newtonian mechanics - Gravitational potential energy of mass between two planets

Suppose I want to launch a rocket from earth to some point $O$ between the center of earth and the center of moon (on a straight line connecting their centers), where the gravitational force of the moon 'cancels out' the gravitational force of the earth (this point is located at $\approx 54 R_E$ from the center of earth where $R_E$ is the radius of earth). I want to know how much energy I should spend in order for the rocket to get there (neglecting the atmosphere and the rotation of the earth around its axis). So, I know that this is basically the difference between the potential energy at the start point and at the end point of the destination. However, $O$ is located not only in the gravitational field of the earth, but also in the gravitational field of the moon. And it seems that I cannot neglect the gravitational potential energy of the body at the moon's gravitational field. So my question is - how can I combine these two? How can I calculate the total GPE of the body in two (or even more) intersecting gravitational fields?

Answer

Gravitational Potential is a scalar quantity so can be added algebraically directly for both(or more) bodies.

Also GPE is just Gravitational potential times mass. $$E=\underbrace{\big(\sum P\big)}_{\text{due to all bodies in vicinity}}\times m$$

Now , rest of your aproach is allright ! Continue using this.

lagrangian formalism - Use partial or covariant derivatives when deriving equations of a field theory?

I feel like this question has been asked before but I can't find it. would the Euler Lagrange equation for, say, the standard model Lagrangian be $$\frac{\partial L}{\partial \phi}=\partial_\mu \frac{\partial L}{\partial (\partial_\mu \phi)}$$ Where $\phi$ is whatever field is an question and $\mu$ is (I believe) being summed from 0 to 3. Or, ist the correct equation $$\frac{\partial L}{\partial \phi}=D_\mu \frac{\partial L}{\partial (D_\mu \phi)}$$ Where $D$ is the covariant derivative of the theory. My intuition tells me its the second eqn but i just wanted to be sure, and I think I once saw someone say that the two were equivalent.

homework and exercises - How to show that the charge conjugation reverses the charge of a state?

How to show that the charge conjugation operator reverses the charge(s) of a (fermionic or bosonic) state?

Let us consider a spin-$\frac{1}{2}$ fermionic state of momentum $\textbf{k}$ and spin projection $s$ given by $a_s^\dagger(\textbf{k})|0\rangle$. The normal ordered charge operator associated with the symmetry $\psi\to e^{iq}\psi$ is given by $$Q=q\int d^3\textbf{p}\sum\limits_{s=1,2}\Big[a_s^\dagger(\textbf{p})a_s(\textbf{p})-b_s^\dagger(\textbf{p})b_s(\textbf{p})\Big]$$ where $a^\dagger, b^\dagger$ are respectively the particle and antiparticle creation operators. It is a trivial matter to show that $$Qa_s^\dagger(\textbf{k})|0\rangle=qa_s^\dagger(\textbf{k})|0\rangle~~ \text{and}~~ Qb_s^\dagger(\textbf{k})|0\rangle=-qb_s^\dagger(\textbf{k})|0\rangle.$$

Now, the charge conjugation operator is given by $$C=i\gamma^2\gamma^0.$$ I want to show that $$Q(Ca_s^\dagger(\textbf{k})|0\rangle)=-q(Ca_s^\dagger(\textbf{k})|0\rangle)$$ or something like $$Q(Ca_s^\dagger(\textbf{k})|0\rangle)=b_s^\dagger(\textbf{k})|0\rangle.$$

Am I on the right track? If yes, any hint how to act C on the state $a_s^\dagger(\textbf{k})|0\rangle$?

Answer

Yes, you are on the right track.

The rest of this answer is about the distinction between the charge conjugation operator $C_{op}$ and the charge conjugation matrix $C_{mat}$. If this misses the point of the question, let me know and I'll revise (or delete) the answer.

The field operator $\psi$ is an array of operators — a Dirac spinor whose four components each act as an individual operator on the Hilbert space. A Dirac matrix $\gamma^\mu$ does not act on the Hilbert space; it only mixes the components of the field operator $\psi$. When we write an expression like $\gamma^0\psi|0\rangle$, we are mixing matrix notation (it is a column matrix in which each component is a whole state-vector) and operator notation (because each component of $\gamma^0\psi$ is a different operator acting on $|0\rangle$).

The charge conjugation operator $C_{op}$ exchanges states of the form $$ \int d^3x\ \sum_n f_n(x)\psi_n(x)|0\rangle \tag{1a} $$ with states of the form $$ \int d^3x\ \sum_n g_n(x)\psi^*_n(x)|0\rangle, \tag{1a} $$ where $f$ and $g$ are ordinary functions with the same number of components as $\psi$, and where $\psi^*_n$ denotes the operator adjoint of $\psi_n$. (If the components of $\psi$ are $\psi_n$, then the components of $\psi^\dagger$ are $\psi^*_n$.) We want $C_{op}$ to exchange these because $\psi$ has components that create antiparticles (and components that annihilate particles), and $\psi^\dagger$ has components that create particles (and components that annihilate antiparticles).

The key point is that the charge conjugation matrix $C_{mat}$ is used to define the charge conjugation operator like this: $$ C_{op}\psi C_{op}^{-1} \sim C_{mat}\psi^* \tag{2} $$ where $\psi^*$ denotes the component-wise operator adjoint of $\psi$, without any matrix transpose. Again, this equation mixes matrix notation and operator notation. On the left-hand side, $\psi$ is a column-matrix, and the operator transformation $C_{op}\cdots C_{op}^{-1}$ acts on each individual element of this matrix. On the right-hand side, $\psi^*$ is again a column matrix, the transpose of the row matrix $\psi^\dagger$, and $C_{mat}$ is a square matrix of ordinary numbers; this matrix product mixes the components of $\psi^*$ with each other.

The matrix $C_{mat}$ can be expressed as some combination of Dirac matrices (as shown in the question), but an explicit expression for the operator $C_{op}$ requires more. We could presumably express it as some combination of field operators (or creation and annihilation operators), but usually we just define it through its effect on the field operators, as in equation (2). After writing $\psi$ in terms of creation/annihilation operators, equation (2) tells us how $C_{op}$ affects them, too.

The matrix $C_{mat}$ should be chosen so that the effect of $C_{op}$ on the current $q\overline{\psi}\gamma^\mu\psi$ is equivalent to reversing the sign of the charge $q$, because this is the current that couples to $A_\mu$ in the Lagrangian. According to this requirement, different representations for the $\gamma$-matrices lead to different representations for the matrix $C_{mat}$. One common representation leads to $C_{mat}\propto\gamma^0\gamma^2$, as stated in the question.

When the matrix $C_{mat}$ is chosen to satisfy that requirement, the operator $C_{op}$ will do something like $$ C_{op}a^\dagger C_{op}^{-1} \sim b^\dagger \hskip2cm C_{op}a C_{op}^{-1} \sim b $$ as anticipated by the question. Then, since the vacuum is invariant under $C_{op}$, we have $$ C_{op}a^\dagger C_{op}^{-1}|0\rangle = C_{op}a^\dagger |0\rangle \sim b^\dagger |0\rangle. $$ The fact that $C_{op}$ reverses the charge of the state is guaranteed by the fact that it reverses the sign of the charge density operator: $$ C_{op}QC_{op}^{-1}=-Q \hskip1cm \text{with } Q\sim q\overline{\psi}\gamma^0\psi = q\psi^\dagger\psi\sim q(a^\dagger a - b^\dagger b). $$ I glossed over a lot of details in this answer. The main point is that $C_{op}$ and $C_{mat}$ are two different things — related, but different.

thermodynamics - Why doesn't hydrogen gas exist in Earth's atmosphere?

The root mean square velocity of hydrogen gas at room temperature is:
Gas constant: $R=8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}$
Molar mass of hydrogen gas: $M=2.02\times10^{-3}\ \mathrm{kg/mol}$
$$\begin{align} v &= \left(\frac{3\times8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}\times300}{2.02\times10^{-3}\ \mathrm{kg/mol}}\right)^{\frac12}\\ &= 3356.8377\ \mathrm{m/s}\\ &= 3.356\ \mathrm{km/s} \end{align}$$
The escape speed of Earth is $11.2\ \mathrm{km/s}$, which is larger than the root mean square velocity of hydrogen gas. But still, hydrogen gas doesn't exist in Earth's atmosphere. Why? Have I made any mistakes in my calculations?

electromagnetism - What's the core difference between the electric and magnetic forces?

I require only a simple answer. One sentence is enough... (It's for high school physics)

electromagnetism - Is a suit that hides a soldier's heat signature fundamentally possible?

I recently played "Crysis", a game where the protagonist wears a suit that allows the player to hide both himself and his heat signature. Then I watched Iron Man 3, where a kid suggests that Tony Stark should have implemented retro reflection panels in his suit.

So I'm thinking, well, as is the nature of things, people are going to be pursuing this sort of thing in real life too, pretty soon. But I'm trying to figure out whether a suit can contain a person's heat signature without emitting the heat somewhere. Is such a thing fundamentally possible to do, without over-heating the person within?

Answer

There are some nice ideas in other answers, but they are overseeing some basics. Let's do some thermodynamics. The efficiency of a thermal engine is bounded by the Carnot efficiency:

$$\eta \le 1 - \frac{T_c}{T_h} $$

Where $T_c$ is the temperature of the cold end and $T_h$ the heat source. Assuming we are in a cool environment, $T_c=0 C$, $T_h=37 C$, so:

$$\eta_C \simeq 12\%$$

Whatever you do to convert your body heat into something useful, you will still have to pump out 88% of what you produce. And this is a very optimistic estimation, given that you have to make it portable. Way to go.

Let's look at the heat sink option. A human body has a surface of roughly $2 \mathrm{m}^2$, so it radiates about $10^3 \mathrm{W}$, but we are getting in around $600 \mathrm{W}$ from our surroundings (again, assuming a temperature of 0 C). If we leave base with a heatsink of iron cooled with liquid nitrogen (76 K) we have:

$$\frac{c_e \Delta T}{P}=\frac{449\frac{J}{K kg} 234 K}{400 W} = 262 \mathrm{\frac{s}{kg}} \simeq 4 \mathrm{\frac{min}{kg}}$$

So if we manage to redirect all our excess heat to the sink we will finish it in a few minutes.* A full hour of invisibility would require loading an extra 15 kg of weight. Other metals have a better specific heat ratio, but expect no wonders here.

How to actually become IR invisible

In order to become truly invisible, what we need is to dissipate all our heat via conduction. The way I would do it (but I am not an expert), is a multi layered system, where the inner layers are designed to give away heat, while the outer layers have to allow for ventilation while shielding the IR emited by the dissipation layer. The peak emission for a human body is in the range of $ 9 \mu m$, so a metallic mesh of that width should be a pretty good shield.

Now, you "just" have to figure out how to build a cloth with that metal embedded, while being able to endure hard conditions. But that is mere Engineering, not Physics.

In this scenario, all our heat is being carried away by the air around us, but that will not create a strong signal, as it is not dense. Also, turbulence will quickly mix it with the surrounding air, cooling it instantly.

---

*- The specific heat varies with temperature, but as a ballpark it works.

nuclear physics - Age of the Earth and the star that preceded the Sun

One of the great unheralded advances made in the history of science was the ability to determine the age of Earth based on the decay of isotopic uranium. Based on the apparent abundance of uranium in the early Earth, what conclusions can be drawn about the star that preceded the Sun?

Answer

Dear Humble, the cosmic nucleosynthesis (first three minutes) only produced hydrogen, helium, lithium, and beryllium. All heavier elements came from slow fusion inside living stars and, especially the heaviest ones, originate from dying stars.

Regular fusion-related processes inside stars (see the list at the bottom of the answer) only produce elements up to $Z=70$ or so: I included the periodic table for your convenience. The even heavier elements, especially gold and uranium, were produced by three extra processes

• s-process


• r-process


• rp-process

The s-process depends on the existence of elements in the iron group. An extra neutron may be absorbed (probably coming from reactions inside red giants), increasing $A$ by one, and if an unstable element is produced in this way, a neutron in the nucleus beta-decays by emitting an electron. Additional neutrons may be absorbed and the process may continue. Nuclei in the "valley of beta stability" can be produced in this way. "S" stands for "slow".

On the contrary, the r-process is "rapid". The neutrons are absorbed in a similar way but in the cores of supernovae. The seed nucleus is usually Ni-56. The rp-process, which may occur in the neutron stars and elsewhere, is also rapid, "r", but the particle is that absorbed is a proton, therefore "p" in "rp". Logically, unlike the previous two, it produces nuclei on the "proton-rich side" of the stable valley.

The uranium we observe on the Earth probably comes from all these processes - and from many stars - there is arguably no "the star" that preceded the Sun. In particular, our Earth hasn't orbited any other star before the Sun because it is as old as the Sun, at least this is what is believed. The hydrogen used by our Sun couldn't have been "recycled" and it began to burn soon after the sufficient collapse - it couldn't have been recycled from elsewhere. The heavier elements were recycled from many places. There was probably no "permanent region" that was inheriting the brand "Solar System". These issues were discussed yesterday:

How many times has the stuff of the Sun been recycled? How many times has the "stuff" in our solar system been recycled from previous stars?

Let me mention that it's not a problem for the heavy material to spread across large distances of the Cosmos. For example, an exploding supernova shoots most or all the matter by the speed 1% of the speed of light. So in 400 years, the material from the Sun - if it went supernova (it won't) - reaches Proxima Centauri and in less than 100 million years, it may reach almost any point in the Milky Way. Even the Solar System is moving at speed of 0.1% of the speed of light which is enough to move matter by light years in thousands of years. It's silly to imagine that the material had to wait on the same place from an "ancestor star", being saved for some humans on some Earth.

It may be useful to list all processes of stellar nucleosynethesis, not only those linked to uranium:

pp-chain / CNO cycle / α process / Triple-α / Carbon burning / Ne burning / O burning / Si burning / R-process / S-process / P-process / Rp-process

rotational dynamics - Angular momentum and Rotation

Is Rotation a necessary condition for angular momentum? I mean can two bodies under translational motion in particular directions have a total angular momentum that is not zero?

Answer

You do not have to have rotation to have angular momentum. And you do not have to have two bodies.

One body moving with constant speed in a straight line has nonzero angular momentum around any point not on that line!

In Newtonian physics, angular momentum is $\mathbf{r}\times\mathbf{p}$. It is much more general than something involving circular orbits or rotating objects. And it is just as important as the momentum $\mathbf{p}$ because it is conserved just like $\mathbf{p}$ is.

sun - Could a mirror array used for thermal solar power plant double as a telescope at night?

I was looking at the Ivanpah Solar facility and it occurred to me that the large array of mirrors could double as some sort of telescope array at night. The climate in the desert would be ideal for this, and there would be no power generated at night anyway.

Answer

Any reasonably flat piece of sort-of reflective metal will function perfectly well as a heat collector, but would not be terribly suited to do astronomical observations with.

In principle, you could probably pull it off.

But it would require a lot more accurately shaped mirrors, with a lot better quality reflective surface. Also, there's good reasons many large telescopes have domes around them, and are located on the tops of high mountains. Also, more complex electronics would need to be attached, different collectors need to be built, data must be transferred to inhabited parts of the world, etc. All this drastically increases the cost of the array, which usually already is one of the main prohibitive factors in getting such an array off the ground in the first place.

You could compromise and settle for a telescope that produces very low quality images in far infrared, but with awesome light gathering power. Perhaps it is even possible to use some smart computational techniques to combine a couple of garbage images into a single high-quality one, and still do reasonable science with it. It's at least worth a PhD research.

So in short, yes, I think it's possible. But I expect you can build a good solar array and a good large telescope for less money than this combination.

electromagnetism - What is the quantum state of a static electric field?

This is something that I've been curious about for some time. A coherent, monochromatic electromagnetic wave is well described by a coherent state $|\alpha\rangle$. The quantum treatment of the interaction between the field and matter then reduces at mean-field level (i.e. neglecting fluctuations) to the usual description of a classical external field acting on quantum matter, so long as $\alpha\gg 1$.

I want to know: does there exist a similar quantum state description for a DC field? For example, the electric field in between two capacitor plates. The expectation values of the field operators in such a state should of course reproduce the classical field strength. If this state (which may not be a pure state) cannot be written down, then I would be curious to know why.

(Feel free to consider, say, a bosonic scalar field rather than vector fields if that makes things simpler.)

Answer

At first glance, what you are describing sounds a lot like squeezed coherent state. However, the more I think about it, what you need is to act the displacement operator $D(\alpha)$ on the coherent state and pick (the real part of) $\alpha$ such that the field fluctuates around some value $E_0$ rather than 0. The displacement operator is $$D(\alpha) = e^{\alpha a^\dagger - \alpha^* a},$$ which you can re-write in terms of the real and imaginary parts of $\alpha$ as $$D(\alpha) = e^{ \sqrt{2} \operatorname{Im} \alpha \, i Q + \sqrt{2} \operatorname{Re} \alpha\, i\frac{d}{dQ}}.$$ or $$D(\alpha) = e^{ -\frac{i q_0}{\hbar} P + \frac{i p_0}{\hbar} Q}.$$ As you know, $P$ is the generator of translations in $Q$ space, so the displacement operator translates the state in $PQ$ space (i.e., phase space)

The simplest way to see this is to consider the wave-function of a SHM in its ground state: $\psi(x) = A e^{- \frac{m\omega}{2\hbar}(x-x_0)^2}$. The value $x_0$ represents the point about which the oscillator oscillates, and is conventionally taken to be zero, because we pick our coordinate origin to coincide with the minimum of the potential $V(x)$. However, nothing in principle stops us from writing down such a state for which $x_0$ isn't zero. The state will just not be an energy eigenstate.

Edit:

The problem with the visual you are having of that coherent state fluctuating around zero is that you are using the free-field Hamiltonian. However, if you have capacitor plates with charges on them, then from the point of view of the EM field, you will have to places sources $A_0 J_0$ in your Lagrangian which will change your Hamiltonian. In that case, the minimum for the potential for the fields will no longer lie at zero, but at some other value.

neutrinos - Do small-angle coherent scattering experiments really see coherent effects over arbitrarily large distances?

After integrating over all possible outgoing angles, the total cross-section of coherent elastic scattering from a fixed target of characteristic length $L$ scales like $L^4$. Does this mean, given a beam of sufficiently small angular dispersion and detectors capable of sufficiently fine angular resolution, there are coherent effects over macroscopic distances arbitrarily longer than the wavelength of the beam?

Consider a target of $N$ identical atoms located at positions $\vec{x}_n$ (where $n = 1,\ldots, N$) which is bombarded by a flux of incoming particles, which we'll call neutrinos. Set $\hbar = 1$ and let $\vec{k} = 2\pi/\lambda$ be the momentum of incoming neutrino, where $\lambda$ is the de Broglie wavelength. Let $f(\theta)$ be the scattering amplitude for a neutrino on a single free atom so the differential cross section is \begin{align} \frac{d \sigma_0}{d\Omega} = \vert f(\theta) \vert^2 \end{align} for one atom. As is befitting neutrinos, assume that $f(\theta)$ is very tiny so that the Born approximation is applicable. In particular, we can ignore multiple scattering events.

For many scattering processes the cross section of the target $\sigma_\mathrm{T}$is just $N$ times the cross section of each atom: $N \sigma_0$. However, for elastic scattering of extremely low energy neutrinos from the relatively heavy nuclei in atoms, the very long wavelength of the neutrinos means the various nuclei contribute coherently to the cross section. When $\lambda \gg L$, where $L$ is the size of the target, one has $\sigma_\mathrm{T} = N^2 \sigma_0$ rather than $N \sigma_0$ [1]. For general $k$, the total cross section is [2] \begin{align} \frac{d \sigma_\mathrm{T}}{d\Omega} = \left\vert \sum_{n=1}^N f(\theta) \, e^{-i \vec{x}_n \cdot (\vec{k}-\vec{l})} \right\vert^2 \end{align} where $\vec{l}$ is the momentum of the outgoing neutrino [3] and $\theta$ is the angle between $\vec{k}$ and $\vec{l}$. Because it's elastic, $|\vec{l}|=|\vec{k}|=k$.

A property of equation (1) is that for $L \gg \lambda$ there is a large suppression of scattering in most directions because the phase in the exponential tends to cancel for the different atoms in the sum. The exception is when $\vec{l}$ is very close to $\vec{k}$ (i.e. low momentum transfer, very slight scattering), because then the phase of the exponent varies very slowly from atom to atom. This means that for large targets, the vast majority of the scattering is in the forward direction.

Now restrict to the case $A \lesssim \lambda \lesssim L$, where $A$ is the typical atomic spacing. What's initially confusing about this is that if we ask for the total cross section by integrating over $\hat{l}$, we find for large $L$ that [4] \begin{align} \sigma_\mathrm{T} = \int_\Omega d \hat{l} \frac{d \sigma_\mathrm{T}}{d\Omega} \sim \frac{N^2}{L^2 k^2} = \rho^{2/3} \lambda^2 N^{4/3} \end{align} where $\rho = N/L^3$ is the number density of the atoms. This means that, for fixed density and fixed neutrino momentum, the total cross section grows faster that than the number of atoms in the target---even over distance scales much larger than the neutrino wavelength. In this sense the coherent scattering effects is non-local over large distances.

The story I have been told is that this is resolved by incorporating the realities of a real-world detector. For any traditional experiment, there is always a minimum forward acceptance angle $\theta_0$ that it can detect. Particles which are scattered at smaller angles are indistinguishable from unscattered particles in the beam. Indeed, if we let $\tilde{\sigma}_\mathrm{T}$ be the detectable cross section scattered at angles greater than $\theta_0$ for any fixed $\theta_0>0$, we find \begin{align} \tilde{\sigma}_\mathrm{T} = \int_{\theta > \theta_0} d \hat{l} \frac{d \sigma_\mathrm{T}}{d\Omega} \sim \frac{N^2}{L^4 k^4} = \rho^{4/3} \lambda^4 N^{2/3}. \end{align} This accords with our intuition. Growing like $N^{2/3}$ is the same as growing like $L^2$, i.e. there is complete cancellation in the bulk of the target, and the only significant scattering is from the surface (which scales like $L^2$).

Is this all there is to say? Can we potentially see scattering of particles (with atomic-scale wavelength) which demonstrates coherent contributions from target atoms separated by meters? Are there any other limiting factors besides a finite mean free path of the incoming particle (which breaks the Born approximation) and the angular resolution of the detector?

For positive answer, I would accept (a) anything that points to a reputable source (textbook or journal article) which explicitly discusses the possibility of coherent effects over arbitrarily large distances or (b) an argument which significantly improves on the one I have made above. For negative answers, any conclusive argument would suffice.

It has been argued to me that the Born approximation is invalid for small momentum transfer $\vec{k}-\vec{l}$, because the approximation requires the energy associated with this transfer to be much larger than the potential (which cannot be the case for $\vec{k}-\vec{l}=0$). This seems to explicitly conflict with textbooks on the Born approximation which state things like

For any potential $V$ there is a $\bar{\lambda} > 0$ such that the Born series (9.5) converges to the correct on-shell $T$ matrix for all values $\vec{p}$ and $\vec{p}'$, provided $|\lambda| < \bar{\lambda}$.

["Scattering Theory: The Quantum Theory of Nonrelativistic Collisions" (1972) by John R. Taylor]

(Here, $\lambda$ is the coefficient of the expansion.)

Is there any validity to this objection?

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[1] For macroscopic $N$, this can be a stupendous boost. This was responsible for some (misplaced) optimism in the '70s and '80s that relic neutrinos might be detectable.

[2] This form also appears in many less exotic places than neutrino physics, like neutron and X-ray scattering.

[3] $d\Omega$ is the differential over the possible directions $\hat{l}$.

[4] This behavior is independent of the details of the geometry. The $1/3$'s come from the integrals over 3 spatial dimensions.

stability - Why is the orbital resonance of the Galilean moons stable?

It is well known that the orbits of Ganymede, Europa and Io are in a 4:2:1 resonance. Most online sources (including but not limited to Wikipedia) say that such an orbital resonance, along with the 3:2 resonance, is "stable and self-correcting", but fail to explain why this is so.

The textbook Fundamental Astronomy says that this phenomenon is due to "tidal forces" but does not elaborate further. I presume it refers to the tidal deceleration which causes the orbits of the moons to evolve outwards, which, other sources say, caused the moons to eventually enter into resonance, but this also does not explain why the resonance is stable.

I am aware of a similar question here but I'm more interested in stability rather than instability in this case.

In short, why is the orbital resonance of the Galilean moons stable, and how is it different from other cases of orbital resonance that are unstable? I don't mind (and would prefer) if the answer is mathematical in nature.

fourier transform - Converting from frequency domain to time domain

If I have an equation of a signal in the frequency domain given by

$L(x)=\frac{1}{\pi}\frac{\frac{1}{2}\Gamma}{((\omega-\omega_0)^2+(\frac{1}{2}\Gamma)^2)}$

and I want to convert this to the time domain, can I do this using the inverse Fourier transform? I think the result should be an exponential, but when I tried to calculate this on mathematica I got a strange result.

thermodynamics - Wasn't the Hawking Paradox solved by Einstein?

I just watched a BBC Horizon episode where they talked about the Hawking Paradox. They mentioned a controversy about information being lost but I couldn't get my head around this.

Black hole thermodynamics gives us the formula $$S ~=~ \frac{A k c^3 }{4 \hbar G}=4\pi k\dfrac{M^2}{M_P^2}.$$

where $M_P$ is the Planck mass. And we also have Einstein's famous $E = M c^2$, which mean that mass can be turned into energy, right? Hence information is either lost or it is preserved, but now in energy-form instead of mass-form.

I can't understand why radiation from black holes would be any different than an atomic bomb for example, where mass is also turned into energy?

Answer

Radiation normally contains subtle correlations. For all practical purposes you can't use it, but it's there. Hawking radiation is, according to the theory, perfectly thermal and does not contain any more information than the temperature itself. The problem is that then the process of black hole evaporation is not reversible, in principle. Unlike all other processes that we know of (which might be irreversible in practice, but are reversible in principle). That irreversibility (which implies non-unitarity) is incompatible with quantum mechanics. That is the problem in a nutshell. It is really a conflict between quantum mechanics and semi-classical general relativity.

There are many more things to be said but I get the impression you haven't done a lot of reading about this and details would be rather pointless. I suggest you browse around for a bit with that starting point in mind.

thermodynamics - Maxwell's Demon - laser cooling

There’s an interesting article on Scientific American that tells how to cool individual atoms to within a very tiny fraction of Absolute Zero.

It uses two laser beams acting on a very cold rarified gas in a magnetic trap. Initially, all of the atoms are in one state. The first laser beam (call it red) only affects atoms in a second stable state. A second laser (call it orange), offset and parallel to the first of a different color is turned on. This laser switches atoms to the previously mentioned second stable state when it hits an atom.

These excited atoms cannot pass the first laser beam (the excited atoms rebound from this beam). This leads to a concentration of the excited atoms on the side of the second beam. All of the atoms will eventually cross the two beams and move into the smaller area, being compressed without raising their temperature (supposedly). The lasers are then turned off, the gas is allowed to expand (and cool) into its original volume and you end up with a lower temperature gas.

I’ve left the following question there, but haven’t gotten an answer (I keep getting notices that the comments have been added to, but even when I clear my browser cache, or use a different browser, I still don’t see them). So here is my question. Can you help me understand?

electromagnetism - Electromagnetic waves in vacuum

If there is no source then also there is electromagnetic waves described by Maxwell equation. how if there is no source then existence of EM waves. What gives energy to this EM waves. Is it vacuum fluctuation or something else?

pressure - Effects of making a hole in a mercury barometer

In a mercury barometer, if we make a hole in the portion above the level of mercury, then the mercury level in the column drops and ultimately all the mercury in the column goes into the reservoir below or if it's capacity is not large enough it overflows.

The following image shows a different situation, when we make a hole below the level of mercury in the column:

The pressure at $A$ is zero and it gradually increases as we move down the column. The pressure at $C$ and $D$ are same as they are in the same horizontal level, and it is equal to the atmospheric pressure $P_{atm}$ From this reasoning, we can say that the pressure at $B$ is lesser than $P_{atm}$.

So, I concluded that the mercury doesn't come out through this hole.

But, what will happen after this step? The reason, why I got this doubt is described below:

We know from Pascal's Law that a pressure applied to a particular section of a fluid gets transmitted to all regions without any loss. Before we made the hole, the force exerted by the mercury column at $B$ was balanced by the contact force provided by the walls of the column. But, after we make a hole, the force exerted on the hole by the atmosphere is larger than the internal force. So, the pressure at all the points within the fluid must increase by $P_{atm}$.

Will the level of mercury in the column rise above $A$? Will it eventually touch the topmost part of the vertical column, as pressure at the topmost level of mercury must increase by $P_{atm}$ which was initially $0$? What will happen to the mercury in the reservoir? Will all the mercury in it will be sucked in by the column till the mercury level touches the upper part of the column? Doesn't this seem counterintuitive?

Kindly explain what will happen to this system. Is it correct to apply Pascal's Law in this way? Are my reasonings correct?

^{Image Courtesy: My Own Work :)}

Answer

The simple force diagram that we rely on to prove that we can measure air pressure with this device gets far more complicated when you start drilling holes.

Consider doing this with water. Water behaves reasonably intuitively. If you drill a hole in the end of a water barometer like this, what you expect will happen. Water will pour out of the hole, and then all the water will drain from the tube.

The reason for this has less to do with the top and bottom of the barometer and far more to do with the top and bottom of the hole. The pressure differences there are sufficient to permit a water/air exchange, letting air into the tube.

Mercury, however, has an extraordinarily high surface tension. It loves to attach to itself and cares much less to interact with air or the edges of the tube (for most tubing materials).

Now our model is more complex. What you will find is that the forces on the mercury near the hole are affected not just by air pressure but by the surface tension as well. The mercury will try to keep the surface area of the mercury/air interface as small as possible, preventing air from going in or mercury from leaking out.

The result? The barometer will probably remain intact. The surface tension effects will provide the additional forces needed which the air pressure does not, and you'll see no significant change.

This, of course, depends on the size of the hole. Make the hole too big, and the surface tension will no longer be able to hold the mercury in.

By the way, mercury spills are bad news. I highly recommend not drilling into mercury barometers. Its really nasty stuff and needs to be disposed of properly.

orbital motion - Calculating the eccentricity of an exoplanet

I'm wondering how to calculate the eccentricity of an exoplanet by its radial velocity vs. phase graph. To clarify my question I will take an exoplanet called WASP-14b 2 as an example (http://exoplanets.org/detail/WASP-14_b).

A plot of the radial velocity of the star vs the phase is displayed in the upper left corner. I am wondering how I could possibly calculate the eccentricity of the exoplanet using this graph (or some other values given in the original measurements). I found a few ways to calculate the eccentricity:

$$ e = \left | \mathrm{e} \right| $$

This uses the eccentricity vector which is calculated using this formula:

$$ \mathrm{e} = \frac{v \times h}{\mu}-\frac{r}{\left | r \right |} $$

The problem here is that this formula needs the specific angular momentum vector and the position vector, which I do not know given only the measurements. However, there is another way to calculate the eccentricity:

$$ e = 1 - \frac{2}{(r_a/r_p) + 1} $$

where $r_a$ is the radius of the apoapsis and $r_p$ the radius of the periaosis. These values are not known using only the measurements, but I believe it should be possible to calculate them by taking the integral of the sine function (radial velocity vs. phase). This would give me the position of the star at any given moment. The problem is that I cannot find the exact points displayed in the graph anywhere, let alone a sine function that would fit them.

When I do get an integral of the function I still have to create one for the planet itself, since this describes the movement of the star. I am able to calculate the mass of the planet using the following formula:

$$ r^3 = \frac{GM_{star}}{4\pi^2}P_{star}^2 $$

which gives me the distance between the star an the planet. Next I can calculate the velocity of the planet using:

$$ V_{PL} = \sqrt{GM_{star}/r} $$

And after that I can calculate the mass of the planet using this formula:

$$ M_{PL} = \frac{M_{star}V_{star}}{V_{PL}} $$

But this is where another problem comes up as a Wikipedia article on Doppler Spectroscopy states: "Observations of a real star would produce a similar graph, although eccentricity in the orbit will distort the curve and complicate the calculations below."

Where do I find the corrected calculations and how can I possibly calculate the eccentricity of this planet using only these values ($M_{star}$ and the plot, of which I cannot find the exact points)?

Additional sources: http://adsabs.harvard.edu/abs/2009MNRAS.392.1532J

Answer

There are a number of options if you want an off-the-shelf solution to fitting RV curves. Perhaps the best free one is Systemic Console.

However, it is not too hard to do something basic yourself.

First define some terms:

$\nu(t)$ is the true anomaly - the angle between the pericentre and the position of the body around its orbit, measured from the centre of mass focus of the ellipse.

$E(t)$ is the eccentric anomaly and is defined through the equation $$\tan \frac{E(t)}{2} = \left(\frac{1+e}{1-e}\right)^{-1/2} \tan \frac{\nu(t)}{2}$$

The mean anomaly $M(t)$ is given by $$M(t) = \frac{2\pi}{p}(t - \tau),$$ where $p$ is the orbital period and $\tau$ is the time of pericentre passage.

"Kepler's equation" tells us that $$M(t) = E(t) - e \sin E(t)$$

Finally, the radial velocity is given by $$V_r(t) = K\left[\cos(\omega + \nu(t)) +e \cos \omega \right] + \gamma,$$ where $K$ is the semi-amplitude, $\gamma$ is the centre of mass radial velocity and $\omega$ is the usual angle defining the argument of the pericentre measured from the ascending node.

OK, so the problem is that the radial velocity does not depend explicitly on $t$, but rather on $\nu$. So what you do is the following:

1. 
   

   Choose values for $K, \gamma$, $\omega$, $\tau$, $p$ and $e$; these are your "free parameters that describe the orbit. The closer you can get your initial guess, the better.

   
   



2. 
   

   You use these parameters to predict what the radial velocities would be at the times of observation of your RV datapoints. You do this by calculating $\nu(t)$ using the equations above. Start with the second equation and calculate $M(t)$. Then you have to solve the third equation to get $E(t)$. This is transcendental, so you have to use a Newton-Raphson method or something similar to find the solution. Once you have $E(t)$ then you use the first equation to find $\nu(t)$. Then use 4th equation to calculate $V_r(t)$ at each of your datapoint times.

   
   


3. 
   

   Calculate a chi-squared (or similar figure of merit) from comparing the predicted and measured values of $V_r(t)$.

   
   


4. 
   

   Iterate the values of the free parameters and go back to step 2. Continue till your fit converges.

   
   

quantum field theory - Charge conjugation in Dirac equation

According to Dirac equation we can write, \begin{equation} \left(i\gamma^\mu( \partial_\mu +ie A_\mu)- m \right)\psi(x,t) = 0 \end{equation} We seek an equation where $e\rightarrow -e $ and which relates to the new wave functions to $\psi(x,t)$ . Now taking the complex conjugate of this equation we get

\begin{equation} \left[-i(\gamma^\mu)^* \partial_\mu -e(\gamma^\mu)^* A_\mu - m \right] \psi^*(x,t) = 0 \end{equation} If we can identify a matrix U such that \begin{equation} \tilde{U} (\gamma^\mu)^* ( \tilde{U} )^{-1} = -\gamma^\mu \end{equation} where $ 1 =U^{-1} U$.

I want to know that, why and how did we do the last two equation. More precisely, I want to know more details and significance of the last two equations.

Answer

The Dirac equation for a particle with charge $e$ is $$ \left[\gamma^\mu (i\partial_\mu - e A_\mu) - m \right] \psi = 0 $$ We want to know if we can construct a spinor $\psi^c$ with the opposite charge from $\psi$. This would obey the equation $$ \left[\gamma^\mu (i\partial_\mu + e A_\mu) - m \right] \psi^c = 0 $$ If you know about gauge transformations $$ \psi \rightarrow \exp\left( i e \phi\right) \psi $$ (together with the compensating transformation for $A_\mu$, which we don't need here), this suggests that complex conjugation is the thing to do: $$ \psi^\star \rightarrow \exp\left( i (-e) \phi\right) \psi^\star $$ So it looks like $\psi^\star$ has the opposite charge. Let's take the complex conjugate of the Dirac equation: $$ \left[-\gamma^{\mu\star} (i\partial_\mu + e A_\mu) - m \right] \psi^\star = 0 $$ Unfortunately this isn't what we want. But remember that spinors and $\gamma$ matrices are only defined up to a change of basis $\psi \rightarrow S \psi$ and $\gamma^\mu \rightarrow S \gamma^\mu S^{-1}$. Possibly we can find a change of basis that brings the Dirac equation into the form we want. Introduce an invertible matrix $C$ by multiplying on the left and inserting $ 1 = C^{-1}C $ (note that $C$ is the more common notation for your $\tilde{U}$): $$ \begin{array}{lcl} 0 &= & C \left[-\gamma^{\mu\star} (i\partial_\mu + e A_\mu) - m \right] C^{-1} C\psi^\star \\ &= & \left[-C\gamma^{\mu\star}C^{-1} (i\partial_\mu + e A_\mu) - m \right] C\psi^\star \end{array}$$

Note that if we can find a $C$ which obeys $-C\gamma^{\mu\star}C^{-1} = \gamma^\mu$ then $C\psi^\star$ makes a perfectly good candidate for $\psi^c$! It turns out that one can indeed construct $C$ satisfying the condition and define charge conjugation as $$ \psi \rightarrow \psi^c = C\psi^\star $$

You can see this more explicitly in terms of two component spinors in the Weyl basis: $$ \psi = \left( \begin{matrix} \chi_\alpha \\ \eta^{\dagger}_{\dot{\alpha}} \end{matrix} \right) $$ (the notation follows the tome on the subject). The charge conjugate spinor in this representation is $$ \psi^c = \left( \begin{matrix} \eta_\alpha \\ \chi^{\dagger}_{\dot{\alpha}} \end{matrix} \right) $$ So charge conjugation is $$ \eta \leftrightarrow \chi $$ This representation explicitly brings out the two oppositely charged components of the Dirac spinor, $\eta$ and $\chi$, and shows that charge conjugation acts by swapping them.

To recap: we want to define a charge conjugation operation so that given a $\psi$ with some electric charge $e$, we can get a $\psi^c$ with charge $-e$. Complex conjugating the Dirac equation gets us there, but the resulting spinor $\psi^\star$ is in a different spinor basis so the Dirac equation is not in standard form. We introduce a change of basis $C$ to get the Dirac equation back in standard form. The necessary conditions for this to work are that $C$ is invertible (otherwise it wouldn't be a change of basis and bad things would happen) and $-C\gamma^{\mu\star}C^{-1} = \gamma^\mu$.

gravity - Why does my apple not weigh 500 tons?

Related to this question: What is potential energy truly?

$E=mc^2$ - energy equals mass. So, if an object has gravitational potential energy relative to another object, it should have additional mass to represent that extra energy.

So, if I have an apple in my hand, it should have quite a large amount of gravitational potential energy relative to a black hole at the edge of the observable universe. I did a quick calculation of how much this gravitational potential energy would be (using Newton's Law of Gravitation) and it came out to be the equivalent of around 500 tons (and that's just for one black hole!).

So, why does my apple not weigh a lot more than it appears to?

EDIT: Here's my calculation of the GPE:

Assumptions:

1. The apple has mass $m_a=0.3 kg$


2. The black hole has a mass equal to 1000 solar masses, $m_b=2\times10^{33} kg$


3. Distance between the apple and black hole is 60 billion light years

Obviously $GPE=mgh$ can't be used, because the gravitational acceleration is not constant and varies with distance, so it has to be integrated. I used Newton's Law of Gravitation to provide the acceleration:

$$ F=\frac{G{m_1}{m_2}}{h^2} $$

A small change in height produces a small change in energy, which can be used to integrate:

$$ \begin{eqnarray*} \delta E&=&F \delta h\\ \int_0^EdE&=&G{m_a}{m_b}\int_1^R\frac{1}{h^2}dh \end{eqnarray*} $$

I'm integrating the force starting at 1m from the center of the black hole, as the force goes to infinity at $h=0$. It's a little crude, but it should provide a conservative estimate.

$$ \begin{eqnarray*} E&=&G{m_a}{m_b}(1-\frac{1}{R})\\ E&\approx& G{m_a}{m_b}\;\;\;\;\;\;(R>>1)\\ G&=&6.673\times 10^{-11}\;\;N(\frac{m}{kg})^2\\ G{m_a}{m_b}&=&4\times 10^{22}\;\;Nm\\ E&=&mc^2\\ \Delta m&=&\frac{4\times 10^{22}}{(3\times 10^8)^2}\\ &\approx&444,000\;\;kg \end{eqnarray*} $$

So, a little short of 500 tons, but still pretty big!

Answer

You are reading the formula the other way round. With $$\tag{1}E = mc^2$$ we mean that the energy associated to the mass $m$ of a particle at rest is $E=mc^2$. In oher words, (1) strictly holds for a particle at rest, not interacting with anything else.

What about all the potential energy that a particle has? That must be included on the right hand side of (1). So the version of (1) including potential energy would be $$ \tag{2} E = mc^2 + E_g,$$ with $E_g$ denoting all the gravitational potential energy contributions that you want. So you can now replace $E_g$ with your calculated gravitational potential energy (and then of course all the gravitational energy contributions coming from the interaction of the apple with the rest of the universe... good luck with that).

Now it is crucial to understand that (as also pointed out by Phoenix in the comments) energy is always defined up to a constant, and what physically matters in the dynamic of a particle (or an apple) is the change in the energy between two states (say the apple at point A and the apple at point B). So $E_g$ can be as big as you want (it can be infinite for that matter), and that would not change anything in the dynamic of the apple. This is way it is not generally included, unless the gravitational energy plays a role in the calculations. You could think of a more general version of (1) as being something like $$ \tag{3} E = \sqrt{m^2c^4 + c^2p^2} + E_{gravitational} + E_{em} + E_{whatever} + C, $$ where $E_{em}$ keeps count of the electromagnetic interaction of the object with the rest of the universe (if there is one), $E_{whatever}$ keeps count of all the other possible interaction you can think of, and $C$ is a constant that you can always choose arbitrarily. Obviously, writing (1) like (3) would be in most cases useless, and hence only the relevant contributions are included.

So to conclude, the mass in $F=ma$ keeping count of relativistic corrections is just $$ \tag{4} m = \gamma m_0 \equiv \frac{m_0}{\sqrt{1-v^2/c^2}},$$ where $\gamma$ is the Lorentz factor, $m_0$ the rest mass used in (1), and $v$ the velocity of the particle. Einstein's formula is meant to tell you the amount of energy that you would obtain by completely disintegrating the mass of a particle, not as a way to calculate the inertial mass of an object.

protons - Is there any Violation of conservation of mass in positron emission?

In positron emission, a proton decays into a neutron, electron, and neutrino. Since the mass of a proton is less than that of a neutron, does that mean that energy is converted into mass in the reaction? Beta decay#β+ decay says that some of the binding energy goes into converting a proton into a neutron, but that's not really what I'm looking for. So, does the extra mass result from the conversion of energy into mass?

adiabatic - Speed of a process in thermodynamics

As per my understanding, all adiabatic processes carried out in absence of an adiabatic wall need to be fast enough so that there is no heat exchange with the surrounding. Similarly, all isothermal processes are slow so that thermal equilibrium can be established with the surroundings in order to maintain a constant temperature in the system.

But then since isothermal processes are slow, how are isothermal irreversible processes possible?

Is my notion that 'irreversible processes are fast' wrong?

Answer

There are some irreversible processes that can be slow. However, when we refer to an isothermal irreversible process, what we really mean is that the boundary of the system with its surroundings are maintained at a constant temperature during the process, even if the process is fast. So, although the temperature at the boundary is constant, the temperature distribution within the system is not spatially uniform, and irreversible heat conduction is occurring within the system. Also, typically in an isothermal irreversible process, the boundary temperature is assumed to be held at the initial uniform temperature of the system in its initial thermodynamic equilibrium state, and the temperature of the system again becomes uniform at the boundary temperature in the final equilibrium state of the system.

geometry - What is the physical meaning of a dot product and a cross product of vectors?

My teacher told me that Vectors are quantities that behave like Displacements. Seen this way, the triangle law of vector addition simply means that to reach point C from point A, going from A to B & then to C is equivalent to going from A to C directly.

But what is the meaning of a product of vectors? I cannot imagine how a product of displacements would look like in reality. Also, how do we know whether we need the scalar (dot) product or a vector (cross) product?

thermodynamics - Relation between heat capacities

$C_p$ here means specific heat at constant pressure and $C_V$ at constant volume. My book says that $C_p$ is "generally" greater that $C_V$ because at constant pressure a part of heat given maybe used for expansion whereas at constant volume all the added heat produces a rise in temperature. The term "generally" has been used because substances generally expand with increase of temperature at constant pressure but in a few exceptional cases there may be contraction. After a few pages the relation $C_p-C_V=R$ is derived I know $R=8.3$ so this means $C_p = C_V+R$ or $C_p>C_V$. so according to this relation $C_p$ is always greater than $C_V$ but the book claims that this is not always true!! How is it possible?? Why are the two statements contradicting each other??

Answer

The equation $C_p - C_v = R$ will have been derived for an ideal gas. For any other substance the relationship between $C_p$ and $C_v$ will be more complicated. However in the vast majority of cases materials expand as they get hotter so if the pressure is kept constant the material will do work as it expands. That means $C_p$ must be greater than $C_v$ even though the difference will no longer simply be $R$.

However materials do exist that have a negative thermal expansion coefficient i.e. the material contracts as it gets hotter. In this case if the pressure is kept constant the material will have work done on it and $C_p$ will be less than $C_v$. These materials are special cases and they are few and far between. Nevertheless such materials exit.

fluid dynamics - How does this spray gun work?

This is the critter spray gun.

The liquid in the jar is at atmospheric pressure (there's a vent hole connecting it to the atmosphere). A fast jet of air flowing right over the end of the pickup tube sucks the liquid up the tube and into the air stream. How exactly does this fast stream of air creates suction? It's released in the atmosphere, so shouldn't it also be at atmospheric pressure?

rotational kinematics - Can the direction of angular momentum and angular velocity differ?

While studying rotational mechanics, I came across a section where it mentioned that angular momentum may not necessarily be parallel to angular velocity. My thoughts were as follows:

Angular momentum ($L$) has the relation $L=I\omega$ where $\omega$ is angular velocity and $I$ is the moment of inertia, so following this relation, it seems they should be in the same direction. Why are they not?

Answer

Consider a thin rectangular block with width $w$, height $h$ resting along the xy plane as shown below.

The mass of the block is $m$. The mass moment of inertia (tensor) of the block about point A is

$$ {\bf I}_A = m \begin{vmatrix} \frac{h^2}{3} & -\frac{w h}{4} & 0 \\ -\frac{w h}{4} & \frac{w^2}{3} & 0 \\ 0 & 0 & \frac{w^2+h^2}{3} \end{vmatrix} $$

This was derived from the definition (as seen on https://physics.stackexchange.com/a/244969/392)

If this block is rotating along the x axis with a rotational velocity $$ \boldsymbol{\omega} = \begin{pmatrix} \Omega \\ 0 \\ 0 \end{pmatrix} $$ then the angular momentum about point A is

$${\bf L}_A = m \Omega\,\begin{pmatrix} \frac{h^2}{3} \\ -\frac{w h}{4} \\ 0 \end{pmatrix} $$

As you can see, there is a component of angular momentum in the y direction. The angular momentum vector forms an angle $\psi = -\tan^{-1} \left( \frac{3 w}{4 h} \right)$

In the figure below you see the direction of angular momentum, and the circle about which the center of mass is going to orbit due to precession.

conservation laws - Why current in series circuit is the same?

I have read in the internet that the charges do not have any other path to go and they must go through the same in a series circuit,hence the current is same.

It was quite convincing but what confused me was: "A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element. Resistors act to reduce current flow..."(according to the Wikipedia). This means that the resistors slow down the rate of flow of charges. By definition, electric current is the rate of flow of charges. Then must not the current be reduced in a resistor even when the amount of charge is same?

How was the definition and the formula of work derived? Is it the best possible?

Work done is defined as the dot product of force and displacement.

However, intuitively, should it not be the product of force and the time for which the body is acted upon by the force (force * time) because while time is independent of force applied, displacement is not.

Were these formulae (for work and energy) actually derived based on some physical understanding or are they just constructs to understand forces better?

Answer

Work done is defined as the dot product of force and displacement. ... should it not be the product of force and the time

Were these formulae (for work and energy) actually derived based on some physical understanding or are they just constructs to understand forces better?

Neither of the two. Most formulas and definitions have an historical motivation. The issue is too complicate to fully discuss it here. There are lots of definitions which are not 'rational' in science, as the first explanations of phenomena where based on misconceptions and some are the cornerstones on which a skyscraper has gradually been erected.

You surely know the names of the orbitals of an atom: s, p, d, f. They are derived from the description by early spectroscopists of certain series of alkali metal spectroscopic lines as s harp, p rincipal, d iffuse, and f undamental. It would be more simple and rational to call them in any other way: a, b, c, d, or 1, 2, 3, 4 or , even simpler, to identify them with $l$, the angular momentum quantum number: 0,1,2,3. It would be rather easy and unpainful to change these definitions (but also physicists are subject to the 'force of inertia': see the comment by John Rennie here), whereas it's extremely difficult to alter the fundamental definitions/ derivations, just like changing the left-hand traffic in England: you ought to change all road signs overnight and, what is worst, scrap all LHD vehicles.

You may find a full and detailed description of how energy was first discovered and neglected in my post here and feel proud that you have had the same ideas of Leibniz.

Leibniz had suggested the more logical, rational and natural integration on time. Now, as you surely know, that is not possible anymore since $F * t$ is defined as momentum.

Another historical reason/ justification related to this is that the concept of energy was understood very late, suffice it to say that, to date, it is not yet considered a fundamental concept and has no own unit: you probably know that the SI has seven base units and energy is not among them. It is really amazing and disconcerting: *the most important concept in the whole universe is a [derived unit]*(http://en.wikipedia.org/wiki/SI_derived_unit) derived in various ways from derived units.

The term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis as "weight lifted through a height", which is based on the use of early steam engines to lift buckets of water out of flooded ore mines. The SI unit of work is the newton-metre or joule (J). (wiki)

Another reason is that the definition was modelled upon the force of gravity, which was for a long, long time the only force that was understood and math- described, that definition, $F*s$, of course, is appropriate to describe it: if you want lift 1 kg by 2m you spend energy that's double the energy to lift 1Kg by 1m. This peculiar definition produced also the 'paradoxical' [see here] (The physical definition of work seems paradoxical) consequence that, in some circumstances, you have spent a lot of 'energy calories' but you have really spent no 'energy work'

But, coming back to the derivation of the formula, after all 'integration on time' is not even necessary as there is not really anything much to be integrated, it is the simple geometric formula you need to find the side of a square: if you consider the unitary mass the real, 'true' relation between entities reveals itself in all its stunning semplicity: $$ v = p = \sqrt{E}$$ We might even drop the 'k' since not only $E_k = W$, but it is also same energy as thermal, and EM etc. energy.

Physicist are probably accustomed to it and do not bother, but it is really puzzling to see the energy of light being related in some way 'equivalent' to the dot product of $F *s$. It would be sensible to replace/integrate this definition distinguishing between work done (energy spent) and work done on something (mechanichal work), and call difference wasted energy/work: $W_d = W_{mech} + W_w$

You'll find more details about wasted energy and wasted energy in a torque at the question linked

I hope this is enough to appease your disconcert.

field theory - What is the definition of soliton?

What is the definition of soliton? I've encountered this name in different situations like when the topic discussed is about QFT, fluid dynamics or optics, but I cannot find a general definition. I've understood it's a solution of a non-linear wave equation, but I didn't find more.

Can you explain me what is a soliton?

PS. I'm undergraduate and I've completed only introductory courses of EM and QM.

Answer

What is the definition of soliton? I've encountered this name in different situations like when the topic discussed is about QFT, fluid dynamics or optics, but I cannot find a general definition. I've understood it's a solution of a non-linear wave equation, but I didn't find more.

Can you explain me what is a soliton?

A soliton is a self-reinforcing single wave, which moves at a constant velocity, whilst maintaining its shape. Solitons represent solutions to a large class of weakly nonlinear dispersive partial differential equations, which are associated with physical systems. Solitons are caused by a cancellation of nonlinear and dispersive effects in the medium. (The term "dispersive effects" refers to a property of certain systems where the speed of the waves varies according to frequency).

Dispersion and non-linearity can combine and result in permanent and localized wave forms. Solitons are a result of the non-linear Kerr effect: the refractive index of a material at a given frequency depends on the wave's amplitude. If the pulse has just the right shape, the Kerr effect will precisely cancel the otherwise disruptive dissipative effect and the pulse's shape will not change over time.

Simulation of collision between 2 solitons. Image source: Scholarpedia. org

Source: From Wikipedia Korte-de Vries

The Korteweg-de Vries (KdV) equation is the basis of model equations of nonlinear waves.

KdV equation is $u_t + u_{xxx} + 6uu_x= 0$

Consider solutions in which a fixed wave form (given by $f(X))$ maintains its shape as it travels to the right at phase speed c. Such a solution is given by ${\displaystyle \phi }(x,t) = f(x − ct − a) = f(X)$. Substituting it into the KdV equation gives the ordinary differential equation

${\displaystyle -c{\frac {df}{dX}}+{\frac {d^{3}f}{dX^{3}}}+6f{\frac {df}{dX}}=0,}$

or, integrating with respect to X,

${\displaystyle -cf+{\frac {d^{2}f}{dX^{2}}}+3f^{2}=A}$

where $A$ is a constant of integration. Interpreting the independent variable $X$ above as a virtual time variable, this means $f$ satisfies Newton's equation of motion in a cubic potential. If parameters are adjusted so that the potential function $V(f)$ haslocal maximum at $f = 0$, there is a solution in which $f(X)$ starts at this point at 'virtual time' −∞, eventually slides down to the local minimum, then back up the other side, reaching an equal height, then reverses direction, ending up at the local maximum again at time $∞$. In other words, $f(X)$ approaches 0 as $X → ±∞$. This is the characteristic shape of the solitary wave solution.

More precisely, the solution is

${\displaystyle \phi (x,t)={\frac {1}{2}}\,c\,\mathrm {sech} ^{2}\left[{{\sqrt {c}} \over 2}(x-c\,t-a)\right]}$

where sech stands for the hyperbolic secant and a is an arbitrary constant. This describes a right-moving soliton.

From Scholarpedia.org. Below is a summary of just some of the solitary waves and soliton types found in many disciplines, not only in physics, but for example biology and mathematics.

Since the discovery of solitary waves and solitons, a menagerie of localized pulses has been investigated in both one dimension and multiple spatial dimensions , though one must be nuanced when considering what constitutes a solitary wave (or even a localized solution) in multiple spatial dimensions. Many localized pulses have been given a moniker ending in "on" for conciseness, although they do not in general have similar interaction properties as solitons. The most prominent examples include the following:

Envelope Solitons : Solitary-wave descriptions of the envelopes of waves, such as those that arise from the propagation of modulated plane waves in a dispersive nonlinear medium with an amplitude-dependent dispersion relation.

Gap solitons: Solitary waves that occur in finite gaps in the spectrum of continuous systems. For example, gap solitons have been studied rather thoroughly in NLS equations with spatially periodic potentials and have been observed experimentally in the context of both nonlinear optics and Bose–Einstein condensation.

Intrinsic Localized Modes (ILMs) : ILMs, or discrete breathers, are extremely spatially-localized, time-periodic excitations in spatially extended, discrete, periodic (or quasiperiodic) systems. (At present, it is not clear whether analogous time-quasiperiodic solutions can be constructed for general lattice equations.

q-breathers : Exact time-periodic solutions of spatially extended nonlinear systems that are continued from the normal modes of a corresponding linear system.

Topological Solitons : Solitons, such as some solutions to the sine–Gordon equation, that emerge because of topological constraints. One example is a skyrmion, which is the solitary-wave solution of a nuclear model whose topological charge is the baryon number.

Other examples include domain walls, which refer to interfaces that separate distinct regions of order and which form spontaneously when a discrete symmetry (such as time-reversal symmetry) is broken, screw dislocations in crystalline lattices, and themagnetic monopole.

Vortex Solitons : A term often applied to phenomena such as vortex rings (a moving, rotating, toroidal object) and vortex lines (which are always tangent to the local vorticity). Coherent vortex-like structures also arise in dissipative systems.

Dissipative Solitons : Stable localized structures that arise in spatially extended dissipative systems. They are often studied in the context of nonlinear reaction–diffusion systems.

Oscillons: A localized standing wave that arises in granular and other dissipative media that results from, e.g., the vertical vibration of a plate topped by a layer of free particles.

Higher-Dimensional Solitary Waves : Solitary waves and other localized (and partially localized) structures have also been studied in higher-dimensional settings.

Numerous generalizations of the above examples have also been investigated, as one can consider chains of solitons, discrete analogs of the above examples (such as discrete vortex solitons), semi-discrete examples (such as spatiotemporal solitary waves in arrays of optical fibers), one type of soliton "embedded" in another type, solitary waves in nonlocal media, quantum solitary waves, and more, including their involvement in the periodic motion of Newton's Cradle

thermodynamics - How do blankets keep you warm?

I heard a famous physicist (was it Feynman?) argue that blankets do not keep you warm by trapping heat but by trapping air next to the body. Is this true?

Answer

It may be worth pointing out that blankets also (surprisingly) act as (thermal) radiation shields. This is the reason that "emergency blankets" can sometimes be found in survival kits that appear to be nothing more than thin shiny plastic. But they really make a difference in the amount of heat lost by a warm (37 °C) body on a cold night (cloudless sky - assume 0 °C).

For a body with an area of 30 cm x 180 cm facing the sky, the area is roughly $0.5 m^2$. Assuming an emissivity of 0.3 (just picking a number), the heat loss is given by

$$E = \epsilon \sigma (T_1^4-T_0^4) = 53 W/m^2$$

Or 25 W for the human I just mentioned. That is a not insignificant amount of heat... especially when you consider that the basal metabolic rate ("doing nothing" which is a good approximation of sleep) is around 60 W. And that's not counting the heat you will lose by breathing (heating cold air, and filling it with vapor).

Heating cold air (still going with 0 °C as our baseline):

250 ml per second, heat capacity 1020 J/kg/C, $\Delta T = 37 C$, you get about 12 J

Evaporating water:

Saturated vapor pressure of water at 37 °C around 47 mm Hg, and breathing about 250 ml per second (900 liter per hour) with an effective fraction of 47/760 per volume of water, this takes another 25 W.

So surprisingly, these three mechanisms result in similar amounts of heat loss - and protecting yourself from radiative heat losses is indeed significant. Because of this, a good blanket (which will reflect some of that heat back to you) is indeed "keeping heat in".

The above underlines that the most significant form of heat loss is evaporation. A good blanket stops circulation, and will keep the air near your body "moist". This will slow down the rate of evaporation, helping you stay warm. Stopping the air from circulating also stops it from carrying away "heat" - but the amount of heat carried by moist air is significantly greater than "just air", as the above example demonstrates.

There is more to this question than meets the casual eye...

cosmology - Expanding universe - Creation of Space

Is the expansion of the space between the galaxies caused by stretching of existing space or the creation of new space?

The fact that the energy content remains constant, and is therefore not being diluted, would suggest to me that it is not being stretched and therefore must be the result of the creation of additional space.

If this is so, are there any theories as to how space is created.

Answer

The unit of the Hubble constant is 1/sec. and it is determining the percentage by which space is expanding each second. The Hubble constant is not constant in time, for this reason it is called also Hubble parameter.

For getting an idea of space expansion you have to imagine a cubic grid throughout the whole space where the grid/ the number of cubes does not change, but each cube is increasing in size. It is similar to a sort of "stretching" of the size of the cubes.

Space includes dark energy whose energy density is constant and does not dilute. This is why in an advanced stage the universe is/ will be dominated by dark energy, which is prevailing/ will prevail over matter and radiation.

Local assemblies of objects in space such as galaxies are practically not concerned by space expansion because their gravitational forces are holding them together even if space is expanding

Space expansion has been observed in the form of redshifting of radiation from very far objects, but at my knowledge there is no current theory explaining how space is created.

quantum field theory - Is gravity a tensor?

All other fundamental forces are mediated by vector bosons. The Higgs boson is a scalar boson & the interaction it mediates isn't called as a force. A force is a vector in the usual description. But the hypothesized graviton is a tensor boson. Would that mean gravity is not a usual force but may be a generalization of the concept of force into a tensor? (Einstein field equation relates Einstein tensor & stress-energy tensor & therefore sort of relates curvature & energy. And gravity is curvature).

statistical mechanics - What is the maximum temperature for a photon?

Assuming that the maximum temperature for particles is $10^{32}$ kelvin (i.e., the Planck temperature) and the electroweak symmetry breaking is below $10^{15}$ kelvin, What is the maximum temperature for a photon? For example, if we could heat a photon to above $10^{15}$ kelvin, would it remain a photon or would it transform into some other type of particle such as the mathematically modelled electroweak $W_1$, $W_2$, $W_3$, or $B$ bosons?

Also, for an optional bonus point question, Would string theory or M-theory make a big difference for these calculations?

Answer

This turns out to be more complicated than you (probably) thought. The electroweak transition that produces the photon, $Z$ and $W^\pm$ is due to an interaction with the Higgs field, but it depends on the energy of the Higgs field.

At the risk of oversimplifying, a quantum field has states of increasing energies. There is a lowest energy state, the vacuum state, and then adding energy excites the field to higher energy states. A quantum field is excited by exchanging energy with other quantum fields, so for a field to have a higher energy means all the fields it interacts with will also have a higher energy. In effect this is just thermal equilibrium. A hot object sustains its temperature because it is continually exchanging energy with its surroundings. We have a system in thermal equilibrium with a well defined temperature.

So we can talk about a temperature for the Higgs field, and all the other fields it interacts with, by considering the equilibrium system in which all the quantum fields are excited and all continually exchange energy with each other. At the risk of oversimplifying yet again, the Higgs field is symmetric at high temperature and asymmetric at low temperature, and it is the transition from the high to low symmetry state that causes electroweak symmetry breaking and the conversion of the $W_1$, $W_2$, $W_3$ and $B$ bosons to the photon, $Z$ and $W^\pm$. The temperature at which this occurs is the $10^{15}$K that you refer to in your question.

The point of all this is that a single very high energy photon does not constitute a high temperature any more than a single very high energy gas molecule constitutes a high temperature gas. So if we consider a single photon and ramp up its energy then we don't expect anything special to happen as we reach then pass the electroweak transition energy. The photon will just carry on being a photon. As we approach the Planck energy then we may well need to start considering stringy effects but right now we have no concrete understanding of physics at these energies so it's impossible to say what happens.

The situation is very different if you consider a system of many particles, e.g. the debris from a high energer collider experiment, and consider a system in which the average photon energy is high. In this case as the energy approaches the electroweak transition energy that energy is shared with all the field including the Higgs field, and as a result the Higgs will be in its high energy symmetric state and the photons will transform back to combinations of the $W_3$ and $B$ bosons.

quantum field theory - Virtual Particles real? Virtual particles create a universe?

I am reading the book of Lawrence Krauss "A universe out of nothing", where he explained that the vacuum is not empty. It is a boiling brew of virtual particles that come out of their existence. And they can create a universe, even space.

But virtual particles are inner lines in a feynman-diagram. They converge in a vertex and so on. They are off-Shell.

I am a layman and I cannot grasp clearly about this stuff.

My question are:

1. 
   

   Is the vacuum empty or is it not?

   
   


2. 
   
   

   Are there particles in the vacuum and can they create a universe?

   
   


3. 
   

   But when virtual particles are just a mathemacical "trick" to calculate something, what does Lawrence Krauss mean?

   
   


4. 
   

   What is the matter about the vacuum?

   
   

quantum mechanics - Finite Square Well Inside an Infinite Square Well

Ok here's a potential I invented and am trying to solve:

$$ V(x) = \begin{cases} -V_0&0a \\ \end{cases}$$

and $V(-x) = V(x)$ (Even potential)

I solved it twice and I got the same nonsensical transcendental equation for the allowed energies:
$$ \frac{-k}{\sqrt{z_0 - k^2}} \frac{e^{2kb} + e^{2ka}}{e^{2kb} - e^{2ka}} = \tan(b \sqrt{z_0-k^2}). $$

, where $k=\sqrt{-2mE}/\hbar$ and $z_0 = 2mV_0/\hbar^2$

The problem is that when I take the limit as $b→a$ (the ordinary infinite square well) I get a division by 0.

So is there something fundamentally wrong with trying to solve this potential? Is it wrong to have an Infinite potential and bury some of it under the 0 (negative potential) ? Note: I am solving it for negative energies (bound bound states?).

EDIT by CZ, for readability and analytic continuation of l.h.side, cf. Gilbert et al.: $$\frac{-k}{\sqrt{z_0 - k^2}} \coth(k(b-a)) ~. $$

Answer

The potential you have written is perfectly legitimate, let's do the derivation from scratch: $$ V(x) = \begin{cases} -V_0&0a \\ \end{cases}$$

and $V(-x) = V(x)$ the Hamiltonian will commute with the parity operator $[\hat{H},\hat{\mathscr{P}}] = 0$, therefore the eigenstates of the energy will be eigenstates of the parity operator too. Also, we can solve the problem in the interval $[0,a]$ with $b0$ (you can easily check that if you started from the condition to get the bound states inside the finite well you would get an impossible condition for the infinite one at the point $x=a$)

Region $[0,b]$:

The Shrodinger equation is $$-\frac{\hbar^2}{2m}\nabla^2\psi-V_0\psi= E\psi$$ with $V_0>0$ in one dimension the solution will be: $$\psi_+(x)=C_1 cos(\omega x)\\ \psi_-(x)=C_1 sin(\omega x)$$ we will work with parity $+$. for comodity. With $\omega=\sqrt{\frac{2m}{\hbar^2}(V_0+E)}$.

Region $[b,a]$:

Here the Shrodinger equation is: $$\frac{\hbar^2}{2m}\nabla^2\phi= E\phi \\ \phi(a)=0$$ notice that $\phi(a)=0$ is the foundamental condition that "tells" the particle that there is an infinite well, and this condition is also the reason that $E>0$ infact if $E<0$ you would have $\phi_+= A_1 \cosh(kx)$ which only has immaginary solutions for $\phi(a)=0$ or the useless $k=0$. The solution for $E>0$ here will be: $$\phi_+(x)=A_1 cos(k x)\\ \phi_-(x)=A_1 sin(k x)$$ with $k=\sqrt{\frac{2m}{\hbar^2}E}$, now since $\phi_{\pm}(a) = 0$ we have that $k$ must be: $$k_+=\frac{(2n+1)\pi}{2 a} \\ k_-=\frac{n\pi}{a} $$.

Now we have to match the solutions in $x=b$. (we have to match the derivatives too since the general solution must be $C^2$)

$$\psi_+(b)=\phi_+(b) \\ \psi'_+(b)=\phi'_+(b)$$ this leads to (for the parity even eigenstates):

$$C_1 cos(\omega b) = A_1 cos(k b)\\ C_1 \omega sin(\omega b) = A_1 k sin(k b)$$ The solution reads: $$\tan(\omega b) = \frac{k}{\omega} \tan(k b) $$ which can be written as : $$\tag 1 \omega \tan(\omega b) = \frac{(2n+1)\pi}{2a} \tan\left( \frac{2n+1}{2a}\pi b \right)$$

Now, if you do $b \rightarrow a$ you get that $\tan\left(\frac{2n+1}{2}\pi\right) \rightarrow \infty$ which means that $cos(\omega b)=0$ which means that $\omega = \frac{2n+1}{2 b}\pi $ (also $k$ vanishes) which is the usual energy for the infinite well.

The limit $b \rightarrow 0$ is consistent too. Consider equation $(1)$ written as $$\omega=k \frac{\tan\left(kb\right)}{\tan\left(\omega b\right)}$$ now take the limit $b \rightarrow 0$ on both sides $$\ \omega = k\cdot \frac{k}{\omega}\longrightarrow \omega = k$$ and you are left with $k = \frac{(2n+1)\pi}{2 a}$ which is the usual infinite well energy.

Universal reference frame based on the speed of light/special relativity

Everything I've read has said that there is no universal static reference frame, but based on my understanding of special relativity I don't understand why it can't be determined.

Here is an experiment I can think of that would seem to determine "universal static", what am I missing? Assuming 1 dimensional space for a second, our reference frame (a spaceship) is moving and an indeterminate speed $V_0$. We take two probes with atomic clocks and send one in each direction at some fraction of the speed of light ($V_0+V_p$ and $V_0-V_p$). After a period of time, the probes return at a slower speed than they left (taking advantage of the fact that time dilation effects increase exponentially with velocity). Upon returning, the clock in the probe sent in the positive direction ($V_0+V_p$) should show less time past than the clock on our spaceship, and the probe sent in the opposite direction ($V_0-V_p$) should show that more time has past than the clock on our spaceship, this tells us which direction we are moving with respect to "universal static", and with repeated experiments we can determine our exact velocity relative to it.

quantum mechanics - 7/2 versus 9/2 for diatomic heat capacity

Question

I calculated the classical heat capacity of a diatomic gas as $C_V = (9/2)Nk_B$, however the accepted value is $C_V = (7/2)Nk_B$.

I assumed the classical Hamiltonian of two identical atoms bound together as $$ H = \dfrac{1}{2m}( |\bar{p}_2|^2 + |\bar{p}_2|^2)+ \dfrac{\alpha}{2} |\bar{q}_1-\bar{q}_2|^2. $$ I calculated the partition function of $N$ particles as $$ Z = \left( \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} e^{-\beta H} ~d^3q_1~d^3p_1~d^3q_2~d^3p_2 \right)^N \propto V^N T^{(9/2)N}. $$ I calcuated the heat capacity as $$ C_V = \dfrac{\partial }{\partial T} \left( k_B T^2 \dfrac{\partial \ln(Z)}{\partial T} \right) = \dfrac{9}{2}k_BN. $$

Why does the classical argument fail?

Classical Derivation

The partition function is \begin{align} Z &=& \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta H(\bar{q}_1,\bar{q}_2,\bar{p}_1,\bar{p}_2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1 ~d^{3}p_2 \right)^N \\&=& \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta ((|\bar{p}_1|^2+|\bar{p}_2|^2)/(2m)+\alpha |\bar{q}_1-\bar{q}_2|^2/2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1 ~d^{3}p_2 \right)^N \end{align} A useful gaussian integral \begin{align} \int_{-\infty}^{\infty} e^{-\gamma (x-x_0)^2}dx = \sqrt{\dfrac{\pi}{\gamma}} \end{align} The partition function can be evaluated using separated integrals \begin{align} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_1|^2} ~d^{3}p_1 = \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_2|^2} ~d^{3}p_2 = \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \end{align} and \begin{align} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha |\bar{q}_1-\bar{q}_2|^2/2 } ~d^{3}q_1 ~d^{3}q_2 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 \iiint_{-\infty}^{\infty} ~d^{3}q_1 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \end{align} The last set of integrals are improper integrals. One has to take the limit as the space approaches infinite containment. In that limit, integrating one set of variables $d^3q_2$ approaches the limit of a finite Gaussian term, while the other $d^3q_1$ approaches the diverging value of the total volume of the gas.

The partition function is \begin{align} Z &=& \left( h^{-6} \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \right)^N \\&=& \left( h^{-6} \left(k_B T \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} V \right)^N \\&=& \left( h^{-6} \left(k_B \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} \right)^N V^N T^{9N/2} \end{align}

homework and exercises - Showing Wien's Displacement Law from Wien's Law

Does anyone know how I would show that $\lambda * T$ is constant, using only Wien's Law? That $\rho(\lambda,T) = 1/\lambda^5*f(\lambda T)$ I differentiated, but all I could get was $\lambda T = 5f(\lambda T)/f'(\lambda T)$, which I don't think means it's necessarily a constant.

newtonian mechanics - Does rolling friction increase speed of a wheel?

I know there are a lot of questions about rolling friction, but I couldn't find any that addressed my specific problem.

Note the following picture where the wheel rotates clockwise. The rolling friction force $\vec{F}_R$ points in the same direction as the rotation.

Normally, rolling friction inhibits the movement (speed $\vec{v}$) of the system. However, since the force is pointing in the "backward" direction, it looks like it is increasing the speed of the wheel. Therefore, it should accelerate the system. This seems unintuitive to me. Can someone please clarify this contradiction?

Thanks!

Update:

Based on the comments and answers, I have reconsidered the diagram and terminology. I mean rolling resistance and no static friction. The deceleration force caused by deformation (and other reasons). In the diagram above, this force is divided into its x- ($\vec{F}_r$) and y- ($\vec{N}$) components.

$\vec{v}$ is not independent of $\omega$ for pure rolling, since $v=R\cdot\omega$. To me, it looks like the resistance $\vec{F}_R$ that slows down the system accelerates $\omega$. And because of the kinematic connection between $\omega$ and $\vec{v}$, the latter must also increase. That's illogical, but I don't understand the reason. Maybe some comments or answers have already addressed that, but I don't understand :( This is really confusing.

Answer

You are right on saying that the rolling resistance increases the angular speed. But I think your confusion is mainly because

$\vec{v}$ is not independent of $\omega$ for pure rolling, since $v=R\cdot\omega$. To me, it looks like the resistance $\vec{F}_R$ that slows down the system accelerates $\omega$. And because of the kinematic connection between $\omega$ and $\vec{v}$, the latter must also increase.

Whatever you have written is correct, but for clarification, we will consider two cases. One where a net external force is acting on the system, and one where it isn't.

With external force (Gravity in this case)

Consider a sphere on an inclined plane, with the upper half of the plane as smooth, and the lower half with sufficient friction for rolling.

Here, as usual the body will have an increase in its speed while sliding on the frictionless surface from point B to point A. But after point A, the speed will decrease, but angular velocity will increase. This seems to go exactly opposite to $v=R\cdot\omega$. Right? The reason for this is that pure rolling hasn't started yet - it is transitioning, so this formula is not yet applicable. Here, the resistance is doing negative work for translation, which is exactly equal to the positive work on rotation(you may use equations of mechanics to prove this).

In short, here the resistance is extracting some energy from translation and gives it to rotation, until they are in such a way that the condition for pure rolling is satisfied.

After point A(once pure rolling has started), whatever you are saying is valid, as obviously, the sphere will speed up while rolling down the plane.

Without an external force and considering deformations

We have our good ol'blue sphere rolling towards right on the ground as shown here.

Here, a portion of the sphere is in contact with the ground, and the points on this portion will existence the force due to ground. In the figure, the red arrows indicate the forces, with their length proportional to the magnitude. The forces at the front will be greater than the forces towards the back. This is also clear according to this which says:

The resulting pressure distribution is asymmetrical and is shifted to the right. The line of action of the (aggregate) vertical force no longer passes through the centers of the cylinders. This means that a moment occurs that tends to retard the rolling motion.

To analyze the forces, we can assume an equivalent force $F$, which will produce the same effect as all these forces combined, and will act as shown below:

Here, the torque can be given as $F.d$ which will act against the rotating motion(You can use simple geometry to understand why it will act against. If any issues with that, feel free to ask for clarification in comments). Further, this force can be divided into two components to analyse the resultant acceleration. The vertical component $N$ will not contribute towards linear acceleration as it will be balanced by the weight, whereas the horizontal component $R$ will act as shown, and contribute towards retardation, opposing the velocity.

In the end, it can be concluded that both angular velocity and linear velocity are decreasing due to the deformations.

We did not use the components to analyze the torque because both would produce torque in opposite directions, and there comparison might not be very obvious.

Apart from these, the following also contribute towards rolling resistance

• 
  

  Material of the sphere (or any other rolling body) : Compare a steel ball and a soft rubber ball

  
  


• 
  

  Type of ground (or whatever surface it is rolling on) : Compare a sharing rink and sand

  
  
  


• 
  

  Mass of the sphere : More the mass, more is the resistance

  
  


• 
  

  Diameter of sphere : More the diameter, less is the resistance

  
  


• 
  

  Adhesion effects

  
  



• 
  

  Hysteresis effects

  
  

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Some links I found useful: