optics - Theoretical penetration limit for evanescent waves

Consider a problem in classical electrodynamics, when a monochromatic beam experiences total internal refraction when traveling from a medium with $n>1$ to a medium with refractive index $1$ - see schematic below. Using Fresnel equations one gets the penetration depth $$d = \frac{1}{\sqrt{k_x^2+k_y^2-(\tfrac{2\pi}{\lambda})^2}},$$ where $k_x$ and $k_y$ are the perpendicular components of the wave vector, and $\lambda$ is the wavelength (in the vacuum).

At least in theory, it is possible to have an evanescent wave of an arbitrary penetration depth $d$. However, in such case one needs to use a plane wave, thus a wave of unbounded spatial size. For a beam with a finite variance $\langle x^2\rangle$ (and $k_y=0$ to reduce the problem to two dimensions) there seems to be a relation that $\langle d\rangle/\sqrt{\langle x^2\rangle}$ is lesser than a constant.

The main questions: is there any strict bound in the form of $$\text{a measure of penetration depth}\leq f(\text{transversal beam size},n)$$ (perhaps in the style of Heisenberg uncertainty principle, or using other moments of $x$, $y$ and $d$)?

nuclear physics - Would a matter-antimatter explosion cause fallout?

I know matter and antimatter annihilation release a lot of gamma rays which are considered ionizing radiation if I am not mistaken. But what if the explosion happened on the surface of the earth, would the the material taken into the fireball cause fallout afterwards ?

My question in another form, what causes fallout ? and is ionizing radiation capable of radiating materials for a long time ?

Answer

In a fission bomb, the fallout consists of fission-decay fragments, which are nuclei that can have long enough half-lives to be transported by winds. Fusion bombs are basically the same idea, because they use fission triggers.

and is ionizing radiation capable of radiating materials for a long time ?

In theory, yes, e.g., exposure to neutrons in reactors can be used to intentionally produce radioactive isotopes. In practice, although nuclear bombs must produce this kind of artificial transmuation of the surrounding matter (e.g., they do emit neutrons), I think there isn't enough of this kind of process to contribute noticeably to the fallout.

Matter-antimatter annihilation from a hypothetical macroscopic explosion would produce the same particles as proton-antiproton annihilation in microscopic quantities in accelerator experiments. You get high-energy (~100 MeV) gammas, medium-energy (e.g., 511 keV) gammas, pions, muons, and neutrinos. The neutrinos fly off harmlessly and undetectably into outer space. Matter is nearly transparent to the high-energy gammas; the downward-emitted ones are absorbed somewhere underground. The medium-energy gammas are absorbed in nearby matter. The pions and muons are unstable and decay quickly into stable particles such as electrons. Nothing long-lived is produced.

general relativity - What is the most natural way of attaining gravity as the gauging of the Lorentz group?

Note: I don't know much about QFT, aside from some basics, I am a student and first and foremost a student of GR.

As far as I am aware, the fact that interactions are gauge theories allow some pretty natural ways of deriving them. For example, for QED, if one is given a matter field $\phi$ (matter here can be bosons as well), whose lagrangian is invariant under global $U(1)$ transformations, then demanding that this lagrangian should be invariant under $U(1)$ transformations that depend on spacetime points as well naturally leads to the introduction of an $U(1)$ gauge connection, and then conjuring up a gauge-invariant lagrangian for the gauge connection too gives one a system of Maxwell's equations being coupled to the dynamics of a matter field.

Since we know from classical electrodynamics that only charged particles interact electromagnetically, this also gives a natural way of telling which particle is charged. A particle is charged if the corresponding matter field admits a global $U(1)$ symmetry.

For example the complex scalar field $$\mathcal{L}_{KG\ (C)}=\partial_\mu\phi^\dagger\partial^\mu\phi-m\phi^\dagger\phi$$ is charged but the real version of the same thing isn't.

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Now, assume that one does not know about Riemannian geometry or general relativity, but because all other interactions arise via a gauge principle like in QED, this person tries to give rise to a theory of gravity the same way.

One knows that gravity interacts with everything, so one cannot use some fancyful gauge group $G$, because every matter field needs to know this symmetry. One fundamental requirement of relativistic field theories is that the action should be Lorentz-invariant, so the only Lie group that satisfies this requirement is the Lorentz group, therefore one needs to gauge the Lorentz group.

The Lorentz group is however an "external" symmetry group, not an "internal" one, eg. Lorentz transforms are related to spacetime geometry.

It is clear that the usual way of doing stuff, namely to replace $\partial_\mu$ with some gauge connection $D_\mu$ won't work here. For example in SR, the coordinates $x^\mu$ form a 4-vector. What is the meaning of a position dependent transform $\Lambda^\mu_{\ \nu}(x)x^\nu$? Nothing.

Of course, if one does know Riemannian geometry and GR, one can see that you need to give up flat spacetime one way or another. The preferred way to do that (preferred in the sense of following the example of QED or QCD) is to make Lorentz symmetry "internal" and have every Lorentz tensor $S_{\mu\nu...}$ replaced with some "section" $S_{a,b...}$ and provide a function $\theta^a_{\mu}$ that relates the "internal" and "external" spaces to one another (a vielbein, basically, but I am using its interpretation as a solder form here) and also replace the volume element $d^4x$ with the invariant volume element $\det(\theta)d^4x$ and all "spacetime" indices $\mu,\nu$ should be interpreted as referring to a general coordinate system, and of course, one needs to introduce a gauge connection $D_\mu=\delta^a_b\partial_\mu+\omega_{\mu\ \ b}^{\ a}$ instead of $\partial_\mu$, but these steps do not follow naturally from the idea of making Lorentz transformations local.

Question: Assuming one does not know GR or differential geometry, but has the bright idea of creating a gauge theory of gravity by gauging the Lorentz group, is there any "natural" way of performing this gauging that leads to a consistent field theory of gravity that couples to everything?

If so, is this necessarily unique? I guess not, since it is not a fundamental requirement to have the gauge connection determined uniquely by the vielbein (Einstein-Cartan theory for example).

But is it necessary to have a vielbein actually appear? Is it possible to arrive at this purely from the idea of gauging the Lorentz group?

binding energy - Obtaining isotope stability

For a given isotope, one can obtain the binding energy using the semi-empirical mass formula.

For example, has a binding energy of 1782.8 MeV. From this information, how can the likelihood of the isotopes stability be found?

Answer

From the binding energy given experimentally, using precise QM calculations or using a given formula, one should first check for "stability in particles", if the binding is negative, you will of course not have stability.

Then the next thing, if you have a formula, is to check for each type of stability. For example, to check for stability against a given fission, calculate the binding energy of the fragments, obtain the new energy and compare.

To check for, let's say a beta minus, replace the nucleus (A,Z) in the formula by (A,Z+1), obtain the binding energy, the new total energy (you can safely neglect the mass of the neutrino and even the one of the electron in most cases) and compare.

For your specific example, this is a bit tricky because a large variety of decay channels are potentially allowed.

Edit

Another way to proceed is to look at a binding energy per nucleon or mass excess against A diagram:

We can see that 237Np is far on the right as a binding energy per nucleon smaller that the most stable elements like 56Fe.

One can then conclude that 237Np can potentially decay to a more stable element to increase it's binding energy (although these decay, that can be alpha decay can have excessively small probability and a time constant excessively long).

electromagnetic radiation - Why don't absorption and emission lines cancel out in our Sun?

I was looking at this answer on why absorption lines and emission lines don't cancel out:

An experiment shining light on the material and looking at the reflected spectrum will see absorption lines at those frequencies, because the relaxation of the excited electrons will emit back radiation all around randomly, whereas the reflected spectrum is at a specific angle.

However it is not totally convincing to me. The photons that get emitted towards the center of the Sun (by the electrons in higher energy states that absorbed a photon earlier) would get absorbed again when they hit a different electron (and then re-emitted) and would eventually make it out. So why don't absorption and emission lines cancel out in our Sun?

Answer

I think that this is a very good question.

In my answer I will only mention the formation of one of the absorption lines, the 589 nm of sodium, and I will call the photon associated with that wavelength a "sodium photon".
What I will try to explain with reference to that particular wavelength of light will be true for all the other wavelengths for which absorption occurs.

The schismatic layout of a standard demonstration of absorption and emission lines is shown below.

At position $A$ one would see an absorption spectrum and at position $B$ an emission spectrum and the "re-radiated in all directions" explanation works very well.

The difference with the Sun is that the "sodium flame" envelopes the Sun and the rate of "sodium photons" emerging from the Sun is less than the rate of emergence of photons with comparable wavelengths.

I think that the OP is asking "Where do the re-radiated sodium photons go?"

The fact is that the rate at which sodium photons escape from the "sodium flame blanket" around the Sun (the Sun's outer layers) is smaller than the rate at which photons close in wavelength escape.

So in effect that outer layer of the Sun is fairly opaque to sodium photons.

As the sodium photons which are being produced in the internal layers of the Sun progress through the outer layers of the Sun they get absorbed and re-radiated so the net rate in the forward (away from the Sun) direction decreases and there is a flux of sodium photons heading back towards the Sun.
Here they interact with the hotter "inner" layers of the Sun and do not necessarily emerge again as sodium photons, their wavelength is changed.
They are thermalised (I cannot think of a better wording).

Those sodium photons enter regions within the Sun where they they are in excess of what might expect for the distribution of wavelength for the temperature of those inner layers.
Interactions within those layers reduce that excess number of sodium photons, so they cease to be sodium photons.

So the net effect is that the "sodium flame blanket" around the Sun sends back towards the Sun sodium photons which are then transformed into photons of other wavelengths.

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gauge theory - Why are the "coupling constants" constant?

The coupling constants (in the gauge theory) fix an inner product on the lie algebra of the gauge group and we use it to define strength of the fields. we are using ad-invariant inner products which are determined by some numbers. In other words, the set of all ad-invariant inner products form a space of more than one dimension and to fix a specific inner product we need to choose some numbers which are the coupling constants. This is the story which happens over each point of the space-time. Mathematically one can produce a theory in which these numbers (and so inner product) changes from a point to another (similar to Riemann metric on a general Riemannian manifold). In other words we can have coupling fields rather that constants.

What is the physical reasoning which disallow us having such a theory (with coupling fields)? And what would be the physical implications if the nature follows such a model?

Answer

You may always promote "couplings constants" (charge, mass, etc...) to fields. Now, as a physicist, you need to make some contact with reality. So you have to tell why and which field you are using (for instance the Higgs field (up to a constant), which has a $SU(2)$ charge, is used to replace a constant mass coupling in the interaction $m (\bar e_R e_L + \bar e_L e_R)$, this is because the left electron has a $SU(2)$ charge while the right electron has not), and how an experiment can test your hypothesis.

Finally, even coupling constants, in a Quantum field theOry, are not really constant, and depenD on the energy scale $e = e(\Lambda)$

resource recommendations - Good math books for physicists

In his first lesson (transcripted in "Tips on Physics"), Feynman talks about math for physicists in a very cool and practical way. And at the end of the section he talks something like "so the first thing to do is to learn to learn derivative, integral and algebra" (I don't know how much precise I'm being because I've read it in Portuguese). I would to know if there is some book that deals with math as Feynman did it in this lesson (respecting formalities, but teaching how to use the practical rules)? Also, someone have any recommendations for algebra book (college level)?

Answer

Higher maths for beginners is ana amzing little book on all the subjects you mentioned, written by one of the fathers of Soviet nuclear bomb, and theoretical phsyicists.

On math physics, the best introductory test is Elements of applied math physics, it has dufferential equations and complex analysis and other cool topics. Unfortunately, it may not have English version.

The comprehensive analysis text is Fundamentals Mathematical Analysis. It's a Russian textbook, but it's old school, i.e. very readable.

Another must have book is Differential Equations and Calculus Variations.

The best reference on PDEs is PDE by Bitsadze, I consult it all the time, it's very thin, and chapters are mostly self-contained.

All these books were used by Physics students, I can guarantee that.

quantum information - Approximate cloning of a qubit, given multiple starting copies

Suppose I'm given several clones of a qubit in a pure unentangled state. That is to say, I'm given the state $(a \left|0\right\rangle + b \left|1\right\rangle)^{\otimes n}$. My goal is to make $d$ clones of the qubit; to expand the input state so that it approximates the target state $(a \left|0\right\rangle + b \left|1\right\rangle)^{\otimes n+d}$.

This can't be done exactly because of the No Cloning theorem. On the other hand, if $n$ is very large, then we can use tomography to get a pretty good approximation of $a$ and $b$. Once we have that, we can start churning out as many copies as we want.

I'm looking for papers/websites/answers giving bounds and on how accurately and efficiently this can be done, as well as explicit techniques for doing it and how well they perform.

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For example, suppose we know that $a$ and $b$ are constrained such that $a = \sin \theta$, $b = \cos \theta$, and $0 \leq \theta \leq \pi/2$. Then a very simple $n$-to-$n+d$ cloning procedure is to count the given qubits that are ON in the computational basis and estimate $\theta$ from that (note that we're conditioning operations on the count, not measuring the count). So, given $\psi(t) = (\cos t) \left|0\right\rangle + (\sin t) \left|1\right\rangle$, we get something like:

$\psi_{\text{target}} = \psi(\theta)^{\otimes n+d} = \sum_{l=0}^{n+d} \left| n+d \atop l \right\rangle \cdot \cos^{n+d-l} \theta \cdot \sin^l \theta$

$\psi_{\text{made}} = \sum_{k=0}^{n} \left| n \atop k \right\rangle (\cos^{n-k} \theta) \cdot (\sin^k \theta) \cdot \psi(f_n(k))^{\otimes d}$

$\psi_{\text{target}}^* \cdot \psi_{\text{made}} = \sum_{k=0}^{n} \left( n \atop k \right) \cdot (\cos^{2n-2k} \theta) \cdot (\sin^{2k} \theta) (\cos^d \left( \theta - f_n(k) \right))$

Which, for $n=d=10$ gives this plot w.r.t. $\theta$:

Since the inner product stays above 0.8, the resulting state is basically at most ~35 degrees off of the true state (which is pretty good in a $2^{20}$-dimensional complex vector space).

But presumably there are processes that perform better than this, and without restricting the state space as much.

Answer

R.F. Werner, Optimal Cloning of Pure States describes the optimal procedure for cloning multiple copies of the same pure state (which, remarkably, for pure states is independent of the figure of merit).

Abstract: We construct the unique optimal quantum device for turning a finite number of $d$-level quantum systems in the same unknown pure state $\sigma$ into $M$ systems of the same kind, in an approximation of the $M$-fold tensor product of the state $\sigma$.

neutron stars - Complex structures bound by nuclear forces ( nuclear molecules ?)

This is very hypothetical question. Consider all chemistry (even bio-chemistry) is just physics of valence electrons. Because solutions of Schroedinger equation for system of electrons and nuclei are highly anisotropic and directional ( like sp$^3$-hydridization ) you can at the end form structures like proteins and DNA.

The only stable structures bound by nuclear forces we currently know are around 250 nuclei of stable isotopes ( Am I right ?). All of them are basically just spherical clusters of nucleons (only few halo nuclei deviate from liquid drop model but just a little).

But is there any (even very speculative) possibility to form something more complex (perhaps like nuclear polymer), or is it completely exclude by laws of nature?

Possibly it is just matter of kinetics? Almost all nucleosynthetic processes (both natural and human made ) create new nuclei in excited state by harsh collisions. It is natural that such nuclei has a lot of excess energy to overcome activation barriers and create most stable but boring spherical clusters. It's a bit like doing chemistry in hot gas - you never create more complex molecules. Maybe if there would be an hypothetical method how to prepare such nuclear structures in gentle way ( cool, not excited ) it would be quite stable?

And even if it is not possible in normal conditions on earth, maybe such complex nuclear structures may exist on the surface of neutron stars, in phase transition from normal matter to nuclear matter. Is that what is called nuclear pasta?. ( I would like to read something )

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By the way - is there any good book which discuss nuclear structure form the perspective close to chemist ( quantum chemist, chemical physicist ) - i.e. discussing everything in terms of "nuclear orbitals" (and it's shape) in analogy to atomic orbitals and molecular orbitals.

Answer

Yes, there are phases of matter that correspond to your question that exist inside neutron stars and are known collectively as nuclear pasta.

The inner crust of a neutron star has densities $>4 \times 10^{14}$ kg m$^{-3}$ and consists of a lattice of extremely neutron-rich nuclei, surrounded by degenerate gases of neutrons and ultrarelativistic electrons that behave like fluids. The equilibrium composition and structure is found by forming the sum of the energy densities of these components and then minimising it with respect to the number densities of the nuclei, neutrons and protons, and the numbers of neutrons and protons in the nuclei, subject to baryon and charge conservation.

The energy density sum must include the rest mass energy of the nuclei, which is a function of the number of protons and neutrons, their binding energy plus modifications that account for the presence of the neutron fluid surrounding the nuclei that reduces the surface energy term and a negative term due the binding energy of the nuclear lattice.

Above densities of $\simeq 3\times10^{16}$ kg m$^{-3}$, nuclei probably start to lose their separate identities and spherical nature, forming elongated rods or planes of nuclear material -- nuclear pasta. Initially the nuclei are thought to group into "spaghetti" and then into "lasagna", where the sauce is provided by the gaseous free neutrons. As densities exceed $10^{17}$ kg m$^{-3}$, it is thought these roles reverse; the gaseous neutrons form into spaghetti like structures sourrounded by nuclear material and then finally, shortly before dissolution, the crust consists of nuclear material enclosing spherical bubbles of neutron gas.

Here is a schematic picture, due to Newton et al. (2011) that shows how the pasta phases develop with increasing density. As I show in some more detail below, the departure from an ordered lattice of spherical nuclei is expected to occur on the nuclei fill more than abut 1/10 of the volume.

A few details

An approximate understanding of the process arises by considering the Bohr-Wheeler condition for the fission of nuclei. $$ E_{C}^{(0)} > 2 E_S\, , $$ where $E_{C}^{(0)}$ is the self-Coulomb energy of the nucleus and $E_S$ its surface energy term. When the Coulomb energy of the nucleus satisfies this inequality then it becomes energetically feasible for the nucleus to split.

$$ E^{(0)}_{C} = \frac{1}{4\pi \epsilon_0} \int_{0}^{r_N} \frac{q(r)}{r} \, dq \, , $$ where $r_N$ is the nuclear radius. If we assume the total charge $q=Ze$ and the charge (due to $Z$ protons) is spread uniformly in the nuclear volume, such that $q(r) = Ze(r/r_N)^3$ and hence $dq = 3Ze (r^2/r_{N}^{3})$, then this integral shows that $E^{(0)}_{C} = (3/5)(Z^2e^2/r_N)$.

However, in terms of the energtics of the gas as a whole, the nuclei are not isolated, but sit in a neutral Wigner-Seitz sphere, accompanied by $Z$ electrons distributed pseudo-uniformly, where $q_e(r) = -Ze(r/r_0)^3$, is the charge distribution of the electrons, and $r_0$ is the radius of the neutral Wigner-Seitz sphere. From this, it can be shown that the total Coulomb energy is $$ E_C = \frac{Z^2 E^2}{4\pi \epsilon_0}\left( \frac{3}{5r_N} + \frac{3}{5r_0} - \frac{3}{2r_0}\right)\, . $$ If we write the fraction of the volume occupied by the nuclei as $f = (r_N/r_0)^3$, we can rearrange this to give $$ E_C = E_{C}^{(0)}\left( 1 - \frac{3}{2} f^{1/3} \right)\, . \tag{1} $$

Now since $E_{C}^{(0)} \propto Z^2/r_N$ and $r_N \propto A^{1/3}$, where $A$ is the number of nucleons in each nucleus, then for a certain value of $Z/A$, the Coulomb energy per nucleon $E_C/A \propto A^{2/3}$. The surface energy term is proportional to the area of the nucleus, so $E_S/A \propto A^{-1/3}$. If we minimise the sum of these two terms with respect to $A$ as follows $$ \frac{\partial (E_{\rm tot}/A)}{\partial A} = \frac{\partial }{\partial A} \left( \alpha A^{2/3} + \beta A^{-1/3} \right) = 0\, \, $$ where $\alpha$ and $\beta$ are just arbitrary constants of proportionality. Solving this gives $2\alpha A^{2/3} = \beta A^{-1/3}$ and hence $2E_C = E_S$ at equilibrium.

If we put this into the Bohr-Wheeler condition and use equation (1) for $E_C$, then we find that $f > 1/8$. In other words the nuclei become unstable to fission and deformation once they fill about one tenth of the total volume, which corresponds to densities of $\sim 3 \times 10^{16}$ kg m$^{-3}$.

waves - Red color has largest wavelenght and violet minimum (in the range of visible light). then why does violet light appears reddish? RED + BLUE = VIOLET

My question is simple. Green light is more similar to red light than violet, then why is violet reddish and green not? in the language of frequencies and wavelengths, red and violet should contrast each other. Then why do they don't?

Gravity vs Gravitational Waves

I thought I had a reasonable understanding of relativity, the speed of light speed limit, and how this stuff related to gravity. Then I read through all the answers/comments for this question:

How does Zumberge's 1981 gravitational measurements relate to gravitational waves?

And now I'm more confused than ever. Here's the opening for the (currently) most upvoted answer:

"This represents a major misunderstanding of what a gravitational wave is. The effect presented is simply the semi-static gravitational field at earth due to the earth, moon and sun. It is predicted by Newtonian gravity. There is no 'wave' that propagated, it's the instant positions of the 3 bodies that change over 1 day (and over 1 year also). "

So... my understanding of relativity prohibits the existence of a "static gravitational field" based on the "instant positions" of some masses. If it were allowed, that would imply that information about gravity traveled at infinite velocity which violates relativity. All changes in gravitational fields must propagate at a maximum of the speed of light.

Fundamentally though, I guess my big confusion is that it seems like everyone in that thread keeps arguing about some difference between "measuring gravity" and "measuring gravitational waves". Is measuring gravity equivalent to measuring magnetism around a permanent magnet? Is measuring gravitational waves equivalent to a digital camera capturing photons?

edit diving a little deeper

There's more context in the linked question, but I guess to begin with, I always assumed that "regular gravity" was "transmitted" via virtual particles of some type of gauge boson. So I always assumed that "gravity waves" then must refer to "real" particles in the same way that photons are real particles. Furthermore, afaik, there's no possible way to detect a static magnetic field using a mechanism which could also detect a photon. That's the whole point of a virtual photon.

So in terms of the magnet vs camera analogy, that doesn't make any sense to me in the context of the linked question. Let me bring some of that background here.

"This is the SAME gravitational wave effect measured by the LIGO researches recently

This is not a gravitational wave, this is a measurement of tides caused by the movement of a single detector within the earth-moon dimple. If there had been a second detector, it would have been clear that these tides do not propagate across the earth at the speed of light, but at the speed of the movement of the moon."

"LIGO actually detects, then filters out, this local gravitational wave

This uses the term "gravitational wave" incorrectly. It is more correct to say, "LIGO actually detects, then filters out, this local gravitational tidal noise."

According to the magnet/camera analogy, LIGO shouldn't even be able to detect tidal effects. It's fundamentally the wrong type of particle. Also, if gravity waves are the "real" equivalent of a force carrying particle, then there would be no need for a second detector. You don't need two cameras to take one picture... That doesn't even make sense.

general relativity - Spacetime Torsion, the Spin tensor, and intrinsic spin in Einstein-Cartan theory

In Einstein-Cartan gravity, the action is the usual Einstein-Hilbert action but now the Torsion tensor is allowed to vary as well (in usual GR, it is just set to zero).

Variation with respect to the metric gives:

$$R_{ab}-\frac{1}{2}R g_{ab}=\kappa P_{ab} \quad (1)$$

where $P_{ab}$ is the canonical stress energy tensor. Variation with respect to the torsion tensor ${T^{ab}}_c$ gives:

$${T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d = \kappa {\sigma_{ab}}^c \quad (2)$$

where ${\sigma_{ab}}^c$ is the Spin Tensor.

By contracting that equation, I can see that if the spin tensor is zero, the Torsion tensor is identically zero:

$${T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d = \kappa {\sigma_{ab}}^c = 0$$ $$ {g^b}_c({T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d) = 0$$ $$ {T_{ab}}^b + {g_a}^b{T_{bd}}^d - {g_b}^b {T_{ad}}^d = 0$$ $$ {T_{ab}}^b + {T_{ad}}^d - 4 {T_{ad}}^d = 0$$ $$ {T_{ad}}^d = 0$$ $${T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d = 0 \quad \Rightarrow \quad {T_{ab}}^c =0$$

My understanding is that:

The spin tensor and the stress energy tensor, are defined entirely in terms of whatever matter Lagrangian we add to the theory. Therefore, from above, the equations in vacuum are exactly the same as normal GR (so solving for outside matter, only the boundary conditions with the matter could be different).

Assuming my understanding up to here is correct, my line of questioning is:

1. 
   

   How is the Spin tensor (and hence Torsion) related to the concept of material with intrinsic spin?

   
   


2. 
   

   Hopefully answered by 1, but if the matter has zero intrinsic spin, yet we have an extended "spinning" body, is the spin tensor still zero (as I'd consider that orbital angular momentum then)?

   
   
   


3. 
   

   Does this mean Einstein-Cartan predictions are identical to normal Einstein GR if (and only if) the instrinsic spin of any particles and fields in the theory are zero?

   
   

Answer

Please let me first refer you to the following review by I. L. Shapiro, which contains a lot of theoretical and phenomenological information on spacetime torsion. The answer will be mainly based on this review.

In the basic Einstein-Cartan theory, in which the antisymmetric part of the connection is taken as independent additional degrees of freedom, the torsion is non-dynamical: (apart from a surface term which does not contribute to the equations of motion).

In the following, only minimal coupling to gravity will be considered (in which the flat metric tensor is replaced by the full metric tensor and the derivatives are replaced by covariant derivatives). There is a huge number of suggestions for nonminimal couplings in most of which the torsion becomes dynamical.

The torsion contribution to the gravitational part of the Lagrangian is quadratic in the torsion components, please see Shapiro equation: (2.15), where the additional terms to the torsion can be more economically expressed using the contorsion tensor whose components are linear combinations of the torsion tensor:

$$ K_{\alpha\beta\gamma} = T_{\alpha\beta\gamma} -T_{\beta\alpha\gamma}-T_{\gamma\alpha\beta}$$

A scalar field minimal coupling to gravity, does not require covariant derivatives because the covariant derivative of a scalar is identical to its ordinary derivative, thus the scalar field Lagrangian does not depend on the torsion, therefore in the case of a scalar field coupled to gravity, the torsion remains sourceless, and the its equations of motion imply its vanishing.

When the Dirac field coupled to gravity with torsion, the Lagrangian can be written in the following form

$$\mathcal{L} = e^{\mu}_a\bar{\psi}\gamma^a (\partial_{\mu} -\frac{i}{2}\omega_{\mu}^{cd}\sigma_{cd} )\psi + e_{\mu a} K_{\alpha\beta\gamma} \epsilon^{\mu\alpha\beta\gamma} \bar{\psi}\gamma^a \gamma_5 \psi$$

($e$ are the vielbeins and $\omega$, the torsionless part of the spin connection, they both do not depend on the torsion).

Taking the variation with respect to the contorsion components, we obtain an algebric equation of motion for the contorsion tensor:

$$ K^{\alpha\beta\gamma} = \frac{\kappa}{4}e_{\mu a} \epsilon^{\mu\alpha\beta\gamma} \bar{\psi}\gamma^a \gamma_5 \psi$$

($\kappa = 8 \pi G$). The last term in the Lagrangian is just of the form:

$$\mathcal{L_K} = K_{\alpha\beta\gamma} \sigma^{\alpha\beta\gamma}$$

Where $\sigma^{\alpha\beta\gamma}$ is the intrinsic spin part of the Noether current corresponding to the local Lorentz symmetry:

$$M^{\alpha\beta\gamma} = x^{\alpha}\Theta^{\beta\gamma}-x^{\beta}\Theta^{\alpha\gamma}+\sigma^{\alpha\beta\gamma}$$

$\Theta$ is the stress energy tensor. This example shows that for the Dirac field, the torsion source is the spin tensor.

As can be observed, the torsion couples axially to the Dirac field. This type of coupling is known to produce anomalies. A careful analysis shows that in the baryonic sector, the same anomaly cancellation criteria of the standard model lead to the cancellation of the axial anomalies due to torsion as well but not in the leptonic sector. This is one of the difficulties of this theory. One possible solution is to absorb the torsion contribution into the definition of the axial current. In contrary to the gauge and photon fields, where this contribution is not gauge invariant, since the torsion field is not a gauge field, this redefinition seems possible. This seems also consistent with the Atiyah-Singer index theorem which states that the anomaly density must be equal the Pontryagin class which is a topological invariant, while torsion can be introduced without altering the topology.

There is another difficulty related to the torsion coupling coming from the fact that the torsion couples only to the intrinsic part of the fields:

In the case of gauge fields such as the Maxwell field. The intrinsic spin is not gauge invariant and only the sum of the spin and the orbital angular momentum is gauge invariant. Thus, although the minimal coupling leads to a coupling of the torsion to the intrinsic spin, the gauge invariance is lost. The following recent article by Fresneda, Baldiotti and Pereira reviews some of the suggestions to overcome this problem.

quantum mechanics - Prove that angular momentum commute with the hamiltonian of a central force

I'm trying to prove that $$[\hat{L}_i,\hat{H}]=0$$ for $\hat{H}$ the hamiltonian of a central force $$\hat{H}=\frac{\hat{p}^2}{2m}-\frac{\alpha}{r}.$$

I'm getting this:

$$[\hat{L}_i,\hat{H}]=[\hat{L}_i,\frac{\hat{p}^2}{2m}-\frac{\alpha}{r}] = [\hat{L}_i , \frac{\hat{p}^2}{2m}] - [\hat{L}_i , \frac{\alpha}{r}].$$

Already prove that the first one is zero (it's a known result too), but the second one, I dont know what to do with it, there is no way for me. I was doing something like this:

$$[\hat{L}_i , \frac{\alpha}{r}] = [\epsilon_{ijk} r_j p_k , \frac{\alpha}{r}] = \epsilon_{ijk}r_j [p_k ,\frac{\alpha}{r}] + \epsilon_{ijk} [r_j , \frac{\alpha}{r}]p_k.$$

Obviously, the last term is zero, but the other one, I do something that has no sense to be zero. Have any idea for this? I'm I doing wrong? All of this is to prove that Laplace-Runge-Lenz operator commute with hamiltonian of a central force.

quantum mechanics - On the atomic level, how is incandescent light structured?

I want to know from the smallest possible originating structures how the light I see generated from heat is made by atoms themselves.

grid deduction - Double feature: In cold blood

_{This puzzle is part 5 of the Double feature series (first part here). The series will continue in "Double feature: Legal trouble".}

---

Rules of Nurikabe¹

• Shade some cells in the grid.


• Numbered cells are unshaded.


• Unshaded cells are divided into continuous regions, all of which contain exactly one number. The number indicates how many cells there are in that unshaded region.


• Regions of unshaded cells cannot be adjacent to one another, but they may touch at a corner.


• Shaded cells must all be orthogonally connected.


• There are no groups of shaded cells that form a $2\times2$ square anywhere in the grid.

Across
2. Fine threads of Nazis holding top-tier currency (7)
5. Turn over some journalism lab meat preserves (7)
7. Ancient Greek city moving to China (7)
9. Engulfed in boiling liquid (3)
11. Drive in a nail at the end of hotel's staircase support (8)
13. Taking in incomplete additional dose of sugar (8)

Down
1. Source of blue dye from Manila (4)
2. Tuscan city welcoming new brownish colour (6)

3. Law exam returned by business intelligence teacher (5)
4. To start off, Sukarno is Indonesia's first president (4)
6. Foolish spectator's heart becoming extremely lonely (7)
8. Document written up from Latin manuscript (5)
10. Three lost in the middle of sandbar region in Germany (4)
12. Treatments for infectious disease (4)

_{¹ Paraphrased from the original rules on Nikoli.}

Solve both puzzles to answer the question: What is found in blood?

Answer

Solution, with cryptic clue explanations:

So the answer is ANIMAL MITOCHONDRIA.

homework and exercises - Pressure and altitude

I am going to ask a simple question, for sure.

The pressure with respect to the altitude is given by this formula

Where

• sea level standard atmospheric pressure p0 = 101325 Pa


• sea level standard temperature T0 = 288.15 K


• Earth-surface gravitational acceleration g = 9.80665 m/s2.


• temperature lapse rate L = 0.0065 K/m


• universal gas constant R = 8.31447 J/(mol·K)


• molar mass of dry air M = 0.0289644 kg/mol

( from Wikipedia )

In addition to this, we have $L = \frac{g}{C_p}$ where $g = 9.80665\ m/s^2$ and $C_p$ is the constant pressure specific heat $ = 1007\ J/(Kg\ K)$

Understood and thence the above formula can be written in this simple way:

$$\boxed{P = P_0\left(1 - \alpha h\right)^{\beta}}$$

where

$\alpha = \frac{g}{C_p\ T_0} \approx 3.3796\cdot 10^{-5}\ m^{-1}$

$\beta = \frac{C_p\ M}{R} \approx 3.5081971$

On the other side we learnt from the elementary physics that the formula for the pressure is also given by

$$\boxed{P = P_0 + \rho g h}$$

Where $\rho$ is the air density $(1.23\ kg/m^3)$.

The question

First thing first: I assumed that

$$h = h_1 - h_0$$

is that correct? I mean it's the difference between two heights (maybe from a table and the ground, just to say).

Since the two formulae seems quite different, I tried with a numerical example in calculating the pressure in two points, with a difference of height about $0.18$ m and I got a really similar result.

Since the first formula is more technique, I think it's the correct formula but I would like to understand if one could pass from the first to the second or vice versa somehow.

Also I would like to know if there are cases in which I can use only the second formula or only the first formula!

grid deduction - Adderlink: Almost symmetrical

Rules:

Hey, this looks like slitherlink! In fact, it is two slitherlinks added together. So both of the two slitherlinks have clues in exactly the same locations as this grid below. Furthermore, a 2 means that there is either a 2 in one of the small grids and a 0 in the other, or a 1 in each. It cannot be a two in one grid and no number in the other. And it was symmetrical (except for the column on the right) until I killed ambiguity...

And in CSV:

- 2 - - - 2 - -
0 1 - 1 - 1 0 2
- - 2 - 2 - - -
2 1 1 3 1 2 1 2
- - - 5 - - - -
1 - 2 - 2 - 1 -
5 - - 2 - - 5 -
- - 1 - 1 - - -

Answer

1. The 2 in the upper left corner offers a starting point. A 2-0 split eliminates whichever fork holds the 2. Therefore it must be a 1-1 split.

2. The combinations for the blue shaded 5 and the 1 are locked as 13 and 02, since 03 cannot exist on an edge.

3. A 0 is impossible in the blue shaded cell, so the 1 must go on the left grid. The same is true of the red shaded cell 1 two cells to the right.

4. Because the red shaded cell in the right grid cannot be 2, some x's can be placed, showing that 1 is that cell's maximum. Looking at the left grid, if we tried 2, it would require that the blue cell also be 2 to prevent a premature loop, which is impossible because the blue shaded cell on the right grid must have at least 1. Therefore the red cells are a 1-1 split.

5. The blue cells must have at least 1, requiring that the red cells must also have at least 1. Both red and blue cells must be 1-1 splits.

6. The blue shaded 5 must be a 2-3 split. The x's to the lower left of this cell on both grids indicate that whichever side holds the 2 must have exits out of the lower right and upper left corners. The red cell must have the 1 on the grid where blue is 2. On the other grid, where blue is 3, there are only two exitable corners, upper right and lower right. In both grids, an exit from the 5 cell must be on the lower right. The dashed line can be drawn in both grids.

If the green cell on the right grid were 1, every scenario would result in an isolated strand or premature loop. The green cell on the right grid can't be 0 either, so it must be 2.

7. The dotted line on the left grid would produce a strand that can only exit at the circled node, producing a premature loop. The blue cell on the left grid cannot contain the 1, as it would result in immediate loops around the red cells.

8. On both grids, a 03 combination in the red cells produces a contradiction, because no more edges can be placed on the blue cells. The combinations of these cells must therefore be 13 and 02. 02 cannot be the combination in the right grid, because it would create an isolated strand.

9. Some standard slitherlink deductions produce the dashed edges on the lower right corner of each grid. The red and blue cells cannot both be 0 in either grid, so they must have opposite splits. Because they cannot be edges in either case, x's can be placed on edges of the red and blue cells.

If the dotted path exists one one grid, it must exist in the other as well. Since one of the green cells must be 0, the dotted paths can be eliminated.

10. Standard slitherlink deductions produce the dashed paths on the right grid. An x can be placed in the top edge of the blue cell in the left grid because it would produce an isolated loop. As a result, the right grid must have the 1 in the red cell, which gives us a 1-1 split on the blue cell.

11. The blue cells must be a 2-0 split, with the dotted lines appearing in one of the grids. In both grids, the top strand has two possible exit nodes, which have been circled. In both grids, the lower node creates a premature loop because of the maximum of 1 for the red cell. As a result, the dashed paths and 1-1 splits on the green shaded cells in both grids can be determined.

12. We can expand the dotted path that will appear in one of the grids, and as a result, place some x's in the blue cells. In both grids, the lower right strand only has one exit node (circled), necessitating the dashed paths. If the dotted path were on the left grid, neither side could have a 2 in the red shaded cell. Therefore the dotted path is on the right grid.

13. The blue cell cannot contain a 1 in the left grid, resulting in the dashed lines. A 1 in the red cell on the left grid would create isolated strands, so the 1 must be on the right.

14. The dashed path leads to the dotted path as the only possible way to close both loops.

fluid dynamics - Bulk transport of an aerosol by a blast wave

I'm trying to model the behaviour of aerosols in response to blast waves. (This is for visual effects.)

Intuitively, if a flare left a smoke trail above a field and then a bomb detonated elsewhere in the field, one would expect the smoke to be dispersed, to a degree varying with the bomb's energy (among other things.)

Studying blast waves, it has come to my attention that the associated overpressures have both a static pressure and a dynamic pressure component, as seen in this graph:

If I understand correctly, static pressure is due to thermodynamic motion while dynamic pressure is due to net motion of the medium.

The dynamic pressure component is fairly easy to understand: it's basically a wind, and figuring out how something moves in response to it should be straightforward.

The static pressure part is more confusing. I'm assuming this is ordinary longitudinal wave propagation without mass transport, similar to sound. Kind of like shoving one person in a lineup and only the 'shove' itself is transmitted.

One thing I find confusing is that static pressure is a directionless (scalar) quantity, so how does a pressurized region transmit energy only in the direction of propagation? Why doesn't the 'shove' just scatter? (I almost asked this as a separate question.)

Rather than worry about it, my first question is, would the static pressure component of a blast wave even effect any bulk transport of an aerosol?

general relativity - Covariant derivative and Leibniz rule

I read the Wikipedia page about the covariant derivative, my main problem is in this part:

http://en.wikipedia.org/wiki/Covariant_derivative#Coordinate_description

Some of the formulas seem to lead to contradictions, I assume I'm making some mistakes.

Here are some formulas from that page.

They define the Covariant derivative in the direction $\mathbf e_j$, denoted $\nabla_{\mathbf e_j}$ or $\nabla_j$ so that:

$\nabla_{\mathbf e_j} \mathbf e _i = \nabla_j \mathbf e _i = \Gamma^k_{\ \ ij}\mathbf e_k$

And define it so it obeys Leibniz' rule.

They then go on to show that

Where it seems they used

$\nabla_{\mathbf e_i} u^j = \frac {\partial u^j}{\partial x^i}$

But then later they define here: http://en.wikipedia.org/wiki/Covariant_derivative#Notation

$\nabla_{\mathbf e_i} u^j = \frac {\partial u^j}{\partial x^i} + u^k \Gamma^j_{\ \ ki}$

1) Is this a misunderstanding of mine or a problem in Wikipedia?

Also instead of the definition:

$\nabla_j \mathbf e _i = \Gamma^k_{\ \ ij}\mathbf e_k$

I saw in other places the Christoffel symbols defined so

$\partial_j \mathbf e _i = \Gamma^k_{\ \ ij}\mathbf e_k$

2) Is the covariant derivative of basis vectors the same as the regular derivative of a basis vector?or are these just two different definition of the Christoffel symbols?

Another contradiction I saw is that they write the following formula:

in the end of the section "Coordinate Description"

where you add here a Gamma for each upper index and subtract a Gamma for each lower index according to the rule written there.

According to this it seems to me that:

$\nabla_j \mathbf e _i = \partial_j \mathbf e _i - \Gamma^k_{\ \ ij}\mathbf e_k$

Which is also inconsistent with how they defined the covariant derivatove

3) Is this a contradiction or a confusion of mine?

Thank you very much, sorry it's so long

If it's a problem I can break the question up into two questions or something

Answer

1) The confusion comes from an omission of parentheses in these notations. In the first case we do indeed have $$\nabla_{\vec{e}_i}(u^j) = \frac{\partial u^j}{\partial x^i},$$ since $u^j$ is just one non-specific component of $\vec{u}$. In the second case, they mean to take the component after differentiating the tensor: $$u^j{}_{;i} = (\nabla_{\vec{e}_i} \vec{u})^j = \frac{\partial u^j}{\partial x^i} + u^k \Gamma^j_{ki}.$$ I am using arrows instead of Roman type to indicate vectors in order to emphasize which things are full vectors (which may have subscripts, e.g. $\vec{e}_i$ is the $i$-th vector in your basis) and which things are components.

2) There should only be one set of Christoffel symbols. In what context was this the definition?

Also, covariant derivatives reduce to partial derivatives on scalars.

3) The confusion here comes from the use of $i$ in $\vec{e}_i$ as a label on which basis vector is being used, rather than on which component of a given vector is in place. Think of $\vec{e}_i$ as one symbol, such as $\hat{x}$ or $\hat{y}$. (This is indicated by the Roman as opposed to Italic typeface in the question, which again I've switched to an arrow to draw attention to the vector nature of the symbol.) We use lower subscripts so they don't interfere with the upper superscripts that would label the components. That is, $\vec{e}_i$ has components $e_i^0$, $e_i^1$, etc. As an object whose components are indexed with upper indices, one uses a positive Christoffel term: $$(\nabla_j \vec{e}_i)^k = \partial_j e_i^k + \Gamma^k_{jl} e_i^l.$$ Note that $e_i^k = \delta_i^k$, which is a constant and therefore has vanishing partial derivative. Contracting the Christoffel symbol with the Kronecker delta in the second term leaves only $\Gamma^k_{ji}$, as expected.

mathematical physics - When does the $n$th bound state of a 1-D quantum potential have $n$ maxima/minima?

In Moore's introductory physics textbook Six Ideas that Shaped Physics, he describes a set of qualitative rules that first-year physics students can use to sketch energy eigenfunctions in a 1-D quantum-mechanical potential. Many of these are basically the WKB formalism in disguise; for example, he introduces a notion of "local wavelength", and justifies the change in amplitude in terms of the classical particle spending more time there. He also notes that the wavefunction must be "wave-like" in the classically allowed region, and "exponential-like" in the classically forbidden region.

However, there is one rule that he uses which seems to work for many (but not all) quantum potentials:

The $n$th excited state $\psi_n(x)$ of a particle in a 1-D potential has $n$ extrema.

This is true for the particle in a box (either infinite or finite), the simple harmonic oscillator, the bouncing neutron potential, and presumably a large number of other 1-D quantum potentials. It is not true, however, for a particle in a double well of finite depth; the ground state, which has a symmetric wavefunction, has two maxima (one in each potential well) and one minimum (at the midpoint between the wells).

The following questions then arise:

1. 
   

   Are there conditions can we place on $V(x)$ that guarantee the above quoted statement is true? For example, is the statement true if $V(x)$ has only one minimum? Is the statement true if the classically allowed region for any energy is a connected portion of $\mathbb{R}$? (The second statement is slightly weaker than the first.)

   
   


2. 
   

   Can we generalize this statement so that it holds for any potential $V(x)$? Perhaps there is a condition on the number of maxima and minima of $V(x)$ and $\psi_n(x)$ combined?

   
   

I suspect that if a statement along these lines can be made, it will come out of the orthogonality of the wavefunctions with respect to some inner product determined by the properties of the potential $V(x)$. But I'm not well-enough versed in operator theory to come up with an easy argument about this. I would also be interested in any interesting counterexamples to this claim that people can come up with.

Answer

I) We consider the 1D TISE $$ -\psi^{\prime\prime}_n(x) +V(x)\psi_n(x) ~=~ E_n\psi_n(x) .\tag{1}$$

II) From a physics$^{\dagger}$ perspective, the most important conditions are:

1. 
   

   That there exists a ground state $\psi_1(x)$.

   
   


2. 
   

   That we only consider eigenvalues $$ E_n ~<~\liminf_{x\to \pm\infty}~ V(x). \tag{2}$$ Eq. (2) implies the boundary conditions $$ \lim _{x\to \pm\infty} \psi_n(x)~=~0 .\tag{3}$$ We can then consider $x=\pm\infty$ as 2 boundary nodes. (If the $x$-space is a compact interval $[a,b]$, the notation $\pm\infty$ should be replace with the endpoints $a$ & $b$, in an hopefully obvious manner.)

   
   
   

Remark: Using complex conjugation on TISE (1), we can without loss of generality assume that $\psi_n$ is real and normalized, cf. e.g. this Phys.SE post. We will assume that from now on.

Remark: It follows from a Wronskian argument applied to two eigenfunctions, that the eigenvalues $E_n$ are non-degenerate.

Remark: A double (or higher) node $x_0$ cannot occur, because it must obey $\psi_n(x_0)=0=\psi^{\prime}_n(x_0)$. The uniqueness of a 2nd order ODE then implies that $\psi_n\equiv 0$. Contradiction.

III) Define

$$ \nu(n)~:=~|\{\text{interior nodes of }\psi_n\}|,\tag{4}$$

$$ M_+(n)~:=~|\{\text{local max points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)>0\}|,\tag{5}$$

$$ M_-(n)~:=~|\{\text{local max points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)<0\}|,\tag{6}$$

$$ m_+(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)>0\}|,\tag{7}$$

$$ m_-(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)<0\}|,\tag{8}$$

$$ M(n)~:=~|\{\text{local max points for }|\psi_n|\}|~=~M_+(n)+M_-(n), \tag{9}$$

$$ m(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)\neq 0\}|~=~m_+(n)+m_-(n), \tag{10}$$

$$\Delta M_{\pm}(n)~:=~M_{\pm}(n)-m_{\pm}(n)~\geq~0.\tag{11} $$

Observation. Local max (min) points for $|\psi_n|\neq 0$ can only occur in classical allowed (forbidden) intervals, i.e. oscillatory (exponential) intervals, respectively.

Note that the roles of $\pm$ flip if we change the overall sign of the real wave function $\psi_n$.

Proposition. $$ \Delta M_+(n)+\Delta M_-(n)~=~\nu(n)+1, \qquad |\Delta M_+(n)-\Delta M_-(n)|~=~2~{\rm frac}\left(\frac{\nu(n)+1}{2}\right).\tag{12} $$

Sketched Proof: Use Morse-like considerations. $\Box$

IV) Finally let us focus on the nodes.

Lemma. If $E_n

Sketched Proof of Lemma: Use a Wronskian argument applied to $\psi_n$ & $\psi_m$, cf. Refs. 1-2. $\Box$

Theorem. With the above assumptions from Section II, the $n$'th eigenfunction $\psi_n$ has $$\nu(n)~=~n\!-\!1.\tag{13}$$

Sketched proof of Theorem:

1. 
   

   $\nu(n) \geq n\!-\!1$: Use Lemma. $\Box$

   
   


2. 
   

   $\nu(n) \leq n\!-\!1$: Truncate eigenfunction $\psi_n$ such that it is only supported between 2 consecutive nodes. If there are too many nodes there will be too many independent eigenfunctions in a min-max variational argument, leading to a contradiction, cf. Ref. 1. $\Box$

   
   

Remark: Ref. 2 features an intuitive heuristic argument for the Theorem: Imagine that $V(x)=V_{t=1}(x)$ belongs to a continuous 1-parameter family of potential $V_{t}(x)$, $t\in[0,1]$, such that $V_{t=0}(x)$ satisfies property (4). Take e.g. $V_{t=0}(x)$ to be the harmonic oscillator potential or the infinite well potential. Now, if an extra node develops at some $(t_0,x_0)$, it must be a double/higher node. Contradiction.

References:

1. 
   

   R. Hilbert & D. Courant, Methods of Math. Phys, Vol. 1; Section VI.

   
   


2. 
   

   M. Moriconi, Am. J. Phys. 75 (2007) 284, arXiv:quant-ph/0702260.

   
   

--

$^{\dagger}$ For a more rigorous mathematical treatment, consider asking on MO.SE or Math.SE.

quantum field theory - Quantization surface in QFT

What does the Quantization Surface mean here?

Reference: H. Latal W. Schweiger (Eds.) - Methods of Quantization

Answer

The word quantization surface is not standard terminology. It apparently refer to a (generalized) Cauchy surface. A (generalized) Cauchy surface is a hypersurface on which the initial conditions are given for a well-posed initial value problem. Phrased differently, for given initial conditions on the Cauchy surface, there exists a unique solution for the evolution differential equation in an appropriate bulk region. (For more information, see also hyperbolic PDE and causal structure on Wikipedia.)

Here the evolution parameter of the system is usually referred to as time, although it doesn't have to be actual time. It could also be light cone time $x^{+}$, etc. Similarly, the word initial does not need to refer to actual time.

The evolution differential equation could be e.g. the Schrödinger equation, the Klein-Gordon equation, the Maxwell's equations, the Navier-Stokes equations, the Einstein field equations, the Arnowitt-Deser-Misner (ADM) equations of GR, etc.

In the context of quantization of a classical field theory, it is usually meant that the evolution problem is given in Hamiltonian form. Traditional Hamiltonian formulation (as opposed to manifestly covariant$^1$ Hamiltonian formulation) has a distinguished spacetime coordinate, which then plays the role of evolution parameter.

--

$^1$ There are various manifestly covariant Hamiltonian formalisms. See e.g. De Donder-Weyl theory; this Phys.SE post; and the following reference: C. Crnkovic and E. Witten, Covariant description of canonical formalism in geometrical theories. Published in Three hundred years of gravitation (Eds. S. W. Hawking and W. Israel), (1987) 676.

rocket science - At what fraction of the speed of light have people traveled?

I'm guessing that, this would be someone in a rocket or something... When they hit their top speed, at what fraction of $c$ are they traveling?

Answer

Maximum velocity attained by the Apollo spacecraft was 39,897 km/h which is $3.6\times 10^{-5}$ times the speed of light...

quantum field theory - Relation of $Q^2$ with the distance of interaction

In a process with only one Feynmann diagram at leading order such as $e^- \mu ^-$ scattering:

the photon propagator is \begin{equation} \frac{-i g_{\mu \nu}}{q^2}, \end{equation} with $q^2 = (p_3 - p_1)^2$. Since $q^2 <0$, we sometimes define the positive value $Q^2 = -q^2$. Apparently, it seems ok to relate $Q^2$ just as a quantity related to the exchange of energy and momentum between the particles.

But when reading about the running coupling constants in Halzen & Martin's book, they show that the running coupling constant of QCD has a small value for increasing $Q^2$, and the author relates high $Q^2$ to small distance interactions (i.e. asymptotically freedom) and small $Q^2$ to long distance interactions.

I cannot understand how the value of the exchanged 4-momentum relates to the interaction distance. Could someone shed some light in the subject?

Answer

In a sense it is based on the heisenberg uncertainty principle, HUP

In every interaction the four momentum transfer is made up of the three momentum transfer and the energy transfer. Looking at large Δp of the Q^2 transfer within the limits of the HUP the Δx gets smaller.

mathematics - Fastest way to collect an arbitrary army

I am looking for solution of this puzzle:

There is a kingdom with a square shape with sides of length 1. The castle is located at the center of the square. At the castle the king lives under the protection of one soldier. There are other soldiers, but each soldier is at his own house. Houses are spread throughout the area of the kingdom, and everyone is aware of where they are.

A war has been declared with a neighboring kingdom, and the king needs to collect all the soldiers at the castle urgently. To do this, he sends the soldier at the castle with the news to gather troops from the other places. Each soldier who learned the news, if necessary, can participate in the gathering. They can travel at the speed of 1.

What is the minimum time in which the king can collect all solders, regardless of their number and initial placement? (Number of soldiers is finite).

I see that the time should satisfy:

$$T_{min} \geq 2+\sqrt{2} = 3.4142...$$ This is the minimum time to gather solders from 4 houses in square's corners. $T = 2+\sqrt{2}$ feels to be the minimum: somehow with help of new solders everyone in a middle zone should be gathered while the very first solder goes via 3 corners.

But I have no idea how to find a generic algorithm for $T_{\mathrm{min}}$, which will achieve such a time for arbitrary number of houses.

P.S. Please be aware when you formulate your algorithm and prove it that there are so many possibilities that it is easy to miss something. I tried many times already, feeling that I've almost got it. But my solutions always had some small unclearness, and when I investigated the "proof" more explicitly and carefully I would find a gap in it...

P.P.S. I tried the following strategy:

1. When solder decides to go to some house he "reserves" it.


2. Once a solder is free (has delivered the news to another house) he takes the closest house from all unreserved and goes there. If there are several closest houses he chooses at random.

It doesn't work.

cosmology - Other explanation for cosmological redshift?

I'm interested if any of the following explanations have enough predictive capability to explain the observations we see today. The claim is that the Universe is not expanding, and that red-shift of light is caused by:

Though a couple of these are simply ridiculous, I'm still interested. Keep in mind that I am not a creationist, but am looking for reasons to see if any of these are valid or invalid.

Answer

Stellar motion would imply we're in the place where the Big Bang occurred while everything else is speeding away from us. I can see how this would be a worthwile explanation for a creationist, but it doesn't really make much sense building just on physics.

Doppler Effect has the same issue - it assumes that every star we can see moves in a speed proportional to distance with respect to us. That's extremely Sun-centric, and doesn't really work if you accept that e.g. stars in our galaxy orbit around its center, including us.

Gravitation of course does have effect on wavelength, but it's by far not enough for common stars, much less planets. And of course, the effect is only at work when you're in the gravitational well - this would imply that the Sun is again at the center of everything, with everything else orbitting around it.

Photon interaction - never been observed, doesn't really play well with accepted theories of electro-magnetism and photons. With the exception of - the expansion of space itself. Yes, photons are losing energy, that's why they're red-shifted. Even though there is a certain preference for photons to fill in similar states, this mostly manifests as a tendency for self-collimation. Interacting photons would not explain why the red-shift is correlated with distance either.

The others work in a similar fashion. Slowing of light would have huge effects on every interaction in the universe - which would mean that even things like water wouldn't be stable over time, they would change their behaviour rather a lot. Combined with the proposed age of ~6000 years for the universe, with humans from the very beginning, this is just absurd. The assumptions made by theory of relativity make this even more crazy - in relativity, speed of light is basically the maximum speed of propagation of information. So this hinges on older models of the speed of light, independent of the space-time itself.

Galaxies spiraling towards the Earth was already handled in the Gravitational red-shift explanation and the Doppler effect.

All in all, you can see why a creationist-proponent would like those theories. They mostly work on the basis that we are at the center of everything, and everything else revolves around us. It's the good old Earth-centric solar system model again, just dressed in the universe this time.

The problem is, addressing every single hypothesis put forward by people lacking in basic science practice and knowledge is fighting windmills. You can always imagine sillier and sillier explanations, and when people finally get tired of responding, you'll just say "See? The scientists have no explanation for that!" It's an uphill battle, and somewhat pointless, really.

Don't forget that when you propose an alternative theory/hypothesis, you need to explain everything the old theory did. If we find out that photons travel at lower or higher speeds than the speed of light, it might put relativity out of the question, but we'll still have to find out why it works almost all the time we've checked. It's possible to "overthrow" a well estabilished theory, but most often, you're only adding to what's already there. For example, Newton's law of gravity is wrong, but it's not thrown out outright - it's simply explained differently, and broadened to explain new observations. Most of the time, it works well enough. If it starts contradicting your observations (like the famous precession of Mercury issue), you start refining - in this case, Einstein pretty much redefined the whole universe. But it didn't make the old theory wrong outright - just incomplete. Creationist hypothesis tend to ignore this completely - they focus on one pet theory, and ignore the other explanations that need to be done to maintain a consistent model.

experimental physics - Near-room temperature DIY cloud chamber?

Conventional design of DIY cloud chamber uses IPA and cooling down to -26..-33°C (sometimes even lower, down to dry ice temperature).

What could be changed in design approach to shift operating point closer to room temperature? -15°C is much easier to maintain with single-stage TEC compared to -26..-30°C. And something around 0°C could be easily achieved in large volumes. On the other hand - heating (if needed) is much easier than cooling.

For me it is not clear why optimal temperature is so low, and hence unclear whether I should pick liquids with lower or higher boiling point (acetone / water or other non-toxic solvents).

Answer

If you prefer to avoid low temperatures, maybe you should use the original approach to supersaturation - pressure change (see e.g. https://www.citycollegiate.com/wilson_cloud_chamber.htm).

EDIT (5/26/2019): The OP asked for a "steady-state" solution in their comment. Such a solution is described in R.S.I. (Review of Scientific Instruments), vol. 10, March 1939, p. 91, http://hep.ucsb.edu/people/hnn/cloud/articles/ALangsdorfRSI10_91_1939.pdf). It includes heaters, vaporizer, and a refrigerator, which can apparently operate at ice (not dry ice) temperature. See also the literature review there.

quantum field theory - Representations of the Conformal Group in terms of the Poincare Group Reps

The Conformal group contains the Poincare group. Typically, if you take a representation of a group and then look at it as a representation of a subgroup, the representation will be reducible. I often hear that CFT's cannot have particles, and I have some understanding of this since $P_\mu P^\mu $ is not a Casimir of the conformal algebra. However, I would think reps of the conformal group should still be reducible representations of the Poincare group, and thus have some particle content.

Is it known how to decompose representations of the conformal group into reps of Poincare? Can we understand it as some sort of integral over masses that removes the scales of the theory? Are there any significant differences between reps of Poincare appearing in reps of the conformal group and the usual representations we are familiar with from QFT?

I'd appreciate any information or a reference that treats this thoroughly from a physics perspective.

special relativity - Does the speed of light have a range of speeds due to medium-dependency?

An EM particle-wave propagates in a vacuum at a constant speed $c$, independent of the source/receiver but dependent on most mediums it moves through. Since the vacuum is a medium, and if logically permissible, all mediums affect the speed of light, these affects of a medium can:

1. 
   

   alter the speed of light.

   
   


2. 
   

   possibly alter the distance light has to travel between source/receiver.

   
   
   


3. 
   

   subsequently alter the interval of time necessary to travel that distance.

   
   

Therefore the vacuum will affect the speed of light, and the speed light is measured to move at will depend on its medium, the vacuum. If each medium has a refractive index measuring this affect on light speed in a medium, then in a vacuum it would be at $c$. Is it possible that light has a true speed beyond what we observe in a vacuum? Also, since light travels the shortest distance/in the shortest possible interval of time through any medium, despite changes of lights speed in that medium,: does the speed of light actually slow down in a medium or just appear to slow down because of path deviations caused by a medium that alter distances thereby altering time intervals?

mathematics - How to get the sum 30 by choosing 3 numbers from these sequence?

I have a sequence of numbers - 1,3,5,7,9,11,13,15 I want to get the sum 30 by choosing 3 numbers from this sequence of numbers . The numbers can be repeated. Thank you

Answer

my choice is

11+13+3! = 11+13+6 =30

orbital motion - How intense a magnetic field would it take to keep an hypothetical iron-made moon orbiting around it?

The intention of the question is to provide an example of the weakness of gravity.

I imagine a horseshoe magnet located at the Earth's centre (remove the Earth), and a ferromagnetic moon. How intense would the magnetic field need to be to keep such a moon in orbit at the same distance that the moon is from the Earth?

everyday life - Why does blowing on a candle put it out but sucking doesn't?

Alternatively, why does the force created by blowing out air feel so much stronger than the force created by sucking in air?

Ok, so forget the human factor involved in blowing out candles. Consider a vacuum cleaner with a suction end and a blower end. Anyone who has tried it out notices the blower end creates a much stronger force than the suction end, despite the discharge being (more or less) equal at both ends.

Why does this happen?

Answer

In a blower, the air is directed along the axis of the blower as it exits, creating a high-pressure narrow cone. Exit pressure can also be multiple times of atmospheric pressure.

At a sucker entry, the low-pressure zone is fed by a much wider angle of atmospheric air at atmospheric pressure. Additionally, the underpressure can at most be 1x atmospheric pressure. therefore the inflow has an upper limit to its velocity.

Energy conservation in General Relativity

I understand that energy conservation is not a rule in general relativity, but I'd like to know under what circumstances it can still be possible. In other words, when is it possible to associate a potential energy to the gravitational field, so that the energy is constant in the evolution of the system?

Here are some examples, is there a convenient way to define energy in these scenarios?

• Just a system of gravitational waves.


• A point mass moving in a static (but otherwise arbitrary) space-time. Equivalent (if I'm not mistaken) to a test mass moving in the field of a second much larger mass, the larger mass wouldn't move.


• Two rotating bodies of similar mass.

Overall, I'm trying to understand what keeps us from associating a potential energy to the metric. When we break the time translation symmetry of a system by introducing an electromagnetic field, we can still conserve energy by defining an electromagnetic potential energy. Why can't we do the same when we break TT symmetry by making space-time curved?

Answer

There are a few different ways of answering this one. For brevity, I'm going to be a bit hand-wavey. There is actually still some research going on with this.

Certain spacetimes will always have a conserved energy. These are the spacetimes that have what is called a global timelike (or, if you're wanting to be super careful and pedantic, perhaps null) Killing vector. Math-types will define this as a vector whose lowered form satisfies the Killing equation: $\nabla_{a}\xi_{b} + \nabla_{b} \xi_{a} = 0$. Physicists will just say that $\xi^{a}$ is a vector that generates time (or null) translations of the spacetime, and that Killing's equation just tells us that these translations are symmetries of the spacetime's geometry. If this is true, it is pretty easy to show that all geodesics will have a conserved quantity associated with the time component of their translation, which we can interpret as the gravitational potential energy of the observer (though there are some new relativistic effects--for instance, in the case of objects orbiting a star, you see a coupling between the mass of the star and the orbiting objects angular momentum that does not show up classically). The fact that you can define a conserved energy here is strongly associated with the fact that you can assign a conserved energy in any Hamiltonian system in which the time does not explicitly appear in the Hamiltonian--> time translation being a symmetry of the Hamiltonian means that there is a conserved energy associated with that symmetry. If time translation is a symmetry of the spacetime, you get a conserved energy in exactly the same way.

Secondly, you can have a surface in the spacetime (but not necessarily the whole spacetime) that has a conserved killing tangent vector. Then, the argument from above still follows, but that energy is a charge living on that surface. Since integrals over a surface can be converted to integrals over a bulk by Gauss's theorem, we can, in analogy with Gauss's Law, interpret these energies as the energy of the mass and energy inside the surface. If the surface is conformal spacelike infinity of an asymptotically flat spacetime, this is the ADM Energy. If it is conformal null infinity of an asymptotically flat spacetime, it is the Bondi energy. You can associate similar charges with Isolated Horizons, as well, as they have null Killing vectors associated with them, and this is the basis of the quasi-local energies worked out by York and Brown amongst others.

What you can't have is a tensor quantity that is globally defined that one can easily associate with 'energy density' of the gravitational field, or define one of these energies for a general spacetime. The reason for this is that one needs a time with which to associate a conserved quantity conjugate to time. But if there is no unique way of specifying time, and especially no way to specify time in such a way that it generates some sort of symmetry, then there is no way to move forward with this procedure. For this reason, a great many general spacetimes have quite pathological features. Only a very small proprotion of known exact solutions to Einstein's Equation are believed to have much to do with physics.

quantum mechanics - What Planck units are limits?

Some Planck units, like time, length, or temperature, describe a physical maximum or minimum, at least approximately: you can't get hotter than the Planck temperature, measure anything smaller than Planck time or length, etc. Others, like the Planck charge, Planck momentum, or Planck energy, seem to have no associated maxima. Which units are of what type, and is there a reason that some are limits while others are in the 'middle' of a spectrum of possibilities? Are there limits to physical units which are distinct from the associated Plank unit?

gravity - Why can't General Relativity be written in terms of physical variables?

I am aware that the field in General Relativity (the metric, $g_{\mu\nu}$) is not completely physical, as two metrics which are related by a diffeomorphism (~ a change in coordinates) are physically equivalent. This is similar to the fact that the vector potential in electromagnetism ($A_\mu$) is not physical. In electromagnetism, the equations can be written in terms of physical (i.e. gauge invariant) quantities -- the electric and magnetic fields. Why can't Einstein's equations similarly be written in terms of physical variables? Is it just that nobody has been able to do so, or is there some theorem/argument saying that it can't be done?

EDIT: Let me rephrase: Prove/argue that there is no explicit prescription that can be given which would uniquely fix coordinates for arbitrary physical spacetimes. I.e., show that there is no way to fix gauge in the full theory of general relativity (unlike in E&M or linearized GR where gauge can be fixed).

quantum field theory - The superconformal algebra

1. 
   

   How does one derive the superconformal algebra?

   
   


2. 
   

   Especialy how to argue the existence of the operator $S$ which doesn't exist either in either the supersymmetric algebra or the conformal algebra?

   
   



3. 
   

   What is the explanation of the convention of choosing If $J _ \alpha ^\beta$ as a generator of $SU(2) \times SU(2)$ and $R^A _B$ and $r$ are the generators of the $U(m)$ R-symmetry ( $m \neq 4$)? How does one get these $R$ and $r$?

   
   


4. 
   

   One further defines $R_1, R_2$ and $R_3$ as the generators of the Cartan subalgebra of $SU(4)$ (the $N=4$ R-symmetry group) where these are defined such that $R_k$ has $1$ on the $k^{th}$ diagonal entry and $-1$ on the $(k+1)^{th}$ diagonal entry.

   
   

Using the above notation apparently one of the $Q-S$ commutation looks like,

$\{ S^A_\alpha , Q^\beta _B\} = \delta ^A_B J^\beta _\alpha + \delta ^\beta _\alpha R^A_B + \delta ^A_B \delta ^\beta _\alpha (\frac{H}{2} + r \frac{4-m}{4m})$

I would like to know how the above comes about and why the above equation implies the following equation (..each of whose sides is defined as $\Delta$)

$2\{ Q_4 ^- , S^4_ - \} = H - 3J_3 + 2\sum _{k=1} ^ 3 \frac{k}{4} R_k$

I can't understand how to get the second of the above equations from the first of the above and why is this $\Delta$ an important quantity.

{Is there any open source online reference available which explains the above issues?

special relativity - Why doesn't $E = mc^2$ contradict the conservation of mass principle?

I was watching this very interesting video: The mathematics of weight loss

and at 12:35 the presenter says: "(...) people think that you can turn atoms into energy. It's one of the founding principles of modern chemistry: you cannot turn atoms into pure energy. It's called the conservation of mass."

Einstein's famous equation immediately came to my mind: $$E = mc^2.$$

Isn't that equation saying you can turn mass into energy and vice versa? Doesn't that contradict the conservation of mass principle then?

quantum mechanics - Does photon interference violate causality?

Suppose a hydrogen atom jumps down an energy level and emits a photon and that photon later goes through a double slit and gets absorbed by one atom of a screen making it jump up an energy level. Doesn't the photon have an extreme uncertainty in position and exist as a wave that travels in all directions until it suddenly collapses to have an uncertainty in position of about the diameter of an atom and gets completely absorbed by a single atom in the screen. That seems to violate causality because it's guaranteed that the screen will detect at most one photon from that atom so the detection by one atom sends the information to another atom faster than light to not detect it.

One possible explanation is that the universe follows de Broglie–Bohm theory and despite the fact that information can travel faster than light, it can never be sent in such a way to create a paradox. That in turn might be because those laws are a simulation by the fundamental laws which is that the universe is a Conway's game of life so it could not have simulated a universe with laws that allow for a paradox.

Another possible explanation is that the universe follows a different theory where it was already determined ahead by a hidden variable which atom will absorb the photon before the atom actually absorbs it. Maybe the photon emits a classical wave which stays a classical wave which gets gradually absorbed classically and the probability density of any atom on the screen jumping up an energy level at any time is completely determined by the rate of absorption of the classical wave so the jump down in energy level of the hydrogen atom actually can causing two atoms in the screen to jump up an energy level. Maybe it can actually be determined ahead which atoms will jump up in energy level but it seems random because the universe is chaotic. If that were the case, we would never be able to tell that the universe doesn't follow the accepted theory because we couldn't tell which jump ups in energy level of an atom in the screen came from which jump downs in energy level of a hydrogen atom. Just because a theory is consistent with observations doesn't mean the universe actually follows that theory.

faq - Book recommendations

Every once in a while, we get a question asking for a book or other educational reference on a particular topic at a particular level. This is a meta-question that collects all those links together. If you're looking for book recommendations, this is probably the place to start.

All the questions linked below, as well as others which deal with more specialized books, can be found under the tag resource-recommendations (formerly books).

If you have a question to add, please edit it in. However, make sure of a few things first:

1. The question should be tagged resource-recommendations



2. It should be of the form "What are good books to learn/study [subject] at [level]?"


3. It shouldn't duplicate a topic and level that's already on the list

Related Meta: Do we need/want an overarching books question?

newtonian mechanics - What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?

As the question states, what is our current best machine for translating falling gravitational potential energy, such as a large weight, into launching a smaller projectile vertically? A lever? A trebuchet? What sorts of efficiency levels have been achieved?

What I'm looking for is the best way to translate:

$$ g * h * m_{\textrm{weight}} * \textrm{efficiency} = \frac{1}{2} * m_{\textrm{projectile}} * v^2. $$

Here's an example: I want to launch a 10kg mass at 100m/s using only gravitational energy (no chemical rockets, no rail guns, etc.). What's our most efficient machine to do this, and what percentage of the energy of the falling weight will actually be transferred to the projectile? This would get me an idea of the size of the weight and how much it has to fall to accomplish this. The only machine I know of that does this on a large scale and for decent size masses is a trebuchet, so this might be a good starting point unless there are others I'm unaware of.

Answer

This question was asked four months ago, but none of the existing answers mentions trebuchets.

To my knowledge the trebuchet design is the only design that is purely mechanical. Other projectile throwing devices store elastic energy and on release transfer that to the projectile.

So a trebuchet it is.
(In effect 'lever' and 'trebuchet' are synonymous. A trebuchet gets its leverage by being a lever (pun intended)).

It may be worthwhile to research the noble art of pumpkin chunkin'. There's a town in the US where there is a yearly pumpkin chunkin' contest that also features a trebuchet category.

Can a trebuchet deliver close to 100% efficency?
For one thing, frictional losses can be kept quite low, that's good.
The problem is this: to throw the projectile the lever arm must be accelerated to a large angular velocity. After the projectile has been released the lever arm is swinging violently. That kinetic energy of the lever arm is energy from the counterweight that was not transferred to the projectile.

The trebuchet design involves trade-offs. A longer lever arm gives the potential for faster throws, but a longer arm also has a bigger moment of inertia.

An ideal trebuchet would transfer all of its kinetic energy to the projectile, so that right after releasing the projectile it would just sit there, nearly motionless. I'm not aware of any trebuchet design that achieves that.

To reduce the inefficiency the mass of the lever arm must be as low as possible. That is what the pumpkin chunkers do: they make the lever arm as flimsy as they dare. For every throw they stand well back, as their machines tend to self-destruct when the trigger is pulled.