{NMI}{2R} \cdot \left(
\left(1+\left(\frac{x+\frac{a}{2}}{R}\right)^2\right)^{-3/2}
+\left(1+\left(\frac{x-\frac{a}{2}}{R}\right)^2\right)^{-3/2}
\right)$$
$N$ = Number of wire coils
$M$ = Magnetic constant
$R$ = Radius of coil
$x$ = Distance from mid-point of coils
$a$ = Distance between coils (in this case $2R$, $R$, and $R/2$)
$I$ = Electric current
| HINT: The error propogation can be calculated for a relation $Z\pm\Delta Z=(A\pm\Delta A)^n(B\pm\Delta B)^m$ , and $Z\pm\Delta Z=\frac{(A\pm\Delta A)^n}{(B\pm\Delta B)^m}$
$$\frac{\Delta Z}{Z}=n\frac{\Delta A}{A} +m\frac{\Delta B}{B}$$
The propogation for $Z\pm\Delta Z=(A\pm\Delta A)(B\pm\Delta B)$ is:
$$\Delta Z=\Delta A+\Delta B$$
Only quantities that have some error in them should be substituted into this propogation.
| {
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Is polarization complementary along its different axes? Is polarization complementary along its different axes -- much like the spin of a particle is -- thus implying that the uncertainty principle holds for polarization measurements on these different axes?
| The answer is yes, in the sense of mutually unbiased bases.
Polarization can be expressed in terms of $2\times 2$ Pauli matrices, and the eigenstates of the Pauli matrices are mutually unbiased, meaning that, if $\vert \pm\rangle_i$ are the two eigenstates of the Pauli matrix $\sigma_i$, then eigenstates of any of the other two Pauli matrices have an equal probability of being measured when the state of the system is prepared in $\vert \pm\rangle_i$. In more mathematical term, the equality
$$
_k\langle \pm\vert\pm\rangle_i=\frac{1}{2}\, ,\qquad i\ne k
$$
holds independently of $i$ and $k$.
Within the context of complementarity one would state that, if you know everything about polarization in the direction $i$, you know nothing about the polarization in any of the other orthogonal complementary directions, since the possible outcomes in the complementary directions are equiprobable.
| {
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My hairs on my hand stand straight whenever I move my hand to TV screen and feel vibration and sound on TV screen Whenver I try to reach my hands near the screen of my TV . My hairs on my hands get straight and more importantly I can hear and feel a sound and kinda vibration . Please mention if this sound duplicate. My purpose is simply know the reason behind those strange sound and vibration that I could feel.
| Is your television tube-based or LCD?
Tube-based TVs work by shooting electrons at the front of the display. Most of these get recycled back to the back of the tube using a wire that connects on a pad that's just behind the front, under the case where you can't see it. However, the front of the display is not perfectly connected and some static builds up.
LCD's don't do this, but they have a plastic sheet on the front that can hold a static charge from the surroundings.
In either case, you are feeling static.
| {
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Interacting term in Jellium model I have a question about the deduction of interacting term in Jellium model. In the text book Condensed Matter Field Theory ed.2 Alexander Altland, Ben Simons, pg.52. Author gives the expression of e-e interaction without deduction details:
$$
V_{ee}=\sum_{k_1,k_2,q} a^\dagger_{k_1-q \sigma_1} a^\dagger_{k_2+q \sigma_2} \frac {e^2}{q^2}a_{k_2 \sigma_2}a_{k_1 \sigma_1}
\tag 1
$$
I try to get this form via inserting the Fourier expansions of each term into the "classical" Coulomb interaction expression
$$
V_{ee}=\int dr_1dr_2 \ a^\dagger_{r_1 \sigma_1} a_{r_1 \sigma_1} \frac {e^2}{r_{12}} \ a^\dagger_{r_2 \sigma_2} a_{r_2 \sigma_2}
\tag 2
$$
with
\begin{align}
a^\dagger_{r} & = \sum_{k} e^{-ikr}a^\dagger_k \\
a_r & = \sum_k e^{ikr}a_k \\
\frac {1}{r} & =\sum_q e^{iqr}\frac {1}{q^2}
\end{align}
When I have done this, I got an expression with different order of operators. This is mine:
$$
V_{ee}=\sum_{k_1,k_2,q} a^\dagger_{k_1-q \sigma_1} a_{k_1 \sigma_1} \frac {e^2}{q^2} a^\dagger_{k_2+q \sigma_2} a_{k_2 \sigma_2}
\tag 3
$$
When $q \neq k_1-k_2$ or $\sigma_1 \neq \sigma_2$, eq.3 can be re-ordered thus becomes identical to eq.1. However, if $k_1=k_2+q$ and $\sigma_1=\sigma_2$, the re-order cannot be done as before since $[a_{k_1 \sigma_1},a_{k_1 \sigma_1}^\dagger]=1$. Therefore, I cannot reach eq. 1?
So, what's wrong? Any suggestions are appreciated.
| Suppose that you have two quartic terms with different orderings, like $a^\dagger a^\dagger a a$ and $a^\dagger a a^\dagger a $, assuming that total momenta and Lorentz spins etc. of the oscillators sum to zero. As you say, if you try to re-order one of them you "wrong" terms via the commutator $[a^\dagger,a]=1$, like
$$g \sum \ a^\dagger a^\dagger a a \sim g \sum a^\dagger a a^\dagger a + g \sum a^\dagger a.$$
The "wrong" term you get just renormalizes the coupling in front of the mass term
$$\int dr\, a_r^\dagger a_r$$
that you have in your Hamiltonian. So it's nothing to worry about.
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How does light 'choose' between wave and particle behaviour? Light exhibits wave behaviour in phenomenon such as interference but particle behaviour in the photoelectric effect. How does light 'choose' where to be a wave and where to be a particle?
| The light doesn't chose. You as an experimenter chose which observable you want to measure, and thus which operator you use. Such measurement will result in a wavefunction collapsing into one of the eigenstates of this operator.
E.g. a positiin operator will give you a position, i.e. particle.
A momentum operator will give you a momentum, i.e. wavelike object.
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Why is the ratio of gravitational force and the inertia to resist it 1? Is there a deeper meaning behind how things of different mass fall at the same acceleration? It feels so perfectly balanced...
| This is a priori an assumption.
In Newtonian physics, they are not assumed to be equal but because empirically they are, we often take them to be the same. For example, Newton's second law says $\vec{F}=m_i\vec{a}$ and Newton's law of gravitation says $\vec{F} = -(GMm_g/r^2)\hat{r}$. The standard (high school) calculation assumes $m_i=m_g$ from empirical results but as stated, the two laws do not say they are the same.
In general relativity, however, it is assumed that gravitational mass and inertial mass is the same: this automatically leads to the result that every object accelerates the same way under gravity, since it is simply a statement that every (point) particle (with no non-gravitational forces acting on them) follows spacetime geodesics. This is closely connected to weak equivalence principle. The fact that general relativity remains the currently best explanation (verified by experiments) for gravitation can be taken as evidence that they should be equal: if deviations were found by ultra-precise experiments, then one will have to forgo general relativity.
Note that unlike Newtonian gravity where the equality is not justified by the framework itself (Newton's laws did not specify they have to), general relativity tells us that we should take them to be equal as an explanation for how gravity works. The way test particles follow geodesic equation in curved spacetime can only make sense if test masses's (inertial/gravitational) mass do not enter the picture. That's why one cannot simply "geometrize" electromagnetism (at least not by itself, as in the case of Kaluza-Klein which geometrizes electromagnetism and gravity).
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How is this derivation of a field transformation, in Weinberg's QFT book, performed? I am reading Weinberg's book Quantum theory of fields.
Could you explain to me the following things? Vol.1, page 60 (transcribed from this image):
To first order in $\omega$ and $\epsilon$, we have then
\begin{align}
U(\Lambda,a) \left[\tfrac12\omega_{\rho\sigma}J^{\rho\sigma} -\epsilon_\rho P^\rho\right] U^{-1}(\Lambda,a)
& =
\tfrac12(\Lambda\omega\Lambda^{-1})_{\mu\nu} J^{\mu\nu}
\\ & \quad - (\Lambda\epsilon - \Lambda\omega\Lambda^{-1}a)_\mu P^\mu
\tag{2.4.7}
\end{align}
Equating coefficients of $\omega_{\rho\sigma}$ and $\epsilon_\rho$ on both sides of this equation (and using (2.3.10)) we find
\begin{align}
U(\Lambda,a) J^{\rho\sigma} U^{-1}(\Lambda,a) & = \Lambda_\mu^{\ \rho}\Lambda_\nu^{\ \sigma} (J^{\mu\nu} - a^\mu P^\nu + a^\nu P^\mu)
\tag{2.4.8}\\
U(\Lambda,a) P^\rho U^{-1}(\Lambda,a) & = \Lambda_\mu^{\ \rho} P^\mu
\tag{2.4.9}
\end{align}
How are we equating the coefficients? How we find the formula (2.4.8)?
The previous formula that the chapter refers to is
$$ (\Lambda^{-1})^{\rho}_{\nu} = \Lambda_{\nu}^{\rho} = \eta_{\mu\nu}\eta_{\rho\sigma}\Lambda_{\sigma}^{\nu} \tag{2.3.10}$$
| Equation (2.4.9) follows easy:
The coefficient of $\eta$ on the LHS of (2.4.7) is $-U P U^{-1}$. And on the RHS it is $-\Lambda P$ (here we don't have to worry about the indices). Set these two equal and (2.4.9) follows.
The coefficient of $\omega$ on the LHS of (2.4.7) is $U J U^{-1}$. The coefficient on the RHS is $(\frac 1 2 \Lambda {\Lambda}^{-1}J-\Lambda {\Lambda}^{-1} a P)$.
I'll leave it to you to put in the indices after which you can use (2.4.10) and obtain (2.4.8), though I have the feeling that this is where you are stuck. Try harder! I'm not supposed to give you the full answer.
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The Meaning of Electromagnetic 'News' in Griffiths Book In the Introduction to Electrodynamics book, by David J. Griffiths, 4th edition, page 60, the author makes the following statement:
"it is not the position, velocity, and acceleration of Q right now that matter: electromagnetic "news" travels at the speed of light, so what concerns Q is the position, velocity, and acceleration Q had at some earlier time, when the message left."
In Example 7.9, page 320, he also makes the same statement:
"electromagnetic "news" travels at the speed of light, and at large distances B depends not on the current now, but on the current as it was at some earlier time (indeed, a whole range of earlier times, since different points on the wire are different distances away)."
*
*In fact, I am struggling with the idea of "electromagnetic news" and its meaning.
*In addition, what do we mean by "at some earlier time," because Griffiths always make this statement throughout his book?
*How could electromagnetic waves depend on quantities, such as the current in Example 7.9, took place "at some earlier time?"
Any help is much appreciated. Thank you so much.
| Imagine that there was an explosion.
You are $3\times 10^8\,\rm m$ (=$1$ light second) away from the explosion.
You will receive "news" of the explosion ie see the flash of light associated with the explosion, one second after the explosion actually happened.
Here the use of the word "news" has the implication of a record of an event that has happened in the past.
So what is affecting you, the light arriving from the explosion, originated from an event, the explosion, which happened one second earlier in time.
| {
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Lorentz Transformation: Message sent before finish line As she wins an interstellar race, Mavis has a “hooray” message sent from the back of her 300m long ship as she crosses the finish line at v=0.6c. Stanley is at the finish line and at rest relative to it. He claims the message was sent before she crossed the line.
I understand how to get the answer using the Lorentz transformation. However, I am having trouble conceptually understanding why he observes the message before she crosses the finish line.
| I assume that the signal is sent when she actually crosses the finish line in the ship's reference frame (the issue of how they know when to send the signal is irrelevant).
Now, imagine that there is someone in the center of the ship that, at a time before the ship reaches the finish line, sends a signal to both, the front and the back of the ship, and that this signal reaches the ends at exactly the same time than the ships reaches the finish line. In the ship's reference frame these two events are simultaneous, they coincide with the ship reaching the finish line and with the time the message was sent.
But what does an observer stationary with the finish line sees? He sees that the light ray that moves towards the front takes a larger time to reach the front than the light that travels towards the back takes to reach the back. This is because to him, the front of the ship moves in the same direction than the light, whereas the back of the ship moves towards the ray that moves backwards, reaching him before the front ray reaches the front of the ship.
| {
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If an electric field passing through a dielectric medium, back into the original medium, is it "back to normal"? Suppose there is an electric point charge causing an electric field E in a medium with a dielectric constant $\epsilon_1$. You can calculate the scalar potential $\phi$ at a given distance $r$, as well as the gradient field $E$.
Now imagine you insert a dielectric medium with a different $\epsilon_2$ somewhere across that distance (like a piece of glass, a plastic board, etc (as long as its not a conductor like sheet metal). I am aware there will be refraction etc., but apart from that there won't be an effect on the electric field beyond that material, right? Meaning the scalar field will be the same strength as if the material 2 was not there? The only effect would be the refraction which would cause a parallel shift of the electric field, right?
Is this some kind of natural law or so? Does this effect have a name?
| Andreas Schuldei makes a great point about the parallel plate capacitors and different media between the plates.
On the micro level there are some important factors. An applied field distorts the electron cloud around atoms in molecules yielding a dipole moment. However, this happens among adjacent molecules/atoms. So if you have two molecules stacked on top of each other, with electrons being pointed in the same direction, the negative part of the resulting dipole will overlap with the positive part of the dipole below, cancelling out. Consequently, the net bound charge difference ends up being on the surface of the media.
With equal and opposite bound charge collected on the surfaces of a media, they cancel each other out so the media below that doesn't "see" the effect of the bound charge.
| {
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What had Feynman meant when he told nobody understands Quantum mechanics? What do we mean by understanding Quantum mechanics? What had Feynman meant when he told nobody understands Quantum mechanics?
What do we mean by understanding Quantum mechanics?
| He certainly meant understandable in terms of principles of classical mechanics (points of matter with masses, subjected to forces in a three-dimensional Euclidean space and satisfying deterministic evolution laws), since we understand very well quantum mechanics. That is evident from the fact that nowadays most technology is strongly based on QM so we can handle practically and, before, theoretically all notions and theoretical constructions of the quantum world. Exactly as we do with classical physics. In that sense we understand both.
The idea that classical physics is more understandable is an illusion in my opinion: we are only much more familiar with it and its theoretical description than quantum physics and relativistic physics.
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Cooling effect caused by evaporation Evaporation is said to cause cooling effect because it absorbs energy from surroundings to change its phase from that of a liquid to gas. I am in doubt as to why would the surroundings be ready to give up energy more than the water molecules. Wouldn't 'lower temperature of the water molecules in comparison to the surroundings' or other such conditions which might involve specific heat capacity or conductivity etc. be mandatory to ensure that the heat is transferred from the surroundings to the water molecules and not the other way round ?
| Evaporation requires the liquid molecules to overcome the intermolecular attractive forces and escape to the surroundings. Only the most energetic molecules, near the surface of the liquid, have enough kinetic energy to overcome the attractive forces. As these highly energetic molecules escape from the liquid (i.e., evaporate), the average kinetic energy of the molecules in the liquid decreases and temperature is nothing but a measure of the average kinetic energy of the system. Hence, temperature decreases with evaporation.
Now, for more number of liquid molecules to have energy enough to overcome the intermolecular forces, the average energy of the liquid must be high. The liquid can gain more energy by absorbing energy from its surroundings. But according to the second law of thermodynamics, this is possible only if the temperature of the liquid is lower than that of its surroundings. Such a flow of energy occurs until the liquid and its surroundings are at the same temperatures (or until all the liquid evaporates).
So obviously, evaporative cooling is possible only if the liquid is cooler than its immediate surroundings. A very good example of that would be sweating, in which the sweat absorbs energy from our body and evaporates.
Edit:
When you perspire the sweat will be at the same temperature as that of the body. However as the more energetic molecules escape the liquid, the average energy (and hence temperature of the liquid) decreases and this is turn causes the sweat to absorb more heat from the body.
| {
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When air is ionised, which component gases are actually ionised? Air is a mixture of O2, N2 etc. Which gases are ionised at STP, and is there a different level of ionisation for different gases?
| The most common ionizations are of oxygen and nitrogen. Nitrogen ionizes first (at lower levels of energy input) and yields a deep blue color. Oxygen ionizes at a slightly higher energy level and adds a pinkish-white color to the plasma. Once the plasma heats up due to current flowing through it, it begins to emit a black-body spectrum and becomes white.
| {
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Will tsunami waves travel forever if there was no land? If there was no land for tsunami waves to collide with, can the waves travel around the globe for forever?
| Waves keep going forever, in a way
As others pointed out already, waves tend to lose energy.
However, what will (theoretically) happen is that at any point in time the wave will lose a fraction of its energy, but never truely 100%. So though you will soon reach a point where random noise will make it practically impossible to detect a wave, the effect of the wave will never truely be gone.
So, while you will not notice it and definitely won't call it a tsunami, the wave can be seen as going on forever!
| {
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How to calculate pumping pressure for backfilling of a borehole? I am trying to calculate the pumping pressure needed to backfill a borehole with geothermal grout. The hose is pulled up at the same rate as the borehole fills, so pumping pressure will get lower as the hole is filled. The borehole is initially filled with water and is filled bottom to top.
The way I am calculating it now is like this:
With:
Pb: pressure at B (start of the hose going into the borehole).
i: Percentage of the borehole filled with grout).
ρ: Density.
g: acceleration due to gravity.
y: depth of the borehole.
τw: friction loss from hosewall.
D: diameter of the hose.
f: Friction coefficient.
vg: velocity of the grout in the hose.
Problem is: I don't know if this is correct. Can somebody validate this or come with the right way to calculate this?
| A force balance on the grout column gives $$P_B+\rho_{grout}gy(1-i)-P_A-4\frac{\tau_w y}{D}=0$$This neglects any acceleration of the grout fluid and any drag caused by the upward motion of the hose. For the water, the force balance is $$P_A=\rho_{water}gy(1-i)$$ So, $$p_B=-(\rho_{grout}-\rho_{water})gy(1-i)+4\frac{\tau_w y}{D}$$So the density difference between the grout and the water reduces the pressure at B, and the grout fluid drag flow increases the pressure at B.
| {
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Deriving classical Hall effect from quantum Hall effect I'm interested in the derivation of the classical Hall effect coefficient, given in cgs by $$R_{H}=-\frac{1}{nec},$$ where $n$ is the electron number density, $-e<0$ is the electron charge,and $c$ is the usual, ubiquitous velocity in Physics, from the fact that QHE provides the quantum of electrical conductance $$g=\frac{2e^{2}}{h},$$ where $h$ is Planck's constant, and the 2 comes from spin degeneracy.
Is there a convenient way to go from the quantum to the classical case for this problem?
| I think you cannot derive classical case from the quantum case. $g=\text{filling factor}\cdot\frac{2e^2}{h}$ occurs at very high magnetic fields where Landau levels start filling and current is carried only by edge states. In classical regime, magnetic fields are low and Landau levels haven't started filling and there is no edge states
| {
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Can heat increase cutting power? I'm writing a book in which the protagonist has a sword that she closes in lightning. (It doesn't melt because it is made of a special material.) Could the heat generated by the superhot plasma allow the sword to cutt through metal?
| An average lightning bolt is at $30,000 K$ and contains $10^9 J$ of energy and the average sword has a weight of $1.5 kg$. For calculation purposes let us assume that the sword is made of carbon steel which has a specific heat capacity of $490 J/kg K$ and let us also assume that all the energy of the lightning bolt gets transferred to the sword. We can calculate the temperature change of the sword using the equation,
$$Q = mc\Delta T \implies \Delta T = \frac{Q}{mc} = \frac{10^9 J}{490 J/kg K . 1.5 kg} \cong 1.36 \times 10^6 K $$
So your sword (made of special material) should be at a temperature much higher than the melting points of most known metals and contains enough energy to cut through standard metal items.
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No hair theorem and Killing tensors I have 2 questions regarding Killing Tensors :
*
*A practical question is how to guess whether a spacetime has Killing tensors or not. We can guess some simple Killing vectors by looking at the isometry of the metric components. Is there any such simple intuition behind finding Killing tensors for a given metric?
*The no hair theorem states that for static spacetimes we would only have three conserved quantities - Charge, Mass and Angular momentum. What about the conserved quantities from the Killing tensors? Does the no-hair theorem take Killing tensors into account?
|
A practical question is how to guess whether a spacetime has Killing tensors or not.
This is very hard and in general case open problem, however for some situations (such as vacuum stationary and axially symmetric solutions) there are techniques. See e.g. this paper for the description of an algorithm (Cartan–K̈ähler or prolongation-projection method) with several examples. Since the calculations are quite involved, a practical first step would be to check for some numerical criteria of integrability for the geodesics in such a spacetime (such as plotting Poincaré sections).
The no hair theorem states that for static spacetimes we would only have three conserved quantities - Charge, Mass and Angular momentum. What about the conserved quantities from the Killing tensors?
No-hair theorems like this one, restricts the number of parameters characterizing a spacetime (vacuum or non-vacuum with a fixed matter content) with given properties (stationary, with an event horizon). Killing tensors allow us to write conserved quantities for geodesics in a given spacetime, so they offer a parametrization of completely different problem.
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Normalize accelerometer data based on non-central location I am trying to "normalize" accelerometer data for a sensor that is not centered on a car. In the image below, you can see how centripetal force changes based on location.
In my case, I only have one sensor on the right side of the vehicle. For logistical reasons, I can only have one sensor and it must be on the right hand side. As I turn, the force is more exaggerated turning left compared to turning right.
Assuming I know the exact dimensions of the car and the exact location of the sensor, is there a way to normalize the data as if the accelerometer were centralized (in other words, left and right turns would have identical signatures)? Assume there is no noise, that we are pre-smoothing the signal.
| With only the readings of one accelerometer you're out of luck.
The basic reason is that a particular reading could be due to a high-speed, large-radius turn or a low-speed, small-radius turn and the two cases call for different corrections.
If you have secondary data—such a ground velocity (speedometer speed is only approximately correct for this but you could make do) or angular velocity (from a compass or gryoscope)—then you can work something out.
Of course, for smooth driving behaviors you could also get enough secondary data from GPS, but in that case you don't actually need the accelerometer at all.
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Why are orbits $1.5r_{s} < r < 3r_{s}$ unstable around a Schwarzschild black hole? The "Orbital motion" section of the Wikipedia entry corresponding to Schwarzschild metric reads:
A particle orbiting in the Schwarzschild metric can have a stable circular orbit with $r > 3r_s$. Circular orbits with $r$ between $1.5r_s$ and $3r_s$ are unstable...
*
*What does justify this instability?
Some follow-up questions:
*What would change if one consider a big object, say, a soccer ball, instead of a particle?
*Does the same result hold for rotating (Kerr) black holes?
| Regarding your first question:
In general relativity the energy of a light object moving around a spherically symmetric mass can be written as:
$E=mc^2(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2((1-\frac{2GM}{rc^2})^2(\hat{r}\cdot\hat{v})^2+(1-\frac{2GM}{rc^2})|\hat{r}\times\hat{v}|^2)}}}) $
The strange part above is because in the Schwarzschild solution the local velocity of light is different in the radial direction and transverse to the radial direction. For pure circular motion this expression reduces to:
$E=mc^2(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1--\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}}}) $
This can be rewritten as:
$E=mc^2(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}})$
For pure stable circular motion we also have, just as classically, $v=\sqrt{GM/r}$ so we can write:
$E=mc^2(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{GM}{rc^2}}})$
From the last expression it can be seen that infinite energy is needed to stay in a circular orbit at the photon radius $r=\frac{3GM}{c^2}=1.5r_s$. By taking the derivative of the last expression with respect to r we see that it has a minimum for $r=\frac{6GM}{c^2}$. This means that as long as you are "higher up" than $r=\frac{6GM}{c^2}=3r_s$ circular orbits require less energy to sustain as you go closer to the black hole. However, closer to the black hole than that radial distance, circular orbits require more and more energy to sustain the closer you go to the black hole, and that is why they are unstable.
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Is throwing dice a stochastic or a deterministic process? As far as I understand it a stochastic process is a mathematically defined concept as a collection of random variables which describe outcomes of repeated events while a deterministic process is something which can be described by a set of deterministic laws. Is then playing (classical, not quantum) dices a stochastic or deterministic process? It needs random variables to be described, but it is also inherently governed by classical deterministic laws. Or can we say that throwing dices is a deterministic process which becomes a stochastic process once we use random variables to predict their outcome? It seems to me only a descriptive switch, not an ontological one. Can someone tell me how to discriminate better between the two notions?
| As you commented, rolling classical dice is a deterministic chaotic process, and thus impossible to model in enough detail to predict which face will be up, unless you have very accurate input data.
More accurate than is plausible for dice rolled by hand, unless maybe you're collecting it with high rez / high-speed cameras or other 3D motion capture system, and have carefully measured the physical properties of the dice and the surface.
A simpler model of the same thing is that it's just random which way up they land.
There might be some scope for predicting which way the face is rotated, and significantly more scope for predicting where the die lands / dice land on a flat table, especially if the table is soft so the dice don't bounce a lot.
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Does a particle exert force on itself? We all have elaborative discussion in physics about classical mechanics as well as interaction of particles through forces and certain laws which all particles obey.
I want to ask,Does a particle exert a force on itself?
EDIT
Thanks for the respectful answers and comments.I edited this question in order to make it more elaborated.
I just want to convey that I assumed the particle to be a standard model of point mass in classical mechanics. As I don't know why there is a minimum requirement of two particles to interact with fundamental forces of nature,in the similar manner I wanted to ask does a particle exerts a force on itself?
| This answer may a bit technical but the clearest argument that there is always self interaction, that is, a force of a particle on itself comes from lagrangian formalism. If we calculate the EM potential of a charge then the source of the potential, the charge, is given by $q=dL/dV$. This means that $L$ must contain a self interaction term $qV$, which leads to a self force. This is true in classical and in quantum electrodynamics. If this term were absent the charge would have no field at all!
In classical ED the self force is ignored, because attempts to describe have so far been problematic. In QED it gives rise to infinities. Renormalisation techniques in QED are successfully used to tame the infinities and extract physically meaningful, even very accurate effects so called radiation effects originating from the self interaction.
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Photoelastic modulator vs. Acousto-optic modulator I am trying to understand the salient differences between photoelastic modulators (PEMs) and acousto-optic modulators (AOMs).
Both of these devices are based on the photoelastic effect - that is, under mechanical stress, a medium's optical properties (e.g., refractive index) change. PEMs are generally used as tune-able phase retarders (refractive index along one axis changes compared to the orthogonal direction), whereas AOMs induce a periodic index variation, yielding effects similar to Bragg diffraction. AOMs are often said to utilize the acousto-optic effect, but that is a subset of the photoelastic effect (not clear on what the distinction is).
Am I correct to assume that the same underlying principles apply to both the PEM and AOM? The only difference is that in the PEM, we have a standing (sound) half-wave, whereas in the AOM we have a travelling wave with many cycles present in the medium, yielding spatially-periodic modulations? Is the direction of applied stain another important difference between the two?
| I think you are correct that in an AOM we have a traveling wave of phonons. I know experimentally that laser diffracted in the same(opposite) direction as the RF input will be up(down)-shifted in frequency. This makes sense if one thinks about the momenta of the photon and phonon adding or subtracting. Most AOMs also consist of an RF transducer bonded on one side of the crystal with a piece of rubber damping material placed at the other end, which is also often cut on an angle to reduce RF reflections from the opposite crystal face.
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"source": "stackexchange",
"question_score": "3",
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Is there a temporal difference between planets due to the sun's gravitational field? since the Sun generates a gravitational field it also generates gravitational time dilatation. Hence, time further from the Sun should pass quicker than in its proximity.
Can we, therefore, say that the time on Mercury is different from the time on Pluto?
Do space probes take into account this difference? For instance, was 'New Horizons' adjusted for the time dilatation during its trip to the Kuiper's belt?
And if there is a difference, does not this rise a paradox, for planets have been generated in the same geological era but then have different relative time? [this actually works also on Earth itself, for Wikipedia reports that there is a difference of 39 h between sea level and the top of the Everest].
| Time dilation is a concept from special relativity. GR doesn't have a concept of time dilation, except in the special case of a static spacetime, which is the only case where the metric can be derived from a scalar potential. An observer on the surface of a planet is orbiting the sun, so the spacetime they're in is not static. All GR really ultimately says is that when you integrate the metric along a world-line, you get the proper time. The result is typically not describable in terms of a static time dilation.
Do space probes take into account this difference? For instance, was 'New Horizons' adjusted for the time dilatation during its trip to the Kuiper's belt?
What is actually observable is a Doppler shift, and yes, this is taken into account for space missions. A cleaner, simpler example to correctly demonstrate the concepts you have in mind is GPS. You can find careful treatments of this online.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can you change the wavelength of light keeping frequency constant and can you do the opposite as well? Can you change the wavelength of light keeping frequency constant and can you do the opposite as well?
I understood the basics but please don't hesitate to go deeper into the concept. Also, If you happened to have an elegant explanation please drop it here if you can.
| Why does frequency remain constant?
The reason that light does not change its frequency is because of conservation of energy. Light is made up of quanta (small discrete particles). The energy of light is proportional (or equal if you use appropriate units) to its frequency.
$$E=nh\nu$$
Where $n$ is number of quanta. When this light ray goes to a different medium the energy must be conserved. Hence the frequency remains same. The relation between the new and the old wavelength can also be calculated using this relation.
Is the opposite possible?
To my knowledge the opposite is not possible for light rays. Since the speed of light is constant for every observer in a medium the frequency and wavelength must change. It is not possible to keep wavelength constant while changing the other two.
Hope this helps.
| {
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Why does electrical resistivity have units of $\Omega \cdot \mathrm{m}$ rather than $\Omega \cdot \mathrm{m}^3 ?$ Electrical resistivity has units of $\Omega \cdot \mathrm{m} .$ However, since resistivity can be described as the resistance of a unit cube, shouldn't the units therefore be $\Omega \cdot \mathrm{m}^3$ instead?
I ask after seeing this question to which the answer is apparently $\left(\text{D}\right) :$
$
\text{Resistivity can be described correctly as:} \\
\hspace{1em}
\begin{array}{cl}
\mathbf{A} & \text{resistance of a unit length.} \\
\mathbf{B} & \text{resistance per unit area.} \\
\mathbf{C} & \text{resistance per unit volume.} \\
\mathbf{D} & \text{resistance of a unit cube.}
\end{array}
$
| Resitivity can be thought of as resistance of a unit cube, but for a unit cube, the resistance doesn't work out to $\text{material constant} * \text{volume}$.
Instead, restivity ($\rho$) is given by $\rho = \frac {RA}{L}$ (where $R$ is resistance, $A$ is area and $L$ is length of material) or to rearrange in terms of net resistance $R = \frac {\rho L}{A}$.
We can see that this suggests that resistance will increase with length, but decrease with area. This should make sense, because if you send the same current through a wider area, it should experience less resistance, and if you push the same current through a longer path of resistance, it should experience more net resistance over the path.
You can see from that relationship that the units will work out to resistance per unit length, not per unit volume.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is a hole better for creating a toroidal vortex than a pipe? Does a hole work better for creating a toroidal vortex, where the fluid starts to turn as it exits, than a pipe, where the flow inside is more laminar. Some high power vortex guns have a cone shaped barrel. So, is a cone better?
| enter link description here
Hope this image help you!!!
The thing here assume the pipe only one side, gas is moving similar to given image.
If it is a cone then you can see that almost all the gas entering leaves out. Hence the power is also increased.
| {
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Viscous flow through a concave tube Consider gravity-driven incompressible flow through a concave tube with known radius $R(z)$, where pressure is atmospheric $P_a$ on both ends.
How does the viscous pressure distribution $P(z)$ compare with that of purely inviscid flow?
I think both flows should definitely have a pressure drop, so that $P(z) < P_a$ throughout the tube. Furthermore, I think the viscous flow should result in a larger pressure drop to balance the shear stresses that don't exist in inviscid flow. I therefore think the pressure distributions for viscous and inviscid flow should look something like this:
Is this qualitatively correct?
| In the case of viscous flow, I gave the answer here. Viscous laminar flow during expansion of the nozzle will come off the walls with the formation of a jet and return flow zones - see Figure 1.
In the case of an inviscid continuous flow, the distribution of velocity and pressure is shown in Fig.2. In this case, the pressure distribution on the axles is similar to what @Drew drew. In both cases, a parabolic velocity profile is specified at the input.
Align the pressure in the viscous flow at the inlet and outlet of the pipe by subtracting $\rho gz$. The distribution of the velocity of the flow does not change, and the distribution of pressure on the axis takes the form similar to that painted @Drew.
Now we need to calibrate the two solutions so that we can coordinate the pressure on the axis. First, we can calibrate by the water flow rate at the pipe inlet by setting the same velocity profile. Such inviscid flow is shown in Fig.4, where $\Delta P=P-\rho gz$.
Finally, it is possible to compare two pressure profiles on the axis for viscous (red) and inviscid flow (blue) with the same velocity profile at the inlet to the pipe - Figure 5.
However, if we reduce the Reynolds number to $Re=90$ and create a non-separable flow, then the pressure distribution will be similar to what @Drew suggested - see Figure 6 (right).
| {
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Could one theoretically use the expansion of the universe to travel through it? At least in one direction? Could one theoretically use the expansion of the universe to travel through it? At least In one direction?
That’s it that’s my question.
I’m not a physicist but I do get ideas.
I also wonder if one could theoretically ride the magnetic fields in the universe and if space is like water could it be possible to “slap it and ride the wave”?
| The expansion of the universe is happening in all points in our surroundings, all three dimensions at a time. All points in three dimensional space were at the beginning of the universe's one point ( classically, ignoring quantum mechanics) in the beginning of time.
Think of the surface of an expanding balloon, where all points are going away from each other, an ant on the surface would see no specific direction of expansion in its two dimensional universe on the surface. It would be traveling whether it wanted to or not.The expansion on the balloon is not "going" anyplace, for the ant to use it as a transport mechanism. It is more complicated in the four dimensional space time but similar.
There are forces that keep the atoms and molecules and planetary systems from expanding with the space time expansion, but that is another story.
| {
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Why are solar panels kept tilted? I have noticed that, in my country India, most of the solar panels are tilted southward at an angle of ${45}^{\circ} .$ Even on buildings with inverted V-shaped roofs, solar panels are still oriented southward on both the sides of roof.
Research
Many sites suggests that the tilt aids in self-cleaning also another site stated that tilt depends on factor like latitude
My questions:
*
*Why are solar panels tilted southward?
*How is latitude of the location of a solar panel relevant in increasing efficiency?
| Cleaning issues aside, the best orientation of a solar panel is the one that maximizes solar exposure.
It can be fixed or change periodically if a motor mechanism is provided. If it is static, then it can be horizontal assuming the latitude of the location is close to the equator.
Since India is near the equator, there is no reason to tilt southward except the one you mention (cleaning).
For other locations: same rules, maximize sunlight unless there are issues regarding cleaning. So if you go to the northern hemisphere, the farther north you go, the more the Sun will spend time southward in your sky, so you want to incline your panels accordingly. In the southern hemisphere, this is just the opposite. If you own a car, the panels can be horizontal on the roof or if you can move them and camp, the rules above apply.
Also the farther you are from the equator, the more the planet tilt will have an effect on seasonal variation, meaning even less Sun during the winter, or more during summer than during the rest of the year, i.e. a position farther or closer to the vertical, respectively. If the panels are used occasionally or can be moved, you want to aim at the sun anyway.
The basic principle is that you make sure your panels face the Sun as much as possible during the period of time of your interest. Since the Sun spends as much time in the western as in the eastern part of the sky, whatever the latitude, unless for example there is a building producing shadows that would require you to compensate, it should not tilt either westward nor eastward.
| {
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"url": "https://physics.stackexchange.com/questions/479515",
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What happens when in space a basketball flies and hits another resting basketball? We have the following situation:
In space, a basketball flies and hits another resting basketball.
What happens next?
a) Both balls fly in the original direction (but slightly slower) = momentum conservation (inelastic collision)
b) First ball stops, second ball flies in original direction with originally speed = momentum conservation (elastic shock) (newton's cradle)
c) Both balls repel each other and fly in different directions = action-reaction
And again the same with metal balls?
| The word you are looking for is the Coefficient of Restitution.
The coefficient of restitution$(e)$ decides whether the collision is elastic or inelastic. It depends on the surfaces of contact between the balls. The value of $e$ is generally between $0$ and $1$. With $0$ representing completely inelastic, and $1$ representing completely elastic.
*
*If $e=0$, then the balls would stick together at contact. They would move together with a velocity lower than that of the original in the same direction. The law of conservation of momentum will be valid here.
*If $0<e<1$, then a partially elastic collision will take place. According to the masses, the balls could move in the same direction, with different velocities. Or move in opposite directions and still satisfy the law of conservation of momentum.
*If $e=1$, then the collision is completely elastic. The incoming ball might rebound and the second ball might move in the original direction. All this happens in accordance with the laws of conservation of momentum and energy. In a special case, if the masses of the balls are equal, then the first ball could stop, and the second one moves in the same direction with the same velocity.
In short, the coefficient of restitution tells us how much loss energy loss occurs to heat. The law of conservation of momentum is valid in all three cases. But only for $e=1$ is the energy not lost to the surrounding as heat.
| {
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Interpreting the conserved charge in scalar QED In scalar QED, applying Noether's theorem for internal global symmetries results in a Noether current that is dependent on the gauge because of the presence of the covariant derivative.
$$j_\mu=-i(\varphi^*\partial_\mu\varphi-\varphi\partial_\mu\varphi^*)-2eA_\mu\varphi^*\varphi.$$
When integrating the current to get the conserved charge, the term dependent on the gauge doesn't obviously seem to cancel out.
My question is how do you interpret this conserved charge that depends on the gauge field? For example, when you take a free complex scalar field, the conserved charge is very simply interpreted as the difference between the number of particles and anti-particles. However, I can't find a similar simple interpretation when it is coupled to a gauge field.
| The total current is gauge invariant, because the first term is also not gauge invariant, because of the partial derivative and not covariant derivative. In fact the current is just like for free charged scalar just replacing partial derivative by covariant derivative.
As intuition to the appearance of the second term, now the matter current does change, and only combination of some current attributed to the field, as happens for momentum and energy for example
| {
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Kinetic Molecular theory ideal gases, deriving the equation $PV = \frac{1}{3}mN C^2$ The problem I have with this proof is how to acquire the fraction $\frac{1}{3}$. In theory we were taught by taking into consideration a rectangular vessel which obviously has three paths which a particle can take when colliding between three faces.nevertheless we still apply this equation to spherical vessels. which have infinite equi-probable paths.
I'm convinced that there is a more generalized way of acquiring the constant(my guess is by integration but i'm clueless as to what is integrated )
Please describe the derivation clearly with a mathematical perspective.
| (a) I won't reproduce a complete proof (See for example Jeans: Kinetic Theory of Gases), but will outline how the factor of 1/3 enters, even for a container of arbitrary shape.
If we consider a small 'patch' of the container wall, it is straightforward to show that the total momentum of molecules hitting that patch per unit time is proportional to $\overline {u^2},$ the molecules' mean square velocity component normal to the patch.
If $\overline {v^2}$ and $\overline {w^2}$ are the mean square velocity components in the two directions orthogonal to each other and to the normal, then we can easily show that the mean square speed of the molecules is given by
$$\overline{c^2} = \overline{u^2} + \overline{v^2} + \overline{w^2}$$
Assuming no direction is favoured,$$\overline{u^2} = \overline{v^2} = \overline{w^2}$$
So $$\overline{u^2} = \tfrac 13 \overline{c^2}.$$
(b) A delightful elementary derivation of $pV= \frac13 Nm \overline{c^2}$ considers a gas in a spherical container and assumes that molecules bounce off the walls as elastic spheres. Orthogonal components are not considered; rather the factor of 1/3 arises almost magically, as the quotient of numerical coefficients in the formulae for the volume and area of a sphere!
(c) It might seem as though the factor of 1/3 arises quite differently in (a) and (b). But on a deeper level there is a common origin: the 3-dimensionality of space!
| {
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Energy density in electrostatics - definition Background:
It can be shown that the potential energy of charge distribution can be calculated by
$U = \frac{1}{2}\int_V \rho(r')\Phi(r') d^3r' $
by means of integration by parts and the poisson equation $\Delta\Phi(r) = 4\pi \rho(r)$ the integral can be rewritten
$U = \frac{1}{2}\int_V \rho(r')\Phi(r') d^3r' = \frac{1}{8\pi}\int_V (\nabla \Phi(r')) (\nabla \Phi(r') )d^3r' = \int_V \frac{|E(r')|^2}{8\pi}d^3r'$
The integrand of the integral can be interpreted as the energy density $u$:
$u(r) = \frac{|E(r)|^2}{8\pi}$
Considering an example of a homogeneously charged solid sphere with radius $R$:
$\rho(r) = \rho_0 \Theta(r-R)$
where $\Theta(r)$ is the heavyside-function.
With $Q=\frac{4}{3}\pi \rho_0 R^3$, this results in the following potential:
$\Phi(r) = \frac{Q}{R} \left(\frac{3}{2}-\frac{r^2}{2R^2} \right)$ for $r<R$
$\Phi(r) = \frac{Q}{r}$ for $r>R$
The radial electric field is:
$E(r) = \frac{Qr}{R^3}$ for $r<R$
$E(r) = \frac{Q}{r^2}$ for $r>R$
With this in mind we can calculate the energy density:
$u(r) = \frac{Q^2r^2}{8\pi R^6}$ for $r<R$
$u(r) = \frac{Q^2}{8 \pi r^4}$ for $r>R$
Integrating this will result in the total potential energy:
$U = \int_0^{\infty} 4\pi r^2 u(r) dr = \frac{3Q^2}{5R}$
Now to my question:
Starting from the first integral, why can't the energy density be defined as followed:
$u(r) = \frac{1}{2} \rho(r')\Phi(r')$
When considering the example above this would equate to:
$u(r) = \rho_0 \frac{Q}{R} \left(\frac{3}{2}-\frac{r^2}{2R^2} \right) = \frac{3Q^2}{4\pi R^4} \left(\frac{3}{2}-\frac{r^2}{2R^2} \right)$ for $r<R$
$u(r) = 0$ for $r>R$
Integrating over this we get the same energy (as expected)!
$U = \int_0^{\infty} 4\pi r^2 u(r) dr = \frac{3Q^2}{5R}$
How can there be two different ways of defining a energy density $u(r)$. Where have I made a mistake?
Help will be apprechiated!
| You just showed that $\frac{1}{2}\rho(r')\Phi(r') = \frac{|E(r')|^2}{8\pi}$. (EDIT: as @Fabian said in the comments, it is actually that $\frac{1}{2}\int_V \rho(r')\Phi(r') d^3r' = \int_V \frac{|E(r')|^2}{8\pi}d^3r'$)
So, it does not matter whether you define $u(r) = \frac{1}{2} \rho(r')\Phi(r')$ or $u(r) = \frac{|E(r)|^2}{8\pi}$. As both lie in the same volume integral, each can be interpreted as the mean the same thing.
| {
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Write electromagnetic field tensor in terms of four-vector potential How can we know that the electromagnetic tensor $F_{\mu\nu}$ can be written in terms of a four-vector potential $A_{\mu}$ as $F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$? In the vector calculus approach, this is not really hard to see (under reasonable 'smoothness' conditions on the fields), but I would like to know how one would see this in the four-vector approach.
More specifically, how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$.
| The Bianchi identity $\mathrm{d}F~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.
| {
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Pressure in irrigation water pipe I would like to know the pressure at the bottom of the pipe which I use for irrigation purpose. The land is step cultivated and the pipe goes slanting for the length of 140 meters and the top height would be 45 meters, water is pumped out from borewell using 7.5 hp submersible pump. I would like to know the pressure at the bottom of the pipe (Surface of the land, borewell depth can be ignored) when the water reaches the top.
| I gave in my comment :
use pressure = density * gravity * height. google this as there are several calculators available on the web.
Using 1000kg/m^3 for density, 9.81m/s^2 gravity and 45m
Pressure = 1000 * 9.81 * 45 = 441450 pa
converted to atm is 4.41
Different slightly to the result in the link, as I assumed 1000kg/m^3 for the density.
So, here is one link, out of many possible:
enter link description here
| {
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Approximation of multiplicity when Ideal gas low density is applied $\frac{M !}{(M-N)!} \approx M^{N}$ Our lecturer today mentioned how a piston's head being at equal pressure maximised the multiplicity of states.
He mentioned the following:
If I have a fixed number of particles $N_A$ on left and $N_B$ on the right, and the whole system has a fixed total volume of $M$, we can say the movable piston partitions the total volume into $M_A$ and $M_B$ lattice sites respectively on each side, and therefore
$$M_{total} = M_{A}+M_{B} = constant$$
Hence if we want the volumes of the two sides that maximise the multiplicity function,
$$\Omega(N, M)=\frac{M !}{(M-N) !} \frac{1}{N !}$$
This all made perfect sense.
The next step is where I have confusion.
As we are dealing with an ideal gas, the densities are so low that $\frac{N}{M} << 1$ hence
$$\frac{M !}{(M-N)!} \approx M^{N}$$
I understand the approximation as there are fewer molecules N than there are sites M for an ideal gas, but I fail to understand how he managed to reach the conclusion above from the previous equation above that.
I tried to solve this using the approximation:
$$x! = \left(\frac{x}{e}\right)^x$$ but this failed to reach me to the correct working in order to prove the $\approx M^{N}$ approximation.
I managed to get:
$$\frac{M^M}{(M-N)^{M-N}}\times \frac{1}{N^N}$$
but I suspect this is deviating from the main way of reaching the approximation.
How is the approximation achieved?
| $$
\frac{M !}{\left(M-N\right)!}
~=~
\begin{alignat}{10}
M
& \times & \left(M - 1\right)
& \times & \left(M - 2\right)
& \times & ~\cdots~
& \times & \left(M - N + 1\right)
& \times & \left(M - N\right)
& \times & \left(M - N - 1\right)
% & \times & \left(M - N - 2\right)
& \times & ~\cdots~
% & \times & 2
& \times & 1
\\[-25px] \hline
& &
& &
& &
& &
& & \left(M - N\right)
& \times & \left(M - N - 1\right)
% & \times & \left(M - N - 2\right)
& \times & ~\cdots~
% & \times & 2
& \times & 1
\end{alignat}
$$
$$
\begin{align}
&~=~
M
\times \left(M - 1\right)
\times \left(M - 2\right)
\times ~\cdots~
\times \left(M - N + 1\right) \\[10px]
% &~=~ \prod_{i=M-N+1}^{M}{i}
\end{align}
$$
Since $\frac{N}{M} \ll 1 ,$ then $M \gg N,$ and $M \gg 1,$ so
$$
\begin{align}
\frac{M !}{\left(M-N\right)!}
&~=~
M
\times \underbrace{\left(M - 1\right)}_{\approx M}
\times \underbrace{\left(M - 2\right)}_{\approx M}
\times ~\cdots~
\times \underbrace{\left(M - N + 1\right)}_{\approx M} \\[10px]
&~\approx~
\underbrace{M \times \cdots \times M}_{N~\text{times}} \\[10px]
&~=~
M^N
\,,
\end{align}$$
so
$$
\frac{M !}{\left(M-N\right)!}
~\approx~
M^N
\,.$$
| {
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Liouville's integrability theorem: action-angle variables For classical dynamical systems, let $I_{\alpha}$ stand for independent constants of motion which commute with each other. 'Remark 11.12' on pg 443 of Fasano-Marmi's 'Analytical Mechanics' suggest that $I_{\alpha}$s can be taken as canonical coordinates.
For a conservative system, the Hamiltonian $H$ is a constant of motion. Let's refer to $H$ as $I_1$. Then $I_1$ becomes one of the canonical momenta. Hence $H$ can be written as $H=I_1$. Application of Hamilton's eqns. of motion implies that only one angle variable $\phi_1$ (corresponding to $I_1$) evolves linearly in time while all others stay constant because
$$
\dot{\phi_i}=\frac{\partial H}{\partial I_i} = 0 ~~~~~~~~~~~~~~~~~\mathrm{for~}i\neq1.
$$
So, is it true that for every Liouville integrable (described here) and conservative system (where Hamiltonian does not depend on time explicitly), Hamiltonian can be written as a function of only one action variable $I_1$ and only one angle variable (corresponding to $I_1$) evolves linearly in time, whereas others stay constant?
| I believe I understand your question. I think Different action-angle variables for a 2D harmonic oscillator is a good example. The 2D oscillator is
$$H = H_1 = \frac12( p_x^2 + p_y^2 + x^2 + y^2)$$
which may be split into
$H = H_x + H_y$
where
$$H_x = \frac12(p_x^2 + x^2)$$
$$H_y = \frac12(p_y^2 + y^2)$$
and then you have one hamiltonian, but you have replaced it with two action variables and it may be solved to obtain two angle variables, $\phi_{H_x}$ and $\phi_{H_y}$, each with constant rates.
I hope that helps.
I think we also need an answer to address the following, which I believe is at the heart of the original question: Suppose I wanted my action variables to be $H$ and $H_x$. When I take the partial derivative of the Hamiltonian ($H=H_1$) with respect to $H_1$, I get 1 so $\phi_{H_1}$ has a constant rate. When I take the partial derivative of the Hamiltonian ($H=H_1$) with respect to $H_x$, I get zero (0), so $\phi_{H_x}$ is constant? Why is this wrong?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Operators acting on a single subsystem within a combined system's state I was reading over combined systems and operators acting on a single system within the combined system, and am confused by the math.
For example, we have individual qubit states for subsystems $A$ and $B$ that, as a combined system, produce the state:
\begin{align}
A\otimes B &= \begin{bmatrix}
a_{0}b_{0} \\
a_{0}b_{1} \\
a_{1}b_{0} \\
a_{1}b_{1}
\end{bmatrix}
\end{align}
We have an operator $O$ that acts on subsystem $A$. To express this on the combined state, it is simply (where $I$ is the identity matrix of the same dimension as $A$ and $B$)
\begin{align}
(O\otimes I)(A\otimes B)
\end{align}
But why is this? I've been reading a text that says this combined operator changes only the coefficients $a_0$ and $a_1$, while leaving $b_0$ and $b_1$ unchanged. But I don't see that. I see this operation as changing whatever the value of the product of $a_i$ and $b_j$ is.
| Welcome to SE! Another way of saying what your text claims is
$$(O\otimes I)(A\otimes B)=(OA)\otimes B.$$
To prove this to yourself, you could write a general $O$, say $O=\begin{pmatrix}w&x\\y&z\end{pmatrix}$, and then just evaluate both sides of the above equation directly for $A\equiv\begin{pmatrix}a_0\\a_1\end{pmatrix}$ and $B\equiv\begin{pmatrix}b_0\\b_1\end{pmatrix}$ (i.e., for the LHS take the tensor product $O\otimes I$ first, then multiply $A\otimes B$ by this, and for the RHS multiply $A$ by $O$ first, then take the tensor product with $B$).
You will find that the equation indeed holds; this is the sense in which $A$ "just acts on" the state of the first qubit.
| {
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Why is baryon number conservation an accidental symmetry I have to write a report surrounding the subject of baryogenesis and I wanted to start this report off with explaining how the first Sakharov condition: Baryon number violation is possible within the Standard Model. I, however, can't seem to find a good explanation as to why in the classical sense the baryon number is conserved and why this is a so-called accidental symmetry.
Most papers I read about baryogenesis state that due to a general Lagrangian being invariant under the gauge group: SU(3) x SU(2) x U(1) that it automatically has an global symmetry around U(1) with which lepton and baryon numbers can be identified. But how?
| I think your misunderstanding is precisely that you think that $U(1)$ gauge symmetry in SM can be associated to any quantum number such as baryon or lepton number. No, it can not. The $U(1)$ coming from $SU(3)_C\otimes SU(2)_L\otimes U(1)_Y$ is related to a quantum number called hypercharge, $Y$
We say that baryon and lepton numbers are symmetries in a 'classical' sense since they are preserved at Lagrangian level. This is, the Lagrangian is invariante under the change
$$
\phi \rightarrow e^{-i\epsilon N}\phi
$$
Where $\phi$ is any field, $N$ can be either baryon ($B$) or lepton ($L$) numbers and $\epsilon$ the group parameter.
Nevertheless, as in QFT books is proven via triangle diagrams you can see that at quantum level (not Lagrangian level) baryon and lepton numbers are not preserved, i.e.,
$$
\partial_\mu j^\mu_{N} = n_{CS} \Rightarrow \Delta N \neq 0 \tag1
$$
with $j^\mu_{N}$ the Noether current related to $N = B, L$ and $n_{CS}$ the Chern-Simmons index. But since this index is the same for $B$ and $L$, from Eq. (1) you can deduce that
$$
\partial_\mu (j^\mu_{B} - j^\mu_{L}) = 0 \Rightarrow \Delta(B - L) = 0
$$
The accidental (which means not pre-impossed in Lagrangian) quantum symmetry is not for $B$ or $L$ but for $B - L$.
| {
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RC circuit: is order of resistance and capacitor important? in a RC circuit is order of resistance and capacitor important?
Typically there is the battery then the capacitor then the resistance, but is it ok if I draw with the battery then the resistance and then the capacitor?
| The two circuits you describe:
are basically the same. For example the change of the current with time when we first connect the battery will be the same. There are some obvious but minor differences, for example in the left circuit when we first connect the battery $V_a$ starts at zero and rises to $V$, while in the right circuit $V_a$ starts at $V$ and falls to zero. Aside from this, when you connect a capacitor and resistor in series it doesn't matter which way round you connect them.
| {
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Quantum Tunneling and Conservation of Energy According to my understanding quantum mechanics, the probability of any particular particle in the universe being at any specific location in the universe is very small but never actually becomes zero. Thus, a little bit of all of us is everywhere.
Let's assume I suddenly quantum tunneled from sea level to the top of Mount Everest. That jump represents a net increase in energy/mass for Earth, which would violate the first law of thermodynamics. My calculations indicate it would have increased the mass of Earth by around 0.0858657µg, unless it is somehow offset some other phenomenon.
It seems implausible to me that energy conservation for a closed system could actually be violated, so how would that net increase in energy/mass be accounted for?
Or does the fact that such a scenario by definition represents such a significant decrease in the entropy of the system mean that an increase in mass should be the expected result?
| Here are the basics of quantum tunneling:
Note, the energy level of the particle in the above simple example does not change.
This holds true for all tunneling scenaria:
There should exist wave functions $Ψ$ , i.e. solutions for the quantum mechanical equations applicable to the problem where the specific boundary conditions have been applied. Then the $Ψ^*Ψ$ computation will give the probability for the particle to be found outside the barrier.
Energy conservation is a universal rule that is incorporated in all wavefunction calculations. Thus the probability of finding the particle in a different energy level than the one in the solution is zero, no matter how complex the solutions and the boundary conditions.
| {
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Why haven't we yet tried accelerating a space station with people inside to a near light speed? Is that something we could do if we use ion or nuclear thrusters?
Wouldn't people in the station reach 0.99993 speed of light in just 5 years accelerating at 1g and effectively travel into the future by 83.7 years?
That would be a great experiment and a very effective way to show relativity theory in action. I mean, the people inside the station would have effectively traveled into the future, how cool is that? Why haven't it been done yet?
| The only current propulsion systems, that I know of, that could achieve high speeds is the same one they've been talking about for decades, setting off nuclear bombs behind a pusher plate. However, I heard somewhere that this would only be feasible for about 10% of the speed of light. And, like in the other answers, you would have to have protection from incoming atoms in space.
| {
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2 windows - will I see the reflections?
I have a question regarding photons nature. Let's say I have a single source of light - regular bulb and the observer - in the same room.
The observer looks through a glass window (normal glass window-nothing special about it) and sees his reflection, but some of the light is passing through the window.
Now I put a second window - as presented on the attached photo. Will the second window reflect some of the light back to the observer, or it will pass 100% of the light forward?
This may be pretty basic but this question sparked a discussion and there where no definitive answer.
| Ok. I am not an expert nor I have enough time to work through an answer comparable to many excellent answers I find here.
Generally your logic is correct, meaning that every time the light goes through a medium like the one you are describing a percentage of the beam will pass through and the rest of it will be reflected.
Naturally, an ideal window will let the beam 100% to pass while an ideal mirror would reflect 100%. Now make sure you understand what I mean by beam.
There are three main ways to think about the nature of light: beams(as in most of early classical geometric optics), waves(modern mostly classical stuff), photons(more quantum stuff). In the problem you are describing, a beam seems like the best way to conceptualize light.
Now, obviously all the views must be correct and consistent. In your problem if you think of it as photons there are statistical formulas(let's say that each photon has a certain probability to pass through or be reflected) that will produce the same result.
To conceptualize these ideas Feynman's QED would be a natural way to go. A more mathematical/professional approach exists in every undergraduate modern physics book.
| {
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Why does thermodynamics use the negative of the Legendre transform? So I see how the negative Legendre transform is very helpful in interchanging dependencies and giving us the four different major thermodynamic potentials, from internal energy to Helmholtz, Gibbs, and enthalpy.
But what I'm unclear on is why we use the negative Legendre transform. Is it more that people derived the four above thermodynamic potentials by physical arguments, and then noticed that you have to use the negative Legendre transform to move between them, or is there some principled physical reason for why the negative Legendre transform makes more physical sense than the usual Legendre transform?
Any thoughts appreciated.
| The point is to preserve the intuition we have about energy. The Helmholtz free energy, Gibbs free energy, enthalpy, and internal energy all can be described as some kind of energy (subject to various other things being held constant) that can be used up to do work, and is minimized at equilibrium. If you added random sign flips, this intuition wouldn't work anymore. It's like defining your $z$-axis to point downward; totally mathematically allowed, but physically confusing.
| {
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Thermodynamic description of few-body systems How large should a system be to become thermal?
Thermodynamic description is well-established for systems with large numbers (say, of order of $N_A\sim 10^{23}$) of constituents. Is there a "lower bound" of sorts, for the number of degrees of freedom $N$ in a system, for which thermodynamic notions such as temperature, entropy, etc., still remain applicable?
Thank you!
| As far as I see it, as it was mentioned in the previous answer - it's all about ergodicity. For example, if we consider a thermodynamic system of $N$ particles, the lower the number of particles in it, the bigger the fluctuations of some physical quantity $f$ ($\langle \Delta f^2 \rangle \sim N^{-1/2}$), and thus the bigger the relaxation time $\tau$ of the system to one of its stationary states. If $\tau$ is comparable with the characteristic time of physical measurements (i.e. the time, over which the averaging is performed), the time average can not equal to ansamble average (which is ergodicity in the simplest form).
Wrapping it up in a simple manner: if a mechanical system demonstrates congruent behavior in some bounded area of its phase space at fixed energy we can apply statistical mechanics formalism, and talk about ergodicity, however definitions of such thermodynamic quantities as temperature can have different meanings.
| {
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How is sound affected by gravity? How would sound be different on a planet with a stronger gravitational field? My reasoning leads to sounds of the same wavelength having higher frequencies - is this correct?
| Sounds travels faster through denser mediums.
In a planet with the same amount of air in the atmosphere, but a larger gravity, the gas would be compressed more tightly and thus be more dense at the surface.
For this reason, the velocity of sound waves would be higher.
For two waves synthesised with the same wavelength, $v = f\lambda \implies f \propto v$. So if the velocity of sound waves on the stronger-gravity planet is higher, then so are their frequencies (for a given wavelength).
| {
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Angular velocity of rotating rod Consider the following system:
Newton's second law for rotational motion:
\begin{equation}\tau=I\alpha \Leftrightarrow rF=\frac{1}{3}mr^{2}\alpha \Leftrightarrow \frac{d\omega}{dt}=\frac{3F}{mr}\end{equation}
Considering RHS constant, we get $\omega=\frac{3F}{mr}t.$
I'm not sure if the angular velocity whould be inverse proportional to the radius (from natural experience I know that pushing farther requiers lower force).
Also what happens if the bar is not fixed and the two opposite forces are acting at the ends of the bar.
Since their sum is $\vec{0}$ there is translational equilibrum and so the axis of rotation is at the $C.M.$ but will the action of the two forces change the angular velocity from the previous situation?
| Consider the FBD of the situation, where the reaction force $R$ below is drawn in a positive sense. Also shown is the center of mass (blue dot), located exactly at $\tfrac{r}{2}$ from the pivot.
*
*Case A, pivoted arm means the linear acceleration of the center of mass is coupled with the angular acceleration $$ \ddot{y}_{\rm cm} = \tfrac{r}{2} \ddot{\theta} $$ and the two (linear + angular) equations of motion being $$ \left. \begin{aligned} R + F & = m\, \ddot{y}_{\rm cm} \\ \tfrac{r}{2} F - \tfrac{r}{2} R & = I \ddot{\theta} \end{aligned} \;\right\}\; \begin{aligned} R & = \tfrac{F}{2} \\ \ddot{\theta} & = \tfrac{3 F}{m r} \end{aligned} $$ when $I=\tfrac{m}{12} r^2$. So yes, rot. acceleration is inversley proportional to $r$ since despite the moment arm being less, the mass moment of inertia is reduced by $r^2$.
*Case B, free arm with equal and opposite $R=-F$ applied on the end of the rod. Same equations of motion, but now $R$ is known, and linear acceleration is not known. The result is $$\begin{aligned} \ddot{y}_{\rm cm} & =0 \\ \ddot{\theta} = \tfrac{r F}{I} & = \tfrac{12 F}{m r} \end{aligned} $$ Again rot. acceleration is inversley proportional to $r$ for the same reason, but the quanity differs from Case A.
NOTE: Just by a cursory look you see in the first case the center of mass accelerating and in the second it isn't. So the two situations differ significantly.
EXTRA CREDIT: If the force $F$ is not applied at the end, but at some other point $d<r$ from the pivot, then the pivot reaction $R$ will be different. Is there a value of $d$ that makes $R$ equal to zero?
| {
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Fisher Information in Statistical Mechanics I am studying the canonical ensemble and it seems to me there is an analogy between derivatives of the partition function, which can extract energy momenta for the system and Fisher score /information.
In partciular we have expressions like $-\partial_\beta log(Z)=<H>$, $\frac{1}{Z}\partial^2_\beta Z=<H^2>$ and finally $\partial^2_\beta log(Z)= Var(H)$ where $Z$ is the partition function $\beta$ the inverse temperature.
Hence I think one could identify the fisher score of the ensemble with the log of ther partition fucntion. Anyway I cannot get further in this analogy nor can get physical meaning and interpretation of this porcess.
Any help would be great!
| There are very useful relations indeed, which I wouldn't necessarily call analogies.
We define a set of thermodynamic variables denoted as {${\theta_i}$} and specify the partition function denoted as $Z(\theta)$. In a typical case where we work with Gibbs measures, we may write
$$lnZ = \psi $$
where $\psi$ corresponds to free entropy. Amusingly, the second derivative of free entropy $$\frac{\partial^2\psi}{\partial\theta_{m}\partial\theta_{n}}$$ yields the thermodynamic tensor, which is identical to Fisher information matrix $F_{mn}$, where
$$F_{mn}(\theta) = \sum_x{p(x)\frac{\partial lnp(x)}{\partial \theta_{m}}}{\frac{\partial lnp(x)}{\partial \theta_{n}}}.$$
This shows how Fisher information is immediately related to our study of statistical mechanics as your question addresses. Identification of Fisher values informs us about how a system behaves in transitions, for example.
| {
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When I do water electrolysis, which water molecules are split apart? If I do water electrolysis I will get hydrogen at one electrode and oxygen at the other.
Is it because a molecule of water somewhere in the middle splits into H and O and then the H and O travel to the appropriate electrodes? Or is it something more complicated?
| Even in "de-ionized" water, some fraction of the water molecules in the bulk will dissociate into $\rm H^+$ and $\rm OH^-$ ions.
In electrolysis, the $\rm H^+$ ions migrate towards the cathode, where they find each other, steal electrons from their surroundings (including the cathode) and form $\rm H_2$ gas. Likewise the $\rm OH^-$ ions migrate to the anode, further dissociate, and form $\rm O_2$ gas. But these migrations are mostly of ions that are already near the electrical terminals. The bulk migration of ions in the fluid, like the "drift velocity" of electrons in a metallic circuit, is surprisingly slow.
| {
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Why electric potential requires absence of acceleration? As of 28/05/2019, Electric Potential is defined in Wikipedia in the article with the same name as:
The amount of work needed to move a unit of positive charge from a
reference point to a specific point inside the field without producing
an acceleration
What is the meaning of "without producing an acceleration"?
| An accelerating charge radiates EM waves that carry energy,
the static electric potential difference does not account for the radiative exchange.
To measure therefore the work done by the static potential difference while moving a charge from one place to another, must be performed so that the charge does not radiate.
| {
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Initial speed is zero and so is power? If I want to accelerate something from standstill to max speed, with a constant force (acceleration and mass don't change), the equation P = F * v would say that in the beginning we use 0 W power. How is that possible? Since power is the rate of transference of energy to the body (J/s), it would appear that that rate should be constant.
Why is it that a fast moving body requires more power to accelerate than a still body? And if we would like to know how much power we need to accelerate that body, should we then use the maximum speed?
| $$ \frac{dv}{dE} = \frac{d}{dE} [(2E/m)^{\frac 1 2}] $$
$$ \frac{dv}{dE} = \sqrt{\frac 2 {mE}} $$
which diverges at $E=0$ (i.e. $v=0$), so the velocity is infinitely sensitive to energy for a body at rest.
| {
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Is my understanding of vectors correct? I recently learned that a vector in mathematics (an element of vector space) is not necessarily a vector in physics. In physics, we also need that the components of the vector on a coordinate transformation as the components of the displacement vector change. So, if my understanding is correct, if $|\mathbf{c}_1|, |\mathbf{c}_2|, |\mathbf{c}_3|,\, \ldots \,,|\mathbf{c}_n|$ are the components of a vector $\mathbf{A}$ and $f$ is the function of transforming coordinates (change of basis), then $$f(\mathbf{A}) = \sum_{i=1}^n{f(\mathbf{c}_i)}$$ where $\mathbf{A} = \sum_{i=1}^n\mathbf{c}_i$.
That is to say, the transformed vector by applying $f$ to it should be equal to the vector formed by the vector components which have been transformed by applying $f$ to them.
Am I correct?
| You are correct in saying they are different. Physics vectors are mathematical vectors, but not necessarily vice versa.
For example, Birkhoff and Maclane "A Survey of Modern Algebra", p162 of the 1953 edition:
A vector space $V$ over a field $F$ is a set of elements, called vectors, such that any two elements $\alpha$ and $\beta$ of $V$ determine a (unique) vector $\alpha+\beta$ as sum, and that any vector $\alpha$ from V and any scalar $c$ from $F$ determine a scalar product $c.\alpha$ in $V$, with the properties
$V$ is an Abelian group under addition
$c.(\alpha+\beta)=c.\alpha+c.\beta , \qquad (c+c').\alpha=c.\alpha+c'.\alpha$ (Distributive laws)
$(cc').\alpha=c.(c'.\alpha),\qquad 1.\alpha=\alpha$
Hence sets of functions form a vector space. So do simple shopping lists. This brings in the dual space, dimensionality and the basis but there is nothing about physical space, tangents, pointing arrows or all the familiar properties us physicists ascribe to a vector.
Physics vectors have the additional property that they can be transformed (by a rotation). Vector equations must remain valid if they are transformed. So if
${\bf A}={\bf B} + {\bf C}$ then ${\bf f(A)=f(B)+f(C)}$. The transformation must be linear. Your result follows, and shows that, if you have a basis $\{{\bf e}_i\}$
and write ${\bf A}=\sum_i c_i {\bf e_i}$ then the function can be written as a matrix multiplication.
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