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Relationship between the constancy of the laws of physics and conservation principles I have heard it said before in passing (I think it was on Star Talk Radio) that there is some specific relationship between the constancy of the laws of physics and conservation principles such as the conservation of momentum and conservation of energy.
The claim, as I remember it, is that the principle of the conservation of momentum is 'equivalent' to saying that the laws of physics are constant throughout space.
Similarly, I heard it said that stating the principle of the conservation of energy is 'equivalent' to saying that the laws of physics are constant throughout all times.
In what way are these claims equivalent? In what way does one follow from the other. Or, am I misremembering some of the details of what was said?
Sorry if this question is a little bit vague.
| You are referring to the theorem of Emmy Noether: "If a system has a continuous symmetry property, then there are corresponding quantities whose values are conserved in time". Any such symmetry refers to the constancy of the laws of nature under this symmetry transformation. For example, if you observe a system in a particular place and then observe exactly the same system in a different place (say, shifted or "translated" by a certain distance), then the behavior of the system should be the same, because the laws of nature do not depend on where exactly things happen (provided no specific differences like different gravity, etc.). Then, according to the Noether theorem, there would be a corresponding conservation law. In case of the space translation symmetry, what is conserved is momentum. In case of the time translation symmetry ( the system behaves tomorrow the same way as today - the laws of nature do not depend on time), the conserved quantity is energy. There are many other symmetries, each producing its own conserved quality, such as angular momentum, electric charge, and others.
https://en.m.wikipedia.org/wiki/Noether%27s_theorem
| {
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Symmetry groups of 3D systems which are periodic in one direction and homogeneous in the other 2 directions What kind of symmetry groups, analogue of 3D crystallographic space groups, can describe the spatial symmetries of 3D systems which have discrete periodicity in only one direction, but are homogeneous in the 2D transverse plane?
What is the classification and the representation theory of such kind of symmetry groups?
| If the system is only discretely periodic in one dimension, the only point groups are the trivial group $C_1$ and the group $Z_2$ generated by reflections about a point.
These, combined with the discrete translational symmetry of periodicity, give the 1D line group.
See https://en.wikipedia.org/wiki/One-dimensional_symmetry_group for more details.
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Is there any experiment that doesn't have a finite countable number of outcomes? There are a lot of random variables in physics that are, in principle, continuous and unbounded (canonical examples: time, position along each direction). In practice, it is my understanding that the number of outcomes of every possible experiment will be a finite integer. That is, the data is always binned, even if only implicitly, at some minimum scale and finite in scope.
Reading the length of something off of a ruler, for example, will always produce a number that is rounded to some nearest rational number, whether that number is a line on the ruler or interpolated from the scale by eye. It will also always only produces either a finite length answer or a lower limit, assuming the experiment has to be carried out in finite time (i.e. using a single ruler you can only translate it a finite number of times to measure something longer than it).
In principle, though, the length of something being measured by a ruler can assume a continuum of values, but this question isn't about the theoretical principle, it's about experiments that are realizable in finite time using finite resources. Is there any example of a real experiment that has an actual infinite number of outcomes in any way? If not, is there a good reason to think that we will never be able to produce such an experiment/measurement?
| All practical physical measurements have a precision associated with them (this is distinct form uncertainty or accuracy). Like the rounding error on your ruler; this limits the number of outcomes to the
$$\frac{\text{Measureable range}}{\text{Precision}} \, .$$
Additionally it is not possible to actually measure an infinite range of results. That's because there is no practical means by which a truely infinte magnitude could be recorded. (Not enough atoms in the universe.)
Consequently we can be sure that we do not have the means to make measurements that are able to return an infinite number of outcomes.
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What would happen to a 10 meter sphere of room temperature water if released into space? Imagine that we had a space station with a relatively large hangar, and we allowed a ball of water to accumulate that had a 10 meter diameter and a water temperature of 20C. While the hangar is pressurized, someone decides to use a (closed loop) rebreather tank to sit in the middle of the sphere and breathe, so they're not dying and they're not exhaling any air into the water (just to keep things simpler).
Someone cycles the airlock and the sphere is now floating in the middle of the hangar in a hard vacuum.
What would happen to the water, and what would happen to the person inside? Would the sphere of water maintain enough pressure on the person that they would be fine, would the water boil off so quickly that it wouldn't be useful for long, or would the water freeze? I see several options, and this is a question I've wondered about for awhile, but I haven't been able to solve it.
| It boils, then the vapor freezes.
It doesn't directly freeze because water is great at holding heat, and the only way to dissipate heat in vacuum is radiation, no convection. However there is no constraint on pressure, and as we know low pressure makes water boil. The process of boiling acts to carry away heat and separates the water into a fine mist which then freezes into flakes.
It's sort of a matter of surface area versus volume, heat loss happens through the surface area but the whole volume is affected by the pressure so it's effects are stronger.
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Does the fact that $j^\mu$ is a 4-vector imply $A^\mu$ is, as argued by Feynman? Let
\begin{equation}
\boldsymbol{\Phi}=\Bigl(\dfrac{\phi}{c},\mathbf{A}\Bigr)
\tag{01}
\end{equation}
the electromagnetic 4-potential. We know that if its 4-divergence is zero
\begin{equation}
\dfrac{1}{c^{2}}\dfrac{\partial \phi}{\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}=0 \quad \text{(the Lorenz condition)}
\tag{02}
\end{equation}
then Maxwell's equations take the elegant form
\begin{equation}
\Box\boldsymbol{\Phi}=\mu_{0}\mathbf{J}
\tag{03}
\end{equation}
where the so called d'Alembertian
\begin{equation}
\Box\equiv \dfrac{1}{c^{2}}\dfrac{\partial^{2} \hphantom{t}}{\partial t^{2}}\boldsymbol{-}\nabla^{2}
\tag{04}
\end{equation}
and the 4-current
\begin{equation}
\mathbf{J}=(c\rho,\mathbf{j})
\tag{05}
\end{equation}
which has also its 4-divergence equal to zero
\begin{equation}
\dfrac{\partial \rho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{j}=0 \quad \text{(the continuity equation)}
\tag{06}
\end{equation}
and is a 4-vector.
The question is : under these conditions is the 4-potential a 4-vector ??? I ask for a proof or a reference (link,paper,textbook etc) with a proof.
EDIT
$^\prime$Mainly Electromagnetism and Matter$^\prime$, The Feynman Lectures on Physics, Vol.II, The New Millenium Edition 2010.
| After searching in Web, in our PSE site and many books, textbooks, papers and so on I end up with this conclusion :
That the electromagnetic 4-potential $\:A^{\mu}(\mathbf{x},t)\:$ is a 4-vector is an assumption.
In $^{\prime}$Quantum Field Theory$^{\prime}$ by Itzykson C.-Zyber J., Edition 1980, we read (in $\S$1-1-2 Electromagnetic Field as an Infinite Dynamical System):
We assume $\:A^{\mu}(x)\:$ to transform as a four-vector field and the lagrangian as a scalar density in order for the action to be a Lorentz invariant.
Also Ben Crowell commented therein Proof that four-vector potential is a valid four vector:
This type of question can't be answered generically. It depends on what assumptions you start from. Someone could choose a logical framework in which the four-vectorial nature of the potential is one of the postulates.- Ben Crowell Sep 20 '17 at 23:06
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Triple Delta Potential in Quantum Mechanics I am facing a problem of Quantum Mechanics, and I gently need your help in continuing to solve it.
The problem is the old usual problem of a particle subject to a potential, which this time has the form
$$V(x) = \alpha \delta(x^3+2ax^2-a^2x - 2a^3)$$
And we need to find the energies and the wave function normalization.
So first of all I used the well known identity for the Dirac Delta Distribution in order to write the potential as
$$V(x) = \alpha \left(\frac{1}{6a^2}\delta(x-a) + \frac{1}{2a^2}\delta(x+a) + \frac{1}{19a^2}\delta(x+2a)\right)$$
By the way, we can take $\alpha = 1$ in case.
From here, a simple sketch of the potential highlights $4$ regions:
$$\begin{cases}
x < -a \\
-a < x < +a\\
a < x < 2a \\
x > 2a
\end{cases}
$$
But my first doubt is: shall I split the second region into two other regions like
$$\begin{cases}
-a < x < 0\\
0< x < +a
\end{cases}
$$
or not?
Also, I attempted to write down the general solution fo the EVEN wave function case, and I got stuck also because of the previous regions question. I think I shall go for
$$\psi_e(x) = \begin{cases}
A e^{-kx} ~~~~~ x > 2a \\
A e^{kx} ~~~~~ x < -a \\
\ldots
\end{cases}
$$
Where the $\ldots$ represent my doubts about how to write the general solution in those cases...
I would really be grateful for any help or clarification about this!
|
From here, a simple sketch of the potential highlights $4$ regions:
$$\begin{cases}
x < -a \\
-a < x < +a\\
a < x < 2a \\
x > 2a
\end{cases}
$$
This is wrong. A delta function $\delta(x + x_0)$ has a peak at $-x_0$, not at $x_0$. You've flipped the sign of $x$.
But my first doubt is: shall I split the second region into two other regions like
$$\begin{cases}
-a < x < 0\\
0< x < +a
\end{cases}
$$
or not?
No, you don't. I guess you're doing this out of habit, because you've seen that done in other problems, but think about it: what is special about $x = 0$? Why should something happen there at all? Why not also split at $x = a/2$ or $x = \pi^\pi a$?
You only need to split the solution at $x = 0$ if the potential actually changes there. In many problems, the potential is set up this way for convenience, but it doesn't happen here.
Also, I attempted to write down the general solution fo the EVEN wave function case, and I got stuck also because of the previous regions question. I think I shall go for
$$\psi_e(x) = \begin{cases}
A e^{-kx} ~~~~~ x > 2a \\
A e^{kx} ~~~~~ x < -a \\
\ldots
\end{cases}
$$
Where the $\ldots$ represent my doubts about how to write the general solution in those cases...
Again, I guess you're trying an even wavefunction out of habit, but this is not right. If the potential is even or odd, it can be shown that your energy eigenstates to be chosen to be even or odd. But the potential you're dealing with here is neither. If you demand your solution is even, you will get no solution at all.
I would really be grateful for any help or clarification about this!
To the left of the first delta function, take a growing exponential. To the right of the last delta function, take a decaying exponential. In both of the two intermediate regions, take superpositions of growing and decaying exponentials.
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If iron can’t undergo fusion, does that mean a black hole is mostly iron? Since stellar fusion can’t progress beyond iron, and a large enough star collapsed into a black hole because an iron core stalled fusion, wouldn’t that mean all black holes are predominantly iron?
| If we are talking about stellar-sized black holes, then the object that collapses to form a black hole will have a high concentration of iron (and other iron-peak elements like manganese, nickel and cobalt) at its core, and it is the core-collapse that begins the black hole formation process, but much more material than this will eventually form that black hole.
It appears, empirically, that the minimum mass of a stellar-sized black hole is around $4M_{\odot}$, but is more typically around $10-15M_{\odot}$. But the extinct core of iron in a pre-supernova star is unlikely to exceed around $1.5-2M_{\odot}$ even for the most massive of supernova progenitors (see for example these slides).
Thus most of the material that collapses into a black hole is not iron, it is actually the carbon, oxygen, silicon neon and helium that surrounded the iron core. Much of the nuclear material will be photodisintegrated into its constituent baryons (or alpha particles) during the collapse. Neutronisation reactions will turn most of the protons in the high density material into neutrons. Even at equilibrium, when densities higher than about $10^{14}$ kg/m$^{3}$ are reached then any remaining nuclear material will begin to transmute into all sorts of weird and wonderful neutron-rich nuclei (as in the crusts of neutron stars) and by the time you reach densities of $\sim 10^{17}$ kg/m$^{3}$ (which is still well outside the event horizon of a stellar-sized black hole), the nuclei will lose their identity in any case, and become a fluid of neutrons, protons and electrons.
A second point to consider is whether it makes any sense to talk about the composition of a black hole. Composition is not one of the things you can measure - these are restricted to mass, angular momentum and charge. The other details are lost from a (classical) black hole.
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Dissapation of photon energy For an incident photon to be absorbed by a material, must it exactly equal a difference in the electron energy levels, or does it just have to be more than one such difference. If more is okay, what happens to the remaining photon energy? Does it continue on as a lower energy photon, if so, the remnant photon might not have enough energy to be absorbed. Would it just continue to travel onward through the material indefinitely? Can one or more electrons be knocked off by a single photon? The
concepts of exact energy levels sounds a bit unnatural, like perfect sine wave. Are there uncertainty bands in these electron energy levels?
| The energy of the incident photon must be equal to the difference in energy levels whether it be absorbed or emitted, when a photon is absorbed, the electron moves up one energy level, when emitted, the electron moves down one energy level.
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How to derive that $pV^k$ is constant in a polytropic process? This is what we did on the lecture:
$$\delta Q=nC dT$$
$$dU=nCdT-pdV$$
$$dU=\bigg(\frac{\partial U}{\partial V}\bigg)_TdV+\bigg(\frac{\partial U}{\partial T}\bigg)_VdT$$
$$dU=\bigg(\frac{\partial U}{\partial V}\bigg)_TdV+nC_VdT$$
$$n(C_V-C)dT=-\bigg(\bigg(\frac{\partial U}{\partial V}\bigg)_T+p\bigg)dV$$
And $\bigg(\frac{\partial U}{\partial V}\bigg)_T=0$ in the case of ideal gas, so:
$$n(C_V-C)dT=-pdV$$
$$n(C_V-C)dT=-\frac{nRT}{V}dV$$
$$\color{blue}{(C_V-C)dT=-\frac{RT}{V}dV}$$
$$\color{red}{pV^k=constant}$$
where $k=\frac{C-C_p}{C-C_V}$
My first question is, how did we get the red one from the blue, or do you know an alternative derivation?
And my second question is, how does this work? For example, what should I do, if I want the $p^V-V^p$ to be constant? How can I get the $k$? It's just a weird example, but I hope you get what I mean.
| The blue line is what is known as a separable differential equation, where we can move all $T$ values to one side, and all $V$ values to the other. So,
\begin{align}
&\left[(C_V-C)\operatorname{d}T=-\frac{RT}{V}\operatorname{d}V\right] \times\left(\frac{1}{T(C_V-C)}\right) \Rightarrow\\
&\int\left[\frac{\operatorname{d}T}{T} = \left(\frac{R}{C-C_V}\right)\frac{\operatorname{d}V}{V}\right] \Rightarrow\\
\ln\left(\frac{T_f}{T_i}\right)&= \left(\frac{R}{C-C_V}\right)\ln\left(\frac{V_f}{V_i}\right) \\
&=\ln\left(\left[\frac{V_f}{V_i}\right]^{R/[C-C_V]}\right) \Rightarrow\\
T_fV_f^{-R/[C-C_V]}&=T_iV_i^{-R/[C-C_V]}.
\end{align}
We conclude that the quatity on either side of the equals signs must be constants for all states related by a particular polytropic process since the final and initial states have no other relationship to them.
Now, for an ideal gas $PV=nRT\Rightarrow T=(PV)/(nR)$ so
\begin{align}
\text{const}&=TV^{-R/[C-C_V]} \\
&=\left(\frac{PV}{nR}\right)V^{-R/[C-C_V]}\\
&=\frac{1}{nR}P V^{1-R/[C-C_V]}.
\end{align}
$nR$ is also a constant for our process, so we can conclude that $PV^\gamma$ is a constant. All that remains is to show that $\gamma$ is the constant in the question, $k$.
\begin{align}
\gamma &\equiv 1-\frac{R}{C-C_V} \\
&= \frac{C-C_V-R}{C-C_V}\\
&= \frac{C-C_P}{C-C_V},
\end{align}
where the last line follows because the molar heat capacities are related by $$C_P = C_V+R$$
because $PV=H-U$, $C_V \equiv n^{-1} \left.\frac{\operatorname{d}U}{\operatorname{d}T}\right|_V$, and $C_P \equiv n^{-1} \left.\frac{\operatorname{d}H}{\operatorname{d}T}\right|_P.$
| {
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Finding the equation of a chain hanging between two points Say I have a connected at either end to two points, $A(x_A, y_A)$ and $B(x_B, y_B)$ of length $l$, where $l \leq \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, how would I go about finding an equation of its shape?
I guess the chain would be of the form $y = \alpha \cosh{(ax + b)} + \beta$, so how do the coefficients relate to the coordinates of the hanging points and its length?
| Taking the solution you propose, you can make $b=0$ since
$cosh(ax+b)=\dfrac{e^{ax+b}+e^{-ax-b}}{2}$
and you can absorb $\beta$ to the remnant of the expansion for the expression. An analogous can be done for constant $\alpha$, and set $\alpha=1$ (Think out: This problem is about two forces acting in one plane, the weight of the chain and the tension of one extreme attached to the ceiling, so it will result on a 2nd order differential equation which will only need 2 constants to determine! as for it is an initial value problem).
Thus, we only need to determine $a$ and $\beta$.
From then, $y(0)$ would be the height the chain has when fully extended and both forces (the tension of one extreme and the weight of the chain) act upon it. From then, $y(0)=\beta$. If the chain is uniform, you can get that in equilibrium:
$l\,T=\beta\,W$
where $T$ would be the tension and $W$ the weight, both per unit length of the chain. From this you get constant $\beta$.
Finally, you can get $a$ when, for example, you set $\dfrac{dy}{dx}=0$ and the chain remains still when hanging.
Also if you are keen on Variational Calculus you may find the way to compute it, you can check chapter 6 from Thornton-Marion's "Classical Dynamics of Particles and Systems" for a good introduction towards it and the same problem.
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Symmetry of ground state electron configuration Take a molecule, whose atoms have a symmetry $S\in \text{O}(3)$ (i.e. S permutes the atoms).
$S$ also acts on the wavefuncion $\psi(x_1,...,x_n)$ of the $n$ electrons in the molecule, by its action on the $x_i$ (maybe the most natural action also permutes the $x_i$, I'm not sure).
Is it true that the electron configuration $\psi_0$ of least energy (the ground state one) is invariant under $S$? If not, when is it/is it not?
| Generally, it need not be the case that the electronic ground state of any given molecule will be invariant under a symmetry transformation $S$ which preserves the overall configuration of the nuclear positions. This can be shown to be the case if the ground state is non-degenerate, but if you have a degenerate ground-state manifold, then the generic case is for the symmetry transformations to swap between the different states within that degenerate manifold.
For an example where this happens, consider the nitric oxide molecule NO, for which the ground state includes one unpaired electron in a pi orbital, which has a degeneracy of 2, i.e. it has a $\pi_x$ and a $\pi_y$ orbitals that congregate electron density in the $x$ (resp. $y$) directions orthogonally to the molecular axis (along $z$). As such, a 90° rotation about the molecular axis will not respect the molecular ground state, but it will change the $\pi_x$ state to the $\pi_y$ state, and vice versa.
More generally, if you have an arbitrary molecule, then one of the first parts of the analysis is to understand its symmetry group $G\leq \rm SO(3)$, and then to understand its representations, using the formal machinery of representation theory within quantum mechanics, as it is the representations of $G$ that will encode the different ways that the symmetry group can act within each degenerate eigenspace. Most molecular physics and quantum chemistry textbooks (or at least, those beyond the pure introductory presentations) devote large sections to this material.
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Sign of work done by friction In Goldstein's classical mechanics (3rd ed.) we read:
"The independence of W12 on
the particular path implies that the work done around such a closed circuit is zero,i.e.
$$\oint \textbf{F}.d\textbf{s}$$
Physically it is clear that a system cannot be conservative if friction or other dissipation forces are present, because $F . d\textbf{s}$ due to friction is always positive and the integral cannot vanish."
My question is: why should the work due to friction be "always positive"? Shouldn't it be nonzero instead?
Also, $F . d\mathbf{s}$ is a typo and should be $\mathbf{F} . d\textbf{s}$ (please let me know if I'm wrong)
| Indeed, it's an errata. You can find a list of them here.
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Tracking Spacetime Events In the linked post: Liouville's Theorem For Spacetime, I indicated the need for tracking the evolution of spacetime events. Is it sufficient to track a spacetime event by placing a particle there with zero initial velocity? You would then identify the spacetime event with the location of the particle. It would need to be a particle with zero mass, so that it could respond to the changes in curvature as quickly as possible. This however runs into trouble with my condition of zero initial velocity.
This seems like a natural choice, since the only way you can measure a spacetime event is by placing a particle there. But without the notion of particles, the notion of spacetime events and how they "move" becomes ambiguous in my mind.
| If a spacetime is globally hyperbolic, it can be foliated by timelike geodesics, ie it is possible that through each point of the spacetime passes a free particle such that no two such particles ever intersect (this can be shown by using the Hamiltonian flow of a Cauchy surface). It can indeed be a way to track spacetime events, although a more important method to do this in the practical case is the intersection of null geodesics with timelike curves - a light ray bouncing between different objects, as this is the most common way to define distances in relativity.
I'm not sure if the global hyperbolicity condition is necessary, but I feel like it might run into some difficulties for instance on the Carter spacetime.
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How do contact transformations differ from canonical transformations? From Goldstein, 3rd edition, section 9.6, page 399 after equation 9.101:
[...] The motion of a system in a time interval $dt$ can be described by an infinitesimal contact transformation generated by the Hamiltonian. The system motion in a finite time interval from $t_0$ to $t$ is represented by a succession of infinitesimal contact transformations which is equivalent to a single finite canonical transformation. [...]
How does the contact transformation differ from the canonical transformation?
| In the 2nd (but not the 3rd!) edition of Goldstein, Classical Mechanics, the word contact transformation appears in its index, and there is a 13 line long footnote on p. 382, which (among other things) states
[...] In much of the physics literature the term contact transformation is used as fully synonomous to canonical transformation, [...]
Concerning canonical transformation, see also this related Phys.SE post.
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Misunderstanding of the functioning of the reflective diffraction grating Suppose we have a sawtooth diffraction grating, as depicted below:
where the angle $\beta$ is the angle of inclination of the 'teeth' of the grating with respect to the plane of the grating and incident plane monochromatic waves normal to the plane of the grating. I am supposed to determine the angle $\theta$ for which the interference pattern for one 'saw-tooth' has a maximum. The diagram in the mark-scheme is as follows:
where the points $A,B$ both belong to the same 'saw-tooth' and the distance between $A$ and $B$ is $d$. The path difference between the two waves, according to the mark-scheme is given by $\Delta = BF - AE = d \sin \beta - d \sin \theta$. My question might seem trivial, but why are the two (BF and AE) not equal? In other words, shouldn't the two parallel incident waves (incident at an angle $\beta$) be simply reflected from the face of the 'saw-tooth' at exactly the same angle, in accordance with the law of reflection? Why even bother defining $\theta$? What am I missing here?
| Your question concerning the law of reflection is a good one, and it still applies here. This is what's known as the 0th (zeroth) order, or specular, pattern. There are other orders, or reflection peaks, that occur at different angles due to diffraction and interference. I think your problem asks you to find those other orders as well. If you google diffraction grating I'm sure you'll find the analysis.
| {
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Incorporate spinor in field equation I got problem understanding the concept as it state
1. Spinors do not work with the principle of General covariance. But how and why?
2. Contracting spinor into the tetrad solves this delemma.
Anybody explaining this concept would be helpful.
| The short answer: spinors transform nontrivially under local Lorentz boosts/rotations.
Spinors do not work with the principle of General covariance.
More precisely, it's the local Lorentz covariance that is the focus point here.
Contracting spinor into the tetrad solves this delemma.
In addition to tetrad $e$, we also need spin connection $\omega$ to write the Lorentz-covariant derivative $D\psi = (d + \omega)\psi$ (usually spin connection $\omega$ is expressed as a function of tetrad $e$ via the zero torsion condition $T = de + e\wedge w + w\wedge e = 0$, which does not fly here since spin currents introduce non zero torsion).
The reality is that tetrad $e = e^I_\mu\gamma_Idx^\mu$ is always there even for nonspinors, only that it's hidden from the plain sight via contracting into the metric:
$$
g_{\mu\nu} = e^I_\mu e^J_\nu \eta_{IJ}.
$$
The thing with spinor action
$$
S_{spinor} \sim \int{i\bar{\psi}e\wedge e \wedge e \wedge (d+\omega)\psi}
$$
is that there are 3 tetrads (or 1 co-tetrad, if you will) and there is no way contracting odd number of tetrads into metric.
| {
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Electric potential concept Imagine having two charged plates, one positive and one negative, and a negative point charge is placed at the negative plate. Let's set the negative plate to zero potential. The distance between the negative point charge and the negative plate is zero, so $V$ is zero in the equation $V=Ed$. However, since $U=qV$, potential energy is zero at this point. This should not be correct because the negative point charge of course has potential energy at this location. What is the conceptual error in this thought process?
| The value of the potential energy doesn't matter, just the change in potential energy. You can set any level of potential to be the zero point. Think of gravitational energy $mgh$, what is $h$? You could measure it from sea level, or you could measure it from the center of the earth. If you're right at sea level, then if you measure your height from sea level your potential energy is 0 but if from the center of the earth you're potential energy is $mg R_{earth}$. The only thing you would really care about, was what you're speed would be if you feel down a $100m$ pit. In that case your kinetic energy would be equal to the CHANGE in potential energy, which would be $mg(100\ \text{meters})$ in both situatons.
In your case you're electron does have potential energy at the negative plate, it has 0 potential energy, but say the positive plate is at $+100$ Volts, it's potential energy is $q(100 \ Volts) = -e(100 \ Volts) = -1.60217662 \times 10^{-17} \ joules$. Negative plate is at $0$, positive is at some negative number, thus the electron moves in the direction of negative potential energy, the positive plate.
| {
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Expectation value of quantum annihilation operator I already know the expectation value of a quantum annihilation operator $\hat{a} $. I want to find the expectation value of quantum creation operator $\hat{a}^\dagger$. Is it equal to following?
$$\langle\hat{a}^\dagger\rangle = \langle\hat{a}\rangle^* $$
$\langle\hat{a}\rangle^*$ is complex conjugate of the expectation value of $\hat{a}$.
| It is helpful here to remember that the expectation value $\langle \hat{a} \rangle$ is taken with respect to a quantum state $|\psi \rangle$ (or a density operator, but let's keep it simple). In this case,
$$
\langle \hat{a} \rangle = \langle \psi | \hat{a} | \psi \rangle
$$Recall now that $\langle \phi | \hat{O} | \psi\rangle^* = \langle \psi | \hat{O}^\dagger|\phi\rangle$. An intuitive picture for this relation comes from the matrix element interpretation of these objects - ie., a linear operator $\hat{O}$ can be expressed as an $N \times N$ matrix where $N$ is the dimension of the Hilbert space. In this case, the 'dagger' operator is the conjugate transpose, so the '$\psi$ by $\phi$'-th element of $\hat{O}^\dagger$ is the complex conjugate of the '$\phi$ by $\psi$'-th element of $\hat{O}$.
Given this relation,
$$
\langle \hat{a} \rangle^* = \langle \psi|\hat{a}|\psi\rangle ^* = \langle \psi | \hat{a}^\dagger |\psi\rangle = \langle \hat{a}^\dagger \rangle.
$$
So indeed the expectation values of $\hat{a}$ and $\hat{a}^\dagger$ are complex conjugates.
| {
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Virial expansion in terms of pressure I'm studying thermodynamics and I found two forms for the virial expansion:
$$pV~=~RT[1+\frac{A_2}{V}+\frac{A_3}{V^2}+\ldots] \tag{1}$$ and
$$pV~=~RT[1+B_2p+B_3p^2+\ldots]\tag{2}$$ my problem is that I can not find the correct procedure to express the coefficients $B_k$ in terms of $A_k$. I just find the answer, that is $$B_2=\frac{A_2}{RT}, \qquad B_3=\frac{A_3-(A_2)^2}{(RT)^2}, \qquad \ldots \tag{3}$$ but I can not find the procedure to obtain those relations (Actually I found a very strange procedure that I didn't understand at all) and I have been trying but I can't solve this problem. Does anybody could help me please?.
| Hints:
*
*Define a "density/inverse volume"
$$\alpha~:=~\frac{RT}{V}, \tag{A}$$
and rescale the $T$-dependent virial coefficient functions
$$ A_k^{\prime}~:=~\frac{A_k}{(RT)^{k-1}} . \tag{B}$$
*Then OP's 2 virial expansions (1) & (2) read
$$p~\stackrel{(1)}{=}~\alpha[1+A^{\prime}_2\alpha+A^{\prime}_3\alpha^2+\ldots] \tag{1'}$$ and
$$\alpha~\stackrel{(2)}{=}~p[1+B_2p+B_3p^2+\ldots]^{-1},\tag{2'}$$
respectively.
*Next, in eq. (2') use the formulas for reciprocal power series:
$$\alpha~\stackrel{(2')}{=}~p[1+B^{\prime}_2p+B^{\prime}_3p^2+\ldots].\tag{2''}$$
*Finally eqs. (1') & (2'') are mutually connected via power series reversion.
| {
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What's the reaction force on a charge moving in a magnetic field? According to newtons third law, all forces occur in pairs. What is the reaction force that the third law predicts when a magnetic force acts on a charged particle moving in a magnetic field?
| In general, Newton's third law is not valid when EM forces due to distant bodies are involved. Two charged moving particles act on each other via electromagnetic forces, but these are not related as action and reaction in Newton's 3rd law; they may even have different directions and magnitudes.
Instead, there is a law of local conservation of momentum, but one has to include momentum of EM field.
| {
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How do fusion reactors deal with blackbody radiation? The plasma of the ITER reactor is planned to be at 150 million K. Using the Stefan-Boltzmann law, setting the surface area as $1000\,\mathrm{m}^2$ (the plasma volume is $840\,\mathrm{m}^{3}$ so this is being generous), and the emissivity as $0.00001$ (emissivity is empirical so I just plugged in an extremely low value) yields a power of $2.87\times 10^{23}\,\mathrm{W}$. It would require somewhere on the order of $10^{35}$ fusion reactions per second just to break even, which clearly is not happening.
How can fusion researchers confine plasmas for several minutes if the blackbody radiation is this extreme? It seems like that with this level of heat, the plasma would just cool down within a few nanoseconds, and everyone in the vicinity would be torn to shreds by gamma rays, but evidently this does not happen. How?
| In ITER the average electron and ion temperatures are about 8 keV and the electron density is about $10^{20}$ m$^{-3}$.
At such a low density, the plasma is optically thin (see for example How large should an optically large fusion reactor be?) and therefore does not emit blackbody radiation.
The primary means of emission will be (optically thin) thermal bremsstrahlung. The link gives an approximate formula to calculate the total power emitted per unit volume.
$$P \simeq 1.7 \times 10^{-38} Z^2 n_i n_e \left(\frac{T}{{\rm eV}}\right)^{1/2}\ {\rm W/m}^3$$
For the parameters above and assuming $Z=1$ and $n_i=n_e$, then $P = 1.5\times 10^{5}$ W/m$^3$. So for 840 m$^3$, the power lost in bremsstrahlung is 12.7 MW; a small fraction of the intended power output of the fusion reactor.
The mean free path for photons due to Thomson scattering by electrons (which will be the primary opacity at those temperatures; or possibly a bit higher for Compton scattering) is $(n_e \sigma_T)^{-1} \sim 1.5\times 10^{8}$ m, or about $10^7$ times the linear dimension of the reactor.
| {
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How does the electric field produced by a simple circuit look? I have not seen anywhere a description of how the electric field looks inside and around a simple circuit. For example let's say we have the circuit shown below. One DC voltage source, two resistors, and a constant current flowing around.
We know that the electric field inside the battery will point from positive to negative, we also know that the field inside the wires is very small and in the direction of the current. Through the resistors there will be a strong field pointing from positive to negative. But in order to maintain the relationship that a closed loop integral of the E field is zero everywhere we must also have a field outside of those circuit elements. I have no idea how this field will look but I have made a crude attempt at sketching it below.
Is this a realistic picture of how the field will look?
| Close but not quite. Here's a very similar circuit from CircuitSurveyor. (I tried to input yours but it wouldn't calculate.) The background grid & flowlines are the Poynting vector $\mathbf S$. For the haphazardly scattered points, blue is $\mathbf E$, green is $\mathbf B$ (always into the screen here), and orange is $\mathbf S$. Note the $\mathbf E$ field near the wires is perpendicular to them, and wraps around circuit components.
Note this visualizer approximates the circuit as a 2D slice of an infinite circuit stack extending through the screen. A real instance would have non-zero $\mathbf B$ (and hence $\mathbf S$) outside the loop.
| {
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Correlators for space-like photon bouncing Consider the diagram below of two detectors-emitters exchanging photons: Thin lines are null-paths between detectors. Detectors are also synchronized in such a way that measurements outside the blue or red windows are ignored. So first detector can only react to photons received in the red window, and the 2nd detector can only react to photons on the blue window.
First detector has some amplitude $\psi_A$ to emit photon from some point $A$ inside the apparatus red window, and second detector has some amplitude $\phi_C$ to emit photon from inside blue window. The key assumption here is that that second apparatus is made such that amplitude $\phi_C$ depends on a photon having being received at some point $B$ inside the measurement blue window. So $\phi_C(B)$ actually depends on past local events in the apparatus.
Now, while the observables and correlators for photons separated by space-like paths like $A-B$ or $C-D$ must commute, Feynmann propagators only vanish exponentially inside the space-like separations, and presumably a measurement of photons at some point $D$ inside the red measurement window of detector 1 are not restricted by space-like commutator rules, presumably the correlations between observations $A$ and $D$ do not need to be zero
Question: Can we expect the correlators between $A$ and $D$ events be zero, or just exponentially small in proportion to the length of
the red and blue reaction windows (and their separation from the lightcones)?
| I have 2 questions plus an optional third, as I’m not sure I’ve fully understood the problem:
1) what does it mean that the probability of second apparatus to shoot a photon from C depends on the detection of a photon in B?
Do you mean that there the emission from the second device is not causal and the second photon “sees” the future of the detection in B and is emitted accordingly?
Or just that the probability of detection in B has some fixed value and probability of emission in C is consequently fixed and will emit accordingly, like a bet on a soccer match?
Is vertical line a time scale such that horizontal lines are same time?
2) why do you say that A-B and C-D are spacelike?
If they’re paths of two photons I would have said that they’re time-like paths;
Am I missing some crucial point?
[ 3) are you assuming flat Minkowski space? ]
Thanks for clarifying
Francesco
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Physical intuition behind Poincaré–Bendixson theorem The Poincaré–Bendixson theorem states that: In continuous systems, chaotic behaviour can only arise in systems that have 3 or more dimensions. What is the best way to understand this criteria physically? Namely, what is is about a space of dimension 1 or 2 that cannot admit a strange attractor? Why does this only apply to continuous systems and not discrete ones?
| There's not enough room for chaos in a 2D flow.
It comes down to the system's solutions being smooth 1D curves in a 2D space: due to uniqueness, these curves cannot cross (they can meet at special points [homoclinic or heteroclinic], but only asymptotically), and that strongly limits the possible end states. Particularly important is the Jordan curve theorem: if a solution closes on itself (forming a loop/cycle), it divides the space into inside and outside regions; and therefore all solutions inside stay inside, so all you can have are fixed points, cycles and spirals.
In discrete systems, the trajectory jumps around from one point to the other, so there's no such restriction and even 1D maps can exhibit chaos.
| {
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How to mathematically prove that point charge and infinitesimal volume charge are same? In electrostatics, while deriving certain elementary equations, I have seen all the books just assuming that point charge and infinitesimal volume charge are same.
How can we rigorously, mathematically and formally prove that point charge and infinitesimal volume charge are indeed the same?
| Maxwell's equations, which are regarded as the fundamental equations governing the evolution of electromagnetic fields, are constructed in terms of charge density and current density. As such, electromagnetism is fundamentally described not in terms of point charges, but a continuous charge distribution.
In the language of distributions, a point charge is described by the Dirac delta function $\delta(\mathbf{r})$. This is defined as the limit of a sequence of charge distributions that are smaller and smaller, but contain the same total charge (and thus, have higher and higher charge density). Taking this limit gives you something like a charge distribution with infinitesimal volume and infinite charge density, but with a finite total charge.
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How are high-energy detectors able to to distinguish between $m_{e}$ and $m_{\mu}$? I had a teacher pose this interesting question yesterday:
Suppose you're running a high-energy scattering experiment at the LHC. For concreteness, let's suppose it's a 2 to 2 scattering event which involves electrons and/or muons.
The theorist uses QFT to compute some cross-section which comes from the amplitude
$$
\mathcal{A}_{p_{1} p_{2} \to p_{3} p_{4} } = F(s,t,u,m_{e},m_{\mu},\ldots)
$$
The amplitude is a function of the Mandelstam variables $s \equiv (p_1+p_2)^2$, $t \equiv (p_1-p_3)^2$ and $u \equiv (p_1-p_4)^2$, as well as the mass of the electron $m_{e}$ and muon $m_{\mu}$ (and some other stuff).
Because we're running a high-energy experiment we obviously have that $s,t,u \gg m_e,m_\mu$, and for this reason the theorist makes the approximation $m_{e} \approx 0$ and $m_{\mu} \approx 0$.
The Question: How is the LHC able to distinguish between an electron and a muon if the theorist makes the approximation that the electron and muon are both massless?
For some reason, the approximation $m_{e} \approx m_{\mu} \approx 0$ is a bad one and the question is why this is. One idea that a colleague had was that the tracks of the electron and muon look different; because of the cyclotron radius $r \sim \frac{cm}{qB} \propto m$ the magnetic fields used in the machine to track the particles coming out of the collision will see the electron spiral more dramatically than the muon.
Any ideas as to other reasons why?
| The usual way to identify an electron vs a muon in a high-energy detector is via their interactions with matter:
- an electron will dump all its energy quickly in an "electromagnetic shower" and quickly stop
- a muon will interact minimally and go far through the material.
In pictures, these look like:
Note that we don't try to measure the rest mass of the electron or muon this way. We know what those are. We just try to identify which kind of particle a particular track is, and then supply the correct mass for the computations.
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Pauli exclusion responsible for "solidity"? I have heard Frank Close say that the reason you can't put your hand through a solid object is the Pauli exclusion principle. However Richard Feynman in his "Fun to Imagine" series attributes it to electrostatic forces.
I have two questions: Firstly, who is correct here (or maybe both)? Secondly, on a classical scale can the Pauli exclusion principle be interpreted as a force? The reason you can't put your hand through a solid object is because of a normal reaction force. So if the PEP is responsible for this it must be creating a force. I have sometimes seen the singularity at 0 in the Lennard-Jones potential interpreted as due to the PEP.
EDIT: I understand that the PEP is not a "fundamental force" carried by force-carrying particles. But it seems to clearly manifest as a force on a classical scale.
| Without Pauli exclusion your hand would go through a glass table or even fuse with it. There would be no chemistry as we know it. It seems unbelievable that the universe is stabilised by something regarded to be a mere consequence of symmetry.
| {
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Why is a fermion field complex? The Lagrangian of a fermion field is
\begin{equation}
\mathcal{L} = \overline{\psi} (i\gamma_{\mu} \partial^{\mu} - m)\psi
\end{equation}
It is said that the fermion field $\psi$ is necessarily complex because of the Dirac structure. I don't quite understand this. Why is the fermion field complex from a physical point of view? A complex field has two components, i.e., the real and imaginary components. Does this imply that all fermions are composite particles? For example, an electron is assumed to be a point particle that does not have structure. How can it have two components if it is structureless?
| Classical fermions in $3+1$ or higher dimensions are not complex. They are Grassmann number-valued. i.e. $\theta_{i}\theta_{j}+\theta_{j}\theta_{i}=0$ for any two fermions $\theta_{i}$ and $\theta_{j}$. There can be no difficulties regarding fermionic fields as complex numbers as long as you don't consider the product of two fields. If you consider them as complex numbers, then there is a thereom saying that fermions live in three representations: 1. complex. 2. real. 3. pseudo-real. For example, the Dirac fermion in 3+1 dimensions can be regarded as the $\mathbb{C}^{2}\oplus\mathbb{C}^{2}$ representation of $SL(2,\mathbb{C})$. But writing the Dirac Lagrangian requires it anti-commutes, otherwise you run into inconsistency.
| {
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Why does this paper use 1/cm for units of frequency? Reading this paper from 1963 $^*$, they use units of cm$^{-1}$ for frequency.
Here is an excerpt:
It doesn't seem like wave number, as they clearly call it frequency. What's going on here?
$^*$ Sievers III, A. J., and M. Tinkham. "Far infrared antiferromagnetic resonance in MnO and NiO." Physical Review 129.4 (1963): 1566.
| A frequency is a number of oscillations, but you need to specify the unit you're refering to : is it oscillations per a given duration (temporal frequency) or per a given length (spatial frequency). Today "frequency" implies a temporal frequency, and we use the word "wave number" for the spatial frequency, allowing us to go away with the adjective. That was not always the case.
| {
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Understanding what it means that gravitation is proportional to the product of the masses Newton's gravitational law states that
$F = G*\frac{M*m}{d^2}$
Intuitively it means that the greater the masses, the stronger the force, but it is more precise than that, it is proportional to the product of the masses not, for example, their sum.
I am not sure how Newton derived that, but I can guess he deduced that if $M_1 = 2*M$, then the force on $m$ should be twice as much.
However, the question is: what if I was able to move mass away from $M$ and add it to $m$? The total mass of the two bodies did not change, but now the force with which they are attracted to each other has changed.
Is there an intuitive explanation of what is going on?
| The most intuitive way of thinking of it seems to be to picture two objects each made up of 100 "particles" (imagine indivisible points of equivalent and constant mass) at a given separation $d$. We can build up to this picture and see why the product of masses makes sense.
First, imagine the simplest picture: 2 particles, $A$ and $B$ at separation $d$. Let's say that $A$ experiences 1 unit of force due to $B$ and $B$ experiences one unit of force due to $A$. Now let's add another particle $C$ that is right next to $B$. The gravitational attraction between $B$ and $C$ is very large but we can imagine it being canceled by some repulsive force of the particles touching. Now $A$ experiences 2 units of force, 1 from $B$ and 1 from $C$. The system of $B$ and $C$ together experiences 2 units of force, 1 from $B$ with $A$ and 1 from $C$ with $A$. So we can see how each particle in a system independently experiences gravitational attraction from each individual particle in the other system.
Now imagine the two objects ($X$ and $Y$), each with 100 particles stuck together, and the two objects at separation $d$. Now each particle in $X$ experiences attraction from every particle in $Y$ (and vice versa). So 100 partices in $X$ each individual experience 100 separate attractions, each of 1 unit of force. Therefore the total attraction is $100\times100=10000$ units of force.
Finally imagine $X$ has 50 particles and $Y$ has 150. The total mass of the two objects is the same as the previous case, but by the same argument $X$ has 50 particles which each independently experience attraction with 150 particles (with each instance of attraction imparting 1 unit of force), therefore $X$ experiences $150\times50=7500$ units of force.
Hopefully it is now more clear why force goes with the product of masses.
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Integrals involving Bose-functions (Computational) In short, I'm looking for some advice/literature how to deal numerically with Bose function.
My physical problem is to calculate a coupled set of Self-energies, thermal loop integrals, self consistently. These integrals involve Bose and Boltzman distributions.
There is some literature for the Anderson model (in NCA) dealing with a similar problem, where instead of Bose function, Fermi function appear.
Here due to the exponential growth of the Boltzman factor in the negative frequency, the involved Greens-function/Spectral-functions must vanish exponentially there. This numerical problem can be simplified by scaling all Spectral-function and Self-energies by the Fermi function (of minus the argument).
When working now with Bose functions the 1/x divergence at zeros is also present.
I would be glad for some advice or literature suggestion, on how numerical calculation involving Bose distributions are done.
| There is a great paper On the Relativistic Bose-Einstein Integrals by Weldon and Haber which I have found useful in the past. It focuses on asymptotic expansions of generic Bose-Einstein integrals in the high-$T$ and low-$T$ limits, but maybe it will be useful for you.
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Relative motion and time When someone reaches to a speed which is close to the speed of light with respect to earth, will he see the things actually moving faster than when he is in low speeds?
| I'm of two minds about answering this because, on the one hand, it's a straightforward homework exercise, but on the other hand, it seems to me that the other two posted answers are at best misleading.
Michele Grosso's answer says that earth clocks tick slowly in the spaceship's frame, but that's not what I think you're asking. I interpret your question to mean this: "If a clock on earth ticks once per second in its own frame, will the light from those ticks reach the spaceship more or less frequently than once per second?".
I'll adopt Árpád Szendrei's setup: The earth is at $(x=0,y-0)$ and the spaceship is at $(1,1)$ moving along the line $y=1$ with speed, say, $1/2$.
If the ship moves rightward, the answer is clear: First, the earth clock ticks slowly in the spaceship's frame, and second, light from each successive tick takes longer to reach the spaceship. So the traveler certainly sees the earth clock tick in slow motion.
If the ship moves leftward, the two effects go in opposite directions, so one needs a bit of calculation.
First, calculate everything in the earth frame:
The earth clock ticks at time $0$ and light from this click happens to hit the traveler just as he passes the point $(1,1)$. This light has traveled a distance $\sqrt2$ so the traveler sees it at $(x=1,y=1,t=\sqrt2)$.
Now the earth clock ticks at time $1$ and light from this tick reaches the traveler at time $T+1$. Thus the interval between the two arrivals is $T+1-\sqrt2$, and the traveler has now reached location $(X,1)$ where $X=1-\left({T+1-\sqrt2\over2}\right)$. Thus the light has traveled a distance $\sqrt{X^2+1}$ in time $T$, giving $\sqrt{X^2+1}=T$, whence $T\approx 1.176$ and $X\approx .619$. Thus the arrival occurs at $(x=.619,y=1,t=2.176)$.
Now Lorentz-transform the two arrivals into the ships frame. The arrival of the first tick at ($x=1,y=1,t=\sqrt2$) transforms to $t'\approx 2.2$ and the arrival of the second tick transforms to $t'\approx 2.87$. Thus the time between the arrivals is $2.87-2.2<1$, which means the spaceship sees the clock speeded up.
I chose to do this with explicit numbers rather than generic locations and velocities, in hopes that it would be more readable. God knows I might have screwed up the arithmetic, but something very like this should be right.
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Can the fact that dark energy increases with volume be explained by classical thermodynamics? Considering adiabatic process in classical thermodynamics, a normal substance with (positive) pressure must do work on its environment in order for the volume to increase by $ dV $(like pushing the walls of a piston) so it's internal energy decreases.
In contrast, in case of an unusual substance with negative pressure, we must do work on it in order for the volume to increase by $ dV $ so it's internal energy increases.
But in the case of dark energy, the volume of space expands due to the repulsive gravitational effect of negative pressure in general relativity without any work applied to it.
I’ve read several sources claiming that the reason dark energy increases with volume can be explained by classical thermodynamics. But I don’t see how it is so since the volume of space increases due to gravitational effects in general relativity, and not from work being done to it like in classical thermodynamics.
Isn’t it illegal to use classical thermodynamics to conclude that the energy of dark energy increases with volume?
|
Isn’t it illegal to use classical thermodynamics to conclude that the energy of dark energy increases with volume?
According to thermodynamics the internal energy $U$ doesn’t change in case of adiabatic free expansion (means outside the piston is vacuum) of an ideal gas, so $dU = o$. What changes is density of the gas, it decreases.
This is much different in general relativity if we consider the accelerated expansion of the universe driven by the cosmological constant. In this case the vacuum density $\rho_v$ remains constant if the volume of the piston increases and hence the internal energy increases which yields $dU = \rho{_v}dV$. Now following thermodynamics the work done by the vacuum pressure is $W = -p{_v}dV$. That clarifies that the vacuum pressure is negative. One could say it is taken from an infite reservoir.
So nothing is illigal, but one has to accept that the energy density of the cosmological constant doesn’t get thinned like matter during expansion.
| {
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How does tilting a bike make it turn sharper? Note that my question is not why do you tilt your bike when on a curve. It's about the reduction in turning radius when one tilts the bike inwards.
Short to-the-point answers are welcome.
| With the bike upright and either no turn or a very slight one the most outer circumference is in contact with the roadway.
When you lean in a sharp turn the inward torque will balance the outward torque created by the centripetal force (preventing tipping over) and the bike moves off the outer circumference and contacts higher up with a leading vector advancing towards the direction of the turn. The portion of the front wheel's rubber after the former lowest point is no longer in contact.
With the rear wheel, which is much fatter and doesn't turn, it's simply the side that pushes and it plays no role in the turning.
| {
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Decomposition of the symmetric part of a tensor The rate of strain tensor is given as $$e_{ij} = \frac{1}{2}\Big[\frac{\partial v_i}{\partial x_j}+ \frac{\partial v_j}{\partial x_i}\Big]$$ where $v_i$ is the $i$th component of the velocity field and $x_i$ is the $i$th component of the position vector. From what I read, I understand that $e_{ij}$ is the rate of strain tensor or the symmetric part of the deformation tensor i.e $\nabla \bf{v}$.
The rate of strain tensor can be decomposed in the following form: $$e_{ij} = [e_{ij} - \frac{1}{3}e_{kk}\delta_{ij}] + \frac{1}{3}e_{kk}\delta_{ij} $$ From what I could gather, $e_{kk}$ can be written as $\nabla \cdot \bf{v}$ which represents the pure volumetric expansion of a fluid element and the first term is some kind of strain which does not encompass volumetric change. Is this correct or is there more to it. What is the correct physical interpretation for it, and why is it useful?
Further more I read that any such symmetric part of tensor can be decomposed into a “isotropic” part and an “anisotropic” part. I am unable to understand Why we can do this and what it represents physically. I would like to have a mathematical as well as a physical understanding for this sort of decomposition. I am very new to tensors and fluid mechanics and would like to have a complete understanding of this. Thank you for the answers.
| There are many different answers to your question (since usefulness is subjective), but here's what I would consider the "main" one.
Very often we assume fluids are incompressible: that is, that the density $\rho$ is constant, and consequently $\nabla \cdot \mathbf{v} = 0$ from the mass continuity equation. By splitting the strain rate tensor $\bf{D}$ into a sum of an isotropic tensor $\mathbf{P}$ and a trace-less deviatoric tensor $\mathbf{S}$,
$$\mathbf{D} = \mathbf{P} + \mathbf{S} = \frac{1}{3}\text{tr}(\mathbf{D})\mathbf{I} + \left(\mathbf{D} - \frac{1}{3}\text{tr}(\mathbf{D})\mathbf{I}\right) = \frac{1}{3}(\nabla\cdot\mathbf{v})\mathbf{I} + \mathbf{S}$$ we can isolate the source of compressibility effects as $\mathbf{P}$ and ignore it in the case where $\rho$ is constant, simplifying constitutive equations considerably.
This can be useful, for example, to give us a straightforward way to mathematically analyze the behavior of fluids in the regime where they become slightly compressible: we know the effects will show up in the strain rate tensor as an extra diagonal term $\epsilon \mathbf{I}$ where $\epsilon \ll 1$, and we can use perturbation theory to see how compressibility propagates into the mechanics.
From a more general perspective, when formulating constitutive laws involving tensors of arbitrary type in classical mechanics, we seek to formulate such laws so that they satisfy objectivity (Galilean transformation invariance). Such laws can only depend on the invariants of tensors, and as a result it's useful to isolate the terms which depend on each individual invariant, of which the trace is one.
| {
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What does it mean if the dot product of two vectors is negative? If the dot product gives only magnitude, then how can it be negative?
For example, in this calculation:
$$W = \vec{F}\cdot\vec{r} = Fr\cos\theta = (12\ \mathrm{N})(2.0\ \mathrm{m})(\cos 180^\circ) = -24\ \mathrm{N\,m} = -24\ \mathrm{J}$$
Why is there a negative sign? What does it tell us?
| The formula you have supplied in your question says it all:
$$\vec F\cdot \vec r=Frcos(\theta)$$
$F$ and $r$ are both positive (vector magnitudes), and so the negative sign comes from $cos(\theta)$
$\theta$ is the angle between the vectors, and $cos(\theta)$ is negative when $\frac {\pi}{2}<\theta<\frac{3\pi}{2}$. This means the two vectors are facing in "opposite directions" (of course not exactly opposite, hence the quotes).
You can think of the dot product as how aligned two vectors are. Its maximum absolute value is just the product of the magnitudes, and the sign indicates if they are facing relatively in the same (positive) or opposite (negative) direction.
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Euler equations in primitive form for a real gas For an ideal gas, it is relatively easy to express the Euler equations in primitive form (variables $\rho$, $u$, $p$), starting from their expression in conservative variables ($\rho$, $\rho u$, $E$).
I did not find any example of such derivation for a general real gas, governed by any equation of state. Is it possible to express the Euler equations in primitive form for any (unknown) real gas (involving the speed of sound somewhere)?
| In general, you can always convert between any two variable sets using the Jacobian that connects them. For example, if we define our conservative variable vector as:
$$ \mathbf{W} = \lbrace \rho, \rho u, \rho E \rbrace^T $$
and our primitive variable vector as:
$$ \mathbf{Q} = \lbrace \rho, u, p \rbrace^T $$
then we want to find the mapping that converts $\mathbf{W}$ into $\mathbf{Q}$. Abusing math notation in ways that make mathematicians cringe, we can say that:
$$ \partial \mathbf{W} = \frac{\partial\mathbf{W}}{\partial\mathbf{Q}} \partial \mathbf{Q} $$
where $\partial \mathbf{W} / \partial \mathbf{Q}$ is the Jacobian matrix. This results in a matrix, which for our system here is:
$$
\frac{\partial \mathbf{W}}{\partial \mathbf{Q}} =
\begin{bmatrix}
1 & 0 & \frac{\partial \rho}{\partial p} \\
u & \rho & \frac{\partial \rho}{\partial p} u \\
E+\rho \frac{\partial E}{\partial \rho} & \frac{\partial E}{\partial u} & \frac{\partial \rho}{\partial p} E + \rho \frac{\partial E}{\partial p}
\end{bmatrix}
$$
where the derivatives of one thermodynamic variable with respect to another, $\partial \rho /\partial p, \partial E/\partial \rho, \partial E/\partial p$, come from your equation of state and can be computed for any equation of state that is analytical.
You can then plug this into the Euler equations:
$$
\begin{aligned}
\frac{\partial \mathbf{W}}{\partial t} &+ \frac{\partial \mathbf{F}\left(\mathbf{W}\right)}{\partial x} = 0 \\
\rightarrow \frac{\partial \mathbf{W}}{\partial \mathbf{Q}} \frac{\partial \mathbf{Q}}{\partial t} &+ \frac{\partial \mathbf{F}\left(\mathbf{W}\right)}{\partial x} = 0
\end{aligned}$$
where the fluxes are still expressed in terms of the conservative variables. If you wanted to derive governing equations for the primitives entirely, there are some additional steps to do to convert the fluxes to primitive variables through the flux Jacobian, which is another matrix constructed using derivatives of the flux vector with respect to the conservative variable vector. I'll leave that as an exercise for the reader.
You can find the Jacobians for a real gas system in pressure-temperature primitive variables in the appendix of this paper, including the flux Jacobians.
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does tension in the string affect its equilibrium? In my textbook (Sears and Zemansky's University Physics), it is written that the vector sum of the forces on the rope is zero, however the tension is 50 N. Then is tension different than the force? And if not, then why force is zero while tension is not?
A body that has pulling forces applied at its ends, such as the rope in Fig 4.27, is said to be in tension. The tension at any point is the magnitude of force acting at that point (see Fig 4.2c). In Fig 4.27b, the tension at the right end of the rope is the magnitude of $\vec{\mathbf{F}}_{M\ on\ R}$ (or of $\vec{\mathbf{F}}_{R\ on\ B}$). If the rope is in equilibrium and if no forces act except at its ends, the tension is the same at both ends and throughout the rope. Thus if the magnitudes of $\vec{\mathbf{F}}_{B\ on\ R}$ and $\vec{\mathbf{F}}_{M\ on\ R}$ are $50\ \rm N$ each, the tension in the rope is $50\ \rm N$ (not $100\ \rm N$). The total force vector $\vec{\mathbf{F}}_{B\ on\ R}+\vec{\mathbf{F}}_{M\ on\ R}$ acting on the rope in this case is zero!
| Each element of the rope is subjected to two forces which are equal in magnitude and opposite in direction so the net force on each element of the rope is zero.
One of the forces on the left side of the element is due to the left hand part of the rope adjacent to the element pulling left and the other force on the right side of the element is due to the right hand part of the rope adjacent to the element pulling right.
At the end of the rope, the rope is exerting a force on the anchor point holding the rope in position and the anchor point is exerting an equal in magnitude and opposite in direction force on the rope.
All the forces in the rope are called the tension.
Overall if the rope is not accelerating (or is massless) the net force on the rope is zero.
You can think of a rope as transferring a force from one position to another and also changing the direction of a force if there is a bend in the rope eg due to a pulley.
| {
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Newbie question: Atom identity. How can you talk about two electrons if electrons are identical? How can you talk about two electrons if they are identical (indistinguible)? Does it make sense to let an electron to have an identity by itself?
If they are on diferent places the place they are is a diference (they are not identical).
The act of labeling, naming them make them distinct. If I say let A be an electron and B another one, then they are distingible by name...(they are not identical)
Edit: if electrons are indistinguible by principle, then you cannot name them properly. A tick after you name one of them it could be any other electron in the universe or not exist anymore. You could say let this be electron A, but a tick after, A means nothing. The most you can say is at some point in time there were A and that "electronism" would remain, but this "electronism" would emerge in diferent form as long as "electronism" holds.
I think that the indistinguibility property makes the interpretation of a particle who has an identity and live forever hard to follow. One need to create strange ideas like "it follows all the paths" or "there are several paralel universes"...
| You can label billiards balls, ping pong balls, marbles and even BBs. But there is no way you can label electrons, even in principle. Think about 2 identical electrons in a head on collision. Before the collision they move in opposite directions along the same straight line. After the collision, 2 electrons move away from each other at some angle to the initial direction less than 180 degrees. How would you be able to tell which electron was which in the final state?
The answer is you can't. So you have to consider the final state as a superposition of both possibilities. Exchange degeneracy.
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A question over Liouville’s Theorem I have some doubts about Liouville theorem, probably its just something conceptual.
So: I know that for a system in which Liouville’s theorem holds, the volume in the phase space is conserved.
But the conservation of the volume does immediately imply the absence of asymptotically stable points.
However, if the Hamiltonian is time dependent and in particular of its time derivative is negative along the phase curves, a system does possess asymptotically stable points.
For that system, Hamilton equations still hold, hence it’s an Hamiltonian system.
But Liouville’s theorem doesn’t hold anymore.
My question: for what kind of systems does Liouville’s theorem hold? For example: the dumped oscillator has one asymptotically stable point.
| Liouville's theorem holds for all Hamiltonian systems.
If your definition of an asymptotically stable point $\boldsymbol{x}^*$ means that trajectories from points $\boldsymbol{x}$ in some neighbourhood of $\boldsymbol{x}^*$ tend to $\boldsymbol{x}^*$ as $t\to\infty$, then
*
*phase-space volume and/or density near $\boldsymbol{x}^*$ are not conserved
*and the system cannot be Hamiltonian (because of Liouville's theorem).
An example of such a non-Hamiltonian system is indeed the unforced damped harmonic oscillator
$$ \ddot{x} + c\dot{x} + k x = 0 $$
with $c\neq0$.
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What does electrical potential at a point mean? From my understanding, potential difference (or voltage) between point A and point B is the difference in electrical potential at the two points. The potential difference is also, the work done per unit charge in moving charges from point A to point B.
But apparently, the potential at one point (e.g point A) is also measured in voltage? How is it possible that at point A, there is work done per unit charge in moving charges from point A to point A?
| In common introductory textbooks, voltage is defined as the electric potential difference (say, between A and B) between two points in space. They say that A and B have a potential difference, and both A and B are at unique electric potentials.
This is completely narrow viewpoint.
Electric potential, electric potential difference, and voltage are all completely synonymous.
In your case, you are right that it doesn't make sense to define a unique potential at point A. When you say the "potential at point A", what you really mean is the difference in electric potential between point A, and some explicit (e.g., circuital ground) or even implied (e.g., infinity) point of reference.
After all, no potential is absolute. When you say a ball "has potential energy $mgh$", you are secretly saying that "the gravitational potential energy difference between its height and the ground is $mgh$". You could also say that its gravitational potential is zero at that height, in which case it would have potential $-mgh$ at the ground.
It's all relative to one's choice of reference point -- and this choice is always made, either explicitly or implicitly.
| {
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How is work transferred to the system recognised? For example, a potato initially at room temperature $25 \sideset{^{\circ}}{}{\mathrm{C}}$ is baked in an oven that is maintained at $200\sideset{^{\circ}}{}{\mathrm{C}}.$
I made potato as the system and the outer surface of the skin as the system boundary. While the oven and the air inside it is the surroundings.
There is a temperature difference between the skin and the air in the oven which is the driving force of heat transfer (temperature difference).
What about work? Is there transfer through work done?
Isn’t the oven working to produce the heat in the oven which is then transferred to the potato? But work is pressure multiplied by the change in volume. However there’s no change in volume of the potato. So does this mean no work is done?
In summary, how do I identify whether work is done to the system or not?
| Well, it is not strictly true that the volume stays fixed. The potato actually gets dilated in the oven, and the change in volume means that a part of the heat transferred has been converted to work (rigorously speaking the work is not pressure by volume because the potato is a solid body, so you should instead consider the internal stresses, but the concept is the same).
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What is the relation between a symmetry and the invariance of the Lagrangian? While proving that homogeneity of space implies conservation of momentum, we use the fact that homogeneity of space means that the Lagrangian of the system remains invariant under translation. Why is it so? I know that homogeneity of space means that if we perform 2 experiments a few metres apart, they will provide same results but how does that imply the invariance of the Lagrangian?
| I'll just add a little to Valter Moretti's excellent answer to explain why in particular momentum conservation is connected to translation invariance. Given the continuous symmetry $\delta\mathbf{q}=\epsilon\mathbf{K}$ with $\dot{\epsilon}=0$, on-shell $$\delta L=\epsilon(\mathbf{K}\cdot\frac{\partial L}{\partial\mathbf{q}}+\dot{\mathbf{K}}\cdot\frac{\partial L}{\partial\dot{\mathbf{q}}})=\epsilon(\mathbf{K}\cdot\dot{\mathbf{p}}+\dot{\mathbf{K}}\cdot\mathbf{p})=\epsilon\frac{d}{dt}\mathbf{K}\cdot\mathbf{p}.$$Thus $L$ is invariant iff $\mathbf{K}\cdot\mathbf{p}$ is conserved. If this works for any constant $\mathbf{K}$, each component of $\mathbf{p}$ is conserved.
| {
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What would happen if total energy of electron in bohr's orbit were positive? I have read somewhere that if electron had positive total energy in orbit it would not follow a closed orbit.
why is that so?
| If the electron had a positive energy, it would mean that the electron would no longer be bound to the nucleus.
Why does the electron have negative energy?
This is because the electron has lost some energy and thus has been successfully attracted by the nucleus. Energy is released which implies that it is negative.
Positive energy with respect to the nucleus would mean that the electron is free since it hasn't lost any energy. This would ultimately mean that the electrons would no longer be revolving in the orbits and they would moving freely in the space.
| {
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Principle of reversibility of light in case of refraction at the critical angle In the case where light is moving from denser medium to a rarer medium, at the critical angle, the refracted ray will be parallel to the surface. Now according to the principle of reversibility of light, if we reverse the direction of light, it will follow the same path. But in this case, if we reverse the direction of light, that is, send a ray parallel to the surface, how will it retrace its path since the ray will never cross the surface?
| Light beam is not an ideal ray and is not perfectly collimated, so, in real life, only its small fraction will be exactly at the critical angle for any given incidence angle.
Another factor, making the transition between refraction and reflection gradual is the dispersion, which makes the refractive index and, therefore, the critical angle, dependent on the frequency. Since the energy of a non-ideal light beam is spread over some finite frequency range, any incidence angle would be exactly critical only for a negligible fraction of light in the beam, even is the beam was perfectly collimated.
This point is illustrated in this video.
Note how the beam gets separated into different colors, each color reaching the boundary at a different incidence angle.
So, a scenario commonly shown on diagrams illustrating total internal reflection, where an incident light beam hit the boundary and then just keeps moving along it, is not realistic.
Therefore, if we direct a beam of light along the boundary between two media with different refractive indices, we are not exactly reversing what happens in the critical angle scenario.
| {
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Arithmetic of Hamiltonian in canonical transformation I have the following Hamiltonian:
$$ \mathcal{H} = \frac{p^2}{2m} + V(q-X(t)) + \dot{X}(t)p, $$
and I make the usual canonical transformation for the momentum:
$$ p \rightarrow p' = p + m\dot{X},$$
and complete the square, which should give the following:
$$ \mathcal{H}' = \frac{p'^2}{2m} + V(q - X(t)) - \color{red}{m\ddot{X}(t)q} - \frac{m\dot{X^2}}{2}.$$
I can get most of this expression apart from the one in $\color{red}{red}$.
This has to come from the cross term $(\frac{(\hat{p} \cdot m\dot{X}+m\dot{X}\cdot\hat{p})}{2m}),$ but I can't get the $q$ to come out.
Any pointers?
| Hint: Try a type-2 generating function $$F_2(q,P,t)~=~\left(P-m\dot{X}(t)\right)q,$$
for a CT $$(q,p,t)~\to~ (Q,P,t),$$ such that
$$K-H ~=~\frac{\partial F_2}{\partial t}~=~\color{red}{-m\ddot{X}(t)q}, \qquad Q~=~\frac{\partial F_2}{\partial P}~=~q, \qquad p~=~\frac{\partial F_2}{\partial q}~=~P-m\dot{X}(t). $$
| {
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Does measured mechanical power and work change between inertial frames? Imagine a car A is accelerating.
Two observers at constant speed, B and C, analyse the change of A's kinetic energy over a same time interval.
B sees A going from 10 to 30 m/s
C sees A going from 100 to 120 m/s
As both see A having the same mass (ignore relativistic effects), and as the kinetic energy depends on the square of the velocity, C sees A having a greater change on its kinetic energy than B does.
This also means C sees that the total amount of work produced by the engine of A's car over that same time interval was greater, and thus its power was greater too.
That's what doesn't make sense for me.
Because, imagine that A's engine was a spiral spring. You can calculate the amount of potential energy released over that interval irrespective to it's velocity, just by knowing the spring properties and its initial and final state.
So, if you were on another referential with a great velocity relative to A, does this mean that the change on A's kinetic energy could be greater than the amount of potential energy released from the string?
| The car is pushing against the Earth. Calculate $K_{tot} = K_{Earth} + K_{car}.$ You will find that $K_{tot}$ changes by the same amount in both frames.
As a simpler example, consider two boxes, masses 10 and 1, attached by a compressed spring. The spring is released, pushing the masses apart.
In frame A, they are both initially at rest, then the larger mass gets a velocity of -1 and the smaller +10. $K_{tot} = \frac{1}{2} 10\times (-1)^2 + \frac{1}{2}1\times 10^2 = 55,$ a gain of 55.
In frame B, they are both initially moving with velocity $v=1,$ so $K_{tot}=\frac{1}{2}11 \times 1^2=5.5.$ After the spring is released, the large mass is at rest, and the smaller mass moves with $v=11,$ so $K_{tot}=0+\frac{1}{2}1\times 11^2=60.5,$ again a gain of 55.
In other words, both frames A and B measure the same gain in $K_{tot},$ which equals the potential energy that was stored in the spring.
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Electric field of an infinite sheet of charge I am trying to derive the formula for E due to an infinite sheet of charge with a charge density of $ \rho C/m^2$. I assumed the sheet is on $yz$-plane. I used Coulomb's law to get an equation and integrated the expression that over $yz$-plane. But, I have not succeeded in deriving the correct expression. The answer I am getting is $0$.
Below is the picture of my work. Kindly, have a look and let me know where did I make mistakes.
In actual, E due to a charge sheet is constant and the correct expression is
E $=\rho/2\epsilon$0 aN , where aN is unit vector normal to the sheet.
| Are you looking to do the integrations by hand? Because
$r = x \hat{x} + y \hat{y} + z \hat{z}$
$r^\prime = y^\prime \hat{y} + z^\prime \hat{z}$
Should yield the correct answer, but the integrations are messy, unless you go to cylindrical coordinates
| {
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The term "Coherence" is overloaded! I have troubles with modern terminology in the field of quantum information technologies. There are a lot of new terms that everyone is using and no-one takes time to explain, even though modern textbooks did not catch up to these yet. Two of them are optical coherence lifetime vs spin coherence lifetime in the context of atomic clocks and quantum information storage/processing: definitions are needed.
*
*Please, notice that I am not asking for a classical definition of coherence. I am rather asking about two very specific terms, listed above in the given context.
*The title was given this way to attract attention and because in the given context the term "coherence" is overloaded and it is a known in the community fact (quantum, optical, quantum information coherence, to name a few).
Thanks to Harry Levine: helpful link
| For a quantum system such as an atom, or rare earth ion in a bulk crystal, the coherence lifetime is defined relative to two states of the system. 'Spin coherence' typically would refer to the situation when the two states are the same up to different nuclear or electronic spin orientations. For example, in a typical atom or ion there are many degenerate ground state levels that are split only by coupling from the nuclear spin to the electron spin (different $F$ levels). The splitting between these states is typically on the scale of < 10 GHz, all within the microwave or radio frequency domain.
However, other types of levels can also be considered: for example, one might consider the coherence between a ground state level and an excited state where an electron is in a higher orbital angular momentum or principal quantum number state. In this case, the energy difference between these states would often correspond to very high frequencies of > 100 THz, which are in the optical domain. Coherence between these types of states would be called optical coherence.
These two different settings are practically very different, in the sense that they are probed either with microwave fields or with laser fields. Their coherence can also be affected by different mechanisms (ie., radiative decay may limit the optical coherence but magnetic field noise might limit the spin coherence).
| {
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Pascal's law - Glass Flask Experiment (doubt)
This image was of my book and written below this image was that when the piston is pushed down, it is observed that liquid flow out with equal pressure from each hole. But, I think that pressure at the lowermost hole should be the most as pressure increases with depth and the lowermost hole is at the greatest depth. Am I wrong? If yes,then please explain me the reason.
| Ur question is right but in a flask the effect of gravity is negligible so the pressure on all points will remain same.
Ur point could be considered while dealing with large volumes under gravity.
I feel it might be helpful.
| {
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Real Lagrangian with complex variable I have a general question concerning real valued Lagrangians that take complex arguments. I have seen in many works of physicists and lecture books where extremal problems are discussed in Lagrangians that are real but depend on complex fields.
But, according to the Cauchy-Riemann equations, a real function $f:\mathbb{C}\rightarrow \mathbb{R}$ is only differentiable, when $f=const.$ since it has no imaginary part, i.e., for $f(x,y)=u(x,y)+iv(x,y)$ (with $v(x,y)=0$) it is
$$
\frac{\partial u(x,y)}{\partial x}=\frac{\partial v(x,y)}{\partial y}=0\qquad \frac{\partial u(x,y)}{\partial y}=-\frac{\partial v(x,y)}{\partial x}=0\, .
$$
Obviously, most Lagrangians are not constants, so I am wondering, how extremal problems, optimisations or Taylor series make even sense? How are these types of calculations consistent with mathematics? And how does one Taylor expand real Lagrangians with complex variables?
| Complex variables in physics are often real differentiable (=smooth) but rarely complex differentiable (=holomorphic/analytic). The simplest example in field theory is a complex scalar field.
| {
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Galilean transformations of velocity If I perform a Galilean boost $$x' = x - vt \\ t'=t$$ between two frames $S$ and $S'$, observers in each frame would disagree on the velocity of a particle because
$$ \frac{dx'}{dt'} = \frac{dx}{dt} - v. $$
Well Galilean transformations preserve the spatial intervals $\Delta x$ and time intervals $\Delta t$, so surely they should preserve velocity
$$ u = \frac{\Delta x}{\Delta t}? $$
There is obviously something going wrong here with my reasoning. I know the spatial interval $\Delta x$ is defined at constant time $t$, but if I was in the Galilean boosted frame $S'$ observing an object moving between two points $x_1$ and $x_2$ in $S$, I would observe that the interval $\Delta x$ between these two points would be a constant over time anyway so I could still conclude that the particle travelled a distance of $\Delta x$ in time $\Delta t$. What is going on?
| You are mixing two things. $dx$ and $\Delta x$. When you say "Gallilean Transformation preserves $\Delta x$ it means when measured by any observer", the relative distance between any two fixed points is same. But that is not the $dx$ in the definition of velocity. Its the change of position of a particle when we measure the distance between that particle and the origin of co-ordinates for that observer. Since the "origin" of the two co-ordinate system are not same(they have a relative velocity with respect to each other) that's why your measured velocity will be different.
| {
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In a vacuum can a cooler body radiate Infrared radiation to a warmer body? I mentioned vacuum, because I want to discount the effects of conduction or convection. I simply want to know if some of the infrared radiation(IR) goes from the cooler body to the hotter body? How does each body know how much to radiate at any particular time? I assume that it ultimately comes down to temperature difference but how does the hotter body know what the temperature is of the cooler body and vice versa? We all know that both bodies will radiate IR at the 4th power of its temperature and obviously they will be eventually in equilibrium with each other, each of them then radiating an equal amount to each other.
| Yes, both bodies radiate, including the cooler one. The warmer one puts out more power in its radiation, proportional to $T^4$ just as you say. The amount of radiation depends on absolute temperature, not temperature difference. But since the colder body has a lower absolute temperature, it radiates less than the hot one and the net result is that heat flows from the warmer body to the cooler one.
| {
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How can a transformer produce a high voltage and a low current? I understand that in ideal transformers, power is conserved. Because of this the product of voltage and current in the secondary winding is a constant.
This means that voltage and current are inversely related, which seems unintuitive because they are directly related by ohms law.
Shouldn't the emf induced in the secondary winding by the alternating magnetic flux be directly related to the current by some constant, such as the resistance of the secondary winding?
I also came across a term known as impedance that seem to be related to the question. Wondering if it is of any relevance.
| Actually the emf induced per turn in both of the primary and secondary windings are equal due to the conservation of energy. Now if you increase the secondary windings with respect to the primary windings then you will get high voltage than the primary circuit for an ideal transformer. And consequently the current in the secondary windings will decrease following the fact underlies in the constancy of product of current and voltage in bother the windings. If any energy dissipative elements are present in those circuit, then you may convert the primary circuit referred to the secondary or vice versa and apply simply kcl and kvl and rigourously solve the resulting instantaneous equations. If the windings are not coupled to an extent such that magnetic flux leakage has become a serious issue , then the calculation will be quite difficult but not impossible.
| {
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Orthonormal Basis integration and Kronecker delta Given that this integral I'm trying to solve is
$$\frac{2}{\pi}\sum^{\infty}_{l=0}\sum^{l}_{m=-l}\int_{r=0}^{\infty}\int_{k=0}^{\infty} R_{nl}(r)b_{lm}(k)j_{l}(kr)k^2 r^2 \int_{\theta = 0}^{\pi}\int_{\phi=0}^{2\pi} Y^*_{lm}(\theta,\phi)Y_{l'm'}(\theta, \phi)\sin\theta \,\mathrm d\phi\,\mathrm d\theta\,\mathrm dk\,\mathrm dr$$
where $R_{nl}(r)$ is the Radial Wave Solution of the Hamiltonian of the Hydrogen atom, $b_{lm}(k)$ is a spectral function that depends on $k$, $j_l(kr)$ is the Spherical Bessel Function and $Y_{lm}(\theta, \phi)$ is the Spherical Harmonics. Given the orthonormal relations of the Spherical Harmonics, the integral should technically then becomes
$$\frac{2}{\pi}\sum^{\infty}_{l=0}\sum^{l}_{m=-l}\int_{r=0}^{\infty}\int_{k=0}^{\infty} R_{nl}(r)b_{lm}(k)j_{l}(kr)k^2 r^2 dkdr \delta_{ll'}\delta_{mm'}$$
I have a very simple question which is how does this integral usually proceed from here with the sums and Kronecker delta given that they are summing over two different indices, do the sums not just collapse to one term where $l=l'$ and $m=m'$?
| The Kronecker Delta in a sum acts like the Dirac Delta function in an integral. So basically you can drop the sums and remove the delta terms. Your integral will be the integrand without the Kroneker Deltas. You technically can make the $l$s and $m$s primed if you wanted to, but that's unneeded primes (unless the context requires the primes).
| {
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Woodworking clamps, does force add up? I was watching a woodworking video about glue, and the guy was clamping two pieces of wood together using a total of 8 clamps. He argued that by doing so he would apply 8 times the maximum force of 150N (a property of the clamp), resulting in 1200N in total.
I think he's wrong. I think the force of 150 N is only working locally where the clamp is and will decline drastically radially from that spot. And so the clamping force on any given spot on the board will never exceed the max. force of the clamp.
Who's right?
| You're both wrong.
Although I would say you are more right than the expert. When you apply a force to an object's surface, the stress (a.k.a pressure) on the object clearly can't be constant across the surface; it must have a maximum closest to the point of contact and then dissipate as distance from the point of contact increases.
A very simple model of the clamping pressure experienced by a piece of wood might look like this:
The units here aren't important (pressure is in unites of 1 clamp). What is important is that clamping stresses add up:
Here is an animation of what happens when seven identical clamps are taken from clamping at the center of a board to being equally spaced across it's length. Total clamp pressure is just the sum of the seven separate clamps. Some observations:
*
*Only when all the clamps are at same location do you get 7 times the pressure.
*The total pressure however is higher (above any one clamp) everywhere.
*The average pressure doesn't change, only the peak pressure. (note it does dip slightly as the end clamps reach the ends).
*The pressure is very uniform when enough clamps are used. This is probably of greater value to a woodworker than maximum clamping pressure.
| {
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Why do vehicles with a higher center of mass roll more easily considering that they have a higher moment of inertia? It is a well known fact that cars with a higher center of mass roll over more easily, but why is this true considering that higher center of mass = higher moment of inertia?
I understand that the higher center of mass = higher torque being applied as the car turns, but this increase in torque is linearly related to the distance from the point of rotation (T = FD), with F being the force applied by the mass of the vehicle as it turns. Moment of inertia, however, is quadratic to the distance of the mass from the point of rotation (I = MR^2), so it seems that moment of inertia increases more than torque does as you move the mass further from the point of rotation, which should always be the tires.
Here's a couple of equations to further show what I am talking about
CoM = 5 meters from point of rotation
Mass = 10 Kg
Centripetal force = 15 N
(I know that these aren't realistic values for a car, but let's use them for simplicity)
Moment of inertia = 10*5^2 = 250KgM^2
Torque = 15*5 = 75NM
With a higher CoM
CoM = 20 meters
Mass = 10 Kg (same as before)
Centripetal force = 15 N (same as before since mass in unchanged)
Moment of inertia = 10*20^2 = 4,000KgM^2
Torque = 10*20 = 200 NM
As you can see the torque increased by much less than the moment of inertia.
| The moment of inertia tells you rate of rotation given a net torque. But for cars, we don't care how fast they tip over, we care if they tip at all. So the question isn't the moment of inertia, but the net torque. Is there something about a high center-of-mass vehicle that makes it more likely to enter a regime where the torques can't be countered?
Lets imagine a car is turning to the left and in the car's rotating frame of reference, it is experiencing a fictitious (centrifugal) force apparently to the right.
The car will not tip if the torque from $F$ does not exceed the torque from the center of mass. So the tip limit is when
$$F y = mgx $$
If mass, width, and forces are kept constant, then increasing the height makes it easier to tip.
| {
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All central forces are conservative forces, but are all conservative forces central forces? I have just been introduced to the concept of central forces, and to the fact that they are per definition conservative forces. I have looked up several examples of central forces (gravity, electric, and spring), but they cover just about all the conservative forces I have ever heard about. Are there any conservative forces that are not central?
There must be, because otherwise there would not be any point in having a subcategory for central forces, yet I cannot find any examples anywhere.
| If $\phi=-xy$, and ${F}=-\nabla \phi=y \hat{i}+x \hat{j}$ is conservative but not central
| {
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Why do we use $m\frac{dv}{dy} = -b(v-v_{ter})$ while determining how the terminal velocity is changing for an object falls down in linear drag force? Why do we use $$m\frac{dv}{dy} = -b(v-v_{ter} )$$ while determining how the terminal velocity is changing for an object falls down in linear drag force . I was Jr Taylor's classical mechanics. In the one dimensional or y directional projectile in linear drag proportional to (instantaneous velocity)^1 . I found this equation. Here drag force is given by $F_{drag} = -b(v - v_{ter})$, $v_{ter}$= terminal velocity. Now while in this equation why do we rule out gravitational force and why? Here $b$ is the constant which is the ratio of drag force and instantaneous velocity.
| Consider the downward direction to be positive. The equation of motion is, $m \frac {dv}{dt} = mg -bv$, where $bv$ is the upward drag force.
The net force will be zero when $v = v_{ter} =mg/b$
Substituting back, we get $ F_{downward} =- b(v-v_{ter})$.
Note that this is the net force (so it becomes zero when the body attains terminal velocity)., and not just the air drag (which is exactly equal and opposite to gravitational force at terminal velocity, making the net force zero)
| {
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The definition of the hamiltonian in lagrangian mechanics So going through the "Analytical Mechanics by Hand and Finch". In section 1.10 of the book, the Hamiltonian $H$ is defined as: $$H = \sum_k{\dot{q_k}\frac{\partial L}{\partial \dot{q_k}} -L}.\tag{1.65}$$
And then author affirms that this quantity is constant and takes the derivative $\frac{dH}{dt}$:
$$\frac{dH}{dt} = \sum_k {\ddot{q_k} \frac{\partial L}{\partial \dot{q_k}} + \dot{q_k}\frac{d}{dt}(\frac{\partial L}{\partial \dot{q_k}}) - \frac{d L}{d t}}.\tag{1.66}$$
Now the book writes: "According to the chain rule for differentiating an implicit function over time": $$ \frac{dL}{dt} = \sum_k {\frac{\partial L}{\partial q_k}\dot{q_k} + \sum{}\frac{\partial L}{\partial \dot{q_k}}{\ddot{q_k}} + \frac{\partial L}{\partial t}}.\tag{1.67}$$
And the summing the second and third gives: $$\frac{dH}{dt} = - \frac{\partial L}{\partial t}.\tag{1.68}$$
Now I don't understand how the third equation is derived and also why is the Hamiltonian $H$ is defined in the way it is in the first equation?
| *
*Ref. 1 is confusingly calling the Lagrangian energy function $H(q,\dot{q},t)$ for the Hamiltonian. We stress that the whole Section 1.10 is purely within the Lagrangian formalism. Eq. (1.68) follows after time differentiation of the definition (1.65) and use of Lagrange equations (1.60).
*In contrast, the function that is usually called the Hamiltonian $H(q,p,t)$ takes the same value but depends on momentum $p$ rather than velocity $\dot{q}$. The Hamiltonian formalism and the Legendre transformation are first explained in Sections 5.3-4.
References:
*
*L.N. Hand & J.D. Finch, Analytical Mechanics, 1998; Section 1.10.
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The work-energy principle for particles reversing direction I've been trying to find an answer to this question, but have really been stumped so far.
The work-energy principle says that work done on a single particle is equal to its change in kinetic energy. Now let's say a particle is moving in the +x direction at constant speed v and we perform work to reverse its direction so that it moves in the -x direction at constant speed v. This clearly requires work, but its change in kinetic energy is zero, because it has the same speed at the beginning and the end.
Sorry if there is an obvious answer to this, but have been bricking my head until now!
| The answer is that it doesn't actually require work. Using the equation $dW=\vec{F}\cdot d \vec{x}$, we can see that the work done on the object is positive if the force is pushing in the same direction as the object is moving, (speeding the object up) while it is negative if the force points in the opposite direction (slowing the object down). In the case of reversing an object's velocity, negative work is done on the object to slow it down to a state of motionlessness, and this is exactly canceled by the positive work done to speed it back up in the other direction.
If you want a picture for this, imagine a ball rolling partway up a steep hill, and then rolling back the way it came. The ball's velocity has been reversed, but the hill is left just the same as it always was.
| {
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Do compact symplectic manifolds play a role in physics? In classical mechanics, the phase space is the cotangent bundle of the configuration space, and it is a symplectic manifold, but not compact.
Do compact symplectic manifolds have physical meaning? Or just of mathematical interest?
| The other two answers do indeed give examples of compact symplectic spaces that can appear in physics. I would like to give some comments about how such things might arise in general, rather than an example thereof.
Starting with an arbitrary Lagrangian system, we construct the conjugate momenta and can construct from the boundary variation of the Lagrangian a 2-form which we will want to interpret as our symplectic form. However, there are no guarantees that this 2-form will actually be non-degenerate. This happens in the case of constraints (and also gauge theory), so this is not exactly an edge case insofar as application is concerned.
Vestiges of this can be seen in Dirac's approach to constrained systems, though he never went so far as working with the symplectic form itself. In the end, the resolution of these difficulties is to perform what's known as a symplectic quotient of the standard non-compact/"flat" phase space. Such quotientings may or may not result in another non-compact space, or it may be mixed in that only some directions become compact.
| {
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Determining conservation of isospin I am trying to understand a worked example. The problem is to determine if all conservation laws are conserved in
$$
\pi^-+n \rightarrow K^++\Lambda^0
$$
The answer says that isospin is conserved, which I don't understand. Isospin for $\pi^-$ is $1$, the isospin for $n$ is $\frac{1}{2}$, for $K^+$ it is $\frac{1}{2}$ and for $\Lambda^0$ is is $0$.
How can it be conserved then? On the LHS we have $1+\frac{1}{2}$ and on the RHS we have $\frac{1}{2}+0$.
In short, I want to know: How to determine if isospin and its projection is conserved.
| The isospin quantum number $I$ is not additive. Isospins combine like angular momenta so in your example you have $I_\pi=1$ and $I_n=1/2$, so the possible values of isospin for the left hand side of your reaction is
$1\otimes 1/2= 3/2\oplus 1/2$, i.e. the output channels must have either isospin quantum numbers $3/2$ or $1/2$.
The isospin projection number is additive; on the left you have the net projection $I_{\pi^-, 3}+ I_{n,3}$, and so on the right you must have the same net projection i.e. $I_{K^+,3}+I_{\Lambda^0,3}=I_{\pi^-, 3}+ I_{n,3}$, where $I_{a,3}$ is the isospin projection for particle $a$ along the isospin-3 axis.
| {
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What direction does electric current flow in when the voltage drop is negative? We were given the below diagram, and we needed to determine whether the unknown component is supplying energy into the system.
$\hspace{200px}$
I thought the unknown component was supplying energy since charge is flowing to a higher potential, but my teacher said it was the other way around.
What does the $`` -2 \, \mathrm{V} "$ label in the diagram mean? Is it:
*
*the difference from $\left(+\right)$ to $\left(-\right);$ or
*the difference from $\left(-\right)$ to $\left(+\right) ?$
I'm sure there is a convention to this sort of stuff but I'm not aware of it yet.
|
I just wanted to know what the -2v actually means.
Operationally, it means that if you place the red (positive) lead of your voltmeter on the $+$ labelled terminal and the black (negative or reference lead) of your voltmeter on the other, the voltmeter will read $-2\,\mathrm{V}$.
This tells you that the $-$ labelled terminal is more positive than the $+$ labelled terminal.
In anticipation of a follow-up question, the $-3\,\mathrm{A}$ means that if you place an ammeter in series with the $+$ labelled terminal of the circuit element such that the red lead is 'to the left', the ammeter will read $-3\,\mathrm{A}$.
This tells you that the current is directed directed out of the $+$ labelled terminal (in the opposite direction of the arrow).
As others have pointed out, with the chosen reference polarity and direction for the voltage and current respectively (passive sign convention), the power associated with the circuit element is
$$P = -2\,\mathrm{V} \cdot -3\,\mathrm{A} = 6\,\mathrm{W}$$
Since the power is positive, power is delivered to the circuit element from the external circuit.
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Is the force which throws a body on a rotating disk outwards centrifugal force? If some object is kept in a radial groove made on a disk and the disk is rotated with say constant angular velocity.
Now as we know that the object moves outwards ie away from the centre of the disk. I want to know which force is it which throws the object outwards.
As centrifugal force is a pseudo force and hence when the observer is in ground frame it shouldn't be present so which force exactly pulls it out.
| No force throws it outwards.
There is also no force pulling it inwards. Therefore it doesn't follow the circular motion. Such a force would have been friction for example. Instead it just continues straight ahead, away from the disk.
All in all, when an object has a speed, then it keeps going at that speed until a force pulls in it. Like a spaceship drifting forever at constant speed effortless. When the object in the disk has an initial speed, and the disk's angular speed is suddenly set too high for the object to follow, then the object will tend to continue straight ahead with this speed. And that means away from the disk.
The fact that it is stuck in a groove may push on it from the sides. But the fact that it moves outwards at all is due to the above describe tendency to continuestraight ahead with constant speed.
"Centrifugal force" is indeed non-existing.
| {
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What is wrong with my deformation gradient calculation? I created two ellipses,$\hspace{150px}$,where the red ellipsis is as the blue one, except translated to the right and rotated by ${30}^{\circ} .$ Using rotation matrix,
$$
\left[
\begin{array}{cc}
x \\[2px]
y \end{array}
\right]
\
\phantom{F} ~=~
\left[
\begin{array}{cc}
cos(30^{\circ}) & -sin(30^{\circ}) \\[2px]
sin(30^{\circ}) & \phantom{}cos(30^{\circ})
\end{array}
\right]
\
*
\left[
\begin{array}{cc}
X \\[2px]
Y \end{array}
\right]
\
$$
Next I tried to calculate deformation gradient by using principal axes(a and b for initial ellipsis,c and d for translated one), and then I decomposed $\vec{c}$ and $\vec{d}$ vectors,$$
\begin{alignat}{7}
\vec{c} &~~=~~ 1.0021 \, \vec{b} &~&+ &~&0.8654 \, \vec{a} \\[5px]
\vec{d} &~~=~~ 0.8654 \, \vec{b} &&- &&0.2505 \, \vec{a}
\end{alignat}
$$and created the deformation gradient as$$
F ~=~
\left[
\begin{array}{cc}
\phantom{-} 0.8654 & 1.0021 \\[2px]
- 0.2505 & 0.8654
\end{array}
\right]
\,.
$$But it is obvious that, this is not the same as the ${30}^{\circ}$ rotation matrix that I had expected it to be,$$
\phantom{F} {\llap{\textsf{rotation matrix}}} ~=~
\left[
\begin{array}{cc}
0.8660 & - 0.5000 \\[2px]
0.5000 & \phantom{-} 0.8660
\end{array}
\right]
\,.
$$
Question: Why aren't the deformation gradient, $F ,$ and and the rotation matrix, $\textsf{rotation matrix} ,$ the same?
| In component form,
$$x=F_{xx}X+F_{xy}Y$$
$$y=F_{yx}X+F_{yy}Y$$
In your example, there is a mapping of 2 vectors:
X = 0, Y= 2 ---> x = 1, y = 1.732
and
X = 1, Y = 0 ---> x = 0.866, y = - 0.5
Solving for the components of the deformation gradient tensor gives:
$F_{xx}=0.866$, $F_{yx}=-0.5$, $F_{xy}=0.5$, and $F_{yy}=0.866$
| {
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Forms of transformation Suppose $O$ is an object to be transformed, and $S$ is the transformation operator. Sometime the transformation is in the form
\begin{equation}
O \rightarrow SO. \tag{1}
\end{equation}
But sometime the transformation is in the form
\begin{equation}
O \rightarrow SOS^{-1}.\tag{2}
\end{equation}
I am confused. I know that there is some difference between these two cases. I just don't know what is the difference? What kind of objects transform in the first way, and what kind of objects transform in the second way? Is there any rule?
| This is a question where the bra-ket notation is truly useful. When $O$ is a column vector, think of it as a ket $\vert O\rangle$. Then the transformation matrix $S$ acts in the usual way
$$
\vert O’\rangle=S\vert O\rangle\, . \tag{1}
$$
Now think of $O$ as an operator, i.e an object of the form $\vert m\rangle\langle n\vert$ that takes one vector into another, where $\vert m\rangle$ is a column vector but $\langle n\vert$ is a row vector.
Applying (1) to $\vert m\rangle$ and $\langle n\vert $ gives
\begin{align}
\vert m‘\rangle &= S\vert m\rangle\, ,\\
\langle n’\vert&= \langle n\vert S^{-1}
\end{align}
so that the operator $O$ now transforms as
$$
O’= \vert m’\rangle\langle n’\vert = S\vert m\rangle\langle n\vert S^{-1}
=S O S^{-1}\, .
$$
This is not a formal proof but it shows (somewhat intuitively) that when the operator $O$ can be represented as a matrix, you can think of $S$ as required to transform the rows of $O$ and $S^{-1}$ as required to transform the columns of $O$.
| {
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Electrical potential with a combination of bodies 1) what would occur with respect to electrical potential if a +1 charged body is brought close to a -2 charged body?
2) a positively charged body next to an earthed conductor
For the 1st one I'm not too sure to make an educated guess. For the second one I'm thinking that the positively charged body will lose potential due to induction while the earthed conductor will have a potential of zero ?
|
1) what would occur with respect to electrical potential if a +1
charged body is brought close to a -2 charged body?
The potential of -2 body will be getting less negative, while the potential of +1 body will be getting less positive. At some point, the potential of the +1 body will become zero and than negative. When the bodies get very close to each other, their potentials will be roughly equal (both negative) and become the same, when they get connected. The final potential will be negative, since the net charge will be -1.
In summary, the potentials of the two bodies will be getting closer to each other as the bodies get physically closer, until they equalize.
2) a positively charged body next to an earthed conductor
Your assumption regarding this case is correct. The general observation above will hold as well, i.e., the potentials of the two bodies will be getting closer to each other as they get closer, but, in this case, the potential of one of them won't be changing.
| {
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Does 'special relativity + newtonian gravity' predict gravitational bending of light? It seems to me that special relativity (SR) already predicts that gravity will bend light rather than this effect being a proof of general relativity (GR). Photons have energy proportional to their frequency and according to $E = mc^2$ they also have a tiny, but non-zero relativistic mass equivalence. I have read the semantic argument that gravity deals with invariant or rest mass, but this should apply to a hypothetical photon at rest, not to real photons at velocity $c$.
I have considered the possibility that the SR effect is much smaller than the GR effect. GR explains the equivalence of inertial mass and gravitational mass as inherent rather than being a puzzling coincidence, but it is true in Newton's gravity, SR, and GR, so the quantitative difference between GR and SR does not seem right.
| As mentioned in tfb's answer, the deflection treating photons with an "effective mass" is half that of general relativity. While photons have zero rest mass in special relativity, one can consider them with infinitesimal mass for the purposes of a Newtonian approximation (a "gravitational force" interacting directly and instantaneously between two masses). Consider a single photon passing a body of mass M at a distance r. The Newtonian deflection angle is: $$\theta= \frac{2GM}{rc^2}$$
In general relativity, the deflection angle is: $$\theta= \frac{4GM}{rc^2}$$
This is because general relativity considers spacetime warpage.
One interesting feature to point out is that in the Newtonian approach, the passing object will experience an increase in speed (potential energy converted into kinetic energy). However, according to relativity, a photon cannot be accelerated past c. Instead, the transfer of this energy causes the photon's frequency to increase as it passes through a gravitational well (and decrease upon escaping); this is the phenomenon of gravitational redshift. The energy of a photon is given by $e=h\nu$, where h is Planck's constant, and $\nu$ is the frequency, thus it can be seen that energy and frequency are directly proportional with Planck's constant as the constant of proportionality.
| {
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What do string theorists mean when they say that little is known about M theory?
*
*It’s often said that we don’t know what M theory is, but what does that actually mean?
*And how long until M theory becomes somewhat of an understood theory?
*Where are we at now with M theory?
| So far, it is known the low-energy limit of the theory. This means that when the energy scale is not high enough to resolve the typical length of the theory, then we will just see point-like objects. Namely, at that energy scale, our system is effectively a set of particles and it will be described by standard quantum field theory. In M-theory, this quantum field theory is 11-dimensional supergravity (see this entry for further details), which contains Einstein's theory.
In addition, it is also known that 11D supergravity, when compactified on a circle, gives rise to the corresponding supergravity that describes the low-energy limit of type IIA string theory.
| {
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Torque acting on car trailers My question: "Which car trailer below will be moved in a straight forward line?"
Relevant info: All forces (represented by the arrows) in the diagram are equal in magnitude. The car trailer diagrams below seem to me very similar in terms of the net torque that acts on each of them. The net torque must be 0 in both, doesnt it? However, according to the answers, only trailer 2 is the one to travel in a straight forward line. Did they make a mistake or am I the one to overlook something?
Note: I received the questions exactly as I present them here from the organization I got them from. Therefore, I suffer from not being able to provide more context and specifications. I am only being able to provide my assumptions based on my preliminary knowledge.
[Please don't pay attention to the size of the wheels]
Wondering about your valuable opinions
| The drawing is not particularly accurate. For instance, although it does not affect the answer, $0.25$m and $0.5$m on the second drawing have about the same length.
Apart from that (without going into the details of the hook-up), it appears that the idea here is that the forces acting on trailer $1$ appear to be symmetric, but the link to the trailer is not in the middle - it is shifted to the left.
The forces acting on trailer $2$ appear to be asymmetric, but, in fact, they are, since the torque from $F_4$ is balanced out by the torque from $F_5$ and $F_6$ and the trailer is hooked up in the middle.
So, it looks more like a trick question.
| {
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Could black holes be formed by highly energetic gravitational waves? Could the gravitational waves released by two merging black holes contain enough energy to produce another black hole?
| In theory, yes, any concentration of energy that is suitably high will form a black hole. In particular, according to general relativity, if any energy is concentrated within about a volume equal to its own Schwarzschild radius or less, it will form a black hole, and given that gravitational waves transmit energy there is no reason this would not apply to them as well. It would effectively be a gravitational version of a so-called "kugelblitz" - a a black hole formed by a similarly-intense and concentrated convergence of electromagnetic radiation. Kugelblitzes are not something that has ever been observed, I believe, and may not have occurred anywhere, since it is very difficult to concentrate such highly mobile forms of energy.
And thus to answer the question here, following on that last point - no it could not, because the gravitational waves are radiating outward, so their concentration is decreasing. The energy of the waves comes from the orbital energy of the black holes - if that were enough to form a black hole then that would effectively mean the black holes had merged already, as they'd now be inside one larger horizon due to it (and thus also to a single large core, presumably) since it's the total energy and momentum in the system that give rise to the gravitational forces (as well as the system's total mass - note that this added mass doesn't belong to each black hole individually but the system collectively).
| {
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How can the solutions to equations of motion be unique if it seems the same state can be arrived at through different histories? Let's assume we have a container, a jar, a can or whatever, which has a hole at its end. If there were water inside, via a differential equation we could calculate the time by which the container is empty.
But here is the thing: through the differential equation, with initial condition, I shall be able to know everything about the container: present past and future.
But let's assume I come and I find the container empty. Then
*
*It could have always been empty
*It could have been emptied in the past before my arrival
So this means I am not able to know, actually, all its story. Past present and future.
So it seems there is an absurdity in claiming that the solution of the differential equation is unique. Where am I wrong?
| Everyone keeps saying all these things but really the issue here is so much more succinct and I think it has a lot more to do with science itself. The purpose of science is to construct models that attempt to predict future behavior based on previously observed behavior. So while yes the differential equation might be able to predict past motion the problem here is that the universe nor its model is necessarily reversible. Nobody has proven nor claimed afaik that any given state of the universe has a unique previous state. In fact, I would claim there isn't such a state. Therefore while your bucket is an analogy I would say that it shows there is unique future behavior and NOT unique past behavior. Of course, there is also the issue that you aren't modelling everything perfectly. There would be evidence to suggest the puddle came from the bucket or whatever such as ripples or the bucket being wet or whatever else would indicate such things.
| {
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How can a particle's position be random and uncertain in quantum mechanics if it is already pre-determined in relativity? In relativity, to my knowledge, the path of an object is described by its worldline in spacetime, and since time is a part of the spacetime geometry, an object's worldline--in a sense--always exists on this “block of time” as I heard the PBS spacetime say. But in quantum mechanics (the Copenhagen interpretation), I’ve heard that a particle will exist in (superposition of being in) multiple places at once until it is observed and the wave function collapses and it has a single position.
So my question is, how could these two viewpoints be reconciled? Apologies if I got anything wrong.
| So when a particle's position is measured, we can say that the particle is really located in some region of space. As time progresses without measurements, the region of where it could be grows according to the uncertainty in its momentum. A first understanding of relativistic QM just makes sure that this region is bounded by the starting region's light cone. relativity is much more concerned with the idea that no information is proven to move faster than the speed of light, than with the idea that nothing is uncertain.
With that said, in the usual description of quantum mechanics there is an instantaneous information propagation which did trouble physicists for a long time, called entanglement. The full pursuit of this feature has convinced most physicists that reality cannot be described in the "local" vocabulary that special relativity would like to use. However we now appreciate that this will never propagate usable information faster than the speed of light: the information is hidden in a correlation between two systems living a great distance apart, and cannot be observed until both measurements are brought back together for comparison.
| {
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What happens if a neutron flies towards a nucleus? Rutherford experiment shows that alpha-particles when they fly towards metal foil sometimes (in minority of cases) can bounce. An explanation proposed was that atoms in fact have positively charged nuclea and majority of space is covered by fields of negative charge caused by electrons. Indeed, these fields must have much smaller charge density so that they (almost) don't affect alpha particles.
However, according to this explanation a neutron when it flies towards a nucleus should not bounce because of electromagnetic force. This is because neutron is an uncharged particle. Gravity is too weak to have any significant effect between neutron and nucleus so they don't merge. Weak forcd probably also is too weak. Therefore we can only consider strong force.
So, when a neutron flies towards a nucleus with high speed, what happens? Does strong force come into effect? Or does it just pass through a nucleus?
| The place to go for neutron scattering data is The Evaluated Nuclear Data Files site, hosted in the US at Brookhaven National Laboratory. There one can get data for a wide range of neutron scattering possibilities, including the cross section vs energy.
Since fission was mentioned in the comments, lets look at U-235. Entering that target nucleus, asking for all neutron reactions, and requesting the cross section (sig = $\sigma$) looks like:
One gets back a long list of possible reactions starting with:
Line one is the total neutron cross section vs energy. Line 2 is the elastic scattering component, and line 3 is the inelastic component. Line 7 is the fission cross section. Things like line 8 are an inelastic neutron scattering through a particular level. Further down one find the (n,p), (n,$\alpha$), and other similar reactions.
Selecting a few of the boxes and hitting 'Plot' up above results in:
So, that is how you find out what neutrons will do.
| {
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Why is it much more difficult to horizontally throw a toy balloon than a football? If you horizontally throw a sphere of radius $R$ it will feel, in this direction, a drag force due to air. Assume the drag is given by Stokes law, $F_D=6\pi\eta R v$, where $\eta$ is the air viscosity and $v$ is the horizontal speed. This force cannot "see" the internal structure of a toy balloon, a football or even a metal sphere. However, anyone who ever played with balls and toy balloons noticed that for the same throwing, the ball will have higher horizontal reach for the same time interval. Just think about someone kicking toy balloons and footballs and the distances reached in each case. How is the resistive force considerably greater for the toy balloon?
Even if we consider a quadratic drag, $bv^2$, I suppose the coefficient $b$ would depend only on the fluid and the geometry of the bodies. Again the drag would be equal.
Another way to put this question: How does the density of the sphere contribute for the resistive force?
| Let's say you get your object to some initial speed $v$ and then let it go so that only the drag force is acting on it (ignoring gravity for now).
Then by Newton's second law:
$$a=\frac{F_D}{m}=\frac{F_D}{\rho V}=\frac 1\rho \cdot \frac {F_D}{V}$$
Where $\rho$ is the density of the object. Since $\frac {F_D}{V}$ depends on the velocity and the geometry, then for two objects with the same geometry and same initial velocity, the object with the smaller density will have the larger acceleration due to the drag force.
We can even use your case to get the velocity as a function of time:
$$\frac{dv}{dt}=-\frac{6\pi \eta R}{\rho V}v(t)$$
$$v(t)=v_0Exp\left (-\frac{6\pi \eta R}{\rho V}t\right)$$
Comparing different plots for various values of $\rho$ gives us something like this:
As you can see, smaller density leads to a faster decrease in velocity.
So your thoughts are correct. The force itself does not care about the insides of the object, only the outer cross section that influences the drag. However, as you can see, this means that more dense objects, which have more inertia (mass), are less effected by this drag force.
| {
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What is state variable and full differential? e.g. in entropy? I am studying basic concepts of entropy and statistical physics. And red a lot what is entropy, and that it is integrative factor of heat; and getting it with the full differential._
Anyway, what I am trying to grasp in all that story, actually to gain a feeling and know what is actually full differential means in this story; and that some variable is state variable (state function)?
For example $dS=\delta Q/dT$.
I know I needed to find integrative factor which is temperature as universal thermodynamics variable. And I need to have on extensive and one intensive variable to get state variable.
But get confused if I actually just start to think about what is variable of state here actually and what does it usually say.
It same here with work where $dV= \delta W/ -p$.
Thanks for help :)
| A function of state is one that depends only on the state of the system, which for a gas means that it's a function of the pressure $p$, volume $V$, and temperature $T$, and not dependent on the path by which the system got there. They're also only defined in equilibrium; if there's no defined $p$, $V$ or $T$, there's no defined functions of them.
For example, internal energy is a function of state - no matter how your gas got to be in that box at that temperature and pressure, it's always going to contain the same amount of energy. No matter how it's compressed, expanded, cooled down, or heated up, if it goes back to the same pressure, volume and temperature, it goes back to the same internal energy.
Work, on the other hand, isn't (for example). The amount of work done on or by a gas depends on the specific path that it takes through pressure, volume and temperature-space. Different thermodynamic cycles will, in general, have different amounts of work associated with them.
| {
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Reducing multi body system to a single body using reduced mass A two body system can be treated as a single body using reduced mass and the motion can be described using one generalized coordinate.
Can this concept be somehow used to reduce a body of say 3 or more particles to a single body?
| The two dof of the system reduces to one because the other one ends up with constant velocity solution and a co-moving reference frame can be constructed that is still inertial. For three bodies, no such co-moving inertial frame exists.
| {
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Cast metal vs. machined metal Why is it, that machined metals, milled out of a primitive shape, tend to bend upon impact rather than crack or shatter? Why do they behave differently to stresses when they are cast into the final shape?
I find this rather confusing. Don't they have the same lattice? Are there more defects introduced when casting it in complex shapes?
| Jon Custer is right- this is a very broad question, about which whole chapters in materials science textbooks have been written. I'll furnish a broad perspective which I hope will get you pointed in the right direction.
First of all, all common metals begin their useful (i.e., engineering) lives as raw solids that were cast from the molten state. But in the case of (for example) a piece of steel bar stock that is being milled to shape in a milling machine, that bar underwent significant post-casting treatment to transform it from a rough billet of as-cast steel into a square chunk of stock: the billet got mechanically deformed while red-hot through a series of incremental squeezing and rolling processes- all of which had the effect of significantly altering the microstructure (grain size, grain orientation, etc.) of the metal
and thereby altering its physical characteristics which include its ability to be bent without fracture or to withstand impact loads without cracking.
Second, those "secondary processes" also furnish the opportunity to cut out and discard things like slag inclusions and voids which occur commonly during the casting process, so that the finished piece of bar stock is usually a lot more defect-free than a cast part and hence more robust.
| {
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If the path integral formulation includes future events, why doesn't that imply retrocausality? I know that such events would cancel out in the math, but if an extreme event were to happen in the future (say a black hole forming or something on that par), would a particle in the present react to it? If not, why?
| The path integral is a broad idea, which comes in several different flavors. In non-relativistic quantum mechanics for one particle, you calculate the propagator matrix element $\langle {\bf x}_f| U(t_f,t_i) | {\bf x}_i \rangle$ by summing $\exp(i\, S[{\bf x}(t)]\, /\hbar)$ over paths connecting $(t_i, {\bf x}_i)$ to $(t_f, {\bf x}_f)$ which travel forward in time, so there's clearly no retrocausality. In relativistic quantum field theory, you typically use the LSZ formalism, which involves integrating $\exp(i S[\varphi(x)] / \hbar)$ with $S[\varphi(x)] := \int_{t_i}^{t_f} dt\, \int d^3 {\bf x}\, \mathcal{L}(\varphi(x), \partial_\mu \varphi(x); x)$ in the limit $t_i \to -\infty$ and $t_f \to +\infty$, so that the incoming and outgoing particles are thought of as far-separated and asymptotically noninteracting. (In practice, this limit is carried out through the use of the "$i\epsilon$ trick" in the denominator of the propagator, which sets the boundary conditions.) In neither case are paths that extend later than $t_f$ considered, so there is no retrocausality. (You do find acausal correlations across spacelike separations, but they cannot transmit acausal influences.)
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Why does light bend towards the normal when passing through a denser medium? When light passes from a medium with less refractive index to a medium with higher refractive index, it bends towards the normal. But why normal . Of course we can take an example of a car moving from the road to the mud in an angle. Its first tire decides the direction. But what about light . It's not a car. It's way small and way too fast.
| This is due to Fermat's principle, which states that light travels between two points along the path that requires the least time, as compared to other nearby paths. Light travels more slowly in a denser medium, and hence will bend more toward the normal.
Why then does light follow this path, and not other paths? This is because light also obeys the principle of least action, and the action for light is proportional to the time for which it travels. Thus, least action requires that the time taken be minimal.
| {
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Half-life of $W$ and $Z$ bosons $W$ and $Z$ bosons should decay through weak interaction. But their half-life is around $\tau = 10^{-25} s$ which is a typical value for particles decaying through strong force (instead of a $10^{-12}-10^{-6} s$ for a weak interaction decaying particle).
Why this can be?
| Yes, it is
correct to say that the W boson is decaying through weak interaction despite its very small lifetime.
The weak interactions were dubbed such in the last century because, at low energies compared to the mass of the W, about 80GeV, the large mass of this particle appeared squared in the propagator of this virtual particle in all weak process amplitudes. This is summarized in the Fermi constant G of the relevant amplitudes. Thus, weak decay widths involve the square of G.
As a result, virtually by dimensional analysis, e.g., μ decay must be of order ${\cal O} (m_\mu^5 /m_W^4)$! Recall the mass of the μ is ~ 0.1GeV, so a thousand times smaller than that of the W. (It is a 2-scale problem: the electron and neutrinos' masses are negligible here.) So, all low-energy weak processes are "cursed" by such a suppression.
As you probably covered/will cover in your particles physics course, the actual small width for μ decay giving it its long, microsecond, lifetime, is
$$
\Gamma_\mu \sim \frac {G^2 m_\mu ^5}{192 \pi^3}=\frac{g^4 m_\mu^5}{m_W^4 \pi^3 ~6144},
$$
where g is the usual EW coupling.
Now, contrast this with the real W decay into μ ν, where there is no propagator suppression:
$$
\Gamma_{W\to \mu\nu} \sim g^2 m_W /48\pi ,
$$
whose order you find again by dimensional analysis: it has to be $m_W$, as it is a one-scale problem---all other scales are negligible.
The ratio of the two, then, amounts to
$$
\frac{m_\mu^5}{m_W^5} ~ \frac{g^2}{128\pi^2}
$$
and so 20 orders of magnitude. And you thought these leaps only happen in astronomy...
You may also marvel at the power of the SM electroweak unification which leads you to such sensible descriptions across 20 orders of magnitude.
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Confusion about how an electron gun works I'm a little unclear about the charge balance aspect of an electron gun. Referring to this diagram and similar diagrams I've seen, what I don't get is wouldn't the target of the electrons have to be connected to the positive anode so that the electrons fired at a target can be recycled if the electron gun is needs to operate continuously? Is the target generally placed on the anode opening so it's connected to the positive?
| I cannot comment on the construction of a specific electron gun, but in a typical CRT, the internal surface of the tube around the screen is coated by graphite, forming the final anode electrode.
This electrode is connected to high positive voltage and one of its functions is to collect electrons arriving to the screen.
So you are absolutely right - any cathode ray tube or electron gun should have some return path for the electrons. It is not shown on your diagram, probably, because this diagram is simplified.
| {
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Is continuum mechanics a generalization or an approximation to point particle mechanics? Newtonian Mechanics is usually presented as a theory of point particles (and forces). My impression of the status of continuum mechanics is that it is mostly taken as an approximate description for certain situations, where many particles are present and where we are interested only in bulk motion. In situations like modeling a string, we see that the continuum limes leads to a very good approximation for the motion of pointlike masses connected by springs under certain conditions. This seems to point in the direction that continuum mechanics is an approximation to point particle mechanics.
But point particles should also be easy to accommodate as a special case in continuum mechanics by admitting delta distributions and the like. Taking this viewpoint, continuum mechanics seems to be a generalization of point particle mechanics.
Therefore my question: Is continuum mechanics a generalization or an approximation to point particle mechanics? Or can it be argued that both are equally valid starting points?
Part of my motivation to ask this question is that I sometimes have difficulties to connect concepts from one viewpoint to the other (see e.g. my question about the point of application of a force).
| It's neither a generalization of, nor approximation to, the classical mechanics of particles, it's simply a different perspective. As described by Walter Noll in discussing the connection in The foundations of classical mechanics in the light of recent advances in continuum mechanics (1959):
It is true that the mechanics of systems of a finite number of mass points has been on a sufficiently rigorous basis since Newton. Many textbooks on theoretical mechanics dismiss continuous bodies with the remark that they can be regarded as the limiting case of a particle system with an increasing number of particles. They cannot. The erroneous belief that they can had the unfortunate effect that no serious attempt was made for a long period to put classical continuum mechanics on a rigorous axiomatic basis.
Worth a read if you have the time.
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Confusion about the meaning of steady current I am trying to learn some elementary EM, but I have some confusion about the basic concepts of steady current.
Suppose I have a wire of uniform cross section area. The current is always flowing from left to right.
I imagine that I can cut a segment of this wire (with the area vectors at both ends parallel to the flow of current) and compute the surface integral $\int\int_S \mathbf{J}\cdot d\mathbf{A}$, where $\mathbf{J}$ is the current density vector, $\mathbf{A}$ is the area vector, and $S$ is the boundary of the segment. I assume that $\mathbf{J}$ does not vary with time.
I believe that the magnitude of $\mathbf{J}$ can depend on its position, so lets say that its magnitude is greater on the right end, as compared to the left end. Because of that, the surface integral $\int\int_S \mathbf{J}\cdot d\mathbf{A}$ should have a non-zero value (the dot products on both ends do not cancel).
However, $\int\int_S \mathbf{J}\cdot d\mathbf{A}$ is precisely the net change in charge out of this segment, i.e., $-\frac{dq}{dt}$.
Now, for a steady current, $\frac{\partial \rho}{\partial t}$ is zero at every point, thus the $-\frac{dq}{dt}$ should also be zero, which contradicts my understanding that $\int\int_S \mathbf{J}\cdot d\mathbf{A}$ can be non-zero.
This is perhaps a very stupid question, but I just cannot figure out what has gone wrong. Any help would be greatly appreciated~
| $\frac{\partial}{\partial t}\rho = 0$ implies $\nabla\cdot\mathbf{J} = 0$. Then, $I = \oint_S \mathbf{J}\,d\mathbf{S} = \iiint_{\Omega} \nabla\cdot\mathbf{J}\,d\Omega = 0$
If you consider $S$ as a cylinder, you will see that the current density does not depend on the cross section you take.
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What is a 'spacelike surface' in relativity? I am studying Noether's theorem in field theory and I am not understanding what spacelike-surfaces mean. I will reproduce the bit of the argument below that contains the term "spacelike-sufaces" in the context I am not understanding.
There will be a conserved ccurrent for each group generator $a$. Each will result in a conserved charge (that is, an integral of motion). To see this, take in spacetime a volume unbounded in the space-like direction, but limited in time by two space-like surfaces $w_1$ and $w_2$. Integrating $\partial_{\mu} J^{\mu}_a=0$ over this volume, we get an integral over the boundary surface, composed of $w_1$, $w_2$ and the time-like boundaries supposed to be at infinity. If we now suppose the current to be zero at infinity on these boundaries, we remain with
$$\int_{w_1}d\sigma_{\mu} J^{\mu}_a=\int_{w_2}d\sigma_{\mu} J^{\mu}_a.$$
In my understanding, spacelike-surfaces are surfaces of constant $t$ (surfaces perpendicular to the time axis in the figure below, like the which passes through the origin), but the text above states to take a spacelike-surface limited in time, which means that my definition of spacelike-surfaces isn't the correct one.
I would appreciate some clarification.
| Your definition of the surface is correct. I think you misread the text. They are talking about a volume that is bounded by two such surfaces "in the time direction" and unbounded "in the space directions".
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What is a complex phase shift? In a complex methods course I am taking, we were given an equation for a particular driven harmonic oscillator where the driving force is trigonometric. I have worked out the math and obtained an equation that tells me that the driving frequency at resonance is the natural frequency multiplied by i. My tutor tells me that this is a 90 degree phase shift, but I don't really understand why. Isn't a phase shift obtained by adding or subtracting 90 degrees? And how can a frequency, which is a measurable physical value, take on imaginary values? I would understand if we were talking about velocity. Because velocity has a direction, addition or scalar multiplication by a real value would not describe a 90 degree rotation of the vector. But frequency is a scalar quantity. What does it mean to have an imaginary frequency?
| If your oscillating function is of the form $e^{i\omega t}$, a phase shift looks like $e^{i(\omega t+\phi)}$, which can be rewritten as $e^{i\omega t}e^{i\phi}$.
Now, recall that $e^{i\phi}=\cos\phi + i\sin\phi$. A 90 degree phase shift corresponds to $\phi=\frac{\pi}{2}$.
Thus,
$$e^{i\frac{\pi}{2}}=\cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = 0 + i = i.$$
So finally we have,
$$e^{i(\omega t+\frac{\pi}{2})}=e^{i\omega t}e^{i\frac{\pi}{2}}=ie^{i\omega t}.$$
So we see that a phase shift of 90 degrees corresponds to multiplication by $i$.
| {
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Speed of sound in a gas and speed of a typical gas molecule Why is speed of sound in a gas less than the average velocity of the gas molecules? Is there an intuitive way to explain this?
| I know the derivation of $v_\text{sound}=\sqrt{\frac{\gamma p}{\rho}}\ $ from macroscopic principles, but I've always thought that there must be a way of deriving it on a molecular level. And I've sometimes wondered about the question being asked here: not so much that $v_\text{sound}\ $ is less than $v_\text{rms}$, as that the two speeds are so close. What follows is speculative and hand-wavy, so please regard it as a possible jumping-off point for further investigation.
As I see it, the sound, perhaps generated by the moving cone of a loudspeaker, superimposes a small velocity on the fast random motion of the molecules. This superimposed velocity is likely to be passed on to other molecules in collisions. But the speed of propagation is almost entirely determined by the random thermal speed of the molecules (since the superimposed 'sonic' velocity is very small by comparison). This accounts for why $v_\text{sound
}\ $ should be close to $v_\text{rms}\ $.
But when one molecule carrying the sonic velocity hits another molecule it has a tendency to pass on that velocity to the molecule it hits, the second molecule is unlikely to leave the collision in the same direction that the first molecule was travelling in, so you'd expect that over a distance, the sound would travel more slowly than the rms speed of the molecules, because, on a microscopic scale, it's not travelling in a straight line.
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What’s wrong with my cloud chamber setup? For my setup, I am using a fish tank, lined with black felt and with one side open for viewing and a slit on a side for the light. I have a pretty bright multi led flashlight. I am using dry ice blocks inside a stryofoam container that is separated from the chamber by a metal cooking sheet. I am using 91% IPA, and taped the bottom of the fish tank to the tray to seal it. I am using some hand warmers to heat the top of the tank.
All that happens (after 4 attempts) is that the IPA begins condensing onto the sheet. I never see mist. I have tried to reposition the light, and have used other lights. It is dark in the room.
| Try and make the black flat metal base which is sitting close to the dry ice (solid carbon dioxide) as horizontal as possible.
If the bottom plate is not approximately horizontal convection currents will be set up within the chamber which will ruin the experiment.
The felt soaked in isopropyl alcohol (IPA) is placed at the top of the chamber.
The sensitive layer in such a cloud chamber is very thin $(\sim \rm mm)$, near the bottom of the chamber $(\sim \rm cm)$ and dynamic in that the alcohol vapour from the top is cooling as it drops and the air becomes supersaturated with alcohol vapour in that sensitive layer.
Once condensed the droplets continue on a downward path so you have a continuous transport of alcohol from the top to the bottom of the chamber.
You will need to wait for several minutes for a stable thermal gradient to be set up within the chamber and also for droplets to form on any dust which is within the chamber.
Do not use a hand warmer as this will introduce unwanted convection currents in the chamber.
I have seen a hot water bottle but initially try it without one.
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Why does work depend on distance? So the formula for work is$$
\left[\text{work}\right] ~=~ \left[\text{force}\right] \, \times \, \left[\text{distance}\right]
\,.
$$
I'm trying to get an understanding of how this represents energy.
If I'm in a vacuum, and I push a block with a force of $1 \, \mathrm{N},$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.
I must be missing something, but I can't really pinpoint what it is.
It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
| When a weight is sitting on the floor, the floor is applying a force to the weight (and vice versa), but no distance. And it should make intuitive sense that there's no work being done.
For your example of a weight in a vacuum: if you push it with force 1N for distance 1m, and then stop pushing, it will move forever at constant speed. If you push another block with force 1N for distance 2m, it will move forever at a higher constant speed. You did more work to it, so it has more kinetic energy.
| {
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what is effective index method which is used to solve modes in rectangular waveguide? I actually want to know the basics of effective index methods for solving the optical modes in slab/rectangular waveguides.
| The effective index method is an analytical method applicable to complicated waveguides such as ridge waveguides and diffused waveguides.
The ridge waveguide, such as shown in Figure Below, is difficult to analyze with simple method since the waveguide structure is too complicated to deal with by the division of waveguide. In order to analyze ridge waveguides, you should use numerical methods, such as the finite element method and finite difference method.
The wave equation for $E_x$ is given by:
$$\frac{\partial^2H_y}{\partial x^2}+\frac{\partial^2 H_y}{\partial y^2}+[k^2n^2(x, y)-\beta^2]H_y=0$$
The basic assumption of the effective index method is that the electromagnetic field can be expressed, with the separation of variables, as:
$$H_y(x,y)=X(x)Y(y)$$
Substituting this above and divided by XY
$$\frac{1}{X}\frac{d^2 X}{dx^2}+\frac{1}{Y}\frac{d^2 Y}{dy^2}+[k^2n^2(x, y)-\beta^2]=0$$
Here we add to and subtract from the equation above the independent value of $k^2n_{eff}^2$
and separate the equation into two independent equations:
$$\frac{1}{Y}\frac{d^2 Y}{dy^2}+[k^2n^2(x, y)-k^2n_{eff}^2(x)=0$$$$\frac{1}{X}\frac{d^2X}{dx^2}+[k^2n_{eff}^2(x)-\beta^2]=0$$
Now we have two equation with one variable that can be applied to x and y boundary and in this way discontinuity of refractive index can be treated simpler.
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What to do with an extra index in the definition of a tensor? I came across this definition of a tensor while reading some vector calculus literature
This definition contains the index $\ell$ in the last term, however the tensor itself only depends on $j$ and $k$. What am I supposed to do with this extra index? Do I sum over it?
Note: The term $\frac{\partial \phi_k}{\partial x_j}$ is irrelevant here; it is just another long expression in terms of the position vectors, although it also contains an $\ell$ within it.
| Yes, generally these sorts of expressions follow the Einstein summation convention. This says that whenever you see an index repeated in a multiplication expression, it means to implicitly sum over that index. So $(x_a - X_a)$ is a vector subtraction, but $x_a X_a$ is an inner product.
This is then made a little bit more rigorous either for skewed coordinate systems, or non-flat geometries like minkowski space. There you define a vector space with upper indices, and a covector space with lower indices. Covectors take a vector and produce a scalar, so whenever you see you the same the symbol for the upper and lower index, you know that they are applying a covector to a vector to create a scalar.
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Why doesn't a charged particle moving with constant velocity produce electromagnetic waves? A charged particle moving with an acceleration produces electromagnetic waves. Why doesn't a charged particle moving with a constant velocity produce electromagnetic waves? As far I understand, the electric and magnetic fields in space will still be time-dependent, if a charged particle is moving with constant velocity, so they could have given rise to electromagnetic waves, but they don't.
Also, why do accelerating charged particles produce electromagnetic waves? What is Nature's intention behind this phenomena?
| Okay, I'll try with a poor but "intuitive" explanation.
According to relativity theories, "it is impossible to tell if you're at rest or moving with constant velocity".
We know that a charge at rest does not emit any wave.
If you were moving at constant velocity and you saw a static charge emitting a wave, you'd think "this charge is not actually static because it emitting waves, so I'm seing static because I'm moving with the same velocity as it, so I am not at rest".
That would violate one of the most basic principles of physics: you cannot tell if the train is moving forward, or the landscape is moving backwards, provided that $\vec{v}$ is constant.
Check that the opposite would lead to the existence of "priviledged observers", or "observers who are at absolute rest". This doesn't make sense.
So we must discard the idea of charges moving at constant velocity emitting waves. It must be accelerated charges, only because there isn't any other option.
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Why isn't Rayleigh scattering a concern in analog communication? The Rayleigh scattering effect applies to 'light' signals, and the scattering of a signal when passing through a material medium, the amount of light scattered is proportional to 1/$\lambda^4$. I am curious as to why(or if) this wasn't considered when we established global communication systems through the various means of modulation. The high-frequency carrier/modulated signal used in say, FM, should visibly have a small wavelength, and since it is an EM wave too(like visible light or others) shouldn't we be concerned about its scattering as much as we are concerned about the consumed bandwidth or SNR etc. ?
I am not sure if I should be posting this in the Stack PE, I request the moderators to kindly direct me. Tag/post edits are welcome. Thank you!
| Rayleigh scattering affects light passing through air, and the light wavelengths range from ~650 nm (red) to ~420 nm (blue) compared to the sizes of the gas molecules in air of ~$4 \unicode{x212B}$ (0.4 nm). That is, the air molecules are about 1/1000 of the wavelength of blue light, which is much strongly scattered than red light.
As red light is quite weakly scattered, and has a wavelength about $1.5 \times$ that of blue light, radio communications signals with wavelengths ranging from metres down to centimetres are barely going to be scattered since those wavelengths are $100,000 \times$ (or more) the wavelength of blue light.
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Why stress always flows through the shorter/stiffer path? Why load has a preference at which way to act? What is the reason behind this preference?
Any link or comment would be welcome
| The majority of the load tends to be carried by a stiffer member. Stress is load per area, so if the stiffer member has a larger cross-sectional area then the stress is not necessarily higher.
As to why... imagine two springs hanging in parallel from the ceiling. Both have an unloaded length of 1 meter. Both are attached to the same point on the handle of a 25kg suitcase.
One spring is stiff. It takes a 20kg mass to extend it 10cm.
The other spring is weak. It only takes a 5kg mass to extend it 10cm.
The 25kg mass will extend the two springs by 10cm. The stiff spring takes 4 times more load than the weak spring.
In summary, stiffness reflects the ratio of load over deflection. If the deflection of two members is the same, then the stiffer one takes a higher load by definition.
| {
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Ray sun heating I recently got a curious question about the sun rays. Is there a material that can absorbs most of them and turn it into pure heat?
For example, we all know that dark material (black t-shirts) get much hot than white material, Although, I've heard that there is some metals that can get much hotter from direct sunlight.
I would like to know, which one can get much hotter from the sun?
| The sun emitts a whole spectrum of wavelengths, so what we'd like in this case would be a material that has a low reflectivity for a large bandwidth. Black t-shirts absorp more heat because they simply absorp the more wavelengths, including most of the visual ones, and that's why they appear black. As we know, white light is a combination of the wavelengths in the visual spectrum so therefore we can conclude that a white t-shirt is reflecting alot of wavelengths (in the visual spectrum). The infrared spectrum (typically heat) is close to the visual spectrum.
The ideal situation for this would be the model of a blackbody. A blackbody is a body that absorbs incoming radiation at a factor $\alpha$ that's between 0 (no absorption) and 1 (full absorption) and then emitts radiation as well with a factor $\epsilon$ between 0 and 1. For you to have a material that retains alot of heat from incoming radiation you'd need a material with a high absorption factor. You can look up example values for a number of materials in the following link https://www.engineeringtoolbox.com/solar-radiation-absorbed-materials-d_1568.html
I hope this answered your question.
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Velocity of separation and velocity of approach Can I have a clear cut dimension or difference between velocity of approach and velocity of separation?
In just simple 1D motion.
Considering two rigid spherical masses of different masses and moving with different velocities.
| I have not seen any formal definition, but this might work.
Let's assume that two bodies, left and right, move along $x$, with positive direction to the right.
Let's define the velocity of approach as positive and the velocity of separation as negative.
Then, the velocity of approach or separation could be determine as $\vec v=\vec v_{left}-\vec v_{right}$.
For instance, if the bodies move toward each other, the velocity of the left body will be positive, the velocity of the right body will be negative, so the difference will be positive, showing that the bodies are approaching each other.
Another example, when the left body is moving to the left at $1$m/s, while the right body is moving to the left at $3$m/s. The difference will be $-1-(-3)=+2$, indicating the the bodies are approaching at $2$m/s.
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What do the points on a Dalitz plot represent I have read that a Dalitz plot is nothing but the plot of $m_{12}^2$ vs $m_{23}^2$, and that the dots correspond to "events". However, this doesn't really tell me anything. My question can boil down to exactly what phase space variable the dots represent. I am currently trying to make a Dalitz plot for the three body decay $K_{long}\to 3\pi^0$. Now, I am able to get the boundary for the Dalitz plot by plotting the constraint
$$\cos(\theta_{12})=\frac{\frac{\left({m_\ell}^2+{m_\pi}^2-{m_{23}^2}\right) \left(-2 {m_\pi}^2+{m_{12}^2}+{m_{23}^2}\right)}{2 {m_\ell}^2}+2 {m_\pi}^2-{m_{12}^2}}{\sqrt{\frac{\left({m_\ell}^2+{m_\pi}^2-{m_{23}^2}\right)^2}{4 {m_\ell}^2}-{m_\pi}^2} \sqrt{\frac{\left(-2 {m_\pi}^2+{m_{12}^2}+{m_{23}^2}\right)^2}{{m_\ell}^2}-4 {m_\pi}^2}} $$
in the $m_{12}^2$ $m_{23}^2$ plane. However, I'm in the dark as to how to populate the inner region. My goal is to be able to answer the following question. Given the constraint above, how do I, for a given ($m_{12}^2$ ,$m_{23}^2$) know whether or not that corresponds to placing a dot there or not.
| Dots on a Dalitz plot should be datapoints (I say "should" because many are drawn with the "SCAT" option in ROOT, which actually fills bins with a number of randomly-placed dots proportional to the bin content). These can come from real or simulated data. Even if you want to draw the functional form of the decay rate (say from the result of an amplitude fit or an amplitude calculated from theory) it's usually easier computationally to generate accept/reject toys than to calculate its value on a fine grid of coordinates.
| {
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Simple question about change of coordinates Suppose we have two coordinate systems (Cartesian and spherical)
$$x^{\mu} = (t,x,y,z)$$
$$x'^{\mu'} = (t',r,\theta,\phi)$$
where $r= \sqrt{x^2 + y^2 + z^2} , \theta = \cos^{-1}(z/r), \phi = \tan^{-1} (y/x)$. My question is, in general, what are the components of a vector $A_{\mu} = (A_t,A_x,A_y,A_z)_{\mu}$ in the primed coordinates? From GR, I believe the answer is $A'_{\mu'} = (A_{t'},A_{r},A_{\theta},A_{\phi})_{\mu'} = \frac{\partial x^{\mu}}{\partial x'^{\mu'}} A_{\mu}$, with the inverse matrix used for upper-index vectors.
If this is the case, in particular it should work for position vectors. That is, $x'^{\mu'} = \frac{\partial x'^{\mu'}}{\partial x^{\mu}} x^{\mu}$. However, applying this transformation gives $x'^{\mu'} = (t',r,0,0)$, not $(t',r,\theta,\phi)$. Am I doing something wrong?
Edit: The second paragraph incorrectly applies the formula I've cited, as pointed out by mike stone.
As for the first question, since we have $x'_r = \sqrt{x_1^2 +x_2^2 + x_3^2}, x'_{\theta} =\cos^{-1}(x_3/x'_r)$,$x'_{\phi} = \tan^{-1}(x_2/ x_1)$, does it follow for any vector $A'_{\mu}$ (for instance, the EM gauge field) that $A'_r = \sqrt{A_1^2 + A_2^2 + A_3^2}$, $A'_{\theta} = \cos^{-1}(A_3/ A'_r)$, and $A'_{\phi} = \tan^{-1}(A_2/A_1)$?
| The GR vector transformation you cite applies to elements of the tangent space at a point. Positions are not vectors in any tangent space, so coordinates $x^\mu$ do not transform as vectors.
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Under what circumstances are general relativistic coordinate transformations physically meaningful? Although the field equations of GR are covariant under arbitrary coordinate transformations, such as the transformation given by Dirac (in Princeton Landmarks pp 34) that eliminate the singularities in the Schwartzschild metric, is it necessarily the case that the new coordinate system, with its singularity free metric, is physically meaningful?
This question comes up because there are transformations that are clearly meaningless. For example, if an observer makes a transformation from his coordinate system to that of his virtual image seen in a mirror, general covariance is preserved, but there is no actual space that can be entered by an observer.
Is there a criterion for deciding when the transformation is strictly virtual in the sense above?
Does the world inside a black hole as described, by say Kruskal coordinates, have the operational meaning of consisting of a world that can be entered and explored by an observer, or is it a strictly virtual construct like the mirror image coordinates?
| As an example of a possible virtual coordinate transformation consider the Schwarzschild metric for r less that the Schwarzschild radius. Then make the space-time switching transformation indicated to get the inside metric:
$$ds^2=\Big(1-\frac{r_s}{r}\Big)\ c^2dt^2-\Big(1-\frac{r_s}{r}\Big)^{-1}dr^2$$$$d\rho = cdt$$ $$dr =cd\tau$$
$$ds^2=\Big(\frac{r_s}{c\tau}-1\Big)^{-1}\ c^2d\tau^2-\Big(\frac{r_s}{c\tau}-1\Big)d\rho^2$$.
$$(dr/dt)(d\rho/d\tau) = c^2$$
Notice that the speed of the same particle in both coordinate systems is related by an inversion transformation using a sphere of radius c.
This means that if the speed of a particle in the r,t coordinates must be less than c, then in violation of relativity, it must be greater than c w/rspt to the $\rho, \tau$ coordinates
Conversly, if $d\rho/d\tau$ is less than c, then $dr/dt$ must be greater.
This seems to imply that the transformation must be virtual because it inevitably leads to a contradiction of relativity. A similar problem can be shown to exist using Kruskal coordinates
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/430180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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