electromagnetism - Explaining Lenz's Law without conservation of energy

I was often told by my professor, using the following example, to demonstrate the relationship of conservation of energy and Lenz's Law.

If you push a conductor into a constant magnetic field. By Lenz's Law, voltage will be induced to oppose the "cause" which then resists the conductor from moving quicker which would be a violation of conservation of energy. In other words, if you push the rod very slightly, and the induced force turns out to push it even further, you generate infinite energy which then contradicts the conservation of energy law.

However, I wasn't happy enough by this answer, even though the fact is that you won't get infinite energy. I was thinking, whether it is something to do with the things happening on the atomic scale. Considering a more trivial example, a magnet moving towards a current loop,

so why would it produce a opposite pole to resist the change?

Is it something to do with the cutting flux of the magenetic field?

Why must a field be created to cancel the effect?

Why can't nothing happen?

Or is it just a matter of fact in nature?

I am finding this hard to understand how this really works. Is it a virtual particle?

Please explain the above phenomena without relating it to energy conservation law.

Answer

According to Lenz's law, polarity of induced emf will be such that it produces the current which opposes the change in magnetic flux which produced it. It directly follows from the law of conservation of energy.

Let's say you have a circular loop of conductor, if you bring the north pole of the magnet towards the loop, current induced in the loop will be anti-clockwise when viewed from the side of the magnet. Remember if current is in the anti-clockwise direction it acts same as the north pole. So, north pole-north pole repel each other. Thus, induced emf is opposing the change in magnetic flux. Therefore, work has to be done in order change the magnetic flux linked with the coil. This work done will be converted into electrical energy.

When you take the magnet away, when viewed from the side of magnet, current in the loop will be clockwise (acts as south pole) whereas magnet side near the loop will be south pole. So, even here south pole-south pole repel each other. Thus, induced emf is opposing the change in magnetic flux. Therefore, work has to be done even here in order to change the magnetic flux linked with the coil. As said above, this work done will be converted into electrical energy.

Thus, in the above two cases energy is conserved.

Let's say that you bring the north pole of the magnet towards the same circular loop, now let current induced be clock-wise (acts as south pole) when viewed from the magnet side. North pole- South pole attract each other, so no work is needed to change the flux linked with the coil. But when flux linked with the coil changes, there is emf induced. Here, electrical energy is produced without any work being done. It violates the law of conservation of energy. Similarly you can consider the other case.

Therefore, from Lenz's law it follows that electrical energy is produced by the expense of mechanical energy. For this to happen, induced emf should oppose change in magnetic flux.

You asked to explain with out using law of conservation of energy, but I felt the above explanation would explain your questions better.

general relativity - How do black holes accrete mass?

Thanks to time dilation, a distant observer watching a man fall in to a black hole will only see him asymptotically approach the event horizon. So how do black holes ever get bigger?

optics - What are the 'types' of parametric down conversion?

I'm looking at photon entanglement, and everywhere in the literature there's a reference to 'type-II' parametric down conversion as a source of entangled photon pairs. I know what parametric down conversion is, and I understand the resulting entangled state of type-II parametric down conversion.

I'm guessing there's some sort of 'type-I' out there, but I can't seem to find a good comparison between the two. What is type-I parametric down conversion? Does it have anything to do with pairs of entangled photons? I feel like I'm missing something.

Answer

As opposed to type II phase matching that produces orthogonally polarized photons in parametric down conversion (PDC), the type I PDC process produces identically polarized photons in the output signal and idler modes (labels $s$ and $i$ below).

Normally the output state from type I PDC is not entangled: to get the required phase matching in the nonlinear material, the pump polarization must be fixed. Both the PDC photons may then either be horizontally or vertically polarized. An often-used trick is to employ two similar nonlinear crystals (placed one after the other with their optic axes orthogonal) and sending a pump with a $45^{\circ}$ polarization. If the crystals are thin enough to simultaneously lie inside the coherence length of the pump, and losses between the first and second crystal are negligible, then a pump photon is equally likely to excite the PDC process in either of the two crystals. In that case, the output state may be approximated as $\propto |H_s,H_i\rangle + e^{i \phi}|V_s,V_i\rangle$ which is an entangled state. The relative phase $\phi$ is a function of the phase matching, thickness of the crystals, etc.

Scalar Field Redefinition and Scattering Amplitude

Consider a field redefinition $$ \phi \rightarrow \phi' = \phi+\lambda \phi^2 $$ Find the Feynman rules for this theory and work out the $2\rightarrow 2$ scattering amplitude at tree level (The result should be zero).

$$ \mathcal{L}_0 = -\frac{1}{2}(\partial^\mu \phi \partial_\mu \phi + m^2 \phi^2) \implies \text{EOM: } \Box\phi - m^2\phi=0 $$ Preforming the field redefinition on the Lagrangian I obtain: $$ \mathcal{L}_0 \rightarrow \mathcal{L}'= \mathcal{L}_0 -2\lambda\phi \partial^\mu \phi \partial_\mu \phi - 2 \lambda^2 \phi^2 \partial^\mu \phi \partial_\mu \phi -\lambda m^2 \phi^3 - \frac{\lambda^2}{2} m^2 \phi^4 $$ I then rewrote the two terms with partial derivatives as $$ -2\lambda\phi \partial^\mu \phi \partial_\mu \phi = -\lambda \partial^\mu (\phi^2 \partial_\mu \phi)+\lambda \phi^2 \Box \phi $$ which gets rid of the total derivative. Then I used the EOM to replace $\Box \phi$ with $m^2 \phi$ to get $$ \mathcal{L}' = \mathcal{L}_0 + \lambda (m^2 - 1)\phi^3 + \lambda^2 \left( \frac{m^2}{3}-\frac{1}{2} \right) \phi^4 $$ Now this looks like a normal scalar field theory with a cubic and quartic vertex. Now it wants a tree level diagram for two to two scattering, so I should put these two vertices in all the combinations that have no loops and result in two in, and two out? Then subsequently work out the scattering amplitude from the said diagrams using the Feynman rules I put together with the above Lagrangian?

Answer

I remember this as a nice problem from chapter 10 of Srenicki's QFT book. Your first expression for $\mathcal{L}'$ should have factors of $m^2$ in the last two terms. You should work with the lagrangian from here and derive the feynman rules and calculate the scattering amplitude. I will sketch this for you below. It is a very nice excercise to check whether you have understood everything up to that point.

You cannot use the EOM as you have done since that was for the original $\phi$ before the field redefinition. After the field redefinition your new $\phi$ (may help you to call it something else, e.g. $\phi \to \psi + \lambda \psi^2$) has a different equation of motion involving quadratic and cubic terms!

As I said above you should just work from your first expression for $\mathcal{L}'$ with the derivatives in and derive the Feynman rules from there. If you go through the whole procedure as is done in the previous chapter (being careful about the derivatives and symmetry factors) for each type of vertex (from each new term in the lagrangian) you will find a whole bunch of new vertex factors. Just calculate the 3 or 4 point function for each individual term and determine the feynman rules that way. Once you get used to how it works you can start to read them off directly from the lagrangian but it is good to go through the whole procedure a few times first. You should find for example, the cubic term with no derivatives gives a vertex joining three lines with a vertex factor of $$-i3! \lambda m^2.$$ The cubic vertex with derivatives has a vertex factor of $$4i\lambda(k_1k_2 + k_1k_3 + k_2k_3)$$ you see that the derivatives act to pull down powers of the momenta from the Fourier transforms of the propagators. It is sort of obvious if you just think about $\partial_\mu$ in momentum space.

So the total vertex joining three lines will just be the sum of these two. Do the same for the four point vertex and then as you say, draw all the feynman diagrams for $\phi\phi \to \phi\phi$ scattering, calculate the total amplitude and you should find if you do everything right that you get lots of nice cancellations and the answer is zero and you will have a nice warm feeling inside :-)

Derivative interactions such as these crop up in non-abelian gauge theories (e.g. the standard model) and gravity so they are important to understand. If you are still having trouble calculating the vertex factors I highly recommend this pdf which goes through in painstaking detail how to derive feynman rules. It follows the treatment by Srednicki too! Check section 5.3 and 5.4 for what you need.

quantum mechanics - Does an electron move from one excitation state to another, or jump?

I'm wondering, when an electron changes state, does it move from one state to another over some (very small) time period? Or does it change from one state to another in no time? If the former, what does it mean for it to be in-between states (for however short a period of time)? If the latter, how does it teleport?

(Does this question make sense?)

How virtual photons give rise to electric and/or magnetic field?

Say a neutron which have magnetic moment despite carrying no charge. Isn't both the electric field and magnetic field consist of virtual photons? So how can the same virtual photons give rise to 2 different properties? Since virtual photon is the excitation of a quantum field how come the same excitation of the quantum field can have different behavior like giving particle electric charge or spin?

special relativity - is SR simultaneity symmetrical?

Alice and Bob's spaceships are converging inertially at 0.99c. Alice knows enough about Bob's clock that she can calculate for any time of her clock what his clock would show "at that time" in her frame. Bob is similarly knowledgeable about Alice's clock. When Alice's clock shows Ta, Alice calculates Tb = what Bob's clock shows at that time in Alice's frame. When Bob's clock shows Tb, Bob calculates Tc = what Alice's clock shows at that time in Bob's frame. Question: does Ta = Tc ?

Answer

Short answer: No. If they did, there would be no relativity of simultaneity.

The simplest way to track the conflict is to add a buoy, say in Alice's frame, at some distance from Alice's own location. Let the buoy carry its own clock, synchronized to that of Alice, such that when Bob passes the buoy location, the buoy clock shows time $T_A$, just like Alice's clock. This means Alice observes 2 simultaneous events at non-coincident locations: her clock beating time $T_A$ at her location, and the buoy clock beating time $T_A$ at its own location.

Now look at Bob as he passes the buoy at his time $T_B$ and buoy time (Alice's time) $T_A$. Relativity of simultaneity means that Bob cannot see both Alice and the buoy marking time $T_A$ at the same moment of his time: spatially separated events that look simultaneous to Alice never appear simultaneous to Bob. So although he does see the buoy clock showing $T_A$ at his time $T_B$, he finds that Alice's clock shows a different time ${\tilde T}_A$.

Fun exercise: Apply the same reasoning to a symmetrical setup using buoys in both Alice's and Bob's frame.

Note following comments: As @dmckee observed, the phenomenon is independent of a particular choice of time origin, in either frame. Alice and Bob may choose their individual $t=0$ marks arbitrarily, yet they always arrive at the same conclusion regarding the specific time intervals. In more formal if obvious terms, time intervals are invariant under time translations in any frame.

quantum mechanics - Is the conserved current probability or charge in Klein-Gordon and Dirac equations?

The conserved currents in KG and Dirac are:

K-G : $j^{\mu}=i(\phi^*\partial^{\mu}\phi-\phi\partial^{\mu}\phi^*)$

Dirac: $j^{\mu}=\bar{\psi}\gamma^{\mu}\psi$

$j^0$ is positive definite in Dirac's but not on KG's. One simple way to see this is noticing that, in KG, the conserved current can be written us $j^{\mu}=2p^{\mu}|N|^2$ where $N$ is a normalization factor. When energy is negative, so is $j^0$.

This problem is solved thinking of $j^{\mu}$ as a charge current and not probability. This makes sense because solutions with positive and negative energy (particles and antiparticles) have opposite charge densities.

How does this interpretation apply to the Dirac current? it is positive definite so it can't change sign for negative energy solutions...

space expansion - What is the universe 'expanding' into?

We say the universe is expanding, and by expanding we mean the distance between objects gets larger over time. We call that "Metric Expansion of the Universe". So far so good. I kind of get the idea about of distances getting larger.

Now, I think of a balloon's surface and the distance between two arbitrary points on the surface gets larger as the metric expansion happens. But, in order for metric expansion to happen, doesn't the universe really expand INTO something. Balloon's surface expands into air so there's no problem imagining it, but how about the universe itself?

Also, do we mean the whole universe or observable universe when we say the universe is expanding? Both maybe?

Edit: Also, I know some multiverse theories that try to explain it, but the idea of universe is expanding has been there before multiverse was even considered, so I guess it can be explained without multiverse theories.

Answer

The balloon analogy is useful in some respects, but it is misleading in one important respect. In the balloon analogy the curavture of the balloon surface is extrinsic while in GR the curvature of the universe is intrinsic.

Extrinsic curvature is easy to understand. The surface of a balloon, or the hills and valleys on a landscape, or (to make a 1D analogy) a railway line are extrinsically curved because there is another dimension external to the surface that allows the surface to curve. We say that our surface is embedded in a manifold with a dimensionality one greater than the surface.

Intrinsic curvature is much harder to understand because it's counter intuitive. I described intrinsic curvature in my answer to Universe being flat and why we can't see or access the space "behind" our universe plane? but let me try a simpler example.

Suppose you watch an ant walking along an elastic rope, and you see the ant changing speed. You would assume the ant is accelerating. But suppose we had stretched some bits of the rope and compressed others:

The dotted lines show equally spaced divisions on the unstretched rope, so when we compress the rope the dotted lines get closer together and when we stretch the rope the dotted lines get farther apart.

The key feature of intrinsic curvature (and GR) is that the ant sees all the divisions as equally spaced no matter how much we stretch the rope. So if the ant crawls one division per second on the unstretched rope it still crawls at one division per second on the stretched rope. So we see the ant moving more slowly at the left end of the rope than at the right end, and we might explain this by saying the ant is being accelerated by some force (like gravity). But actually the ant is moving in an intrinsically curved space.

This is what happens in GR. The curvature of spacetime is like some bits of spacetime being compressed and other bits being stretched, and this is what causes the acceleration that we describe as gravity. There is no external dimension that the universe is being curved in.

You started off by asking about the metric expansion of space. Well this is like the elastic rope being continuously stretched, but the rope is infinite and has no ends. So the rope isn't being stretched into anything - all the stretching is internal. Likewise the universe isn't expanding into anything.

general relativity - Energy-momentum tensor for dust

We all know that the energy-momentum tensor for dust is just $T^{\alpha\beta}=\rho_0v^\alpha v^\beta,$ where $\rho_0$ is the mass density in the dust's rest frame and $v^α$ is the dust's four-velocity. I'm trying to derive the dust energy momentum tensor from the equation $T_{αβ}=-\frac{2}{\sqrt{-g}}\frac{δS_M}{δg_{αβ}}$ but I'm getting the wrong answer.

The action for dust is

$$S=\int -\rho_0\sqrt{-g}d^4x.$$

Thus

$$\frac{\delta S}{\delta g^{\alpha\beta}}=\frac{\delta (-\rho_0)}{\delta g^{\alpha\beta}}\sqrt{-g}-\rho_0\frac{\delta \sqrt{-g}}{\delta g^{\alpha\beta}}.$$

To evaluate $\frac{\delta (-\rho_0)}{\delta g^{\alpha\beta}}$, I define $K_\alpha=\rho_0 v_\alpha$. Then $\rho_0=\sqrt{g^{\alpha\beta}K_\alpha K_\beta}$ and thus $\delta \rho_0=\frac{1}{2\rho_0}K_\alpha K_\beta \delta g^{\alpha \beta}.$ It follows that $$T_{\alpha\beta}=\rho_0 v_\alpha v_\beta-\rho_0 g_{\alpha\beta}.$$

There's an extra term that I can't get rid of. Any idea where I went wrong?

special relativity - Measurement of event time by different observers

Follow up question to Doppler redshift in special relativity

A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. How will the time of the emission of a light pulse as measured by the emitter be different to the time as measured by the observer (how will it be different than the time of the pulse reaching the observer)?

edit: to elaborate slightly, let us say $ t_{E_1} $ is the time of emission as measured by the emitter, $ t_{E_2} $ is the time of emission as measured by the observer, and $ t_O $ is the time of observation of the pulse by the observer. How are these three times different, how would they be measured in practice?

Answer

How are these three times different, how would they be measured in practice?

First, let's see if we agree on what these times are. From what I've read, I believe that:

$t_{E_1}$ is the reading of a clock, co-located and co-moving with the emitter, when the pulse of light is emitted.

$t_{E_2}$ is the reading of a synchronized clock, at rest in the frame of the observer, co-located with the emitter at the time of the emission of the pulse.

$t_O$ is the reading of the observer's "wrist watch", i.e., it is the reading of a clock at rest at the origin of the observer's frame of reference, when the pulse reaches the origin.

If so, then:

$\gamma_v t_{E_1} = t_{E_2}$

(assuming standard configuration etc.)

Intuitively, the moving emitter clock runs slow according to the observer so $t_{E_1}$ is smaller than $t_{E_2}$ by the factor $\gamma_v$.

How is this measured? The observer has a synchronized clock at the spatial location of the emission of the pulse. This clock reads $t_{E_2}$ when the emitter is co-located (and emits the pulse). The emitter's clock reads $t_{E_1}$

---

$t_O = t_{E_2} + \frac{d}{c} = t_{E_2}(1 + \frac{v}{c}) = t_{E_1}\gamma_v(1 + \frac{v}{c})$

Here, d is the displacement, from the origin of the observer's frame, of the emitter when the pulse is emitted at time $t_{E_2}$.

The coordinate time required for the pulse to travel to the origin is $\frac{d}{c}$ but $d$ is just $v\cdot t_{E_2}$

newtonian mechanics - Work-energy theorem and Conservation of energy formula

We have $$W_{net}=\Delta K \quad(work-energy\; theorem)\tag{1}$$ And also $$W_{net}=\Delta K +\Delta U \quad(Conservation\;of \;energy \;formula)\tag{2}$$ How's that happening?

In proof for bernoulli's equation there's a place which they say:$\Delta W=\Delta E $which$ \Delta W$ does not contain work of gravity and however it's related to external or internal energy but I can't understand? You can see the proof here: http://www.4physics.com/phy_demo/bernoulli-effect-equation.html

Answer

This is a nice example of why notation matters.

You know there are two types of forces: non-conservative and conservative forces.

Let's call $W_{cons}$ the work done by conservative forces. Let's call $W_{NC}$ the work done by the rest.

The work-energy theorem states that

$$W=\Delta E_k $$

However, this work is the total work. This can be splitted in two parts: $W=W_{cons}+W_{nc}$. So

$$ \Delta E_k = W_{cons}+W_{nc} $$

Now, we define a quantity called $E_p$ such taht $W_cons=-\Delta E_p$. The minus sign is a convention, but it is important to keep it in mind. Hence

$$ \Delta E_k = W_{cons}+W_{nc} $$ $$ \Delta E_k = -\Delta E_p +W_{nc} $$ $$ \boxed{ \Delta E_k + \Delta E_p = W_{nc} } $$

So your "net" work refers only to non conservative forces in your second equation.

waves - Why does wavelength affect diffraction?

I have seen many questions of this type but I could nowhere find the answer to "why". I know this is a phenomenon which has been seen and discovered and we know it happens and how it happens. But my question is why would wavelength affect the amount of diffraction? I am looking for a very simple logical explanation rather than a complex mathematical answer. Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray? I need an answer that will answer "why" does diffraction depend on wavelength of light.

Image sources: http://www.olympusmicro.com/primer/java/diffraction/index.html

Answer

Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray?

Don't think of bending. Think of diffraction like this: if you have a plane wave incident on a slit, then you can think about the space in the slit as being a line of infinitely many point sources that radiate in phase.

If you are looking straight down the slit, then all those point sources are in phase. There's not much unusual going on here.

However, if you move a bit to the side, then all those point sources aren't in phase. They are, really, but since they are not at equal distances to you, the radiation from each is delayed by a different amount. Depending on your position, the point sources interfere constructively or destructively, and this is what yields the diffraction pattern.

If you look closely at this image, it appears it was generated by an approximation of four point sources in the slit.

Now, the number of these point sources there are, and the maximum difference in phase between them, is a function of the size of the slit, obviously. If the slit is wider, then when viewed from some direction slightly off center, the phase difference from the left-most source and the right-most source will be greater, because the difference in distance between them is greater.

Compare a small slit:

To a bigger slit:

The significance of the size of the slit is apparent, right?

Well, changing the wavelength is equivalent to changing the size of the slit. If we make the slit bigger, and make the wavelength bigger by the same amount, then the difference in distance between the sources is greater, but the rate of change in the wave function is slower, so the phase difference between the two extremes of the slit is the same.

But, if we just make the wavelength smaller, and leave the slit the same, the rate of change in the wave function is faster, which is equivalent to making the slit bigger without changing the wavelength.

Images from Wikipedia

Why is the equivalence principle so important to general relativity?

In its simplest form, equivalence principle states that the inertial mass and the gravitational mass should be the same. This is easy to understand.

But why is it so important to the formulation of General Relativity? To be more specific, I don't understand how the gravitational field equation:

can be derived from this principle.

Answer

A derivation of Einstein's equation isn't why the Equivalence principle is central to GR. The reason that the equivalence principle is central to GR is in the fact that you can represent the gravitational field with a metric tensor at all--you can replace a force equation with a geodesic equation for a test mass precisely due to the fact that the geodesic that that test mass follows (or the "acceleration" felt by a Newtonian mass) is independent of the mass of that test$^{1}$ particle.

The equivalence principle, however, only selects out that one can represent gravity with a metric tensor. There are a great many other so-called "metric theories of gravity" that obey the equivalence principle, but are not general relativity--amongst other things, they will differ in the field equation for the metric tensor, or have extra fields in addition to the metric--the most famous of these is the Brans-Dicke theory, which treats Newton's constant as a scalar field coupled to the metric tensor. Most alternative metric theories have either been experimentally ruled out, or have had their additional fields constrained to the point where their values are consistent with zero (for instance, Brans-Dicke theory has a parameter $\omega$, and tends to GR if $\frac{1}{\omega}\rightarrow 0$. Current data says that $\omega > 4000$, or some similar number.).

$^{1}$Note that this is generally only true if the mass of the test particle is "small" compared to the local curvature of the spacetime, and if it's motion is slow enough to not produce gravitational radiation comparable to its energy. Either of these effects will cause the test mass to perturb the background spacetime, and those effects will both be mass dependent and cause the test mass to not follow a geodesic of the background spacetime. Both of these approximations are true (to great precision, at least) of all of the planets, asteroids and comets orbiting the sun, amongst many other things.

gravity - Gravitation as the source of redshift of light beams

According to Hubble's law, light and other kinds of electromagnetic radiation emitted from distant objects are redshifted. The more distant the source, the more intense is the redshift. Now, the expansion of the universe is expected to explain the redshift and its nearly linear dependence on distance between source and observer. But isn't there an other source influencing the redshift? We know that a light beam passing the Sun is deflected by the Sun's gravity in accordance with predictions made by Einstein's general theory of relativity. This deflection is dependant on a gravitational interaction between the Sun and the light beam. Thus, the position of the Sun is affected by the light beam, though by such a tiny amount that it is impossible to detect the disturbance of the Sun's position. Now, during its journey to the Earth a light beam, originating from a distant source in the Universe, is passing a certain amount of elementary particles and atoms. If the light beam interacts gravitationally with those elementary particles and atoms, affecting the microscopic mechanical properties of the individual elementary particles and atoms at issue, can this interaction be detected as a redshift of the light beam? If so, could we use this gravity redshift to measure the mean density of matter and energy in space? The beginning of the sentence "If the light beam interacts gravitationally with those elementary particles and atoms..." should be interpreted to say "If the light beam interacts gravitationally with those elementary particles and atoms by way of leaving them in a state of acceleration different from their initial state of acceleration...." This clarification seems to necessitate the additional question: "why would a gravitationally interacting object (a cluster of photons) passing another gravitationally interacting object (a mass) leave that mass in the same state as before the passage?"

particle physics - Matter-Antimatter Asymmetry in Experiments?

As I hope is obvious to everyone reading this, the universe contains more matter than antimatter, presumably because of some slight asymmetry in the amounts of the two generated during the Big Bang. This raises the question of whether there are any processes short of the Big Bang that produce more matter than antimatter. That is, is there any known process where a particle collider (or whatever) would convert some energy into matter not through the production of particle-antiparticle pairs but through some process that produced more matter than antimatter? This doesn't need to be restricted to current accelerators-- if there's some mechanism for this that requires impractically high energies, but is based on solid theories (i.e., the Standard Model or straightforward extensions thereof), that would be interesting, too.

I'm fairly certain that the answer is "no," because I know that the matter-antimatter asymmetry is related to CP violation, and I also know that existing measurements of CP violation are not enough to explain the asymmetry. If there were a known way to slam protons together and make more quarks than antiquarks, I wouldn't expect this to still be a mystery. My particle physics knowledge is far from comprehensive, though, so it can't hurt to ask.

(I was briefly confused into thinking that there was such an experiment a while back, but it turned out to just be sloppiness about marking the antiquarks on the part of the people writing about it...)

(This is another question prompted by the book-in-progress, on relativity, this time a single word: I wrote that matter created from energy in particle physics experiments is "generally" in the form of particle-antiparticle pairs. Then I started wondering whether that qualifier was really needed, and thus this question.)

quantum mechanics - Why does Fluorine-19 have a nuclear spin of 1/2?

According to the nuclear shell model, $^{19}F$ has one unpaired proton in the $6$-fold degenerate $1d_{5/2}$ state, which means the orbital angular momentum is $l = 2$ and the total angular momentum is $j = l+ \frac12 = 5/2$. If we follow the rule highlighted by Lubos Motl that nuclear states with higher $m_j$ are filled first, then the one unpaired proton goes in the $m_j = 5/2$ state predicting a nuclear spin of $5/2$. However, we know that $^{19}F$ has a nuclear spin of $1/2$.

Why is that? Is the selection rule claimed by Lubos Motl simply wrong? If so, what is the appropriate way to fill the states and calculate nuclear spin? For example, why does $^{23}Na$ have a nuclear spin of $3/2$?

Answer

Here is a paper on the nuclear structure of this nucleus. It's not spherical, it's deformed. Basically the very first thing you should do when trying to guess the ground-state configuration of a nucleus is to try to find out whether or not it's spherical. If it's deformed, then you're not going to get the right answers by looking at a diagram of energy levels at spherical deformation. For a diagram showing the relevant energy levels in deformed nuclei of this region, see this paper, p. 30. The $d5/2$ shell splits up with deformation into states with definite values of $j_z$ along the axis of symmetry. For a prolate nucleus, the lowest $j_z$ drops the fastest. When you couple this odd particle to the rotational degrees of freedom, you get a rotational band in which the lowest spin equals the single particle's value os $j_z$.

thermodynamics - Work done in adiabatic process

When we try to establish a relation between the pressure and temperature in adiabatic process we come across a equation..

$dU = dq - PdV$

$dq=0$ (Adiabatic process) and,

$dU=C_v.dT$ (Heat capacity at constant volume)

Therefore, $C_v.dT = -PdV$$\tag1$

In this equation we are using heat capacity defined at constant volume but their still is some work done by the system (i.e, $PdV$ is not $0$ or $dV$ not equals to $0$).

The first part of the equation $(1)$ is implying that the volume is constant but the second part is implying that the volume is not constant (if it was there would be no work done).

Then why there is this contradiction ?

Answer

Even though we call $C_v$ the heat capacity at constant volume, what we really mean by the subscript v is that this is the way we measure $C_v$. At constant volume, we can determine the heat capacity of the material by measuring the heat transferred $dQ=dU=C_vdT$.

But this same heat capacity also applies to all other situations for an ideal gas if we recognize that, for an ideal gas, $U$ depends only on temperature, such that $dU=C_vdT$. It is just that, in these other situations (involving work), dQ is not equal to $C_vdT$.

condensed matter - Hollow gold bar

A scammer got a hollow gold bar and fills it with a combination of lead and air, with the same average density as gold.

What's the simplest way of discovering the fraud? I know that x-rays will see it, but are there simple means for analyzing it? (without destroying it)

general relativity - Why does pressure act as a source for the gravitational field?

I'm asking for a qualitative explanation if there is one.

My own answer doesn't work. I would have guessed it's because when a gas has pressure the kinetic energy adds to the rest mass of a given quantity of the gas, so the pressure contribution would be equal to whatever energy density it contributes. But that can't be right. If one had an ideal monatomic gas where the atoms are randomly moving around at non-relativistic speeds, the kinetic energy per volume of the atoms is 1.5 times greater than the pressure, but in chapter 4 of Schutz's book "A First Course in General Relativity" (or any other GR text) he says that rho plus pressure (in units where c=1) plays the role of inertial mass density. In my incorrect view the equation would be rho plus 1.5 pressure

Why is my answer wrong? I'm guessing part of the problem is that the kinetic energy of the atoms is already part of the mass density term--that is, a hot gas of one mole of helium atoms would have a higher mass than a cold gas composed of helium. Then the pressure is tacked onto that, which seems like counting it twice to me, but clearly I'm confused.

cosmology - Does the Universe have finite number of particles?

I read that the number of atoms in the entire observable universe is estimated to be within the range of $10^{78}$ to $10^{82}$.

Does the Universe have finite number of particles? If so, how could it be determined?

Answer

The universe must contain a finite energy - sum of all matter and fields - or the mass-equivalent would collapse within its own gravitation. Said mass-energy is fractionally partitioned among elementary particles and their agglomerates. One then strongly expects there are a finite number of particles including extremely low energy photons and neutrinos.

thermodynamics - At what gap width between two plates does convection not occur?

Does the Grashof number lead to the answer? The Wikipedia article (https://en.wikipedia.org/wiki/Grashof_number) yields an equation for vertical plates $$Gr_L = \frac{g\beta(T_s-T_\infty L^3)}{\nu^2}$$ Could I just solve for $L$, with $Gr_L$ equal to $10^8$ (upper boundary for laminar flow)?

Answer

You're on the right track here. For convection between plates to 'stop' (it won't really stop, just get wicked slow) you want viscosity to be very high relative to buoyancy, so you want low Grashof. Also, in the case of very narrow plates, you'd want to use the gap width as your characteristic length $L$.

$10^8$ is the transition point at which the flow goes from being laminar to turbulent. That's already a fair amount motion. Unit-wise, the velocity should scale as

$$\frac{g \beta \Delta T L^2}{\nu}$$. That'll get you a ballpark idea of the velocities involved.

Also look the Rayleigh number , which will tell you how strong convection is relative to conduction. As that gets towards, one you'll be seeing really negligible flow. (note : it'll be practically the same as Grashof for gases).

And last an empirical correlation using $w$ as gap width and $L$ as height:

$$Nu_w = \frac{hw}{k} =\left(\frac{576}{[Ra_w(w/L)]^2} + \frac{2.87}{[Ra_w(w/L)]^{1/2}} \right)^{-1/2}$$

Good for two plates at the same constant temp (other cases can be found scattered around the web).

Just get $h$ low enough that can say that convection has stopped. Again, you won't stop it, you'll just make it really small.

'Meson-nucleon' scattering in Scalar Yukawa Theory

I'm making my way through a QFT course and at the moment I'm trying to calculate some amplitudes in a toy-model Yukawa theory. The particular scattering I'm interested in is:

$$\phi \psi \rightarrow \phi \psi $$

and I want to do it 'by hand', i.e. using Dyson's formula and Wick's theorem.

My Lagrangian as given here: Lecutre Notes in eqation 3.7

I was looking at a similar example that is solved in the set of notes above. The example is in the section 3.3.3. My equivalent expression for the normal ordered product in 3.47 is:

$$ :\phi(x_1)\psi(x_1)^\dagger \phi(x_2)\psi(x_2): $$

But how do I know which of the $\phi$'s goes to the right? It could be either

$$\psi(x_1)^\dagger\phi(x_1) \phi(x_2)\psi(x_2)$$ or $$\psi(x_1)^\dagger \phi(x_2)\phi(x_1)\psi(x_2)$$ as I'll be using one of them to destroy and one of them to create a meson, but there doesn't seem to be a 'preferred' choice.

On top of that, there's something else I'd like to get a better intuition about.

If I, say, choose the latter one for now, I run into another problem. When I insert an unitary operator and expand my field operators (for readability only the rightmost ones, as I'm not concerned with the remaining ones) and the final state, $a_{k_1}^\dagger b_{k_2}^\dagger |0\rangle$, I obtain:

$\psi(x_1)^\dagger \phi(x_2)|0\rangle \langle 0|\int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}}\left( a_p e^{-ipx_1} + a_p^\dagger e^{ipx_1} \right)\int \frac{d^3q}{(2\pi)^3\sqrt{2E_q}}\left( b_q e^{-iqx_2} + c_q^\dagger e^{iqx_2} \right)a_{k_1}^\dagger b_{k_2}^\dagger |0\rangle $

I can discard the creation operators in my field operators since they kill the vacuum in the unity operator on the LHS, so I'm left with:

$$\psi(x_1)^\dagger \phi(x_2)|0\rangle \langle 0|\int \frac{d^3pd^3q}{(2\pi)^6\sqrt{2E_p}}\left( a_p e^{-ipx_1} \right)\left( b_q e^{-iqx_2} \right)a_{k_1}^\dagger b_{k_2}^\dagger |0\rangle $$

The difference now, compared to the example from the notes, is that where in third line of 3.48 there are two exponential factors in each pair of round brackets, I'll end up with just one (because after a single non-zero commutation among my four creation/annihilation operators the RHS vacuum is exposed to an annihilation operator). Assuming I'm correct up to this point, is this because in this kind of scattering we have a nucleon in our propagator rather than a meson?

Thanks a lot for your help!

EDIT ----------------------

I just had an idea: is the ambiguity in the ordering of $\phi$'s due to the fact that we have two possible diagrams? And we need to account for both cases and as a result our amplitude is twice the amplitude calculated for one particular choice of ordering?

everyday life - Is play-dough liquid or solid?

At room temperature, play-dough is solid(ish). But if you make a thin strip it cannot just stand up on it's own, so is it still solid?

On a more general note, what classifies or differentiates a solid from a liquid?

Answer

Play-Doh is mostly flour, salt and water, so it's basically just (unleavened) dough. There are a lot of extra components like colourings, fragrances, preservatives etc, but these are present at low levels and don't have a huge effect on the rheology.

The trouble with saying it's basically just dough is that the rheology of dough is fearsomely complicated. In a simple flour/salt/water dough you have a liquid phase made up of an aqueous solution of polymers like gluten, and solid particles of starch. So a dough is basically a suspension of solid particles in a viscous fluid. To make things more complicated the particles are flocculated, so you end up with a material that exhibits a yield stress unlike the non-flocculated particles in e.g. oobleck.

At low stresses dough behaves like a solid because the flocculated particles act like a skeleton. However the bonds between flocculated particles are weak (they're only Van der Waals forces) so at even moderate stresses the dough flows and behaves like a liquid. Dough, and Play-Doh, are best described as non-Newtonian fluids.

newtonian mechanics - Proof that the Earth rotates?

What is the proof, without leaving the Earth, and involving only basic physics, that the earth rotates around its axis?

By basic physics I mean the physics that the early physicists must've used to deduce that it rotates, not relativity.

Answer

Foucault pendulum. I don't know how the ancients did it, but it is surely pure classical mechanics.

The animation describes the motion of a Foucault Pendulum at a latitude of 30°N.

differential geometry - Diffeomorphism & Weyl transformations in the 2D worldsheet of string theory and the existence of conformal gauge

D. Tong's notes on string theory, chapter 5 (PDF), feature the following in introducing the symmetries used in the Faddeev-Popov method:

We have two gauge symmetries: diffeomorphisms and Weyl transformations. We will schematically denote both of these by $\zeta$. The change of the metric under a general gauge transformation is $g \rightarrow g^{\zeta}$. This is shorthand for, \begin{equation} g_{\alpha \beta} (\sigma) \rightarrow g_{\alpha \beta}^{\zeta} (\sigma)= e^{2\omega (\sigma)} \frac{\partial \sigma^{\gamma}}{\partial \sigma'^{\alpha}} \frac{\partial \sigma^{\delta}}{\partial \sigma'^{\beta}} g_{\gamma \delta} (\sigma) \end{equation} In two dimensions these gauge symmetries allow us to put the metric into any form that we like — say, $\hat{g}$. This is called the fiducial metric and will represent our choice of gauge fixing.

Is there a proof available that shows that the combined symmetry allows us to "put the metric in any form (locally, of course)?" Is this property restricted to two dimensions?

Answer

Theorem. Every 2D pseudo-Riemannian manifold $(M,g)$ is locally conformally flat, i.e. there locally exist isothermal coordinates.$^1$

Sketched proof of theorem: Given a point $p\in M$. Consider local coordinates $u,v$ in a neighborhood of $p$.

1. 
   

   Generic case $g_{uu}(p)\neq 0$: Do a Weyl scaling such that $g_{uu}\equiv 1$. Then $$g~=~\mathrm{d}u\odot \mathrm{d}u + 2g_{uv}\mathrm{d}u\odot \mathrm{d}v + 2g_{vv}\mathrm{d}v\odot \mathrm{d}v~\stackrel{(2)}{=}~\omega_+\odot \omega_-,\tag{1}$$ where we have defined 2 non-vanishing one-forms $$ \omega_{\pm}~:=~\mathrm{d}u + [g_{uv} \pm \sqrt{-\det(g)}]\mathrm{d}v. \tag{2}$$

   
   
   
   
   • 
     

     Minkowskian case $\det(g)<0$: Then the one-forms $\omega_{\pm}$ are real. There exist locally 2 real integrating factors $\lambda_{\pm}\neq 0$ such that $$\omega_{\pm} ~=~\lambda_{\pm}\mathrm{d}x^{\pm}.\tag{3}$$ Then the metric tensor (1) reads $$g~\stackrel{(1)+(3)}{=}~\lambda_+\lambda_-\mathrm{d}x^+\odot \mathrm{d}x^-.\tag{4} $$ A Weyl scaling brings the metric on light-cone form. $\Box$

     
     
   
   
   • 
     

     Euclidean case $\det(g)>0$: Then the one-form $\omega_{\pm}~=~\omega^{\ast}_{\mp}$ is complex. There exists locally a complex integrating factor $\lambda_+\neq 0$ such that $$\omega_+ ~=~\lambda_+\mathrm{d}z, \qquad z ~=~x+iy, \qquad \omega_- ~=~\omega^{\ast}_+ ~=~\lambda^{\ast}_+\mathrm{d}z^{\ast}.\tag{5}$$
     
     Then the metric tensor (1) reads $$g~\stackrel{(1)+(5)}{=}~|\lambda_+|^2\mathrm{d}z\odot \mathrm{d}z^{\ast}~=~|\lambda_+|^2[\mathrm{d}x\odot \mathrm{d}x+\mathrm{d}y\odot \mathrm{d}y],\tag{6} $$ which is again manifestly real. A Weyl scaling brings the metric on standard Euclidean form. $\Box$

     
     
   
   
   
   


2. 
   

   Special case $g_{uu}(p)=0$:

   
   
   
   
   • 
     

     Euclidean case $\det(g)>0$: Impossible. $\Box$

     
     
   
   
   
   • 
     

     Minkowskian case $\det(g)<0$: Then $g_{uv}(p)\neq 0$. In the Gauss elimination procedure (if we were to bring $g$ on diagonal form) this corresponds to a case of vanishing diagonal element. It is possible to perform an affine coordinate transformation $(u,v) \to (u^{\prime},v^{\prime})$ so that $g_{u^{\prime}u^{\prime}}(p)\neq 0$. Now use the generic case. $\Box$

     
     
   
   
   
   

References:

1. 
   

   M. Nakahara, Geometry, Topology and Physics, 1989; Example 7.32.

   
   
   


2. 
   

   M. Nakahara, Geometry, Topology and Physics, 2003; Example 7.9.

   
   

--

$^1$ And yes, the theorem is not true in higher dimensions, cf. e.g. this related Phys.SE post.

thermodynamics - A physical explanation for negative kelvin temperatures

Just to get the thoughts rolling...

Consider a two state system with discrete energy levels $E_1$ and $E_2$ where $E_2 > E_1$ which contains $N$ particles.

We can easily deduce that the state of maximum entropy is when energy level $E_1$ and $E_2$ each contain $\frac{N}2$ particles.

Based on the (flipped-upside-down) thermodynamic definition of temperature, $$(\frac{\partial E}{\partial S})=T$$ we see that when energy as a function of entropy increases, temperature also increases.

This means, for our two state system, every time a particle populates level $E_2$, there is a corresponding increase in temperature because that population increases entropy.

What about when we go beyond the point of maximum entropy where there are $\frac N2$ particles in $E_1$ and $E_2$?

Well, then we are adding energy to the system, but the entropy is decreasing.

Based on the Boltzmann factor, this can't be possible unless the temperature of the system is negative. That is, $$\frac{P_1}{P_2}=e^{-\frac{(E_1-E_2)}{kT}}$$ which approaches infinity unless one says the system has a negative absolute temperature.

Now we're getting to the point: this increase in energy which is connected to a decrease in entropy is what has been called a negative kelvin temperature. This is most often connected to population inversion in lasers.

Forgive me for saying things you already know.

All of this raises a lot of questions though, the most striking to me is that there does not seem to me to be a good physical explanation for how this can be possible. For instance, when people say that the negative temperature system is "hotter" I get what that means as far as saying that heat will flow from a negative temp. system to a pos. temp system.

I'm fine with that except that it invokes the idea of a fairly macroscopic quantity--hotness--for a purely quantum phenomenon. So, is there any macroscopic physical meaning to this statement that the negative temperature is "hotter"?

Concerning population inversion in lasers: doesn't this mean that the negative temperature required for that laser to exist ought to be so "hot" no mere mortal can handle such a device.

After all, negative absolute temperatures are often explained as being "hotter than infinite temperature." (I've heard this other places, but also see the neg. temp. Wikipedia article for quote.) I understand that again from a mathematical perspective, but what is the physical connection? A statement like that must surely manifest itself in some way that is noticeable beyond the quantum level.

I guess I really want to understand how it could be this "hotter than infinite temperature" system isn't perceptible in the macroscopic realm? I mean, lasers are cool, but they aren't just melting through everything as if they have infinite temperature.

---

My bad for the obvious misuse of the word "physical" when I probably meant something more like "macroscopic". I hope the point is clear anyways.

Thoughts?

Is information propagated across a medium in any other way than waves?

Is information propagated in any other way than waves? Please distinguish "propagation across a medium" from information "storage within stable states of matter", which might difuse or interact chemically.

Information might be stored in stable configurations of matter, which might diffuse, or interact chemically (odor,DNA), but these might be orders of magnitude weaker, in range and dissipation. Is there a domain in physics comparing wave vs non-wave propagation. The two most known are sensory: sound and electromagnetic propagation. I think gravity probes are still searching for waves in this medium.

Why does nature prefer waves for long distance calls? Perhaps because it involves a minimum dissipation of energy?

quantum mechanics - Why the lowest order of matrices in Dirac equation are 4x4 matrices?

Why the lowest order of matrices in Dirac equation (Relativistic Quantums) are 4x4 matrices (and can not be 2x2 matrices)?

How to prove it?

quantum mechanics - How is temperature related to color?

I spent some time studying about temperatures and color of objects. It turns out that as we heat something it turns to red, then yellowish white and if we heat it more it turns to bluish-white.

Like we can say a blue star is hotter than a red star. But why isn't it the same with flames?

Blue flame isn't always hotter than a red flame. It's just the chemistry of it all.

I mean I know the electrons jump from one orbit to another and gets into excited state when heated.

But the questions in my mind are:

1. 
   

   If one object appears blue and another red, that doesn't always mean that the blue object is hotter than red. Is that correct? If yes then how is it so? I am sorry but I'm bit confused over it.

   
   


2. 
   

   With the "Color Temperature" concept on Wikipedia. They say 1,700 K to match a flame. But 15,000–27,000 K for a clear blue poleward sky.

   
   

This confuses me. The sky appears blue, does that mean its hot? But it isn't right? It's colder. I am not sure if I'm able to frame it properly. It's a bit confusing to me.

Is this something like a glowing object vs reflecting object? The sun is white but Earth is blue. But sun is hotter than the earth. It's the sun radiating light but earth is just scattering it.

But the surface temperature of sun is 5778K while the Wikipedia states our sky is about 15000K-27000... I know there's something I'm missing here. I'm hoping someone can tell me what is it.

Answer

1) If one object is appearing blue and other object red. That doesn't always mean blue object is hotter than red? Is that correct? If yes then how is it so? I am sorry but I'm bit confused over it.

Yes it's correct: when an object is glowing blue when burning that means it has more temperature than an object glowing red. It's because an object which is glowing blue has more energy. Energy of an object of temperature $T$ (in kelvins) is calculated using this equation: $E=kT$ where $k$ is Boltzmann's constant. Energy of a light with frequency $f$ is calculated using this equation: $E=hf$ where $h$ is plank's constant. So spectral radiance of emitted light with frequency $f$ from object with temperature $T$ is calculated using $\beta(T) = \frac{2hf^3}{c^2}\frac{1}{e^{\frac{hf}{kT}}-1}$. For derivation See Plank's Law. The higher the frequency of light the more energy it has and it looks 'more blue' and the lower the frequency the less energy and it looks 'more red'.

Here's very good image of "light" (light is part of spectrum we see (visible spectrum), full spectrum is called electro-magnetic spectrum, but for simplicity let's call it just light) spectrum:

Here's a very good diagram of emitted light with wavelength $\lambda$ from an object with temperature $T$ (in kelvins). (NOTE: for light with frequency $f$, wavelength is $\lambda = \frac{c}{f}$ where $c$ is the speed of light.)

2) With the "Color Temperature" concept on Wikipedia. They say 1,700 K for match flame. But 15,000–27,000 K for a clear blue poleward sky.

This confuses me. The sky appears blue, does that mean its hot? But it isn't right? It's colder. I am not sure if I'm able to frame it properly. It's a bit confusing to me.

Is this something like a glowing object vs reflecting object? The sun is white but Earth is blue. But sun is hotter than the earth. It's the sun radiating light but earth is just scattering it.

You can calculate spectral radiance. Spectral radiance is like an (spectral radiance isn't fully intensity, it's just like an intensity) intensity of emitted light with frequency $f$ from object with temperature $T$ (in kelvins) of emitted light from an object with temperature $T$ (in kelvins) using this equation:$$\beta(T) = \frac{2hf^3}{c^2}\frac{1}{e^{\frac{hf}{kT}}-1}$$ Earth is blue because it reflects light, not emits it. But the Sun doesn't reflect light (it's a blackbody, but not perfect), it just emits it, that's why it's hotter.

newtonian mechanics - Pulling apart two interleaved phone books

The TV show Mythbusters had an episode in which they interleaved two phone books and dramatized how hard it was to pull them apart. (A long time ago, a phone book was a book that had an index of phone numbers for everyone in your city. The trick apparently also works pretty well with reasonably thick magazines.) They ended up needing two tanks to pull the phone books apart against the resistance of friction:

http://www.youtube.com/watch?v=AX_lCOjLCTo

http://www.youtube.com/watch?v=QMW_uYWwHWQ

When one of my students described this in class, it strongly violated my carefully cultivated intuition about friction, and I thought maybe the student was getting the description wrong. Actually it seems right.

Why does this work? I found the following online discussion from 2008, the same year in which the episode aired:

http://www.physicsforums.com/showthread.php?t=215857

Posts 8, 27, and 36 seem the most relevant. 36 quotes the result of a calculation, without saying what the calculation was.

Can anyone give a good analysis?

energy - Spacetime and the conservation laws

I'm reading Peter Atkins' book, Galileo's Finger, and in the chapter on energy, he makes the points that the conservation of momentum stems from the shape of space (that it's smooth and not lumpy) and that the conservation of energy stems from the shape of time (that it's smooth and not lumpy). I'm not totally clear on how the shape of spacetime leads to the conservation laws. Could someone elucidate the relationship, in layman's terms?

quantum field theory - Would a spin-2 particle necessarily have to be a graviton?

I'm reading often that a possible reason to explain why the Nobel committee is coping out from making the physics Nobel related to the higgs could be among other things the fact that the spin of the new particle has not yet been definitively determined, it could still be 0 or 2.

This makes me wonder if the spin would (very very surprisingly!) finally be discovered to be 2, this then necessarily would mean that the particle has to be a graviton? Or could there hypothetically be other spin-2 particles? If not, why not and if there indeed exist other possibilities what would they be?

Answer

There are theoretical arguments that a massless spin-2 particle has to be a graviton. The basic idea is that massless particles have to couple to conserved currents, and the only available one is the stress-energy tensor, which is the source for gravity. See this answer for more detail.

However, the particle discovered at LHC this year has a mass of 125 GeV, so none of these arguments apply. It would be a great surprise if this particle did not have spin 0. But it is theoretically possible. One can get massive spin 2 particles as bound states, or in theories with infinite towers of higher spin particles.

special relativity - Wick rotation - time and what else changes?

For aid of example consider two quantities the four-momentum $\tilde P$ and a time-independent four potential $\tilde A$. Now if a wick's rotation was carried out by simply replacing $it$ with $\tau$ then under a Wick's rotation we would get: $$\tilde P'=i \tilde P$$ $$\tilde A'=\tilde A$$ whilst if it was carried out as a rotation by $\pi/2$ in the complex plane of the $0$th component we would get: $$P_0'=-iP_0,\quad A_0'=-iA_0$$ with all other components remaining the same. Which of these (if either) is the correct interpretation of a Wick's rotation - if either? and why?

Answer

1. 
   

   OP's second option is correct: The zero-components $V^0$ of all contravariant $4$-vectors $V^{\mu}$ do Wick-rotate $$V^0_E~=~iV^0_M ;\tag{A}$$ not just time $x^0$ in the spacetime position 4-vector $x^{\mu}$: $$x^0_E~=~ix^0_M. \tag{B}$$ That is how inner products remains invariant $$V\cdot V~=~V^{\mu}~\eta_{\mu\nu}~ V^{\nu} \tag{C}$$ when going from Minkowski$^1$ signature $(-,+,+,+)$ to Euclidean signature $(+,+,+,+)$.

   
   
   


2. 
   

   Similarly, the zero-components $V_0$ of all covariant $4$-vectors $V_{\mu}$ do Wick-rotate in the opposite direction: $$V_0^M~=~iV_0^E.\tag{D}$$ In particular, the zero-component $p_0$ of the energy-momentum 4-covector $p_{\mu}$ Wick-rotate as$^2$ $$p_0^M~=~ip_0^E.\tag{E}$$ (The latter is related to the fact that the Fourier-integral representation $$\delta^4(x)~=~\int_{\mathbb{R}^4} \frac{d^4p}{(2\pi\hbar)^4}~\exp\left(\frac{ip\cdot x}{\hbar} \right)\tag{F}$$ of the Dirac delta distribution cannot be analytically continued to the ambient complexified spacetime: The real integration region can at most be deformed, i.e. the $x^0$ and $p_0$ Wick-rotations must be balanced, cf. this Phys.SE post.)

   
   


3. 
   

   To see how the Wick rotation works in gauge theory, see e.g. this related Phys.SE post.

   
   

--

$^1$ The speed of light $c=1$ is set to one in this Phys.SE answer for simplicity. Concerning the reason for the choice of Minkowski signature, see my Phys.SE answer here.

$^2$ Warning: Traditionally we assign the energy variable $E=p^0$ to behave contravariantly, so that $E_E=iE_M$. However some authors effectively define the energy variable $E=-p_0$ to behave covariantly, so that $E_M=iE_E$ and $E_E$ is effectively the negative Euclidean energy! See also my Phys.SE answer here.

lagrangian formalism - Defining quantum effective/proper action (Legendre transformation), existence of inverse (field-source)?

Given a Quantum field theory, for a scalar field $\phi$ with generic action $S[\phi]$, we have the generating functional $$Z[J] = e^{iW[J]} = \frac{\int \mathcal{D}\phi e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi e^{iS[\phi]}}.$$

The one-point function in the presence of a source $J$ is.

$$\phi_{cl}(x) = \langle \Omega | \phi(x) | \Omega \rangle_J = {\delta\over\delta J}W[J] = \frac{\int \mathcal{D}\phi \ \phi(x)e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi \ e^{i(S[\phi]+\int d^4x J(x)\phi(x))}}.$$

The effective Action is defined as the Legendre transform of $W$

$$\Gamma[\phi_{cl}]= W[J] -\int d^4y J(y)\phi_{cl}(y),$$ where $J$ is understood as a function of $\phi_{cl}$.

That means we have to invert the relation $$\phi_{cl}(x) = {\delta\over\delta J}W[J]$$ to $J = J(\phi_{cl})$.

How do we know that the inverse $J = J(\phi_{cl})$ exists? And does the inverse exist for every $\phi_{cl}$? Why?

inertial frames - Is length contraction in Special Relativity the same as the Doppler Effect?

In my further reading of Special Relativity, the idea of length contraction when travelling at the speed of light is such that the length gets "squished" in the direction of travel. This immediately made me think of the familiar Doppler Effect, with electromagnetic waves travelling at the speed of light, where their wavelength is shifted due to their velocity, i.e red shift.

How does Special Relativity and the Doppler Effect link? Is the Doppler Effect a Relativistic effect, as electromagnetic waves travel at the speed of light?

Answer

Length Contraction involves the apparent spatial separation of two parallel [timelike] worldlines, marking the ends of a stick. The contraction depends only on the relative-speed (but not direction) along the x-axis.

The Doppler Effect involves the apparent spatial separation of two parallel lightlike lines, marking the wavelength... the successive wavefronts of a light wave. The scaled wavelength depends on the relative-velocity, where the approaching case is different from the receding case.

Borrowing from my older reply to Deriving Relativistic Doppler Effect through length contraction

In the source frame, imagine a ruler at rest with a marking at $x=10$.
Interpret this as "where the source says the previous wavefront is located when the source emits the next signal".
Note that this marking has a worldline parallel to the source.
Although the "separation between these timelike-worldlines" is equal to $\lambda_{source}$ in the source frame, these timelike-worldlines are only indirectly related to the source-wavelength [which is the "separation between the lightlike signal-lines"].
In the receiver-frame the "separation between these timelike-worldlines" is given by $OX=\frac{\lambda}{\gamma}=\frac{10}{5/4}=8$, in accordance with length-contraction.
However, this is not the wavelength observed by the receiver---
the observed-wavelength is the "separation between the lightlike signal-lines" given by $OW=20$ in the receiver frame. ($OW=k(O\lambda_{source}=(2)(10)=20.$).

The point is: the observed-wavelength (separation between lightlike signal-lines)
doesn't directly involve length-contraction (involving parallel timelike-lines).

The above spacetime diagram is drawn on rotated graph paper so that it becomes easier to visualize the ticks along the various segments.

Finally....
The Doppler effect occurs for any wave motion... e.g. sound.
For sound waves in Galilean physics, the Doppler factors depend on the velocities of source, receiver, and wind.
However, in special relativity, the Doppler Effect depends only on the relative-velocity between source and receiver.

lagrangian formalism - Derivation of the supergravity action in 11D

The Einstein-Hilbert action of general relativity is uniquely determined by general covariance and the requirement that only second derivatives in the metric appear. Yang-Mills theory can be motivated in a similar way. In the original paper of Scherk, Julia, Cremer there are some arguments given from which they deduced the form of the action. They are only sketched however. Is there a more complete exposition of the derivation in the literature, or possibly even a uniqueness result as in the case of general relativity or Yang-Mills theory?

Answer

A beautifully conceptual derivation of 11d SuGra was given in

• Riccardo D'Auria, Pietro Fré, Geometric Supergravity in D=11 and its hidden supergroup, Nuclear Physics B201 (1982) 101-140 (nLab)

using the excellent supergeometric methods later laid out in their textbook

• Leonardo Castellani, Riccardo D'Auria, Pietro Fré, Supergravity and Superstrings - A Geometric Perspective World Scientific, 1991 (nLab)

condensed matter - Low dim physics: Examples of confinement-deconfinement phases of U(1) gauge theory in 2 dimensions

Please provide some examples of confinement-deconfinement phases of U(1) gauge theory in 2 spacetime dimensions (Low dimwnsional physics).

U(1) gauge theory can be:

pure U(1) gauge theory, or

U(1) gauge theory with matter fields (bosonic/fermionic).

It is the best to provide examples for both cases.

Time in special relativity and quantum mechanics

The time is treated differently in special relativity and quantum mechanics. What is the exact difference and why relativistic quantum mechanics (Dirac equation etc.) works?

Answer

Quantum mechanics can be reconciled with special relativity to make quantum field theory, but there are some awkward things going on in that marriage. SR treats time symmetrically with position, but in quantum mechanics, position is an operator and time isn't. Baez at UCR has a nice discussion of that here: http://math.ucr.edu/home/baez/uncertainty.html

fluid dynamics - Does irrotational imply inviscid?

Let us consider a 2D irrotational flow, such that $\nabla\times\boldsymbol u =\boldsymbol 0$. Defining the stream function such that $\boldsymbol u =\nabla\times\psi \boldsymbol n$ where $\boldsymbol n$ is the unit vector perpendicular to the plane of the motion. The incompressibility condition implies $\nabla^2\psi =0$, so $\nabla^2 \boldsymbol u= \nabla^2 \nabla\times\psi \boldsymbol n=\nabla\times\nabla^2\psi \boldsymbol n=\boldsymbol 0$. From this follows that in the Navier-Stokes equations the term $\nu\nabla^2\boldsymbol u=0$ vanishes, no matter the viscosity.

Now, does this allow us to say that the fluid is inviscid? Or, after considering the nature of the inertia, we can say that:

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  if inertia can be neglected, the equation of the motion is $\nabla p=\boldsymbol 0$.

  
  


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  if inertia is important, the flow is at high Reynolds number.

  
  
  

My problem is: Is every irrotational flow inviscid? This is kind of counter-intuitive. But I think my error is in the "no matter the viscosity"...

Answer

(I try to answer to my own question, after some reflections made with the help of Luboš.)

For an incompressible and irrotational flow, the conditions $\nabla\times \boldsymbol u=\boldsymbol 0$ and $\nabla\cdot \boldsymbol u = 0$ imply, $\nabla^2\boldsymbol u =\boldsymbol 0$. Indeed:

$$\nabla^2\boldsymbol u = \nabla(\nabla\cdot\boldsymbol u)-\nabla\times(\nabla\times \boldsymbol u) = \boldsymbol 0$$

This forces us to write down the Navier-Stokes equation for the motion of the fluid without the viscous term $\mu\nabla^2\boldsymbol u$, no matter the viscosity:

$$ \rho(\partial_t \boldsymbol u + u\cdot\nabla \boldsymbol u) = -\nabla p \ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$

Now, it could seem that this implies the flow is automatically a high-Reynolds number flow (for which we could have written down the same equation, but for a different reason: $\mu=\rho\nu\simeq 0$, and this would have been an approximation). But, even if the viscosity is far from neglectable, we can make another kind of approximation, saying that the inertia, represented by the left-hand-side terms in N-S equation, can be neglected because of $Re=UL\rho/\mu\ll 1$ (this can happen in a lot of situations: microobjects, extra-slow flows, and - of course - high viscosity. In this case, the equations of motion become: $$-\nabla p = \boldsymbol 0\ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$ which are, in fact, the equations we would arrive at if we started by the Stokes equation (for inertia-less flows) for irrotational flows.

Then, an irrotational flow is not necessarily governed by the Euler equation, i.e., it's not necessarily inviscid.

newtonian gravity - How was Newton able to guess that gravitational force is inversely proportional to distance squared?

This question is puzzling me since I learnt about the gravitation law in school. Why did Newton guess/assume that gravitational force is inversely proportional to the square of distance?

Did he verify that experimentally? (I remember hearing that the first experimental verification of the law of gravitation was after Newton's death.)

If the answer to the above question is no, is it for example more plausible to suppose that $F\propto1/r^2$ than to suppose that $F\propto1/r^4$? Did Newton carry out a thought experiment that makes $F\propto1/r^2$ a plausible guess?

So in summary: Why did Newton choose exponent of $-2$ instead of any other exponent? Was it a guess that depended on pure luck or an educated guess?

Answer

For a uniform circular orbit of radius $r$, the acceleration is

$$\tag{A} a~=~ \omega^2r, \qquad \omega~=~\frac{2\pi}{T},$$

where $T$ is the orbital period. Comparing eq. (A) with Kepler's third law

$$\tag{B} T^2 ~\propto~ r^3,$$

we conclude that the gravitational acceleration

$$\tag{C} a~\propto~ r^{-2} $$

is proportional to the inverse square distance $r$.

What does negative energy imply?

I know energy comes in a number of forms and every form of energy is defined uniquely.

But if I had to give a broad definition of energy, I would define it as "ability of a system to perform work."

Using the above definition doesn't make any sense while trying to explain negative energy. So what does negative energy mean?

Answer

If positive work is "work done by the system" then negative work is just "work done on the system". The sign just tells if energy is added to the system or leaving the system.

quantum mechanics - Entangled particles

So we have two particles (A and B) that are entangled.

From what I understand, entanglement isn't destroyed, it is only obscured by subsequent interactions with the environment.

Particle A goes zooming off into outer space.

10 years later, Particle B becomes incorporated into my brain.

The day after that, an alien scientist measures the entangled property on Particle A.

This will have some immediate non-local effect on Particle B won't it?

And since B's state has been altered (in some sense), and it is part of my brain, then my brain state has been altered as well, hasn't it?

Maybe only a tiny amount, obscured by the many environmental interactions that the two particles have been subjected to since the initial entanglement, but in a way that is real and at least conceivably significant.

And if that is true, then to the extent that mental states supervene on brain states, my mental state would also have been altered by non-local effects.

Or is that wrong?

Thanks!

Answer

From what I understand, entanglement isn't destroyed, it is only obscured by subsequent interactions with the environment.

Depends on how you view it. There is an explanation of quantum measurement (called decoherence) in which this is true. I will not be using that explanation in this post because it's unnecessarily complicated.

This will have some immediate non-local effect on Particle B won't it?

Nope. When an alien scientist measures particle A, it does not have an effect on particle B. Specifically, it does not induce any sort of change in particle B that can be detected.

Of course, we would say that the quantum state of particle B (technically: the combined quantum state of the pair AB) has changed, but changes of this nature in the quantum state can't be detected. For example, suppose the entangled state, written in a basis of energy levels, is

$$|\psi_1\rangle = \frac{1}{\sqrt{2}}\left(|1\rangle|2\rangle + |2\rangle|1\rangle\right)$$

so that if particle A has an energy of 1 (in unspecified units), particle B will have an energy of 2, and vice versa. Suppose the alien scientist measures the energy of particle A to be 1. The state of the pair AB becomes

$$|\psi_2\rangle = |1\rangle|2\rangle$$

Then, the next day, when your brain measures the energy of particle B, it will get 2 as the result. But as far as your brain is concerned, the energy of particle B could have just been 2 all along. Or it could have had some arbitrary probability distribution with 2 as one of the possible results. The fact that the state changed from $|\psi_1\rangle$ to $|\psi_2\rangle$ has no effect, and could not have been detected by your brain.

And if that is true, then to the extent that mental states supervene on brain states,

If by that you mean that your thoughts are determined by quantum states, that's definitely false.

thermodynamics - Why does the blackbody radiate even at thermal equilibrium?

Everytime, I read about blackbody, I always get confused at the point where it is written

Under thermal equilibrium conditions , the blackbody radiation depends only on temperature. ..... . At equilibrium , electromagnetic waves bounce around with the walls of the cavity.

What is this equilibrium all about? Why is it important here?? If there is thermal equilibrium, why does the body radiate?? Please help.

Answer

The equilibrium mentioned in that quotation is between the radiation field and the walls of the container. The walls of the container are imagined to be held at some fixed temperature by some (unspecified) means. Under those conditions, the spectrum of the radiation in the cavity is that of a blackbody at the temperature of the walls. The radiation field and the walls exchange energy, but since they are in equilibrium the temperature of both remains the same. The walls radiate and absorb equally.

So far I haven't mentioned anything about radiation that leaves the cavity. In the ideal case of a closed cavity, the discussion is about the walls of the cavity and the field within. One could poke a small hole in the cavity, allowing some of the radiation to escape. In the limit that the hole is infinitesimally small, the disturbance caused by the hole is negligible, and the radiation that leaks out has very nearly a blackbody spectrum. This is a blackbody radiator. Usually the temperature of the blackbody is much higher than the surroundings into which the radiation radiates. Thus the blackbody radiator is not in thermal equilibrium with it's surroundings.

newtonian mechanics - Kinetic energy of an object at rest

Why don't we consider that an object on Earth has kinetic energy as it is also moving at the rotating speed of Earth?

What is the definition of an exchange particle?

After reading through articles, i concluded that a suitable definition is that when 2 particles interact bosons are exchanged between the 2 particles creating a force?

What would a good definition be for exchange particle?

Conservation of Energy and Quantum Fluctuations

Regarding conservation of mass-energy Wikipedia says: "this is an exact law, or more precisely, has never been shown to be violated."

However, regarding quantum fluctuations, Wikipedia says here: "That means that conservation of energy can appear to be violated, but only for small times".

I thought to resolve the two conflicting statements by proposing that the virtual particles from quantum fluctuations are created/destroyed from the "noise energy" of the uncertainty principle. Hence, the energy is already there. Is this correct?

If this is incorrect, how can we resolve the two?

nuclear physics - What is the most stable nuclide of an isobar?

From the semi-empirical mass formula, the mass of an atomic nucleus is $$M\left(A,Z\right)=Zm_p+(A-Z)m_n-\frac{E_b(A,Z)}{c^2}$$ I've been told (at first) that for a given mass number $A$, the most stable nuclide of the isobar is the one with the smallest mass. Then it should be the one with proton number $Z$ such that $$\frac{\partial{M}\left(A,Z\right)}{\partial{Z}}=0$$ i.e. \begin{equation}m_p-m_n-\frac{1}{c^2}\frac{\partial{E_b}(A,Z)}{\partial{Z}}=0\end{equation} but then I had a test where the professor asked for the most stable isobar and he argumented that this was the one for which the binding energy was the highest, meaning only that $Z$ satisfies $$\frac{\partial{E_b}(A,Z)}{\partial{Z}}=0$$ so I got it wrong on the test. Either way the difference is just a (small) constant, so the values $Z$ they both predict are roughly the same for low values of $A$. Let me call ${Z_m}$ the one obtained through minimizing $M$ and ${Z_E}$ the one obtained through maximizing ${E_b}$, then a plot looks like this

The values of $Z$ begin to separate for large $A$, but the good values here *supposedly** are those of $Z_m$, for example, for $A=209$: \begin{align}Z_m(209)&=83.36\approx83\\ Z_E(209)&=82.22\approx82\end{align} because the most stable isobar for $A=209$ *as given in Table A.4 in Shultis & Faw's book** corresponds to $Z=83$ (element $\text{Bi}$).

But both that the binding energy should be a maximum and that the mass should be a minimum makes sense; in the Wikipedia article, for example, the correct one is given as $Z_E$, and I'd even go with the first one, but as I've seen, the best one is the one for minimum mass, so...

How come this happens? Is it a just a fault of the theory (namely the liquid drop model) or what?

Answer

The liquid drop model (LDP) is an approximate description of the mass of nuclei. It is a parametric formula that is fitted to the experimental values. The formula for binding energy is expressed in a similar way and comes from the same assumption. Therefore, both being parametrizations, they are approximate. And indeed if you examine nuclei by nuclei it will fluctuate largely from the experimental values. The small differences you see raise from small differences in the parameters of the formula, due to diversity in the fitting data and ultimately, due to numerical approximations. But bare in mind, non of these formulas have a predictive character, there values are off for most of the nuclei known.

Furthermore, the model itself is known to be approximative since it does not consider many of the features observed in nuclei. This is the ultimate reason why these formulas will give very approximate results, in many cases only valid for order of magnitude.

general relativity - What prevents the accumulation of charge in a black hole?

What prevents a static black hole from accumulating more charge than its maximum? Is it just simple Coulomb repulsion?

Is the answer the same for rotating black holes?

Edit

What I understand from the answers given so far, is that maximum charge is a moving target. You can add charge to a black hole but Coulomb repulsion guarantees that you will do so in a manner than will increase "maximum charge value". Is this correct?

Answer

Coulomb repulsion it is. Specifically, if a black hole has a lot of charge, then particles with a high charge-to-mass ratio will be repelled. Anything that falls in will contribute "more mass than charge," heuristically, keeping the charge-to-mass ratio of the black hole from getting too big.

quantum field theory - Self energy, 1PI, and tadpoles

I'm having a hard time reconciling the following discrepancy:

Recall that in passing to the effective action via a Legendre transformation, we interpret the effective action $\Gamma[\phi_c]$ to be the generating functional of 1-particle irreducible Green's functions $\Gamma^{[n]}$. In particular, the 2-point function is the reciprocal of the connected Green's function,

$$\tilde \Gamma^{[2]}(p)=i\big(\tilde G^{[2]}(p)\big)^{-1}=p^2-m^2-\Sigma(p)$$

which is the dressed propagator.

But, the problem is this: in the spontaneously broken $\phi^4$ theory, the scalar meson (quantum fluctuations around the vacuum expectation value) receives self energy corrections from three diagrams:

$-i\Sigma(p^2)=$ + +

Note that the last diagram (the tadpole) is not 1PI, but must be included (see e.g. Peskin & Schroeder p. 361). In the MS-bar renormalization scheme, the tadpole doesn't vanish.

If the tadpole graph is included in $\Sigma$, and hence in $\tilde{G}$ and $\tilde\Gamma$, then $\tilde\Gamma$ cannot be 1PI. If the tadpole is not included, then $\tilde G$ is not the inverse of the dressed propagator (that's strange, too). What's going on?

Answer

I'm going to give an explanation at the one loop level (which is the order of the diagrams given in the question).

At one loop, the effective action is given by $$ \Gamma[\phi]=S[\phi]+\frac{1}{2l}{\rm Tr}\log S^{(2)}[\phi],$$ where $S[\phi]$ is the classical (microscopic) action, $l$ is an ad hoc parameter introduced to count the loop order ($l$ is set to $1$ in the end), $S^{(n)}$ is the $n$th functional derivative with respect to $\phi$ and the trace is over momenta (and frequency if needed) as well as other indices (for the O(N) model, for example).

The physical value of the field $\bar\phi$ is defined such that $$\Gamma^{(1)}[\bar\phi]=0.$$ At the meanfield level ($O(l^0)$), $\bar\phi_0$ is the minimum of the classical action $S$, i.e. $$ S^{(1)}[\bar \phi_0]=0.$$ At one-loop, $\bar\phi=\bar\phi_0+\frac{1}{l}\bar\phi_1$ is such that $$S^{(1)}[\bar \phi]+\frac{1}{2l}{\rm Tr}\, S^{(3)}[\bar\phi].G_{c}[\bar\phi] =0,\;\;\;\;\;\;(1)$$ where $G_c[\phi]$ is the classical propagator, defined by $S^{(2)}[\phi].G_c[\phi]=1$. The dot corresponds to the matrix product (internal indices, momenta, etc.). The second term in $(1)$ corresponds to the tadpole diagram at one loop. Still to one-loop accuracy, $(1)$ is equivalent to $$ S^{(1)}[\bar \phi_0]+\frac{1}{l}\left(\bar\phi_1.\bar S^{(2)}+\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}\right)=0,\;\;\;\;\;\;(2) $$ where $\bar S^{(2)}\equiv S^{(2)}[\bar\phi_0] $, etc. We thus find $$\bar \phi_1=-\frac{1}{2}\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c. \;\;\;\;\;\;(3)$$

Let's now compute the inverse propagator $\Gamma^{(2)}$. At a meanfield level, we have the meanfield propagator defined above $G_c[\bar\phi_0]=\bar G_c$ which is the inverse of $S^{(2)}[\bar\phi_0]=\bar S^{(2)}$. This is what is usually called the bare propagator $G_0$ in field theory, and is generalized here to broken symmetry phases.

What is the inverse propagator at one-loop ? It is given by $$\Gamma^{(2)}[\bar\phi]=S^{(2)}[\bar\phi]+\frac{1}{2l}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2l}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}, \;\;\;\;\;\;(4)$$ where we have already used the fact that the field can be set to $\bar\phi_0$ in the last two terms at one-loop accuracy. These two terms correspond to the first two diagrams in the OP's question. However, we are not done yet, and to be accurate at one-loop, we need to expand $S^{(2)}[\bar\phi]$ to order $1/l$, which gives $$\Gamma^{(2)}[\bar\phi]=\bar S^{(2)}+\frac{1}{l}\left(\bar S^{(3)}.\bar\phi_1+\frac{1}{2}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}\right). \;\;\;\;\;\;$$ Using equation $(3)$, we find $$\bar S^{(3)}.\bar\phi_1= -\frac{1}{2}\bar S^{(3)}.\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c,$$ which corresponds to the third diagram of the OP. This is how these non-1PI diagrams get generated in the ordered phase, and they correspond to the renormalization of the order parameter (due to the fluctuations) in the classical propagator.