Photon vs electromagnetic waves

Suppose an electron makes a single transition from higher to lower energy level releasing energy. Would that energy be released in exactly one photon equal to $h\nu$? Also, is saying "one photon is released" equivalent to saying "one electromagnetic wave of $\nu$ frequency is released"?

quantum mechanics - What is Quantization?

In classical mechanics you construct an action (involving a Lagrangian in arbitrary generalized coordinates, a Hamiltonian in canonical coordinates [to make your EOM more "convenient & symmetric"]), then extremizing it gives the equations of motion. Alternatively one can find a first order PDE for the action as a function of it's endpoints to obtain the Hamilton-Jacobi equation, & the Poisson bracket formulation is merely a means of changing variables in your PDE so as to ensure your new variables are still characteristics of the H-J PDE (i.e. solutions of the EOM - see No. 37). All that makes sense to me, we're extremizing a functional to get the EOM or solving a PDE which implicitly assumes we've already got the solution (path of the particle) inside of the action that leads to the PDE. However in quantum mechanics, at least in the canonical quantization I think, you apparently just take the Hamiltonian (the Lagrangian in canonical coordinates) & mish-mash this with ideas from changing variables in the Hamilton-Jacobi equation representation of your problem so that you ensure the coordinates are characteristics of your Hamilton-Jacobi equation (i.e. the solutions of the EOM), then you put these ideas in some new space for some reason (Hilbert space) & have a theory of QM. Based on what I've written you are literally doing the exact same thing you do in classical mechanics in the beginning, you're sneaking in classical ideas & for some reason you make things into an algebra - I don't see why this is necessary, or why you can't do exactly what you do in classical mechanics??? Furthermore I think my questions have some merit when you note that Schrodinger's original derivation involved an action functional using the Hamilton-Jacobi equation. Again we see Schrodinger doing a similar thing to the modern idea's, here he's mish-mashing the Hamilton-Jacobi equation with extremizing an action functional instead of just extremizing the original Lagrangian or Hamiltonian, analogous to modern QM mish-mashing the Hamiltonian with changes of variables in the H-J PDE (via Poisson brackets).

What's going on in this big Jigsaw? Why do we need to start mixing up all our pieces, why can't we just copy classical mechanics exactly - we are on some level anyway, as far as I can see... I can understand doing these things if they are just convenient tricks, the way you could say that invoking the H-J PDE is just a trick for dealing with Lagrangians & Hamiltonians, but I'm pretty sure the claim is that the process of quantization simply must be done, one step is just absolutely necessary, you simply cannot follow the classical ideas, even though from what I've said we basically are just doing the classical thing - in a roundabout way. It probably has something to do with complex numbers, at least partially, as mentioned in the note on page 276 here, but I have no idea as to how to see that & Schrodinger's original derivation didn't assume them so I'm confused about this.

To make my questions about quantization explicit if they aren't apparent from what I've written above:

a) Why does one need to make an algebra out of mixing the Hamiltonian with Poisson brackets?

(Where this question stresses the interpretation of Hamiltonian's as Lagrangian's just with different coordinates, & Poisson brackets as conditions on changing variables in the Hamilton-Jacobi equation, so that we make the relationship to CM explicit)

b) Why can't quantum mechanics just be modelled by extremizing a Lagrangian, or solving a H-J PDE?

(From my explanation above it seems quantization smuggles these idea's into it's formalism anyway, just mish-mashing them together in some vector space)

c) How do complex numbers relate to this process?

(Are they the reason quantum mechanics radically differs from classical mechanics. If so, how does this fall out of the procedure as inevitable?)

Apologies if these weren't clear from what I've written, but I feel what I've written is absolutely essential to my question.

Edit: Parts b) & c) have been nicely answered, thus part a) is all that remains, & it's solution seems to lie in this article, which derives the time dependent Schrodinger equation (TDSE) from the TISE. In other words, the TISE is apparently derived from classical mechanical principles, as Schrodinger did it, then at some point in the complicated derivation from page 12 on the authors reach a point at which quantum mechanical assumptions become absolutely necessary, & apparently this is the reason one assumes tons of axioms & feels comfortable constructing Hilbert spaces etc... Thus elucidating how this derivation incontravertibly results in quantum mechanical assumptions should justify why quantization is necessary, but I cannot figure this out from my poorly-understood reading of the derivation. Understanding this is the key to QM apparently, unless I'm mistaken (highly probable) thus if anyone can provide an answer in light of this articles contents that would be fantastic, thank you!

quantum field theory - What are the actual transformation properties of Dirac spinors $u_sigma(p)$?

Let $u_\sigma(p)$ be a Dirac spinor. As far as I know, it transforms under changes of reference frame according to $$ u_\sigma(p)=S(\Lambda)u_\sigma(\Lambda p)\tag{1} $$ where the $\sigma$ label doesn't mix. Why is this? shouldn't the polarisations be frame-dependent? After all, the "spin quantisation axis" is frame dependent.

Put it another way: the relation above is equivalent to $$ U(\Lambda)|p,\sigma\rangle=|\Lambda p,\sigma\rangle\tag{2} $$ without $\sigma$ mixing. To me, one-particle states should transform according to $$ U(\Lambda)|p,\sigma\rangle\stackrel?=\sum_{\sigma'}D_{\sigma\sigma'}(\Lambda)|\Lambda p,\sigma'\rangle\tag{3} $$ though this is obviously not the case. If this were true, then we would have $$ u_\sigma(p)\stackrel?=S(\Lambda)\sum_{\sigma'}D_{\sigma\sigma'}(\Lambda)u_{\sigma'}(\Lambda p)\tag{4} $$ instead of $(1)$.

Question: why don't polarisations mix under Lorentz transformations (in neither $(1)$ nor $(2)$)?

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EDIT

As pointed out by Blazej, the spin components $\sigma$ do mix under Lorentz transformation, and the correct law is $$ u_\sigma(p)=\sum_{\sigma'}M_{\sigma\sigma'}u_{\sigma'}(\Lambda p) $$ for some matrix $M$ (which is actually related to a Wigner's little group matrix, but whose form is not that relevant to me; the important part is that the $\sigma$ components mix, and not what is the actual matrix that mixes them).

My concern is that this is not what I find online: for example, see this answer in physics.SE (last equation). Also, see this wikipedia article. Who is right and who is wrong?

Answer

We can answer this question in more generality: what are the transformation properties for the polarizations associated to a massive field? (I am restricting to massive because the polarizations for massless particles with spin greater or equal than 1 require the discussion of gauge invariance; I'll leave this to another day).

The polarizations can be defined with no reference to the field equation: they are defined as matrix elements for a field between the vacuum and the one-particle states: $$ \langle 0| \psi_\ell(0) | \mathbf{p},\sigma\rangle \propto u^\sigma_\ell(\mathbf{p}) $$ where the proportionality factor is a constant (known as wavefunction renormalization) in the relativistic normalization for the state $|\mathbf{p},\sigma\rangle$. Here the index $\ell$ is the index in the Lorentz representation carried by the field $\psi_\ell$ (for example, $\ell=\mu$ for a 4-vector, $\ell=\alpha$ for a dirac spinor, more generally $\ell=(\alpha,\beta)$ is a pair of indexes in the $(A,B)$ representation of $SO(3,1)\sim SU(2)\times SU(2)$). The polarizations carry also another index, $\sigma$, which represents the spin of the particle. More precisely, it is the little-group index carried by the particle, either the spin or the helicity. This definition immediately tells us how $u^\sigma_\ell$ transforms given that $\psi$ carries a Lorentz representation $$ U(\Lambda)\psi_\ell(x) U^{-1}(\Lambda)=D(\Lambda^{-1})_{\ell\ell^\prime}\psi_{\ell^\prime}(\Lambda x) $$ and the one-particle state transform with respect to the little group with a Wigner rotation $$ U(\Lambda)|\mathbf{p},\sigma\rangle = \mathcal{L}_{\sigma^\prime\sigma}(W(\Lambda),p)|\mathbf{p}_\Lambda, \sigma^\prime\rangle $$ which imply $$ D(\Lambda)_{\ell\ell^\prime}u^\sigma_{\ell^\prime}(\mathbf{p})=u^{\sigma^\prime}_{\ell}(\mathbf{p}_\Lambda)\mathcal{L}_{\sigma^\prime\sigma}(W(\Lambda),p) $$ where $\mathbf{p}_\Lambda$ is 3-vector part of the 4-vector $\Lambda p$. (One way to read this equation is by saying that the polarizations transform on the left under Lorentz and on the right under the little-group transformations: this has to be so such that one can convert the Lorentz indexes of correlations functions of fields into the little group indexes of the scattering matrix $S$ as dictated by the LSZ reduction formula).

This answer your question. But in fact we can say more: those transformation properties are constructive since they allow you to determine explicitly the polarization (and show that they satisfy certain equations, e.g. Dirac for spin-1/2,...) as it was shown long ago in the 60's by Weinberg (see the discussion in his textbook on QFT vol.1 chapter 5). For example, take $k=(m,\mathbf{0})$ (for a massive particle) and apply the canonical Lorentz transformation $L=\Lambda$ that brings it to $p=(E,\mathbf{p})=L\,k$. In this case the Wigner rotation is trivial, $W=1$, and hence $$ u^{\sigma}_{\ell}(\mathbf{p})=D_{\ell\ell^\prime}(L)u^\sigma_{\ell^\prime}(\mathbf{0}) $$ which means that we just need to know them at zero momentum (or, for massless particles with respect to the reference vector used for the little group). Moreover, for an abitrary rotation $\Lambda=R$ we have $W=R$ for any $p$ so that $$ D_{\ell\ell^\prime}(R)u_{\ell^\prime}^\sigma=u^{\sigma^\prime}_\ell(\mathbf{0})\mathcal{L}_{\sigma^\prime\sigma}(R) $$ Taking diagonal the rotations around the $z$ axis, $\mathcal{L}_{\sigma^\prime\sigma}(R_z)=e^{i\sigma\theta}\delta_{\sigma^\prime\sigma}$, the polarization can be extracted by the infinitesimal $z$-rotation $$ D_{\ell\ell^\prime}(J^z) u^\sigma_{\ell^\prime}{\mathbf{0}}=\sigma u^\sigma_{\ell}(\mathbf{0})\,. $$

Let me just give an instructive example: a spin-1 massive state (where $\ell$ is an index in the irrep $(1/2,1/2)$, that is $\ell=\mu=0,1,2,3$ is 4-vector index) has a the 3-dimensional representation of $J^z$ where $D(J^z)_{ij}=-i\epsilon_{3ij}$ and $D(J^z)_{00}=D(J^z)_{i0}=0$ so that the polarizations $$ \epsilon_{\mu}^{\pm}(\mathbf{0})=\frac{1}{\sqrt{2}}(0,\mp 1,-1,0)^T\,,\qquad \epsilon_\mu^0=(0,0,0,1)^T\,. $$ solve the desired equations above. Clearly they satisfy $p^\mu \epsilon_\mu^\sigma=0$, once we boost $(m,\mathbf{0})$ to $p$. Therefore, the matrix element $$ \Psi_\mu(x)\equiv \langle 0| \psi_\mu(x) | \mathbf{p},\sigma\rangle =e^{ipx} \langle 0| \psi_\mu(0) | \mathbf{p},\sigma\rangle \propto e^{ipx} u^\sigma_\mu(\mathbf{p}) $$ satisfies $$ (\square+m^2)\Psi_\mu(x)=0\,,\qquad \partial_\mu\Psi^\mu(x)=0 $$ which are derived rather that used as starting point.

The same can be done for any spin, in particular for the spin-1/2 and see that they solve the Dirac equation. More generally, since the Lorentz group $SO(3,1)\sim SU(2)_A \times SU(2)_B$ the angular momentum is given by $J=J_A+J_B$ which tells us that $$ D(J_A)_{\alpha\alpha^\prime}u^\sigma_{\alpha^\prime\beta}(\mathbf{0})+D(J_B)_{\beta\beta^\prime}u^\sigma_{\alpha\beta^\prime}(\mathbf{0})=\mathcal{L}(J_S)_{\sigma^\prime\sigma} u^{\sigma^\prime}_{\alpha\beta}(\mathbf{0}) $$ where $\ell=(\alpha,\beta)$. In other words, the polarizations are (proportional to) the Clebsch Gordan coefficient for the spin $S$ found inside $A\otimes B$ $$u^\sigma_{\alpha\beta}(\mathbf{0})\propto C^{(S)\sigma}_{(AB)\alpha\beta} $$ Several of the properties of the polarizations come indeed from the unitary condition for these Clebsch Gordan coefficients.

quantum mechanics - Could teleportation ever really happen?

Could any form of matter teleportation every really happen?

If so, what form would it take? Some science-fiction talks about literally sending the atoms from one location to another; others talk of creating a duplicate and destroying the original.

Question originally added here.

Update: there are a couple of comments asking me to specify what I mean by teleportation. I'm talking about stuff of Star Trek - taking an object or person and transporting them almost instantly across great distances (thousands of KMs) by:

1. breaking them up at the atomic (or smaller) level


2. transporting those atomic components (or the information, "recipe" to recreate) and


3. (re-)assembling at the new location.

Answer

I don't think it's possible in the way you mean it:

Option 1

Decompose, move, rebuild

Decomposition is possible as long as we can consider the matter classical. If any quantum state is relevant, it will be automatically destroyed in the process of measuring the state of each atom in order to be able to rebuild later. This will lead to a large increase in entropy (which is relevant in the last step).

The movement of a stream of atoms is possible, but not simple at all, especially if you need to focus it in a spot. The energy required, though is probably very similar to the energy required to transport the matter classically.

Rebuilding is possible (given enough technology, that is), but it would be enormously expensive in terms of energy because of the need of reversing the increase of entropy in the first step, locally.

All in all, it's probably possible, but much more expensive in terms of energy than simply moving the body, so it's not clear to me how useful it would be.

Also, were the matter being transported a living being, there are a number of ethical questions involved: would the person rebuilt be still the same person as the one killed by the first step of the process?

Option 2

Use quantum teleportation

This, as far as I know, requires a quantum entanglement between the mass and a perfect copy of the mass. Although this is probably possible for simple systems, I am not really sure it represents the kind of teleportation you are looking for. It's a teleportation of quantum states, not of matter.

Option 3

Abuse Quantum Mechanics

This is far off, and there's nothing suggesting that it's possible, but if we found a way of interfering radically with the probabilities of quantum mechanical measurements, we could leverage the indeterminacy in the body's position to make it move. It really depends on us understanding better how the wave function actually collapses.

Option 4

Abuse General Relativity

This is also far off, but if we were to open a stable wormhole between two positions then it should be possible to step from one place to another instantly. This merely requires using currently unknown forms of exotic matter to generate the holes, because all the wormholes that we know of, in GR, are actually of really small diameter and also apparently a not very healthy place to be because of extreme gravity, radiation and so on.

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A couple of good resources (divulgative but physically accurate):

Physics of the Impossible: A Scientific Exploration of the World of Phasers, Force Fields, Teleportation and Time Travel

The Physics of Star Trek

experimental physics - Distribution and detection of dark matter

I feel in the dark (no pun intended :),

I'm sure most of you are familiar with this image I pulled from wikipedia:

The caption reads: "3D map of the large-scale distribution of dark matter, reconstructed from measurements of weak gravitational lensing with the Hubble Space Telescope."

I'm wondering what makes us think that we should be able to detect dark matter on Earth and if it's possible that we could be in a region where dark matter doesn't exist, or how do we know that we lie in a region where dark matter lies? Would it be easier to detect dark matter particles in other places?

Can someone shed some light in this for me?

Answer

I was led to your question after answering a related question.

We think that the dark matter in the vicinity of the solar system has a rather smooth distribution and that there actually isn't very much of it - in terms of a density. It is estimated that the local dark matter density is around $\sim 0.01$ $M_{\odot}$/pc$^3$ (Garbari et al. 2012, Bovy & Tremaine 2012) corresponding to only a few $10^{-22}$ kg/m$^3$. For comparison, the density of the interplanetary medium is abut 100 times greater.

The reasons that dark matter is unimportant on solar system scales, but dynamically important on galaxy scales and utterly dominant on galaxy cluster scales is that the total gravitating mass goes up with volume (i.e. the cube of the length scale).

The Galaxy probably has an approximately spherically symmetric distribution of dark matter. Dark matter does not readily "accumulate", either in the disk of our Galaxy or around individual stars or solar systems. It interacts very weakly with normal matter and is primarily influenced by gravity. The Earth's gravity for example is far too small to make a local concentration of dark matter. The local dark matter would be moving in the Galactic potential at speeds similar to that of the Sun around the Galaxy ($\sim 250$ km/s); much faster than the 11 km/s escape velocity of the Earth or the 42 km/s escape velocity of the solar system at the Earth's orbit.

The net outcome of this is that we do expect there to be some dark matter present everywhere on Earth, and that it should have roughly the density expected for its orbital radius in the Galaxy. However, yes, there are places where dark matter could be more concentrated - the centres of large spiral galaxies for example. There are even calculations that suggest that dark matter could be captured in the centres of stars (e.g. Vincent et al. (2015)).

Does light accelerate or slow down during reflection?

After all, it does change direction when reflection occurs. So shouldn't it also accelerate? And since the acceleration cannot increase the speed of light, mustn't it slow down?

Answer

Light does not slow down during a reflection.

Light is a signal disturbance in electric and magnetic fields. These disturbances propagate through space at a fixed speed $c$ in vacuum. The situation is completely analogous, in a mathematical sense, to a wave pulse that is sent along a string. When the pulse encounters a boundary, it flips direction, and may or may not change phase depending on the type of boundary encountered. For good graphical depictions of this phenomenon, visit this page.

If you emit a pulse of light at a distance of 1 meter from a plane mirror, and measure the amount of time it takes for the signal to return, you will find that it is 2 meters / $c$, neglecting refractive effects of the air. In this sense, we say that the light has not slowed down, even though it has changed direction in the middle of its journey.

electromagnetism - Can electromagnetic fields be used to deconstruct and reconstruct atoms?

I was thinking one day and came up with a theory after reading about how scientists were studying anti-matter by using electro magnetic fields to separate matter from the anti-matter they made. It got me thinking would it be possible to use very powerful electromagnetic fields to break down the atomic structure of objects or build things in this way?

Is this atomic reconstruction with electromagnetic fields theoretically possible? That is, is it theoretically possible to use electromagnetic fields generated by a machine to separate the parts of an atom thereby deconstructing an object on the atomic/subatomic level?

I'm not asking about breaking molecular bonds but rather actual atoms apart. If it's possible to break atoms apart with electromagnetic fields, is it also possible to use a similar process to assemble them?

waves - Could sound be considered a kind of renewable energy?

Is sound energy useful as a source for generating electricity? If so, could it be a renewable resource?

Answer

I think you might be a little confused.

The phrases 'renewable energy' and 'un-renewable energy' are used to refer to industrial sources of energy. These industrial sources include Wind, Solar, Wave, and Nuclear power, and traditional fossil fuels (coal, oil, natural gas etc.).

If a source of power is renewable, it is not depleted (used up) when utilised - for example, the wind doesn't run out just because you put a few turbines up.

These industrial sources are not forms of energy themselves - they produce energy (heat energy like in a nuclear reactor or kinetic energy like in wave and wind power), but "solar power" or "wind power" are not a forms of energy. They are sources. This link gives a good overview of the different forms of energy.

Sound energy is associated with the vibrations of matter (sound waves vibrate air particles - that's how you can hear), so is a form of kinetic energy (the energy all things have when they move). It is not an energy source, so it doesn't make sense to call it "non-renewable" or "renewable".

black holes - How does virtual particle explanation of Hawking radiation contradict with consistent loop description?

the picture of virtual particle pairs is categorically not the right way to think about Hawking radiation. Quite obviously it must be wrong, because it is a loop level effect, and loops in QFT have to close, which they don't in this heuristic picture. (in the comment by Edward Hughes from What are the virtual particles generated during the Hawking radiation?)

Can anyone explain how virtual particle explanation of Hawking radiation goes against consistent loop description - why loop is not being closed?

Answer

It is true that vacuum loops of particle antiparticle generation can have a measurable effect on crossections and other measurements , for interactions that already occur,in higher order feynman diagrams.

For example:

Two example eighth-order Feynman diagrams that contribute to the electron self-interaction. The horizontal line with an arrow represents the electron while the wavy lines are virtual photons, and the circles represent virtual electron–positron pairs.

The loops are particle antiparticle pairs, so they have the quantum numbers of the vacuum. The energy for their mathematical existence comes from the line of the real electron, which has an energy and momentum vector on mass shell.

It is a heuristic picture to talk about vacuum loops generating e+e- pairs at the black hole horizon, because the leaving partner (e+ or e-) takes away energy and momentum, which must be given by the ambient particles about the horizon, eventually the energy lost from the black hole. This is a better diagram

Feynman diagram of the Hawking process for particle creation by a black hole. The heavy, zigzag solid lines represent the worldlines of electrons ( − e ) and of a positron (+ e ). The heavy, vertical dashed line on the left represents the event horizon of the blackhole, and the light, vertical dashed line on the right represents an effective emissive surface, for the case of electron emission. Points A and B represent pair creation and pair annihilation, respectively [12]. The “Compton layer” has a thickness on the order of the Compton wavelength, h/mc . The coordinate system being used here is that of a freely falling observer orbiting around the black hole just outside of its horizon.

Here is how Hawking% describes it :

Just outside the event horizon [of a black hole] there will be virtual pairs of particles, one with negative energy and one with positive energy. The negative particle is in a region which is classically forbidden but it can tunnel [emphasis added] through the event horizon to the region inside the black hole where the Killing vector which represents time translations is spacelike. In this region the particle can exist as a real particle with a timelike momentum vector even though its energy relative to infinity as measured by the time translation Killing vector is negative. The other particle of the pair, having a positive energy, can escape to infinity where it constitutes a part of the thermal emission described above.”

The vacuum loops , assumed to exist due to the uncertainty principle, are an easy graphic for the appearance of particle pairs, but not really true, since in order for a real particle to appear energy and momentum must be supplied and there will no longer be a loop.

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%page 202 of refered as ref. 1 here

nuclear physics - Spherical ground state of nucleus

If a nucleus has spin-parity 0$^+$ in its ground state (even-even nucleus), does this necessarily mean that the ground state is a spherical one? Or does this apply only to closed shell nuclei? E.g. compare $^{232}$Th and $^{208}$Pb.

Answer

If a quantum system has angular momentum of zero, then it is necessarily spherical. One can see this as rob mentioned by looking at the Wigner-Eckart theorem and seeing that only scalar moments are allowed. Another way of seeing this is remembering that the total angular momentum operator $J$ is the generator of rotation transformations. For some rotation by an angle $\delta\theta$, we have the rotation operator $$ U(\delta\theta) = e^{i\delta\theta\cdot J} $$ Here the dot means that we choose some axis about which to rotate, and we take the $J$ projection on that axis. The effect of the rotation operator is $$\Psi(r,\theta+\delta\theta) = U(\delta\theta) \Psi(r,\theta)$$

In the case where $J=0$, then we have $e^{i\delta\theta\cdot 0}=1$ for any rotation angle. This means that no matter how we rotate the state $\Psi$, it looks the same, which is exactly what is meant by spherical symmetry.

Now to the question of intrinsic deformation. The way deformation arises is easiest to understand if you think of two particles in a spherical mean field potential. Say we put the first particle into a state with orbital angular momentum $\ell>0$. Because of the spherical symmetry, all projections $m_{\ell}$ have the same energy and we may as well put it in the maximally projected orbit $m_{\ell 1}=\ell$, meaning it's roughly orbiting in the $x$-$y$ plane. It's angular shape looks something like this:

Now we add the second particle. Because the two particles have an attractive interaction, it's now energetically favorable to put this particle in the $m_{\ell 2}=-\ell$ orbit, i.e. orbiting in the $x$-$y$ plane in the opposite direction, minimizing the distance between the two particles. The two particles the form a state with total angular momentum projection $M=m_{\ell 1} +m_{\ell 2} = 0$. (We can ignore spin for now, for simplicity - the main intuitive point is unchanged.) However, this state does not have good total angular momentum $J=\ell_1 + \ell_2$. This is due to the fact that we cannot simultaneously know $m_{\ell 1}$,$m_{\ell 2}$, and $J$. To form a state with good $J=0$, we need to use Clebsch-Gordan coefficients and sum over $m_\ell$ states. $$\Psi(\ell_1,\ell_2,J=0,M=0) = \sum_{m_{\ell_1}m_{\ell 2}} \langle \ell_1 m_{\ell 1} \ell_2 m_{\ell 2} | J=0 M=0 \rangle \Psi(\ell_1,m_{\ell 1},\ell_2,m_{\ell 2}).$$

The physical $J=0$ state is given by the deformed state averaged over all directions, yielding a spherical state.

We could also use the above formula to project out a state $\Psi(\ell_1,\ell_2,J=2,M=0)$. In this case, due to the Clebsch-Gordan coefficient we would be averaging not equally over all directions, but favoring some particular directions. It so happens that the distribution we get is the same as you'd get by rotating the deformed intrinsic state about a particular axis. If you project out a state $\Psi(\ell_1,\ell_2,J=4,M=0)$, then it's as if you rotated if faster.

This is how the picture of a deformed rotor arises. But you should keep in mind that the intrinsic state is not the physical state, just a convenient way to think about it. The "experimental" $\beta_2$ values are obtained by matching electromagnetic transition properties to the predictions of a rigid rotor model.

For a more formal discussion of this, I'd suggest Bohr and Mottelson's Nuclear Structure Volume 2, or Unified Theory of Nuclear Models and Forces by Gerry Brown.

kinematics - How does anything move?

So in order for two things $A$ and $B$ to move apart, for example, relative to each other, $B$ can be set into motion away from $A$. This means that we have to increase $B$'s velocity and therefore the acceleration has to be positive. If the acceleration has to be positive and $A$ and $B$ were stationary before, that means that $B$'s acceleration has to increase from zero and therefore the third derivative of motion with respect to time has to be positive. The third derivative has to increase from zero and etc. etc. Is this just another way to state Zeno's paradox (and thus a dumb question) or does motion really involve an increase of the magnitude of infinitely many derivatives of velocity?

special relativity - What are the mechanics by which Time Dilation and Length Contraction occur?

What are the mechanics of time dilation and length contraction? Going beyond the mathematical equations involving light and the "speed limit of the universe", what is observed is merely a phenomenon and not a true explanation of why time dilates or length contracts. It has been proven to work out, but do we know why? Is it something that happens at a subatomic level?

measurements - How can a beam balance measure mass?

In Newtonian physics, mass is the amount of matter in an object. So, how can a beam balance measure the amount of matter in an object (which is the mass of the object).

Answer

A beam balance does not measure mass; it measures weight.

The gravitational force of attraction between the earth and an object depend on the mass of the object.

$$F = G\frac{Mm}{R^2}$$

where $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object and $R$ is the radius of the earth.

For convenience sake, we define a new quantity called acceleration due to gravity which does not depend on the mass of the object (therefore, has the same value for all the objects).

$$g = \frac{G}{M}{R^2}$$

Using the above simplification, we say:

$$F = mg$$

The beam balance measures the force $F$ exerted by the mass on the beam balance. The beam balance does not directly measure the mass. It uses the fact that the gravitational force on the object is proportional to its mass.

So what if you try to lift the object while it is being measured? You are reducing its weight, right? Does the beam balance read a smaller value?

Yes!

quantum field theory - Bilinears in adjoint representation

Below are two statements from my notes and I am trying to verify them explicitly. In both cases the fields are assumed to transform under the fundamental representation of $O(N)$ -

--'The kinetic term for a Dirac spinor is invariant under the symmetry group $U(N) \otimes U(N)$'.

I considered the case of a Weyl spinor first. This has a kinetic term $i \bar \phi_R \gamma^{\mu} \partial_{\mu} \phi_R$ and if $\phi_R \rightarrow U \phi_R$ then $i \bar \phi_R \gamma^{\mu} \partial_{\mu} \phi_R \rightarrow i\phi^{\dagger}_R U^{\dagger} \gamma_0 \gamma^{\mu} \partial_{\mu} U \phi $. Because $U$ and the gamma matrices act on different spaces, can I just shift the $U$ to the $U^{\dagger}$ and then using $UU^{\dagger}=1$ get the result? The $U(N) \otimes U(N)$ for the Dirac spinors comes about from decomposing a Dirac spinor into its left and right handed components each of which transforms under a 'left handed fundamental representation' or 'right handed fundamental representation' so could write the symmetry group as $U_L(N) \otimes U_R(N)$ (correct if mistaken).

--'If $T_a$ are the generators of $O(N)$, the bilinears $\phi^T T^a \phi$ transform according to the adjoint representation.'

I'm just wondering do generators always transform in the adjoint representation themselves? I read on here in another thread that the adjoint representation can be thought of as the representation anchored at the identity so if anyone could shed some light on this statement that'd be great.

Answer

Yes, the first part of your question is appreciated and answered soundly. The fermion kinetic term splits into two independent parts involving left and right Weyl spinors respectively, so each is independent under a separate U(N) as your wrote down.

The second question is a matter of language. A generator is a matrix with one adjoint index, a here, ranging over the dimension of the Lie algebra, so "universal" as per your question; and two indices each corresponding to its representation, e.g. i,j, the matrix indices, ranging over the dimension of that particular representation. It is an operator of the representation, acting on vectors of it, such as your φ. If your φ is in the fundamental, for instance, T will act on it as $\phi_i\mapsto T^a_{ij} \phi_j$, all in the fundamental. Dotting this with another vector, φ will yield a scalar in the fundamental, your expression, with a loose index a of the adjoint, so, then a vector in the adjoint. Always in the adjoint, regardless of what irrep φ you started with, as long as you used the suitable representation matrices for that generator.

To rotate this vector under O(N), you'll have to act on it with the Lie algebra structure constants, which are the operators T in the adjoint, so, analogously, $\phi^T T^a \phi\mapsto i f_{abc}~\phi^T T^c \phi$. For instance, for O(3) the generators are the familiar vector spin matrices $i\epsilon_{abc}$s.

electromagnetism - Lorenz Gauge Condition in Helmholtz equation

I do not understand why we can apply the Lorenz Gauge Condition in Helmholtz equation. What is its physical meaning? Any help is appreciated.

$\nabla\cdot \vec{A} + \mu_0\varepsilon_0\frac{∂\phi}{∂t} = 0$

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SOLUTION:

We can describe electromagnetic fields as a function of the electromagnetic potentials through the following equations (delayed potential equations):

$\vec{B}=\nabla × \vec{A}$

$\vec{E}=-\nabla \phi - \frac{∂\vec{A}}{∂t}$

We can transform these electromagnetic potentials so that they continue fulfilling the previous equations. Making the following changes we will obtain the same electromagnetic fields but the potentials will lose their physical meaning. These transformations will allow us to solve simpler statements, like Helmholtz's equation.

Then we can rewrite the new potentials like:

$\vec{A'}=\vec{A}+\vec{\alpha}$

$\phi'=\phi+\beta$

As I mentioned before, these transformations will have to satisfy the delayed potential equations so:

$\vec{B}=\nabla × \vec{A'}=\nabla × (\vec{A}+\vec{\alpha})$

$\vec{E}=-\nabla \phi' - \frac{∂\vec{A'}}{∂t}=-\nabla (\phi+\beta)- \frac{∂(\vec{A}+\vec{\alpha})}{∂t}$

If we continue simplifying the following equations we obtain:

$\nabla ×\vec{\alpha}=0$; $\vec{\alpha}=\nabla\lambda$; $\beta=\frac{∂\lambda}{∂t}$

So finally we get:

$\vec{A'}=\vec{A}+\nabla\lambda$

$\phi'=\phi+\frac{∂\lambda}{∂t}$

Conclusion: we can find any electromagnetic potential that satisfies Maxwell's equations. So we will try to find potentials that ease the calculations. It is important to note that these transformations will lead to a loss of potencial's physical meaning. But we do not have to worry about the electromagnetic fields because they will remain the same.

statistical mechanics - Spontaneous symmetry breaking at a finite temperature $T$: How is the state dscribed as a function of $T$?

Consider the equilibrium state of a statistical system with infinite DOF at a finite temperature $T$. For example, a Heisenberg ferromagnet with Hamiltonian $$H=-J\sum\limits_{i,j}\textbf{s}_i\cdot \textbf{s}_j\tag{1}$$ One can see that $\textbf{s}_i\cdot\textbf{s}_j$ is a scalar and therefore invariant under rotation.

However, if the temperature $T$ is greater than a critical value $T_c$, then the equilibrium state respects the symmetry of the Hamiltonian. And if $T

I want to make this picture mathematically precise. Can we describe the equilibrium state, in general, at temperature $T$ irrespective of whether $T>T_c$ or $T

How can I mathematically describe the equilibrium configuration so that I can explicitly see (like in Eqn. (1)) it breaks rotational invariance (symmetry of the Hamiltonian) for $TT_c$?

How Are Quantum Computers Able to Store Any Data at all?

So if qubits can have more than two states, and according to this video, https://www.youtube.com/watch?v=T2DXrs0OpHU you don't know what you get until you actually "open the box", if its all randomness and probability,then how can it store anything? Like, if you tried opening your word document, won't it show up differently everytime you opened it? Sorry if some of my questions seem stupid, I'm a high school student who has just recently gotten interested in quantum computing.

gravity - Electrical force vs gravitational force

Given that the electrical force is so much stronger than gravitational force at atomic levels, why is it that it's the gravitational force between you and the earth that keeps you on the ground rather than the electrical force between you and the earth?

Answer

Gravitational force is what keeps us on the ground rather than electrical force because there's no negative gravitational mass. Electric charge can be neutralized, while gravitational attraction not.

general relativity - If a Kerr-Newman black hole is like a charged, spinning, heavy magnet, what kind of magnet is it like?

I was reading up on De Sitter spaces, which states that the gravitational effects from a black hole is indistinguishable from any other spherically symmetric mass distribution. This makes a lot of sense to me.

I'm now super curious, can we just formulate all the properties of a black hole beyond the limit at which GR is needed in a completely Maxwellian / Newtonian sense? The Wikipedia article on the Kerr-Newman metric seems to indicate so, but the equations are in terms of the GR metrics. This is not what I want, I want a simplification, a limit-case of that math.

Ted Bunn answered a part of my question in Detection of the Electric Charge of a Black Hole. Let me repeat the stupidly simple form of the gravitational and electric field for a black hole beyond the point at which GR is needed.

$$\vec{g} = \frac{GM}{r^2} \hat{r}$$

$$\vec{E} = \frac{Q}{4\pi\epsilon_0 r^2} \hat{r}$$

Correct me if I'm wrong, but these would be a meaningful and accurate approximation in many, in fact, most situations where we would plausibly interact with a black hole (if we were close enough that these are no longer representative, we'd be risking a date with eternity).

Question: Fill in the blank; what would the magnetic field $\vec{B}$ around a black hole be?

Here is why I find it non-trivial: Every magnet in "our" world has some significant waist to it. So here is a normal magnet.

What happens when this is a black hole? Would the approximation for $\vec{B}$ that I'm asking for have all the magnetic field lines pass through the singularity? Or would they all pass through the event horizon radius but not necessarily a single point?

---

I think most people who have understood this already, but ideally the answer would use the 3 fundamental metrics of a black hole. Mass $M$, charge $Q$, and angular momentum $L$. The prior equations for gravity and electric field already fit this criteria. So the answer I'm looking for should be doable in the following form.

$$\vec{B} = f \left( M, Q, L, \vec{r} \right) $$

Answer

It's a little tricky giving a formal answer to this, but here is a sketch:. The electromagnetic potential of the Kerr-Newman hole is given by:

$$A_{a}dx^{a}=-\frac{Qr}{r^{2}+a^{2}\cos^{2}\theta}\left(dt-a \sin^{2}\theta d\phi\right)$$

This field will acquire a magnetic field from the fact that $\frac{\partial A_{\phi}}{\partial r}$ and $\frac{\partial A_{\phi}}{\partial \theta}$ are both nonzero. The problem is that the magnetic field, when re-phrased in terms of vectors and not one-forms, will fall off as $\frac{1}{r^{3}}$. At that point, if we're keeping terms that fall off that quickly, then we need to have a discussion about that asymptotic form of the metric you imply above, because there are terms in the metric that we need to keep, arising from frame-dragging effects of the black hole. If you want, I can go into more detail.

EDIT:

OK, so once we have the vector potential, we can calculate the magnetic field according to the rule: $B^{i}=\frac{1}{\sqrt{\left|g\right|}}\epsilon^{ijk}\left(\frac{\partial A_{j}}{dx^{k}}-\frac{\partial A_{k}}{dx^{j}}\right)$, where $\epsilon^{r\theta\phi}=1$, $\epsilon^{ijk}=-\epsilon^{jik}=-\epsilon^{ikj}$ and $\epsilon^{ijk}=0$ if $i=j$, $i=k$ or $j=k$. So, now, we just plug in the above expression for the spatial components of $A_{a}$, the metric tensor, and turn the crank.

After doing this, and taking the limit that $r$ is larger than everything else, we find that

$${\vec B}=\frac{2Qa\cos\theta}{r^{3}}{\hat e}_{r} + \frac{Qa \sin \theta}{r^{3}}{\hat e}_{\theta}$$

Sensibly, this is zero if either $Q=0$ or $a=0$. And I will once again assert that there are $\frac{1}{r^{3}}$ corrections to the gravitational force that must be taken into account in your high $r$ limit if you are going to keep this magnetic field.

homework and exercises - Incline Plane - Coefficient of friction and kinetic friction

A block of mass 90 kg sits on a slope with an angle of inclination 12° above the horizontal. The coefficients of friction between the block and the slope are µk = 0.03 and µs = 0.2.

a. What is the maximum frictional force that will allow the block to stay in place?

b. At what angle of inclination will the block start to slide down the slope?

My Question: I am not looking for these two questions to be answered, however, the question has caused me to doubt my understanding of static and kinetic friction re: incline planes.

When i calculated the the angle at which the block would begin to slide, using the arc tangent of 0.2 i got an angle less than 12 degrees. However, the question is suggesting that the block is at rest with when the incline plane is at 12 degrees. What is my conceptual error in thinking?

My next question:

Wouldnt the minimum angle of the incline plane (tan inverse of 0.2) be needed in answering the first question? that is, having it placed in the formula for the static friction formula = 0.2 x 90kg x 9.81 x cos of (tan inverse of 0.2)? assuming that tan inverse of 0.2 was found already? Is there something wrong with the question or how am i thinking is conceptual flawed?

My Final Question: Is it that the maximum frictional force that allow the block to stay in place is based on whatever angle the incline plane is set at? So, for example, if the incline plane is lowered to an angle of 5 degrees, the max friction force will be adjusted, even though the block would slide at 11.31 degrees? Is it that if lowered to 5 degrees, that the max friction force to overcome would be different?

In other words: If a block is at rest on the incline plane at an angle of 30 degrees, would the maximum frictional force that would allow the block to stay in place be calculated using 30 degrees, OR would we have to find the minimum angle of the incline plane needed for the block to slide before we can determine the maximum frictional force? That is, find that minimum angle and then substitute it in the formula for fs(max) = UsMgcos (minimum angle needed for block to slide)?

waves - Frequency of the sound when blowing in a bottle

I'm sure you have tried sometime to make a sound by blowing in an empty bottle. Of course, the tone/frequency of the sound modifies if the bottle changes its shape, volume, etc.

I am interested in understanding how the tone/frequency of this sound can be (approximately) calculated from the shape of the bottle. Is there some formula? If you have any references about this please post them in your answers.

Being a mathematician I keep hearing that the principal tone is given by the first eigenvalue of the Laplace operator, and I'd like to understand the physical explanation.

Answer

Yes. This is called Helmholtz resonance or cavity resonance which is an important application in acoustics. You could find larger sets of sample frequencies recorded in coke bottles at Hyperphysics...

In my basic understanding: The air in the bottle exhibits a single resonant frequency. When additional volume of air is blown into the closed cavity, the air will overflow out causing the pressure to decrease inside the bottle. Due to the newly produced low pressure, air outside rushes in. Thus, the air will oscillate into and out of the container for a few cycles at some natural frequency.

                    

Thus, The frequency of sound in such a closed bottle is determined to be $$f_c=\frac{v}{2\pi}\sqrt{\frac{A}{VL}}$$

Edit for symmetry and water add-in: The above formula could be used for an air cavity (as per your question). But in case of a water-filled bottle (If you require an appropriate answer), the experiment becomes somewhat complicated because we've to take the properties of sound in water into account. For example, $v_{air}$ is barely 340 m/s whereas $v_{water}$ is as high as 1484 m/s. It's easy to do if we write down some frequencies using different volumes of water (using the same bottle), and concluding a general relation between frequency & volume. This could be easily achieved through a graph...

                    

This is the $f$ vs. $1/\sqrt{V}$ graph plotted by a bunch of good guys (for a 0.6 liter coke bottle). This curve gives the equation of a straight line which says that $$f=\frac{5184.93}{\sqrt{V}}-30.4$$

forces - Why does the superposition principle work in method of images?

Okay, let there be a conducting sphere having radius $a$ initially charged with $Q$ & insulated. Now, $q$ is brought in front of the conductor at $y$ from the center.

Now, Jackson in his book uses the superposition principle to calculate the field & potential :

If we wish to consider the problem of an insulated conducting sphere with total charge $Q$ in the presence of a point charge $q$, we can build up the solution for the potential by linear superposition. In an operational sense, we can imagine with a grounded conducting sphere(with its charge $q'$ distributed over its surface). We then disconnect the ground wire & add to the sphere an amount of charge $(Q - q')$. ... .To find the potential, we merely note that the added charge $(Q - q')$ will distribute itself uniformly over the surface since the electrostatic forces due to point charge $q$ are already balanced by the charge $q'$.[...]

Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system.

But why doesn't $\left(Q- q'\right)$ exert force on $q'$ on the surface?? On the other hand, superposition principle tells that you can't alter the configuration of either system when superimposing the two systems which means the surface charge density of $q'$ remains the same even when there are $\left(Q- q'\right)$ other charges on the surface?? So, why doesn't $\left(Q- q'\right)$ exert force on $q'$?? If it exerts force on $q'$, then it would change the distribution of $q'$ which can't happen as said by superposition principle. So, can anyone explain why $\left(Q- q'\right)$ wouldn't exert force on $q'$??

Answer

Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system.

I have no idea why you think a zero potential surface has anything to so with anything. While the field of $q$ and the field of $q'$ together make a zero potential surface on the surface of sphere they most definitely exert net force on the charge $Q-q'.$

But why doesn't $\left(Q- q'\right)$ exert force on $q'$ on the surface??

It does. The charges that make up the density of charge $Q-q'$ exerts forces on each other so they exert force on the charges that make up the density of charge $q'$ in exactly the same way.

On the other hand, superposition principle tells that you can't alter the configuration of either system when superimposing the two systems which means the surface charge density of $q'$ remains the same even when there are $\left(Q- q'\right)$ other charges on the surface??

Let's be straight, when you grounded the sphere the external charge $q$ felt a force as if there were a charge $q'$ at the image charge location. And there is a net charge of $q'$ on the surface and the charges that make up the net charge of $q'$ exert forces on each other but the forces on each other add up to zero. But they still feel a force due to the charge $q$ and in fact the total force on all the charge on the surface is equal and opposite to the force the charge $q$ feels due to an image charge at the image location. And the actual force at each point in the sphere points orthogonal to the surface.

That orthogonality is the sole fact we get from the fact that the surface is an equipotential surface. And the force that is orthogonal is the total force i.e. the force due to the other charges on the surface plus the force due to the charge $q$ ... only their sum is orthogonal.

As for superposition, the force due to the uniform spread of charge $Q-q'$ over the surface is a force that points radially outwards as well.

The densities add, the fields add and the forces add. The only reason things don't move is that forces orthogonal the surface don't allow charges to move.

So, why doesn't $\left(Q- q'\right)$ exert force on $q'$??

It does.

If it exerts force on $q'$, then it would change the distribution of $q'$

No it doesn't, because the net force is orthogonal to the surface. And charges are free to move inside a conductor but they are not free to move outside a conductor. So a force orthogonal to the surface simply doesn't move the charges because the conductor itself is capable of keeping charges from leaving the conductor. It is called a work function.

which can't happen as said by superposition principle.

And now I think you misunderstand the superposition principle. It just says that if you add fields and add charges that it is also a solutions. And it does mean that if you add static solutions, you get static solution. If 100% everything were electrostatics then yes you could add like that. But consider the simple case of adding charge $q''$ uniformly spread to the outside of a conducting sphere, no grounds, no external charge. If you kept adding solutions like that over and over again eventually the work function might not be enough to keep such a huge charge density on the surface. Or maybe the field density outside the sphere gets large enough to break down the dielectric that is the air itself outside the sphere.

Just because you add charge and current and fields together doesn't mean that things don't move just because they didn't move on their own. Because the sphere itself or the air itself has to deal with all those things added together and it could have limits to what it can handle.

Superposition is just having a linear equation, add the sources and add the individual solutions and you get a solution. Anything non-linear such as a work function or a dielectric breakdown means the effects might not be static even if the individual things were static. Superposition doesn't mean more than it means.

Why did then Jackson write that the force on $Q - q'$ from $q$ is balanced by $q'$?

The forces due to $q$ and $q'$ are already balanced in the sense that the total field due to both is orthogonal to the surface so the new charge also arranges itself so that its field is also orthogonal to the surface as well which means the new charge is arranged uniformly.

newtonian mechanics - Why does a ballerina speed up when she pulls in her arms?

My friend thinks it's because she has less air resistance but I'm not sure.

rocket science - How could a solar sail be powered 'Upwind' by craft's own (laser) light source?

I've just watched a Discovery Science program - Alien Encounters featuring SETI luminaries such as Jill Tarter, Frank Drake and Nick Sagan. A serious discussion on how to interpret ET messages coupled with a dramatised depiction of the social impact on Earth of an impending visit from ETs. My major point of confusion is that, within our solar system, the visitors head in a sunward direction to Earth powering a craft with a solar sail, powered from behind using very powerful lasers.

I have viewed some topics on Phys.SE on the powering of solar sails, and including 'how massless photons can have momentum'. My problem with the described system is;

1) This sounds like 'pulling yourself up by your bootstraps' - If the photons hitting the sail have sufficient momentum to push the sail along, would there not be an equal and opposite momentum of the photons leaving the laser - and thus cancelling out the thrust?

2) If the solar sail craft is heading sunwards, you are already fighting a 'reverse thrust' from the sun. The program postulated tapping energy sources such as dark energy to power the lasers (OK that's sci-fi at the moment, I admit). So if problem 1. above is wrong, the power source for the lasers would have to be immense as it would to exert more than double the sun's photonics pressure to move forward.

I thought the only sunward movement of a solar sail was possible only by an analog of land-sailing's 'tacking'. I know the TV program was speculative, but I hope this is not received as science fiction, as solar powered space probes are a reality.

To clarify, as I'm getting tangential responses to this. Is the issue I called problem 1. a complete bust? That is, powering a solar sail from behind with your own (laser) light source a non-starter, because of the 'bootstrap' scenario or another more scientific principle?

Attempting to load a screenshot of the craft as requested-

rotation - Coupling between galaxy spin and central black hole spin

What is the relationship between the spin of a galaxy and the spin of its corresponding black hole?

Associated questions:

Do they always have the same axis of rotation?

Do they always spin in the same direction?

Does galaxy angular momentum increase due to frame dragging from black hole, causing continuing increase in galaxy angular velocity?

How does Hawking radiation impact rotational coupling?

Answer

The definitive answer is that we don't know how the spins are connected. And, unfortunately, it might be a pretty long time before we have reliable observational evidence.

Measurements

Black hole spins are very difficult to measure. Two methods have very recently started to be used, but both are approximate - and there is a lot of skepticism about the reliability of their results (especially since they often don't agree). Both methods rely on inferring how close the accretion disk comes to the black hole to infer the Innermost Stable Circular Orbit (ISCO) --- which then implies a spin magnitude and axis (but not which direction along that axis). The first method uses the temperature of the accretion disk (hotter means it comes closer to the BH), and the second method uses broadening of an iron emission line (the broader it is, the faster it is rotating and the more redshifted it is by strong-gravity -- thus the closer to the BH it is).

Both of these methods require fitting noisy data... The main uncertainty seems to be how well the inclination of the system (relative to the observer) can be determined. I.e. if the disk is edge on, could we distinguish that from face-on? How does that affect the measurement?

(Note that both of these methods have mostly been applied to stellar-mass BH systems, within our own galaxy. Massive Black-Holes in other galaxies is much harder.)

Theory

Even [super-]massive black-holes ([S]MBH) are very small compared to the sizes of galaxies they live in; like, 5-8 orders of magnitude smaller. This means it's very hard to simulate/calculate the connection between these size-scales. Specifically, we don't understand very well how MBHs get their gas from the surrounding galaxy. For that reason we don't have very good guesses on how galaxy and BH spins should be related. That being said, both simulations and people's theories suggest the spins should be correlated, but far from directly linked. Because of chaotic effects like galaxy and black-hole mergers, we can say with good certainty that in the 100's of billions of galaxies out there, lots of them will have completely opposite spins. We actually do know of some galaxies that have disk-like populations of stars which rotate in the opposite direction as the bulk of the galaxy... so this suggests the same could happen with BH accretion disks, and thus BH spins.

I can give a couple of definitive answers, however. Both frame-dragging and Hawking radiation from the black-hole definitely have zero effect on the galaxies spin, and evolution. Hawking radiation is completely negligible in all known astrophysical systems, and should only be relevant in micro-black-holes (if they exist[ed]). Frame-dragging is definitely an observable effect in the strong-gravity regime (near the event horizon), but because of the size-scale differences - it will have no way of coupling-to or affecting the galaxy as a whole.

quantum field theory - Can energy be taken out of the QFT vacuum?

There have been recent questions about the vacuum. In my simplified knowledge the vacuum is like a ground state energy level, and also that there might even exist other lower energy levels than the vacuum we find ourselves in. The sea is a soup of created and annihilated pairs of virtual particles with virtual energies and momenta.

In a normal sea on earth, which also represents a ground state of water, energy can be taken out of random waves by the clever construction of valves that allow only one way motion of water. Is it conceivable that a gadget of similar function could be found for the vacuum sea, or is it forbidden by conservation laws?

My intuition tells me that it might be possible if GR is taken into account, but my physics knowledge does not stretch to support this.

Edit : An explanation of why I am asking this question:

Let me expand on the example of the sea. The energy from the waves comes from either tides, i.e. gravitational forces, or wind (temperature differentials). If these were missing the oceans would be like glass representing a unique ground state of the gravitational well of the earth.

In an analogy, a gravitational wave going through the vacuum would be supplying energy to the vacuum sea.

I have been thinking of this analogy ever since cold fusion surfaced and refuses to die out, the most recent one being discussed here too. Approached from nuclear physics orders of magnitude the claims seem preposterous. There are people though who believe they have results of extra energy over input energy, much more than chemical reactions could supply.

This set me thinking on vacuum energy and the analogy with getting energy from the sea. A crystal is a prime candidate for any exploration of such concepts and in all cold fusion "successful" results crystals have been used. Now if the effect depended on the vacuum and how much distorted it was by a passage of a gravitational wave at the time of the experiment, or the exact orientation of the crystal, or the type of impurities in the crystal ( F centers etc) one would expect to get haphazard results, and non repeatable by other experimenters.

Of course this would be the first experimental evidence of gravitational waves :).

Edit 20/7/12

Maybe I should clarify that an acceptable answer in the negative would be one based on conservation laws. I believe that data trumps theory, and next in line are conservation laws,because they are the distillation of an enormous amount of data. Some people seem to think that theoretical definitions can substitute for proof in physics, but physical theories change, solid data do not, and this is physics, not axiomatic mathematics.

Answer

Well, let us be honest here. This question was not supposed to be answered from the very beggining. First of all we don't know how to mix quantum mechanics and gravity. There is no good consistent theory for that. Another thing is that today "the Dirac sea" analogy is not considered to be a very good thing. It is a pre-QFT naiive picture. Finally we are supposed to talk about "waves" in this "sea"... While the "sea" itself is an obsolete analogy... And all that is in a context of non-existing theory... Come on...

Now. There is actually a formal way to answer the question, because the question is about the "QFT vaccuum". And the QFT vacuum have a precise definition. Which basically says that it is "something you cannot take energy from". Actually, we start from that definition to build QFT. So the answer is: "you cannot take energy out of the QFT vacuum by definition".

Maybe we are wrong to start from this definition. Maybe for quantum gravity we need another starting point. But then it wouldn't be a QFT -- it would be a new theory which will have QFT as a limiting case. And in the range of validity of QFT that "formal" answer will hold.

Does light always travel at *c* even when moving through a medium?

The title of the question is a bit misleading, but with limited space it was all I could think of. So, I know that light travels at fractions of c when traveling through a medium; however, as I understand it, the reason for this is because the photons will be absorbed then re-emitted a short period of time later. My question is do the photons travel at c once the atom that absorbed them spits them back out, or do they travel a reduced speed in between their atom absorptions? For example, let's pretend we can track a single photon through a light year of water using an internal tracker on the photon that tells us when the photon "exists," i.e. it's traveling as a photon, and when it "doesn't exist," i.e. it is temporarily absorbed by atom and has yet to be emitted. Let's assume the photon that is absorbed and then emitted by the atom has the aforementioned tracker, regardless if it is the same photon. Now, that tracker is attached to a stationary clock. When the photon "exists," the clock runs, when the photon "does not exist," the clock stops. So, if we were to clock the time it took the photon to travel through the water, basically ignoring the time spent absorbed by the atoms, would it travel the light year in one year? Also, would all photons take the same amount of time? I suppose the fundamental question here is exactly what is it that causes light to slow through a medium. I am assuming the absorbing and emitting by atoms is not the only thing that slows light as I have read some articles where light is slowed to 38 miles per hour (http://news.harvard.edu/gazette/1999/02.18/light.html), and if it was light being absorbed, then the light would be held up in the atoms for the VAST majority of the time spent traveling through the medium.

cosmology - Is inflation theory really dead?

I know the title is little bit challenging but maybe most of you heard about the last BICEP2 paper on February. As I have read about it here and here. My understanding is, BICEP2 results released on March, revealed B-mode polarization and physicists support this observation with inflation theory. But the last paper published that these B modes represent of a cosmic high-dust.

I am not expert in cosmology but the inflation theory makes sense for many physicists. Now is it all dead or new findings twisted a little bit?

Answer

While a B-mode signal in the CMB would be a smoking gun for inflation, inflation is not even under pressure, even if BICEP2 only measured dust.

Inflation explains a number of early universe puzzles neatly. Most importantly it explains why we find an extremely homogeneous CMB. Furthermore, without inflation it is hard to explain that the perturbations on the CMB are correlated between points that seem to never have been in causal contact.

There are many models of inflation out there and most of them (the ones most popular before BICEP2 actually) predict an ubobservably low amplitude for tensor perturbations - which in turn lead to the B-mode signal BICEP2 looked for.

So, while some of the models people cooked up just to encompass BICEP2's unexpectedly large signal might be dead, inflation as a cosmologically well-motivated framework is perfectly fine.

Direct observational evidence beyond B-modes are tough to find. Gravitational waves with a nearly scale-invariant power spectrum are a generic prediction, but we have not even observed these at a single scale yet. However, there are a number of experiments planned within the next decade that aim to put an upper bound on the so-called tensor-to-scalar-ratio $r \lesssim 10^{-3}$, which is the order of magnitude predicted by generic models of inflation.

Very very long edit:

The only kind of signal we can recieve from inflation are of gravitational nature. If one takes the CMB as a given and looks only at cosmic evolution afterwards, the $\Lambda$CDM can explain basically all observations with only a few parameters. No photon signal can come from before recombination, making the CMB the oldest thing we can literally "look" at.

The sucess of inflation is to actually give a mechanism to create the CMB. It is non-trivial to get the CMB to be so homogenous and at the same time to create anisotropies with correlations over space-like intervals.

But note that in inflation, the CMB fluctuations are actually also variations in the metric tensor - they just are in its scalar component ($\sum_i g_{ii}$). So besides these scalar fluctuations there could be vector-like or tensor perturbations in the metric. Direct computations shows that no vector-like perturbations are created in de-Sitter like inflation, leaving tensor perturbations as the only other possible signal.

Now depending on your model of inflation, more or less of these tensor perturbations are created. So you can try to get as much information from two spectra (that are predicted to be nearly scale-invariant):

• The amplitude of one of the signals must be fixed by observation. This is done as the so-called "COBE Normalization" of scalar perturbations.


• The ratio of the amplitude of the two spectra at a fiducial scale. This is what we call $r$, the "tensor-to-scalar ratio". All models of inflation predict $r > 0$, but well-motivated values range from $10^{-4}$ to $0.2$ (where the latter are models more or less tailor-made for BICEP's $r = 0.2$ claim).


• The spectral tilt of each spectrum. Perfectly de-Sitter inflation with a constant Hubble parameter predicts a perfectly scale-invariant spectrum. Actual models are close to this, with only tiny variations of the Hubble parameter during inflation, leaving a nearly-scale invariant spectrum. This observable is called the spectral index, $n_s$ for scalars and $n_t$ for tensors, but measuring $n_t$ is nigh impossible.


• The running of the spectral index, i.e. its value at different scales can also give you information about the inflaton potential. For this you need to measure $n_s$ at different scales; while the CMB allows for measurements of $n_s$, measuring it at other scales than the CMB scale is challeging.


• One can even dream about measuring the running of $n_t$, but this will stay a dream for a long time...

I hope that I could elucidate why we are looking for grav. waves so hard and not for anything else - because there just isn't any other signal that is directly tied to inflation.

acoustics - Extended sound of thunder

Why does the sound of thunder last several seconds even when lightning lasts for only fraction of a second?

Answer

This is because the different parts of a stroke are be very different in distance.

This

Reverberations and echoes of sound last longer than those of light.

is obviously wrong. Which walls or mountains are the reflectors of those echos? Which rooms do the reverberations?

general relativity - How could spacetime become discretised at the Planck scale?

I didn't have much luck getting a response to this question before so I have tried to reword and expand it a little:

In early 2010 I attended this inaugural lecture by string theorist- Prof. Mavromatos entitled 'MAGIC strings'. In it he proposes that some string theory models may violate Lorentz symmetry at the Planck scale resulting in a kind-of foamy spacetime that could be observed by differing arrival times of photons of different energies reaching us from distant astronomical sources. See http://www.kcl.ac.uk/news/events_details.php?year=2010&event_id=2178 or here for one of the papers: http://iopscience.iop.org/1742-6596/174/1/012016

Furthermore, in 'Cycle's of Time' that I read recently, Prof. Sir Roger Penrose mentions (page 203) that Wheeler and others have strongly argued that if we could examine spacetime at the Planck scale we would see a turbulent chaotic situation (from vacuum fluctuations of the quantum fields I suppose) or perhaps a discrete granular one. Penrose goes on to list some other approaches that may suggest how this discrete structure may manifest itself. Loosely transcribed these are: spin foams, casual sets, non-commutative geometry, Machian theories, twistor theory, [EDIT] loop quantum gravity, or strings and membranes existing in some higher-dimensional geometry...

I have studied some QM, introductory QFT and the Standard Model as well as some basic GR but I have no formal experience of string theory. My questions are therefore:

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  What's involved with each of the above approaches? I.e. in what way does the spacetime become discretised? (Particularly in string theory)

  
  
  


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  Are there any other popular(ish) approaches that should be added to the list?

  
  


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  Supplementary query, with GR being a background-independent theory, I fail to see how one can end up with discretised spacetime without it being a pre-defined background onto which a theory of the dynamics would have to be 'bolted-on'??

  
  

Please forgive my ignorance if what I have said is misinformed, all comments and elucidations would be most welcome.

Answer

Let me try to address your questions, even though just the first one seems quite heavy in itself.

1. 
   

   Spacetime Discreteness: let me give you links to references that are relevant to your questions, than i'll make some general comments. An Introduction to Spin Foam Models of Quantum Gravity and BF Theory; Spacetime in String Theory; The quantum structure of spacetime at the Planck scale and quantum fields; Meaning of Noncommutative Geometry and the Planck-Scale Quantum Group; Causal Sets: Discrete Gravity (Notes for the Valdivia Summer School); On the Origins of Twistor Theory — this should get you going. As for string theory and spacetime discreteness, let me say that, in a crude way, the $\alpha$ that appear in the Action in this link Superstring theory, called 'string tension', is basically what 'measures' this.

   
   


2. 
   

   Other approaches not listed: i think your list is fairly complete. But, you didn't list Loop quantum gravity — maybe you were thinking of it, or maybe it fits in one of your named categories: i just thought i'd make it explicit.

   
   


3. 
   

   GR discreteness: to me, this is a subtler question, in the sense that once you have discretized spacetime (for one reason or another), you should not expect that the other [geometric] structures remain 'continuous' — in fact, there's a whole branch of research dealing with 'quantum groups' and 'discretized' (or 'latticized': think computer simulations) theories. The point being that if you discretized all of your ingredientes, you still maintain a certain relation among them (e.g., discrete gauge symmetry, or $q$-gauge symmetry). The bottom line is that you can perfectly define a theory where all ingredients are properly 'discretized', and so it maintains its relevant features (recovering the continuum theory in some limit). As a side note, it's worth seeing that it's possible to discretize theories at the level of differential forms, à la Discrete Differential Forms, Gauge Theories, and Regge Calculus (PDF) (and similar constructions by several other folks). In this sense, many of the relevant properties are kept even after discretization (quite robust method).

   
   
   

I hope this can get this discussion started.

newtonian mechanics - Fluid Pressure: The Collision Model - Approximation or Truth?

Microscopically, the pressure exerted by a fluid on a surface in contact with it is caused by collisions of molecules of fluid with the surface. As a result of a collision, the component of a molecule's momentum perpendicular to the surface is reversed. The surface must exert an impulsive force on the molecule, and by Newton's Third Law the molecule exerts an equal force perpendicular to the surface. The net result of reaction force exerted by many molecules on the surface gives rise to the pressure on the surface.

The above is an extract from Physics by Resnick, Halliday & Krane.

I've a few questions, conceptual in nature, which stemmed from the above paragraph -

1. 
   

   All that is mentioned is that the component of molecule's momentum perpendicular to surface is reversed; nothing is mentioned about its magnitude. If, they wish to tell us that the collision is elastic (as in the case of kinetic theory of gases), why is this a valid assumption? Maybe, the right question to ask is, to what extent is it a valid assumption?

   
   



2. 
   

   According to the aforementioned extract, pressure arises due to collisions between molecules and the surface. It is also a well known fact that pressure in a static fluid increases with depth. How can we explain that using this collision model? I'm confused because, the nature of collisions should be a property of the fluid, and should not vary with depth.

   
   


3. 
   

   Is this model - the one that talks about pressure arising due to collisions, sufficient to explain pressure related phenomena in all possible situations; or is it a mere approximation?

   
   

Answer

The quoted paragraph from the textbook talks about fluids which usually includes gases, liquids, and plasmas. However, it would not be right to say that for liquids (e.g., consider water for concreteness) the pressure is the kinetic pressure $P_k=nkT$. First of all, we know that we can put water under a piston and increase the pressure isothermically at nearly constant density. If the pressure is due to particle collisions then why does it increase without any increase of temperature and density? Furthermore, using the numbers for water at normal conditions, $n=33e27 m^{-3}$, T=300 K, we'd get the kinetic pressure $P_k$ at about 10 million atmospheres, but we don’t see it!

We don't see this huge pressure because it is largely compensated by intermolecular attraction forces. So the total pressure in a liquid is $P = P_k + P_f$, where $P_f$ (negative at normal conditions) is the component of the pressure due to intermolecular forces, strongly dependent on the density. If water is compressed (at a constant temperature) the resulting pressure increase is due to the change of $P_f$.

So, for water compressed under a piston at a constant temperature, the total observed pressure increases; the thermal pressure caused by water molecules bouncing off the surface does not change in this process but the intermolecular forces respond to the compression changing the total pressure.

Given that the thermal pressure in a liquid is almost entirely compensated by the intermolecular forces, one can model a liquid as a large number of slippery almost incompressible balls lumped together, essentially excluding thermal motion from the picture. This model would have the properties of a real fluid (weakly compressible, isotropic pressure, Pascal law, Archimedes law). If we put such a "liquid" in a vertical column then we'd observe that those balls deeper down from the surface are compressed more (because there is a larger weight above them), and a body embedded in this “liquid” deeper would experience a larger external pressure.

Hamiltonian Operator Interpretation of Quantum Anomaly

We can see the definition of quantum anomaly in terms of Lagrangian path integral formulation. What is the definition of quantum anomaly in terms of Hamiltonian operator approach or even more directly in terms of wave functions? What are the characteristics of anomalous Hamiltonians or wave functions?

thermodynamics - Would a pin head heated to 15 million degrees Celsius kill everyone in a 1000 mile radius?

The YouTube video How Hot Can it Get? contains, at the 2:33 mark, the following claim:

A pin head heated to 15 million degrees will kill everyone in a 1000 miles radius.

On what basis can this claim be true? Some of the things I can think of:

1. 
   

   Radiation of the metal as it cools down

   
   


2. 
   

   Energy released in fusion (not sure if this works for an iron pin)

   
   

Would the damage be only to organic matter or will it destroy other structures within that radius?

particle physics - More on matter and anti-matter

1. 
   

   Does every particle that has rest mass also have an anti-particle with which it would annihilate?

   
   


2. 
   

   Does annihilation only occur between like particles? For example what happens if a antineutron (anti u, anti d, anti d) collides with a proton (uud)? What happens if a positron collides with a proton?

   
   



3. 
   

   Since the Tevatron accelerates antiprotons is this more difficult to handle and dump?

   
   


4. 
   

   I've read about WIMP annihilation detection. Why would one assume there is any different proportion of WIMPS to anti-WIMPS than as is for non-WIMP matter (where there is far more matter than anti-matter).

   
   

newtonian mechanics - Intuitively, how can the work done on an object be equal to zero?

To my understanding the work done on an object is defined mathematically as: $$W = \vec{F}\cdot\vec{S}=|\vec{F}||\vec{S}|cos\theta$$ This, I understand. My problem is that I don't understand that if the angle $\theta$ is 90 degrees how can the work done by $\vec{F}$ on the object is zero. For example; say you have a particle and the direction of the displacement is directly to the right, and you also have a force vector acting on the particle that is straight up(like the normal force on a box that is standing on a flat surface). How is it possible that the force vector is not doing any work? Must the particle not take a different route because of the force vector acting upward on the particle, like if you add the vectors together?

There has to be something wrong with my reasoning, but what is it?

Answer

Because it's not any work, but the work done by a force that produces a displacement.

In the scenario you describe, somehow that force is not doing any work on the particle. This could be because the particle is restricted by another force to not go perpendicular and then the sum of forces in the perpendicular direction is zero.

In the second scenario, with the box and the normal force, it's the same. That force doesn't do any work since in the direction of that force there is zero movement. Which is analogous to say that the cosine of the angle between the displacement and such force is 90°.

quantum field theory - Why doesn't a plane wave solution represent a single photon?

Why doesn't a plane wave solution represent a single photon? And what is meant by the quantum-mean field being zero?

EDIT:

This post is an extension to a previous post I made asking about the photon in QFT. I was asked by the person helping me to start a new question so he can respond in the form of an "answer" post and thereby keeping all "answers" on topic. The link is here Quantum State of Photon Question

Answer

Ok so, let's start by defining the photon in the context of quantum theory. A photon is one of the quantum states of definite energy and momentum of the theory of Quantum Electrodynamics. In QED without fermions, a photon of momentum $k$ and spin $s$ is defined a the state

$$ |\gamma, \vec{k},s\rangle=a^{\dagger}(\vec{k},s)\,|0\rangle $$ Where $a^{\dagger}(\vec{k},s)$ is the operator that appears in the Fourier-expansion of the electromagnetic four-potential operator $A^{\mu}(x)$:

$$ A^{\mu}(x)=\int\frac{d^{3}k}{(2\pi)^{3}}\ \frac{1}{\sqrt{2k^{0}}}\ \sum_{s}\ \left\{e^{-ik_{\mu}x^{\mu}}\ a(\vec{k},s)\ e^{\mu}(\vec{k},s)+e^{ik_{\mu}x^{\mu}}\ a^{\dagger}(\vec{k},s)\ e^{\mu\ *}(\vec{k},s)\right\} $$ Depending on the conventions on the normalization of states, you may find different expressions for both the four-potential operator and the normalized state of the photon. In the conventions to which I'm used to, one usually writes $$ |\gamma, \vec{k},s\rangle=\sqrt{2k^{0}}\ a^{\dagger}(\vec{k},s)\,|0\rangle $$ Why it is so depends on the commutation relations between the operators of the theory. The quantization of the electromagnetic field is not an easy subject, so in order to get useful results we shall work under assumptions which would be theoretically justified if one started from the very beginning. First of all, let's assume that the $0$ component of the electromagnetic four-potential can be set to zero: $A^{0}=0$ (this is not the case in QED with fermions, but this assumption can be made in our case, by setting $e^{0}(\vec{k},s)=e^{0\ *}(\vec{k},s)=0$). Second of all, let's state without derivation that $$ [a(\vec{k},s),a^{\dagger}(\vec{k}',s')]=(2\pi)^{3}\ \delta^{(3)}(\vec{k}-\vec{k}')\ \delta_{ss'} $$ $$ [a(\vec{k},s),a(\vec{k}',s')]=[a^{\dagger}(\vec{k},s),a^{\dagger}(\vec{k}',s')]=0 $$ (again, the factor of $(2\pi)^{3}$ depends on the normalization conventions). This being said, you can measure the mean electromagnetic four-potential carried by a photon is the usual way: $$ \langle A^{\mu}(x)\rangle_{\gamma,k,s}=\langle{\gamma,k,s}|A^{\mu}(x)|\gamma,k,s\rangle $$ So let's calculate $$ A^{\mu}(x)|\gamma,k,s\rangle=\int\frac{d^{3}p}{(2\pi)^{3}}\ \frac{1}{\sqrt{2p^{0}}}\ \sum_{\sigma}\ \left\{e^{-ip_{\mu}x^{\mu}}\ a(\vec{p},\sigma)\ e^{\mu}(\vec{p},\sigma)+e^{ip_{\mu}x^{\mu}}\ a^{\dagger}(\vec{p},\sigma)\ e^{\mu\ *}(\vec{p},\sigma)\right\}\sqrt{2k^{0}}\ a^{\dagger}(\vec{k},s)\,|0\rangle=e^{-ik_{\mu}x^{\mu}}\ e^{\mu}(\vec{k},s)\ |0\rangle+\int\frac{d^{3}p}{(2\pi)^{3}}\ \frac{1}{2p^{0}}\ \sum_{\sigma}\ e^{ip_{\mu}x^{\mu}}\ e^{\mu\ *}(\vec{p},\sigma)\ |(\gamma,p,\sigma),(\gamma,k,s)\rangle $$ by using the commutation relations, where $$ |(\gamma,p,\sigma),(\gamma,k,s)\rangle=\sqrt{2p^{0}}\sqrt{2k^{0}}\ a^{\dagger}(\vec{p},\sigma)\ a^{\dagger}(\vec{k},s)\ |0\rangle $$ is a state containing two photons. This way, we find that $$ \langle A^{\mu}(x)\rangle_{\gamma,k,s}=e^{-ik_{\mu}x^{\mu}}\ e^{\mu}(\vec{k},s)\ \langle{\gamma,k,s}|0\rangle+\int\frac{d^{3}p}{(2\pi)^{3}}\ \frac{1}{2p^{0}}\ \sum_{\sigma}\ e^{ip_{\mu}x^{\mu}}\ e^{\mu\ *}(\vec{p},\sigma)\ \langle{\gamma,k,s}|(\gamma,p,\sigma),(\gamma,k,s)\rangle $$ Now of course the inner product between states containing different numbers of photons is zero. So $$ \langle A^{\mu}(x)\rangle_{\gamma,k,s}=0 $$ i.e. the mean four-potential carried by a photon is zero. As a consequence, the electric and magnetic field carried by a photon too must be zero (the derivatives can be brought out of and into the average sign). The direct calculation of the electromagnetic field-strength $F^{\mu\nu}$ is done as follows: $$ \langle F^{\mu\nu}(x)\rangle_{\gamma,k,s}=\langle{\gamma,k,s}|\partial^{\mu}A^{\nu}(x)-\partial^{\nu}A^{\mu}(x)|\gamma,k,s\rangle $$ A direct computation shows that this indeed is zero.

So what does this mean? Does this mean that the photon carries zero electromagnetic field? Of course not. This only means that if you were able to measure the electromagnetic field carried by a photon many times, the point-by-point (i.e. separately in space and in time) average of the field would tend to zero in the limit of infinite measurements, and not because the photon is an oscillating wave: as I said, the average would be zero from point to point. On the other hand, a plane wave potential corresponds to a non-zero, fixed value of the electromagnetic field. Thus by no means you can identify it with a photon. On the other hand, complicated superpositions of states with different number of photons, from zero to infinity, can have non-zero average electromagnetic field. This article explains how one can get a specific classical field configuration from a superposition of photonic states.

homework and exercises - What is the force required to move a raindrop?

I'm doing some research on how much force is required to move a typical raindrop (assuming that it is falling straight down) off to the side of X distance. This is for a school project on creating our own airbrella.

Doing some reasearch it seems a typical raindrop is 4-100MG and falls at about 5-20MPH. Also I would like to know how far the raindrop would be blown for X amount of wind. Assuming the wind is hitting it at a 120 degree angle? (im guesstimating 120 because of the picture below)

Answer

What is the force required to move a raindrop?

The force $F_{(t)}$ that pushes against the drop at time $t$ is

$$F_{(t)} = \frac{cW\cdot A\cdot \rho\cdot v_{(t)}^2}{2}$$

where $cW$ is the drag coefficient, $A$ the cross section area of the drop, $\rho$ the density of the air and $v_{(t)}$ the velocity at any given time, but what you might be asking for is the initial kinetic energy, which is $M\cdot v_{0}^2/2$ where $v_0$ is the initial velocity of the drop.

To calculate how far the drop moves with the kinetic energy you give it initially, I would solve this problem with the ballistic equation, which requires a differential equation if you want to have air resistance:

Code (Mathematica, .nb File)

Here the mass of the drop is 0.004 gram, its initial velocity is 10 m/sec, the initial angle 10° and the height from which it is thrown 1 meter. It is assumed that the drop has 1 mm radius and a cW value of 0.3:

Then the drop would fly 0.67 sec and reach the ground after 4.7 meter.

I might put this into Latex when I have the time later, but in the end you will need a math program anyway since differential equations are hard to solve by hand and in this case there is no analytical solution as far as I know.

If you choose unrealistically high initial velocities you also have to note that the waterdrops would vaporize, and if they are to big they will split up. This is not taken into account in my calculation. More information on drops, their mass and shape (which influences the cW value) can be found here and here.

Edit: put in more realistic values for the drops