particle physics - Frequency of an electron

If frequency is defined as the cycles per time, then what is meant by "frequency of an electron"? If it refers to the rotation of electron around a nucleus, then which phenomenon is considered for a free electron i.e. an electron in a force field?

Is "frequency of an electron" an experimental quantity?

My teacher told me how to calculate the frequency of an electron. We started from finding energy of electron, then difference in energy, then we get this equation according to the Bohr radius of a hydrogen atom and

$$ f = \frac{z^2e^42\pi^2m}{h^3} \left(\frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

Where:

• $z =$ atomic number


• $e =$ charge of proton



• $m =$ mass of electron


• $h =$ Planck constant


• $n =$ orbit number

From the last part of my equation, I am confused. Does the $n_1$ and $n_2$ show that that frequency will be the frequency of energy or electrons?

quantum mechanics - Is the spin-rotation symmetry of Kitaev model $D_2$ or $Q_8$?

It is known that the Kitaev Hamiltonian and its spin-liquid ground state both break the $SU(2)$ spin-rotation symmetry. So what's the spin-rotation-symmetry group for the Kitaev model?

It's obvious that the Kitaev Hamiltonian is invariant under $\pi$ rotation about the three spin axes, and in some recent papers, the authors give the "group"(see the Comments in the end) $G=\left \{1,e^{i\pi S_x}, e^{i\pi S_y},e^{i\pi S_z} \right \}$, where $(e^{i\pi S_x}, e^{i\pi S_y},e^{i\pi S_z})=(i\sigma_x,i\sigma_y,i\sigma_z )$, with $\mathbf{S}=\frac{1}{2}\mathbf{\sigma}$ and $\mathbf{\sigma}$ being the Pauli matrices.

But how about the quaternion group $Q_8=\left \{1,-1,e^{i\pi S_x}, e^{-i\pi S_x},e^{i\pi S_y},e^{-i\pi S_y},e^{i\pi S_z}, e^{-i\pi S_z}\right \}$, with $-1$ representing the $2\pi$ spin-rotation operator. On the other hand, consider the dihedral group $D_2=\left \{ \begin{pmatrix}1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &1 \end{pmatrix},\begin{pmatrix}1 & 0 &0 \\ 0& -1 & 0\\ 0&0 &-1 \end{pmatrix},\begin{pmatrix}-1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &-1 \end{pmatrix},\begin{pmatrix}-1 & 0 &0 \\ 0& -1 & 0\\ 0&0 &1 \end{pmatrix} \right \}$, and these $SO(3)$ matrices can also implement the $\pi$ spin rotation.

So, which one you choose, $G,Q_8$, or $D_2$ ? Notice that $Q_8$ is a subgroup of $SU(2)$, while $D_2$ is a subgroup of $SO(3)$. Furthermore, $D_2\cong Q_8/Z_2$, just like $SO(3)\cong SU(2)/Z_2$, where $Z_2=\left \{ \begin{pmatrix}1 & 0 \\ 0 &1\end{pmatrix} ,\begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix} \right \}$.

Comments: The $G$ defined above is even not a group, since, e.g., $(e^{i\pi S_z})^2=-1\notin G$.

Remarks: Notice here that $D_2$ can not be viewed as a subgroup of $Q_8$, just like $SO(3)$ can not be viewed as a subgroup of $SU(2)$.

Supplementary: As an example, consider a two spin-1/2 system. We want to gain some insights that what kinds of wavefunctions preserves the $Q_8$ spin-rotation symmetry from this simplest model. For convenience, let $R_\alpha =e^{\pm i\pi S_\alpha}=-4S_1^\alpha S_2^\alpha$ represent the $\pi$ spin-rotation operators around spin axes $\alpha=x,y,z$, where $S_\alpha=S_1^\alpha+ S_2^\alpha$. Therefore, by saying a wavefunction $\psi$ has $Q_8$ spin-rotation symmetry, we mean $R_\alpha\psi=\lambda_ \alpha \psi$, with $\left |\lambda_ \alpha \right |^2=1$.

After a simple calculation, we find that a $Q_8$ spin-rotation symmetric wavefunction $\psi$ could only take one of the following 4 possible forms:

$(1) \left | \uparrow \downarrow \right \rangle-\left | \downarrow \uparrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(1,1,1)$ (Singlet state with full $SU(2)$ spin-rotation symmetry), which is annihilated by $S_x,S_y,$ and $S_z$,

$(2) \left | \uparrow \downarrow \right \rangle+\left | \downarrow \uparrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(-1,-1,1)$, which is annihilated by $S_z$,

$(3) \left | \uparrow \uparrow \right \rangle-\left | \downarrow \downarrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(1,-1,-1)$, which is annihilated by $S_x$,

$(4) \left | \uparrow \uparrow \right \rangle+\left | \downarrow \downarrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(-1,1,-1)$, which is annihilated by $S_y$.

Note that any kind of superposition of the above states would no longer be an eigenfunction of $R_\alpha$ and hence would break the $Q_8$ spin-rotation symmetry.

Answer

The set $G$ gives the representation of the identity and generators of the abstract group of quaternions as elements in $SL(2,\mathbb C)$ which are also in $SU(2)$. Taking the completion of this yields the representation $Q_8$ of the quaternions presented in the question.

From the description of the symmetry group as coming from here, consider the composition of two $\pi$ rotations along the $\hat x$, $\hat y$, or $\hat z$ axis. This operation is not the identity operation on spins (that requires a $4\pi$ rotation). However, all elements of $D_2$ given above are of order 2.

This indicates that the symmetry group of the system should be isomorphic to the quaternions and $Q_8$ is the appropriate representation acting on spin states. The notation arising there for $D_2$ is probably from the dicyclic group of order $4\times 2=8$ which is isomorphic to the quaternions.

density - Pressure,Pascals Law

Lets say that i have 2 different types of containers, one cuboidal and another one conical. and lets say both these are massless and fully contain water.

**1.**In the cuboidal one the pressure exerted by liquid column at base is Densitygheight.This multiplied by Area of base will give the force exerted on base.As the container is massless total downwards force is also same.And now by calculating volume and density of liquid , we can find mass of liquid and multiplying by 'g' gives total downwards force.In this case total downward force calculated by both the methods is equal.

**2.**In the conical container, the total downwards force calculated using densitygheight and multiplying by conical container's base area is 3 times greater than the force calculated using volume and density relationship(As the volume is 1/3*height*area).Why is it different, which method is correct to calculate downward force? Why is it so?

Answer

In the conical container, the downforce is certainly the same. It can be found via calculus, taking into account the following:

• Since the inside surface of the container is touching the fluid, the calculation is a surface integral.


• The force on a surface due to a pressure is exerted perpendicular to the surface. The downward component will have to include a $\cos(\theta)$ term. For an upright cone standing on its tip, this will be a constant factor.


• The pressure depends on the depth of water above the point being considered.

Even though the area of the tilted part of the cone is much larger than the bottom of the cube, the total downward force will still be the same.

general relativity - A spin zero graviton?

Why can the graviton not be a spin 0 particle? On a similar note why can it not be a spin 4, spin 6 particle?

Answer

For force carriers the interacting field theory determines the spin. A scalar field yields spin 0; the Higgs is the only example; a vector field yields spin 1, the photon, W, and Z are examples; a tensor field yields spin 2.

Since gravitational field theory requires a tensor field for General Relativity, quantized gravity, in the weak-field, linearized limit, yields the spin 2 graviton.

Higher order spins can be constructed from complex particles, meaning systems with two or more constituents.

For more information see Would a spin-2 particle necessarily have to be a graviton?

virial theorem - Stellar Viscosity in Galaxies

Is there such as thing as the viscosity of stars in a galaxy, along the lines of gravitational attraction between stars changing the dynamics.

If so, how is that put in terms of the Virial Theorem?

mass - How do neutrinos oscillate to more massive states?

Neutrino oscillations show that neutrinos change flavor as they propagate, if I understand correctly, and that this is done by allowing different mass and flavor eigenstates states to mix. Do they have definite mass states when they are measured (or at any time)? If so then how is it possible for a neutrino with a lower mass eigenstate to later have the potential to oscillate into a higher mass eigenstate (or whate makes this not possible)? Is mass and energy conserved by a sudden change in the neutrino's energy?

terminology - What is a dormant black hole?

I was on the worldbuilding SE and saw this question

So I tried a quick google on what it was but there seemed to be a lot of people asking the same question, without answer.

So what is a dormant black hole?

Answer

A black hole "feeds" on its environment, usually. So matter falls into the black hole. Due to angular momentum conservation the matter will start to spiral. At some point it will hit the "centrifugal barrier" and keeps orbiting the black hole. When more matter is accumulated, the black hole will create a massive ring of matter.

Tidal forces will crush the matter, friction will heat the matter. The heat will become so large that the thermal radiation pushes matter away again. Other parts of the matter lose energy and fall beyond the event horizon.

When a black hole has consumed all the matter in its vicinity, it can no longer generate radiation from its belt. At this point you could call it "dormant" since nothing falls into it.

quantum mechanics - Reconciling Wikipedia and textbook descriptions of ladder operator method

I'm trying to reconcile the work in my textbook, Quantum Field Theory and the Standard Model by Schwartz, which I'm finding difficult to follow, with the Wikipedia article for the ladder operator method of quantum harmonic oscillators.

The Wikipedia article proceeds as follows:

The commutation property yields

$$\begin{align} Na^{\dagger} \mid n \rangle &= (a^{\dagger} N + [N, a^{\dagger}]) \mid n \rangle \\ &= (a^{\dagger} N + a^{\dagger}) \mid n \rangle \\ &= (n + 1)a^{\dagger} \mid n \rangle\end{align}$$

and similarly,

$N a \mid n \rangle = (n - 1)a \mid n \rangle$

I understood everything, except the first part:

$$ Na^{\dagger} \mid n \rangle = (a^{\dagger} N + [N, a^{\dagger}]) \mid n \rangle $$

1. How did they get this? What "commutation property" is being referred to for this?

The textbook proceeds as follows:

$$\begin{align} \hat{N} a^{\dagger} \mid n \rangle &= a^{\dagger} a a^{\dagger} \mid n \rangle \\ &= a^{\dagger} \mid n \rangle + a^{\dagger} a^{\dagger} a \mid n \rangle \\ &= (n + 1) a^{\dagger} \mid n \rangle \end{align}$$

2. Here, I don't understand how the author got

$$a^{\dagger} a a^{\dagger} \mid n \rangle = a^{\dagger} \mid n \rangle + a^{\dagger} a^{\dagger} a \mid n \rangle$$

and

$$a^{\dagger} \mid n \rangle + a^{\dagger} a^{\dagger} a \mid n \rangle = (n + 1) a^{\dagger} \mid n \rangle$$

?

I would greatly appreciate it if people could please take the time to clarify this.

Answer

1 -This is simply the definition of a commutator:

$$[N,a^\dagger] := N a^\dagger - a^\dagger N$$

so: $$Na^\dagger = a^\dagger N + [N,a^\dagger]$$

2 - Remember the commutation relation for the ladder operators $[a,a^\dagger] = 1$:

$$a^\dagger a a^\dagger = a^\dagger (a^\dagger a + [a,a^\dagger]) = a^\dagger a^\dagger a + a^\dagger 1 = a^\dagger a^\dagger a + a^\dagger$$

$\quad$ Meaning that: $$a^\dagger a a^\dagger |n \rangle = a^\dagger a^\dagger a |n \rangle + a^\dagger |n \rangle$$

3 - Remember that $a^\dagger a \equiv N$, so:

$$a^\dagger |n \rangle + a^\dagger \underbrace{a^\dagger a}_{N} |n \rangle = a^\dagger |n \rangle + a^\dagger N |n \rangle = (1+n) a^\dagger |n \rangle$$

differential geometry - Geometric mechanics - Symplecticity

I am just trying to wade through literature on classical mechanics and I really don't know where to start, everything is Fibre bundle this or manifold that, and doesn't really ease you in to the topic. I'm sure this is a common question but I would like help with a couple of specific points:

• What would be the main points to take away from the "symplectic structure of phase space". Specifically what does knowing its symplectic do for us? We quote it all the time when talking about Hamiltonian mechanics/Liouville equation/Poisson brackets... etc ... In other words what would I be saying in differential geometry terms when I say " a canonical transformation preserves the symplectic structure?"

I understand this may have been answered before but so far having seen the previous question/answers I'm still struggling. I'm specifically after answers relating to classical mechanics linear/non-linear alike, but not relativity if it can be avoided.

So to summarise: what are the salient points of a geometrical understanding of classical mechanics and what does this do that a basic understanding doesn't!

Answer

The main difference between Hamilton and Lagrange/Newton mechanics is, that it happens directly on the phase space, i.e. any point on your manifold already fully determines the state of your system. Intuitively, you realize this by specifying position and momentum coordinates. On a mathematical level, the world we see is some smooth manifold (a priori not even necessarily Riemann), this is the place where you specify your position coordinates. To specify momentum coordinates, you have to consider points from the cotangent space, therefore the structure where you can fully specify your state is the full cotangent bundle of your original manifold.

However, the cotangent bundle may be regarded as manifold on its own right (with twice the dimension of the original manifold). Furthermore, you can construct a canonical(!) symplectic structure quite easily by using the derivative of the natural projection. The reason, why it makes sense to put the "original manifold" aside and operate only on symplectic manifolds resp. cotangent spaces, is Darboux' Theorem, which basically says that every symplectic manifold is actually locally equivalent to the cotangent space of some manifold. Polemically simplified, one could say "The cotangent bundle is the symplectic manifold".

Lets now take a closer look at the symplectic structure: Locally, it looks like $$ \omega=d\vec{q}\wedge d \vec{p}=dq_i\wedge dp^i $$Using this, you can construct a volume form (up to some constant) as $\omega^n$ where $2n$ is the dimension of your manifold. Especially, you see that this volume is even oriented - locally, you can actually imagine some region as a region in euclidean space. The famous Liouville theorem states, that the Hamiltonian phase flux (i.e. the one-parametric group of the corresponding autonomic system) leaves the symplectic structure intact, therefore it also preserves the volume. This is an important consequence especially in statistical mechanics. However, this is not main the point about the symplectic structure. The important thing is, that the symplectic form together with the Hamiltonian determines the trajectory: Trajectories are defined as integral curves of the Hamiltonian vector field $X_H$, which is uniquely determined by the Hamilton function and the symplectic structure by $$ \omega(\cdot,X_H)=dH $$ Vividly spoken: If you throw some particle onto a symplectic manifold, it will move along the Hamiltonian vector field.

The point now about canonical transformations is, that it has the symplectic form as an integral invariant, or, mathematically speaking, it is a special symplectomorphism. Of course, this yields as consequence, that the local equations governing the trajectories are not affected. You can therefore think of a canonical transformation just as a reparametrization of some region of the symplectic manifold, just like a change from Cartesian to spherical coordinates. In this sense, the major advantage over Lagrange is, that you can simultaneously transform both $p$ and $q$, which of course leads to much easier equations in the end, as you can use canonical transformations to make all coordinates cyclic. This is the basic idea of Hamilton-Jacobi theory.

newtonian mechanics - If Newton's third law is true, why can we sink in sand?

Newton's third law of motion states that every action has an equal and opposite reaction. That is the reason we do not sink into the earth, because when our weight exerts a force on the earth it also exerts an equal and opposite force on us.

But when we stand on quicksand or on fluids we can sink in. How is this possible? Does it not exert an equal and opposite force on us? Or are Newton's laws different in the case of fluids and substances of low densities?

conventions - Is the adjoint representation of $SU(2)$ the same as the triplet representation?

Is the triplet representation of $SU(2)$ the same as its adjoint representation? Where the convention for the adjoint representation used is the one used in particle physics, where the structure constants are real and antisymmetric:

$$ \mathrm{ad}(t^b_G)_{ac} = i f^{abc} $$

I was under the impression that is was, but I see two different forms of the generators in the triplet representations used, one being just the real skew symmetric generators of the $SO(3)$ rotation group, which agrees with the adjoint representation, and the other being:

$$ T^1 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right) \quad T^2 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0\end{matrix}\right) \quad T^3= \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{matrix}\right)$$

These two representations do not agree, I assume that my idea about the adjoint reperesentation of $SU(2)$ being its triplet representation is wrong, but why?

Answer

It is just matter of a missed factor $i\sqrt{2} $ due to different conventions. The antiHermitian matrices $i\sqrt{2} T^k $ can be transformed into the real antisymmetric matrices $L_k $ (which therefore are also complex antiHermitian) by means of a suitable unitary matrix $U$, $$L_k = U i\sqrt{2}\:T^kU^\dagger\quad k=1,2,3\:.$$ This is because both triples of matrices are irreducible representations of the Lie algebra of $SU (2) $ with the same value of the Casimir operator $\sum_k (\sqrt{2} T^k)^2 = -\sum_k (L_k)^2 =2I$ (so that $2= j(j+1)$ with $j=1$ which is the spin of the representation). As is known, up to unitary equivalences there is only one unitary irreducible representation of $SU (2) $ for every value of the spin, essentially due to Peter-Weyl theorem.

newtonian mechanics - Why can some oscillations be modeled by Simple Harmonic Motion, while others cannot?

For some oscillators an increase in the driving amplitude changes the period (frequency) of the oscillation, but the simple harmonic oscillator does not predict this type of behavior. Why?

electrostatics - Why do we have energy loss same for every length of wire in capacitor charging

The loss of energy of a capacitor is independent of length of wire connecting it to voltage source $$(\frac{1}{2}×CV^2)$$So where and when is the coverted into heat?

newtonian mechanics - Derive equation for a cantilever in SHM

I am currently investigating how a hacksaw blade's time period of oscillation changes when I add mass to the end of it or when I change the length it is clamped at.

I found the following equation from an IB worksheet:

$$ T^2=\frac{16M\pi^2x^3}{bEd^3} $$

Where:

• $T$ is the time period for one oscillation;


• $M$ is the mass of the cantilever;


• $x$ is the length of the cantilever;


• $b$ is the breadth of the cantilever and $d$ is the thickness of the cantilever;


• and $E$ is the stiffness of the cantilever.

I've looked around the internet and asked my teachers, but I haven't been able to derive this equation from first principles or the equations given in my syllabus.

If anyone knows where this comes from or would like to try and work it out, I would be very greatful.

optics - Why doesn't a backward wave exist?

Huygens principle says every point of wavefront emit wavelet in all directions. Then why does a back ward wave not exist? Can any expert tell real answer? On different sites I get different and contradictory answer.

I am asking question from my class 12 book. It says " Huygens argued that the amplitude of the secondary wavelets is maximum in forward direction and zero in the backward direction; by making this adhoc assumption, Huygens could explain the absence of backwave. However, this adhoc assumption is not satisfactory and the absence of the backwave is really justified from more rigorous wave theory."

What justification is given by wave theory?

answer explained by Marty green on Is Wikipedia wrong about Huygens-Fresnel Principle?

is different from answer given by Acuriousmind on

How does Huygens Principle incorporate the unidirectional property of a traveling wave?

Which is correct?

newtonian mechanics - Direction of friction when a car turns

When a ball moves to the right, friction acts to oppose the motion, in other words, to the left. However, when a car travels around a bend, the friction acts in the perpendicular direction to the car's velocity and provides the centripetal force. I just cannot understand why friction would act in that direction.

quantum mechanics - Slowing down light in an opaque crystal for a whole minute

I just read about a team of physicists at the University of Darmstadt, Germany, that managed to completely slow down a beam of light that traveled through an opaque crystal (article here).

How is it possible for a beam of light come to a complete stop? In the article they mentioned that they fired a laser at the crystal causing the atoms to go into a quantum superposition. How does this affect the stopping of the light? Also if the uncertainty principle applies to photons (which I do not know if it does), how does this not violate the uncertainty principle if the photons aren't moving?

Non-Newtonian Fluid Stop a Bullet?

I just saw a YouTube video about Non-Newtonian fluids where people could actually walk on the surface of the fluid but if they stood still, they'd sink. Cool stuff.

Now, I'm wondering: Could a pool of Non-Newtonian fluid stop a bullet? Why or why not?

If so, if you put this stuff inside of a vest, it would make an effective bullet-proof vest, wouldn't it?

mass - What is the meaning of "matter" in physics?

What is the meaning of matter in physics? By defining matter in terms of mass and mass in terms of matter in physics, are we not forming circular definitions? Please give a meaning of "matter" in Physics that circumvents this circularity.

Answer

What is the meaning of "matter" in physics?

It doesn't matter. Sometimes matter means "particles with rest mass". Sometimes matter means "anything that contributes to the stress-energy tensor". Sometimes matter means "anything made of fermions". And so on. There's no need to have one official definition of the word "matter", nothing about the physical theories depends on what we call the words.

Discussing this any further is just like worrying about whether a tomato is really a fruit or a vegetable. A cook doesn't care.

electricity - How can I calculate the wave propagation speed in a copper wire?

First of all: I am a computer science student, so I don't have much knowledge of physics. So please keep your answers simple.

I recently learned something about circuit design and its problems (differend kinds of hazards). To model the problems, we introduced a "dead time model" ("Totzeitmodel" in German, I have if it is "dead time model" in English)
We added some dead time to each element of the circuit, but we didn't add dead time to the wires of the circuit. I asked the prof. why we didn't add dead time to the wires. He responded that the signal is moving much faster and you can neglect the time that signals need to pass the wires.

Now I would like to calculate the speed of the signal (is this the wave propagation speed?) for some very simple settings:

• assume we have a copper wire


• the wire is a perfect cylinder with diameter of 1mm


• the current is 2A


• the voltage is 12V

Can you help me with this? Do you need something else to calculate the speed?

---

Notes: I found the wikipedia article Wave propagation speed and some questions on physics.stackexchange.com, but the questions and answers were either too complicated or didn't directly give numbers (like that one)

A little side question: When I think about the electric signal, I imagine some elastic balls. When there is a signal at one end, you push the ball. It gets compressed and expands later, which compresses the next ball a bit and it expands, ... This way, the last ball gets moved and the signal arrives at the end. Do I have to get another thinking-model for simple circuits or will I be able to understand simple circuits with this model in mind?

Answer

To calculate the propagation speed, you need to specify the return current path in addition to the "forward" path. The reason is that the electromagnetic fields that determine the propagation characteristics fill the space between the two conductors. [If you try to calculate the inductance of a single wire, you get an infinite result.]

The filler material between the conductors matters too: its electric polarizability (quantified by the dielectric constant $\epsilon$, which is typically 2-5 times the value for free space $\epsilon_0$) slows down the signal speed. Typically the filler is magnetically neutral, so its susceptibility $\mu$ is the same as for free space.

For a coaxial conductor, the wave speed formula ends up being very simple:

$$ v = \frac{1}{\sqrt{\mu \epsilon}}$$

For a relative dielectric constant ($\epsilon/\epsilon_0$) of 3, one calculates a velocity of 58% of the speed of light.

Finally, your elastic ball analogy is good to zeroth order, but I don't think you can use it to think about propagation velocity. There are two independent (but coupled) fields (electric and magnetic) at play here.

---

UPDATE: It turns out that the geometry of the conductors doesn't matter much; the main determinant of the propagation velocity is the filler material properties. For parallel conductors of arbitrary (but constant) cross-section, the propagation velocity is: $$ v = \frac{c}{\sqrt{\mu_r \epsilon_r}} $$

Here the relative permeability of the filler $\epsilon_r = \epsilon/\epsilon_0$ (typically 2-5) and relative magnetic susceptibility $\mu_r = \mu/\mu_0$ (usually 1), while $c$ is light-speed. So the formula for the coaxial geometry turns out to be quite general (note $c=1/\sqrt{\epsilon_0 \mu_0}$).

As Jaime mentions in the comments below, there will be some additional "internal" inductance due to the magnetic fields within the conductors which will reduce the velocity; that bit is geometry-dependent.

specific reference - What is known about quantum electrodynamics at finite times?

I'm aware that we can describe the time evolution of states/operators (choose your favourite picture) of non interacting quantum fields and that perturbation theory is very effective in computing S matrix elements between free states in the remote past and free states in the remote future. Clearly the non-perturbative description of what's going on at finite times, where the interaction is active is intractible, but my question is - are there simplified toy models (scalar electrodynamics ? reduced numbers of dimensions ?) where we can describe what's happening non perturbatively.

Even if nothing like electrodynamics has been treatable like this, any results on the other "textbook" QFTs (like $\phi^4$) would be interesting.

Answer

A lot is known about QFTs (including QED) at finite time. It is tractable approximately (just like scattering). though in 4D no rigorous treatment is available (neither is there one for scattering).

One can compute - nonrigorously, in renormalized perturbation theory - many time-dependent things, namely via the Schwinger-Keldysh (or closed time path = CTP) formalism.

For example, E. Calzetta and B. L. Hu, Nonequilibrium quantum fields: Closed-time-path effective action, Wigner function, and Boltzmann equation, Phys. Rev. D 37 (1988), 2878-2900. derive finite-time Boltzmann-type kinetic equations from quantum field theory using the CTP formalism.

There are also successful nonrelativistic approximations with relativistic corrections, within the framework of NRQED and NRQCD, which are used to compute bound state properties and spectral shifts. See, e.g., hep-ph/9209266, hep-ph/9805424, hep-ph/9707481, and hep-ph/9907240.

There is also an interesting particle-based approximation to QED by Barut, which might well turn out to become the germ of an exact particle interpretation of standard renormalized QED. See A.O. Barut and J.F. Van Huele, Phys. Rev. A 32 (1985), 3187-3195, and the discussion in Phys. Rev. A 34 (1986), 3500-3501,3502-3503.

Approximately renormalized Hamiltonians, and with them an approximate dynamics, can also be constructed via similarity renormalization; see, e.g.,
S.D. Glazek and K.G. Wilson, Phys. Rev. D 48 (1993), 5863-5872. hep-th/9706149

In 2D, the situation is well understood even rigorously:

For all theories where Wightman functions can be constructed rigorously, there is an associated Hilbert space on which corresponding (smeared) Wightman fields and generators of the Poincare group are densely defined. This implies that there is a well-defined Hamiltonian $H=cp_0$ that provides via the Schroedinger equation the dynamics of wave functions in time.

In particular, if the Wightman functions are constructed via the Osterwalder-Schrader reconstruction theorem, both the Hilbert space and the Hamiltonian are available in terms of the probability measure on the function space of integrable functions of the corresponding Euclidean fields. For details, see, e.g., Section 6.1 of J. Glimm and A Jaffe, Quantum Physics: A Functional Integral Point of View, Springer, Berlin 1987. In particular, (6.1.6), (6.1.11) and Theorem 6.1.3 are relevant.

[The above information was extracted from the Section ''Relativistic QFT at finite times?'' of Chapter B3: ''Basics on quantum fields'' of my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html ]

[Edit October 9, 2012:] On the other hand, a lot is unknown about QFTs (including QED) at finite time. Let me quote from the 1999 article ''Some problems in statistical mechanics that I would like to see solved'' by Elliot Lieb http://www.sciencedirect.com/science/article/pii/S0378437198005172: ''But there is one huge problem that everyone avoids, because so far it is much too difficult to handle. That problem is Quantum Electrodynamics, and the problem exists whether we are talking about non-relativistic or relativistic quantum mechanics. [...] The physical picture that begs to be understood on some decent level is that the electron is surrounded by a huge cloud of photons with an enormous energy. We are looking for small effects, called 'radiative corrections', and these effects are like a flea on an elephant. Perturbation theory treats the elephant as a perturbation of the flea. [...] After renormalizing the mass so that the 'effective mass' (a concept familiar from solid state physics) equals the measured mass of the electron we are supposed to obtain an 'effective low energy Hamiltonian' (again, a familiar concept) that equals the Schroedinger Hamiltonian plus some tiny corrections, such as the Lamb shift. From there we should go on to verify the levels of hydrogen (which, except for the ground state, have become resonances), stability of matter and thermodynamics and all those other good things. But no one has a clue how to implement this program. [...] On the other hand matter does exist and the sun is shining, so the theory must exist, too. I would like to see it someday''

quantum mechanics - Number of Nodes in energy eigenstates

I have a question from the very basics of Quantum Mechanics. Given this theorem:

For the discrete bound-state spectrum of a one-dimensional potential let the allowed energies be $E_1

What is the physical interpretation for the number of nodes in the concrete energy eigenstate? I understand that the probability of finding the particle in the node point is $0$ for the given energy. However, why does the ground state never have a node? or why does every higher energy level increments number of nodes precisely by 1?

Answer

I guess there is not that much to grasp, unless you can really understand dark spots on an electron diffraction pattern. Very roughly explanation would be to interpret wave functions of a particle in a potential well as "standing waves", or as two interfering waves reflected from the walls of the well. Increasing the energy leads to higher harmonics, which leads to additional nodes. Nodes' numbering is the same as in the case of a classical string.

cosmology - Dark matter and dark energy references

I've been looking for questions about dark matter, and I've read some very interesting answers. However, I desire too look into it deeply.

This is not actually a question. I'm asking the community to recommend interesting references to understanding dark matter and dark energy.

I accept all sort of references: notes, books, scientific papers etc.

Let us assume some background on classical physics, thermodynamics and basics about quantum theory.

cosmology - Can we use pressure analogy to understand inflation?

Classically, the expansion of a gas in a container requires the gas to apply outward pressure to the walls of the container. Is it also true for the Universe during inflation? How did the inflation drive the expansion of Universe? I think this analogy is poor because it's not the expansion of a gas in a container. Nothing applies an outward pressure of the "walls" of the Universe rather it is the space that expands.

condensed matter - Why does a fermionic Hamiltonian always obey fermionic parity symmetry?

Consider a closed system of fermions. Any reasonable fermionic Hamiltonian you might think of will always have an even number of fermionic operators in each term. In other words: $[H,P] = 0$ where $P=(-1)^{N_f}$. My question is: why is this? I am not looking for a flowery description in terms of 'fermions are ends of strings and hence can only be created in pairs'. My question is more mathematical in nature: is there some kind of inconsistency in considering a closed system of fermions with e.g. $H = \sum c_i^\dagger + c_i$ ? Indeed it seems that physically we should see fermionic parity symmetry as a gauge symmetry. But is there a mathematical argument to see that we have to see it as a gauge symmetry to keep things consistent?

It would be enough to argue that any fermionic wavefunction has to have a well-defined fermionic parity. (Indeed: if $[H,P] \neq 0$ then time evolution would take a state with a well-defined parity into one without well-defined parity.) But again, I have no good way of arguing this.

Perhaps the resolution is simply pragmatic: we haven't figured out a way of engineering an interaction term which is not bosonic in nature. Or is there a fundamental reason we can't have such an interaction?

electric fields - When work done is taken negative in electrostatics?

Let us say a point charge Q was moved across a potential difference V, then work done would be : QV.

This work is taken negative when done external agent, Please explain when It is negative and positive, also for the general case, if we take change in potential to be V, then by the equation :

When this will give an absolute value.

Answer

If a charge $Q$ is moved from rest by external forces across a potential difference $\Delta V=V_b-V_a$ from point $a$ to point $b$ and ends up at rest, then the net work done by all forces (electrical and external) will be zero by the work-energy theorem: $$W_{ext}+ Q(V_b-V_a) = \Delta K = 0.$$

That means that $W_{ext}=-Q(V_b-V_a)$. So, the work done by the field has the opposite sign of the external work.

If a positive charge is moved in the same direction as the electric field, or from higher to lower potential, the work done by the field will be positive.

work - Demonstration of the existence of a scalar potential for a conservatice force

Mathematically a vector field, $\vec{F}$, is conservative if:

$$\oint_{\gamma} (\vec{F}.d\vec{l})=0$$

Physically, the integral is the same as the work done by a force $\vec{F}$ on a body in a closed path. I intend to demonstrate mathematically that a conservative force assumes a scalar potential, i.e:

$$\oint_{\gamma} (\vec{F}.d\vec{l})=0 \Leftrightarrow \vec{F}=-\vec{\nabla} U$$

where $U$ is the scalar potential.

I know that $$\vec{F}=-\vec{\nabla} U \Rightarrow\oint_{\gamma} (\vec{F}.d\vec{l})=0 $$

(Using Stokes theorem: $\oint_{\gamma} (\vec{F}.d\vec{l})=\int_S \left([\vec{\nabla} \times\vec{F}].\vec{n}\right) dS$. Using Schwarz lemma, we have that $ [\vec{\nabla} \times\vec{\nabla}U]=0. $ So, $\oint_{\gamma} (\vec{F}.d\vec{l})=0.$)

My problem is that I can't demonstrate the inverse, i.e:

$$\oint_{\gamma} (\vec{F}.d\vec{l})=0\Rightarrow\vec{F}= -\vec{\nabla} U $$

I got a more general solution that is different of $\vec{F}= -\vec{\nabla} U. $

How could I do this?

Answer

I'll first show you how to show that $\oint_\gamma F = 0$ implies $F = -\mathrm{d} U$ and then we'll discuss that mathematically the problem lies in the other direction.

1. 
   

   If we know that $\oint_\gamma F = 0$ then we can define the potential $U$ by fixing a point $x_0$ and then setting $$ U(x) := \int_\gamma F$$ for any path between $x_0$ and $x$. This is well-defined since for any other path $\gamma'$ from $x_0$ to $x$ we have that we can form a loop $\ell$ by going first along $\gamma$ and then along $\gamma'$ in the reverse direction so that $$ \int_\gamma F - \int_{\gamma'} F = \oint_\ell F = 0$$ so the potential is well-defined. It's not hard to check that the gradient of $U$ is $F$.

   
   
   


2. 
   

   The converse direction has a subtlety: You have assumed that there is a surface $S$ such that its boundary is $\partial S = \gamma$ for every path $\gamma$. This is only true in simply-connected spaces, but not in general. However, that being a gradient implies all loop integrals being zero is true in all spaces, we simply need to apply Stokes' theorem in its most basic form, the fundamental theorem of calculus. It says that for $\gamma$ starting at $x_1$ and ending at $x_2$, we have $$ \int_\gamma \mathrm{d}U = \int_{\partial \gamma} U = U(x_2)-U(x_1)$$ and for loops $x_1=x_2$, so $\oint_\ell F = \oint_\ell \mathrm{d}U = 0$.

   
   

Now, it's actually important to realize that in non-simply connected spaces, there are curl-free vector fields which are not the gradient of a scalar function. For instance, on $\mathbb{R}^2 - \{(0,0)\}$ we have $$ V(x,y) = \frac{x}{r}\partial_x + \frac{y}{r}\partial_y$$ with $r = \sqrt{x^2+y^2}$ as a curl-free vector field that is not a gradient, and also has a non-zero loop integral: Integrating it along anyloop that goes around the origin once, you get $2\pi$.

This is also of importance physically, for instance, it is the observation underlying the Aharonov-Bohm effect. While the magnetic field is zero everywhere outside the solenoid, the magnetic vector potential is not globally the gradient of a single scalar potential although its curl vanishes, and consequently its integral around a loop around the solenoid is non-zero.

lie algebra - Derivation of the irreducible representations of SO(3)

Is there a way to derive the representations of $SO(3)$ without the usual method with the ladder operators which also gives the ones of $SU(2)$?

The usual way to do these calculations is to start from the commutation relations of the Lie algebra associated with $SU(2)$ (or that of $SO(3)$, which is the same given that $\mathfrak{so}(3) \approx \mathfrak{su}(2)$ as far as I understand) and from there to go throught the ladder-operators-thing to obtain all of the representations of $SU(2)$. Is there another way to derive the representations of $SO(3)$ which is specific of $SO(3)$ and not also applicable to $SU(2)$?

Answer

Be careful. It may be the case that $\mathfrak{su}(2)=\mathfrak{so}(3)$, but it is not the case that $SU(2)=SO(3)$. $SU(2)=\mathrm{Spin}(3)$ and $\rho :SU(2)\rightarrow SO(3)$ is the two-sheeted universal cover of $SO(3)$. It thus turns out that only the integer spin representations of $SU(2)$ factor through $\rho$ to give well-defined representations of $SO(3)$.

Concretely, the spherical harmonics $Y_{\ell m}(\theta ,\phi )$ span an irreducible representation of $SO(3)$ in $L^2(S^2)$. This representation has spin $\ell$ (dimension $2\ell +1$), and this yields all the finite-dimensional irreducible complex representations of $SO(3)$. Note that $\ell$ must be an integer in this context (and so it does not give you all the irreducible representations of $SU(2)$).

quantum mechanics - Heisenberg Uncertainty principle defined as $Delta x Delta p_x geq h$ versus $Delta x Delta p_x geq hbar/2$

At my university, in the during lectures and in the equation sheet for our exams, the formula for the Heisenberg Uncertainty Principle is stated as $\Delta x \Delta p_x \geq h$, for example in one of my lecture notes, the following example is illustrated using this formula

However I know that in my textbook (University Physics by Young and Freedman) and pretty much universally the Heisenberg Uncertainty Principle is stated as $\Delta x \Delta p_x \geq \frac{\hbar}{2}$. Using this formula, we can see the example above is off by a factor of $4\pi$.

Is the formula, $\Delta x \Delta p_x \geq h$, that my university uses a valid formula? If so is just a weaker version of $\Delta x \Delta p_x \geq \frac{\hbar}{2}$?

general relativity - Does the Big Bang need a cause?

Possible Duplicate:
on causality and The Big Bang Theory

Asking here in layman's terms..

When theoretical physicsists discuss the origin of our Universe, the wider consensus appears to be that it originates from a singularity; a position that rests on observations about the apparent expansion of our Universe.

However, the question why singularity itself came into being, still remains.

But this question takes a different angle: Does the Universe as a whole need a cause to exist at all?

If the law of conservation of energy is universally valid, then questions about a beginning become irrelevant since there can't be any by definition and everything boils down to dynamics of interaction. Thoughts?$$\mbox{ }$$

Answer

I don't think I agree with your first sentence. Our simplest theoretical models, based on classical general relativity, say that there was a singularity in the past, but few if any cosmologists take that as a reason to believe that there actually was such a singularity. Rather, the most likely possibility is that those classical models are wrong at early times.

The truth is that we have no idea what happened "at the Big Bang," or even if that phrase is meaningful. There are some theories in which the Universe has existed for infinite time (Google "eternal inflation" for instance), and others in which time started at a finite point in the past.

Conservation of energy arguments don't really help here. For one thing, conservation of energy in the expanding Universe is more complicated than you might initially expect -- there's really no such thing as the total energy of the Universe. For another thing, conservation of energy, all by itself, wouldn't answer the question of whether there was a beginning. (Global) conservation of energy says that the energy at any one time equals the energy at any other time. It doesn't say anything about whether there was an initial time.

general relativity - Does acceleration of time explain gravity (rather than the other way round)?

I have a question about interpreting (explaining, even) the general theory of relativity.

A common interpretation of GR, as I understand it, is to imagine two-dimensional space represented by flatlanders (ants) living on the surface of a trampoline (or stretched elastic sheet). The third ‘hyper-dimension’ from the point of view of the ants is time.

If a heavy object is placed in the middle of the trampoline, it distorts the surface in space-time, so that ‘straight lines’ of motion become geodesics, which are actually curved. A marble rolled across the trampoline curves around the object, because it’s following the shortest path in space-time. So far so good.

A reader here a while back (Why would spacetime curvature cause gravity?) posted the question: so why do two stationery objects move towards each other, and the answer was that in space-time, they are not stationery - they are moving through time. And it’s their space-time lines which still try to take the shortest/straightest route, as illustrated very nicely in a YouTube video linked in one of the answers.

Now I’m interested in getting back to a more intuitive understanding of why an attractive force seems to be exerted on the second object by the first, without getting tied up in the rather complex tensor representations of geodesics etc. At the same time, I want to know WHY a massive object distorts space-time.

Most texts take the curvature of space-time as an axiom (postulate), without pretending to offer any explanation. Note that (as I understand it) the heavy object on the trampoline is only intended to represent or visualise the curvature - there is no suggestion that it’s a simple matter of heavy objects ‘weighing down’ space time.

But I’m thinking, why not take the analogy literally, and follow through on this idea of weighing down?

In the analogy, it seems obvious that a heavy object distorts the surface downwards - it’s a result of the external, real gravity in the analogy (acting in a hyper dimension from the ants’ perspective). And the marble, of course, is simply rolling downhill.

Now as we know from Einstein’s principle of equivalence, gravity equals acceleration. The third dimension in ant-world is time - so, what if we suppose that time is accelerating and hence generating that hyperdimensional force?

For me, the surface of the trampoline is not simply a snapshot, or slice, of space-time, but far more significant - it is the present universe. (As an aside, because of its distortion, the relative ‘hight’ of points in space time (above ground level) varies. Furthermore different observers (ants on the trampoline) project different tangential planes from their standpoints. Do these two effects explain all the results of special relativity?)

Now assume that time moves inexorably forwards, and the surface of the trampoline is moving upwards, indeed accelerating. As we know, acceleration and force are the same thing: so the surface of the trampoline is pushing up against all objects in the universe.

Suppose also that all massive objects have a type of inertia which resists time - then the trampoline surface is retarded, or flexed backwards! That is what causes the distortion.

Now looking at a stationery marble sitting off to the side of the heavy object in the middle, it too is being pushed forwards (upwards) by time (the surface of the trampoline). Only the surface is now inclined - so there is a sideways component of the force, which pushes the marble towards the heavy object.

Simple, obvious, and indeed an inevitable consequence of the assumptions that a) time is accelerating forwards, and b) heavy objects have an inertia which resists the passage of time.

Is this idea complete nonsense, or is it a valid way of interpreting how space-time is bent (and how that generates the appearance of gravity)?

Is this how some physicists already interpret general relativity - am I wrong to suppose that the analogy is not usually intended to provide a reason for the curvature?

It’s really a question for someone very familiar with the mathematical underpinnings of GR - can the curvature of space-time be successfully modelled by an accelerating timeframe and massive objects having an inertia which resists it?

thermodynamics - Temperature at different points in a metal rod during heat conduction

$$ k = \frac{\frac{Q}{t}}{A(\frac{T_1 - T_2}{L})} $$

where k is thermal conductivity of the solid, Q is total amount of heat transferred, t is time taken for the heat transfer, A is area of the cross section, L is the length of the solid and T1 and T2 are the temperatures at the hotter end and the colder end respectively.

According to this formula, when a metal rod is getting heated through conduction, the temperature gradient $$\frac{T_1 - T_2}{L}$$ decides what the temperature will be at different points down the length. However, physics also states that when heat transfer takes place, it goes on until the temperature of the hotter object and the cooler object becomes equal. So, how can heat conduction stop before temperature becomes the same throughout the length of the metal rod without contradicting the basic theory of heat transfer?

Answer

"However, physics also states that when heat transfer takes place, it goes on until the temperature of the hotter object and the cooler object becomes equal."

That's only true in equilibrium.

In this case, we know the system is not in equilibrium because we have heat being added ($Q$). If $Q = 0$ after some amount of time, $\frac {T_1 - T_2}{L}$ will go to $0$.

I believe this even fits with the wording you chose:

"According to this formula, when a metal rod is getting heated through conduction, the temperature gradient decides what the temperature will be at different points down the length."

(emphasis mine) Basically, the gradient can only exist when there is heat being transferred through the rod. When there is no net heat transfer through the rod, the gradient becomes a flat line due to the bar having a uniform temperature.

Another thing to point out is that they aren't talking about the temperature gradient when convection stops. They are talking about the temperature gradient at some time during the heat transfer. This could either be a steady-state conduction or a snapshot of a transient process at one point in time.

lagrangian formalism - Why is Fermat's principle not formulated as principle of least action?

I noticed from the units of $S$ that despite the notational similarity Fermat's principle $$ \delta S= \delta\int_{\mathbf{A}}^{\mathbf{B}} n \, ds =0 $$ is not a principle of least action but a principle of least length. Confusingly one often writes this principle even using a so called "optical Lagrangian" $L= n \frac{ds}{dx_3}$ as $$ \delta S= \delta\int_{\mathbf{A}}^{\mathbf{B}} L \, dx_3 =0 $$ However this Lagrangian doesn't have units of energy as the usual Lagrangian $L= T-V$ has. Instead the optical Lagrangian has no units. So I wondered what is missing to make Fermat's principle into a principle of least action and figured from the units that one has to multiply the principle with some "optical momentum" $p_O$ so $S$ becomes an action: $$ \delta S= p_O\ \delta \int_{\mathbf{A}}^{\mathbf{B}} n \, ds =0 $$ But what should this $p_O$ be? Since we are talking about light we could set $p_O=\hbar k = \frac{h}{\lambda}$. But Fermat's principle is the basis of geometrical optics, which can be derived as the limit of zero wavelength $\lambda \rightarrow 0$ from Maxwells wave equations of light. And a zero wavelength means light of infinite momentum according to $p_O= \frac{h}{\lambda}$. So it doesn't seem to make sense to multiply Fermat's principle with a momentum, because that would yield a infinite value for the action.

So is the reason why we cannot express Fermat's principle as principle of least action the fact that geometrical optics implies an infinite momentum for light?

cosmology - What happens if you travel through space at the same rate that space expands?

Hubble's Law gives the rate at which the universe is expanding, as a function of distance.

Take an estimated value for $H_0$ of 70 (km/s)/Mpc. If we take some star 1 Mpc away from my current position, my understanding is that Hubble's Law says that the star is moving at a rate of 70 km/s away from me.

Now, what happens if I travel in my spaceship at 70 km/s for one second. I have moved 70 km, and the space between me and the distant star has also expanded by 70 km. Am I still the exact same distance away from the star?

Answer

Defining speed and distance in the expanding universe is not trivial. To be clear, let's disregard the gravitational attraction and assume that you start moving toward a remote star with such a speed that you see no redshift. This would essentially mean that your proper distance to the star is momentarily constant and in this sense your speed relative to the star is zero. (The proper distance is measured by the time it takes for light from the star to reach you.) Then the question is, how this situation would evolve in time.

If the universe is flat and expanding without acceleration (such as in the Milne cosmology), then nothing would change. You would always remain at the same proper distance from the star with the redshift of zero. The latest measurements suggest that the universe is flat within a small margin of error. However, the universe is expanding with acceleration described by the cosmological constant attributed to dark energy. In this case, the star would accelerate away from you gradually gaining speed. Currently the amount of acceleration is very small, so your distance to the star would change very slowly and essentially remain approximately the same for some time.

The expansion of the universe in the current $\Lambda\text{CDM}$ cosmology ("Lambda - Cold Dark Matter", where $\Lambda$ is the cosmological constant) is described by the Friedmann equations that depend on the matter content of the universe. Until about 70,000 years from the Big Bang the universe was dominated by radiation, then by matter, and now by dark energy, which is effectively repulsive causing the expansion to accelerate.

The Friedmann equations have an analytic solution for a flat universe dominated by a combination of matter and dark energy. This closely describes the universe since a very young age of about when the Cosmic Microwave Background radiation was emitted. In this solution, the scale factor of the expansion of the universe depending on time is given by

$$ a(t)=\sqrt[3]{\dfrac{\Omega_{m,0}}{1-\Omega_{m,0}}}\sinh^{\frac{2}{3}}{\left(\dfrac{3}{2}\sqrt{1-\Omega_{m,0}}\,H_o t\right)} $$

Where $\Omega_{m,0}$ is the current total matter density, $H_o$ is the current Hubble parameter, and $t$ is the current age of the universe.

The plot of this equation shows the scale factor in blue and its time derivative in red indicating that the speed of expansion has already turned from reducing to increasing. The horizontal axis shows the ratio of the age of the universe to the current Hubble time, currently

$$ \dfrac{t}{t_h}=H_ot\approx 0.99$$

The nature of the dark energy is currently unknown. See Dark Energy and the Accelerating Universe for details.

electromagnetism - What is the importance of vector potential not being unique?

For a magnetic field we can have different solutions of its vector potential. What is the physical aspect of this fact? I mean, why the nature allows us not to have an unique vector potential of a field?

Answer

There is no "physical aspect of this fact". The physical variables are the electric and the magnetic field, not the potentials. Introducing the potential is aesthetically and technically pleasing, but it is not necessary. A gauge symmetry is not a physical symmetry.

The reason you can have a non-unique potential is that every divergence-free field such as the magnetic field has a vector potential whose curl it is, but adding any gradient to that potential still gives the same magnetic field since the curl of a gradient is zero. The equation defining the magnetic vector potential is simply underdetermined.

Note that even the effect that is usually cited as showing the potentials being "physical", the Aharanov-Bohm effect, does not make the potential unique. The quantity that is relevant is the integral of the vector potential $A$ along a closed loop $\gamma$, and if we denote the region inside $\gamma$ as $U$, we have $\int_\gamma A = \int_U B$ by Stokes' theorem, so what really matters here is the flux through the loop, not the specific value of the potential. And one has to close the loop to observe a phase difference (or, well, maybe not always, but the phase is still only dependent on the flux, not on a gauge-variant potential value). In any case, this is a quantum effect. In the classical theory, the potential is definitely not "physical" in the sense of being measureable.

general relativity - Metric of an Evaporating Black Hole

The famous Hawking calculation is done with an assumption that the background is static, i.e. the evaporation doesn't change the mass parameter in the metric. Thus, we simply describe the geometry using the static Schwarzschild (or, generically, Kerr-Newman) metric. But clearly, the evaporation actually makes the geometry non-static and thus, the geometry should actually be described using a non-static metric. I am finding a hard time finding out what metric this is.

I think that even if the Hawking calculation is done within the assumption that the background metric is static, one can safely assume that a spherically symmetric radiation will still be being emitted from an evaporating black hole even during the stages where the static assumption is inappropriate. Thus, the natural guess for a non-static metric that describes the geometry of an evaporating black hole would be the Vaidya metric.

But, as discussed in this answer, an outgoing Vaidya metric describes a metric for which the mass parameter is continually decreasing--but this doesn't describe a black hole geometry, instead, it describes a white hole geometry. Further, as discussed in the same answer, an ingoing Vaidya metric describes a black hole geometry--but with a monotonically increasing mass parameter. Thus, none of the two Vaidya metrics qualify to describe an evaporating black hole.

So, my question is, is there any known metric that can describe a spherically symmetric geometry whose mass parameter decreases with time and the horizon is of the nature that resembles a black hole horizon? If so, then it can be considered as a metric that describes an evaporating black hole.

Edit

I recently read a comment by @JerrySchirmer that the Hawking radiation violates the energy conditions. If so is the case then the argument that an ingoing Vaidya metric has a monotonically increasing mass parameter doesn't work (as this argument relies on the null energy condition). If someone can provide some canonical references in this regard then it would be truly helpful.

thermodynamics - How to calculate temperature of an incandescent bulb filament?

Suppose we have a light bulb, for which we know its power rating, like voltage of $12\mathrm V$, and power consumption of $10\mathrm W$. We also know it's a halogen bulb with a tungsten filament inside. Suppose we also know temperature of the surrounding air.

Is this data enough to compute temperature of the filament? If not, what should also be included? And anyway, how do we find the temperature?

Answer

You really are asking two questions.

First - how do we calculate the temperature:

At the typical temperatures of a halogen bulb, the large majority of heat loss is due to thermal radiation (although there is some conductive loss in a halogen bulb as the bulb is not evacuated). Because of this, the most important factor is the "apparent size" of the filament. I say "apparent" because when you have a tightly wound coil, the parts of the coil facing other parts of the coil don't contribute to a net heat loss.

If you took for example a 5 mm long, tightly wound filament with a mean diameter (after winding) of 0.5 mm, you would have a surface area of approximately 5 * π * 0.5 ~ 8 mm². If you had 10 W of emission, you would use the Stefan Boltzmann law to get the power per unit area:

$$I = \sigma T^4$$

from which we get a temperature of

$$T = \sqrt[4]{\frac{10}{8\cdot 10^{-6}\cdot5.67\cdot 10^{-8}}} \approx 2100 K$$

Getting more accurate numbers is quite hard - there are lots of subtle effects (conduction down the support wires, heat lost to the filler gas, and "true effective area" to name just three).

Second - how do we measure the temperature. For such high temperatures, the disappearing filament pyrometer is very effective: you send a current through a calibrated filament, and compare its color against the color of the object of interest. When the filament "disappears" against the background, the temperatures are matched. Often, filters are used to do the comparison in a narrower range of wavelengths; the result can give resolution down to 10C according to the above linked article. There are obvious problems with this is the emissivity of the object of interest is very different than that of the filament - but if you are trying to determine the temperature of a filament that is not likely to be a problem.

astrophysics - Experimental observation of matter/antimatter in the universe

Ordinary matter and antimatter have the same physical properties when it comes to, for example, spectroscopy. Hydrogen and antihydrogen atoms produce the same spectroscopy when excited, and adsorb the same frequencies. The charge does not make a difference in the potential (regardless if it's generated by a proton or an antiproton) nor in how the positron behaves in this potential (being its mass equal to the mass of an electron)

How can astronomy evaluate if a far galaxy is made of matter or antimatter, given that from the spectroscopy point of view, they behave in the same way? In other words, how do we know that an asymmetry exists between matter and antimatter in the universe?

Answer

To be a little pedantic, nobody has yet done precision spectroscopy of antihydrogen, though the recent success in trapping it at CERN (all over the news this week, paper here) is an early step toward that. It's possible that there are small differences in the spectrum of antihydrogen and hydrogen, though these differences can't be all that large, or they would be reflected in the interactions of antiprotons and positrons with ordinary matter in ways that would've shown up in other experiments.

As I understand it (and I am not an astronomer) the primary evidence for a lack of vast amounts of antimatter out there in the universe is a lack of radiation from the annihiliation. We're very confident that our local neighborhood is matter, not antimatter, which means that if there were an anti-galaxy somewhere, there would also need to be a boundary region between the normal matter and antimatter areas. At that boundary region, there would be a constant stream of particle-antiparticle annihilations, which produce gamma rays of a very particular energy. We don't see any such region when we look out at the universe, though, which strongly suggests that there aren't any anti-galaxies running around out there.

astronomy - How can one get the eccentricity of the orbit of the Sun around center of the Milky Way?

How can one get the eccentricity of the orbit of the Sun around center of the Milky Way? Can it be measured?

Answer

Short answer, no.
The Sun's orbit is non-Keplarian; there are many perturbations and a general unevenness in the motion of the Sun around the Galactic centre. This is a result of non-uniform mass distributions, the galaxy not being a point mass, and the impact the relative motions of neighbour stars has on measuring. Thus, giving a particular eccentricity for the Sun is almost meaningless. For instance, it fluctuates up and down roughly $2.7$ times per orbit and it passes through high density regions which cause major perturbations. This creates instability in any average eccentricity.

Long answer, it is not impossible.
In theory, we could measure it. However, we have two rest frames; local and standard rest. The local rest frame refers to how we can take the average motion of stars within (say) $100~pc$ and use this average to compute our approximate orbital properties. The standard rest frame refers to us using Oort constants/properties and similar things in order to determine our more specific motion around the galaxy based on accelerational perturbations, etc. Both frames have their own advantages and both give slightly different values for our currently computed orbital characteristics. The problem lies with determining the relative weights each might contribute to an eccentricity value.

While the motion of the Sun may be non-Keplarian, we do know that the circular velocity is around $230{km\over s}$ and the peculiar velocity is on order $15{km\over s}$. This leads many astronomers to say that while measuring the eccentricity would be very hard and calculating it would be near impossible, they can say that it is most likely on the order of a few percent. Definitely less than $10\%$, but a value in the range of $e=0.02-0.08$ would be the most likely.

quantum mechanics - State vector vs density operator

We formulate quantum mechanics using language of state vectors. One alternative formulation is possible using density operator or density matrix. Why we are doing this alternative approach? Is the approach of state vector is not sufficient? Is the density matrix approach works more efficiently?

quantum mechanics - Does a photon interfere only with itself?

I sometimes hear statements like:

Quantum-mechanically, an interference pattern occurs due to quantum interference of the wavefunction of a photon. The wavefunction of a single photon only interferes with itself. Different photons (for example from different atoms) do not interfere.

First of all -- is this correct?

If it is correct -- how do we explain basic classical interference, when we don't care about where the plane waves came from?

I heard that there are experiments with interference of two different lasers -- is this considered as a refutation of the statement? If it is -- how should one formally describe such a process of interference of different photons?

Finally -- such statements are usually attributed to Dirac. Did Dirac really say something like that?

cosmology - Why are neutrinos ruled out as a major (or even sole) component of dark matter?

A number of times I have encountered in text-books and articles that neutrinos might contribute only a small fraction to dark matter. The reason has to do with the fact that if all of the dark matter consisted of neutrinos, then small-scale structures in the Universe could not have formed yet, because, as they say, neutrinos "wash out" small fluctuations. However, none of these texts provided a reference to any specific sources explaining in detail what is meant by "washing out". After all, neutrinos are notorious in their weak interaction with baryonic matter, so if there is a small-scale fluctuation of baryons, then how background neutrinos can prevent it from growing further if they practically do not interact with baryons? I guess the question boils down to calculating cross-sections of interactions at specific temperatures. I would appreciate comments and references to sources addressing this particular issue.

Answer

The dark matter energy density of the universe is, at present, thought to be about five times that of the baryonic matter energy density. Meanwhile, the radiation energy density is almost negligible. Matter energy is about 4.5% of the total energy density of the universe. Dark matter makes up about 23%, and radiation is very small at about 0.009%. The number for radiation was calculated including all relativistic particles, including neutrinos. In fact, if you go through and read this link, it details the calculation for the total neutrino energy density and shows that it is thought to be about 68% of the photon energy density. So the 0.009% of the universe that is relativistic particles is not even mostly neutrinos.

My point? There truly is simply not enough neutrinos out there to explain away dark matter as neutrinos. Not only that, but we have clearly already included them in the calculation. Dark matter makes up 22.7% (give or take) of the energy density of the universe. And that is on top of the less than 0.0036% that neutrinos account for. So there's no way that neutrinos could be a major, let alone sole, component of dark matter.

For an overview of the energy densities, see Wikipedia and links therein

To answer your question on "washing out", the Wikipedia article on Dark Matter does a very good job at explaining this. For small scale structure to form, dark matter is required to help gravitationally bind baryonic matter. However, the free streaming length of any candidate particle that accomplishes this must be small. The free streaming length is the distance that the particles move in the early universe from random motions before the expansion slows them down. Primordial density fluctuations provide the seeds for small scale structure to form, but if the free streaming length of the dark matter candidate particle is larger than the scale of the small primordial perturbations, then these perturbations become homogenized (or "washed out") as the particles communicate and equilibrate. Without the perturbations, there is no seed for the small scale structure and, thus, it does not form.

Now you may be wondering why dark matter is needed in the first place for small scale structure to form. After the Big Bang, ordinary baryonic matter had too much temperature and pressure to collapse into structure on its own. It requires a gravitational seed (like giving it a kick-start to get the gravitational collapse going), which means there has to be a perturbation in the density of a colder, less interacting form of matter to provide this seed; that is, a local density of this cold dark matter that is higher than the background value. These perturbations would be formed because of the primordial density perturbations left over from inflation. However, neutrinos are known to have a high free streaming length, thus they would smooth out these perturbations in their own density and you wouldn't get a local high density region that could act as a seed. No seed means no collapse. No collapse means no small scale structure (until it's much too late). Neutrinos are actually the primary candidate for hot dark matter, but they are not a viable consideration for cold dark matter, which is what is necessary to generate sufficient small scale structure formation.

special relativity - The Topology of the Lorentz group in 1+3 dimensions

This is a question of mathematical nature. Let the Lorentz group $O(1,3)$ be defined as a matrix group.

$$\text{O}(1,3) =\{\Lambda\in M_4 (\mathbb{R})| \Lambda^T \text{diag}(+---) \Lambda = \text{diag}(+---)\} $$

One defines the supremum norm on $M_4 (\mathbb{R})$ as:

which turns $M_4 (\mathbb{R})$ into a topological space in the metric topology induced by the norm. Can this supremum norm be made particular for Lorentz matrices? I guess it would, right? It would immediately follow that the Lorentz group is a topological space in the subspace topology of $M_4 (\mathbb{R})$.

Answer

Yes, $\mathrm{O}(1,3)$ is a topological subspace of $M_4(\mathbb{R})$ under this construction.

That said, there's nothing special about the topology induced by supremum norm in this regard; any subset $S$ of a topological space $X$ inherits a natural subspace topology from its "parent" space. You could define some weird topology on $M_4(\mathbb{R}) \sim \mathbb{R}^{16}$ which is a product topology between $n$ copies of $\mathbb{R}$ with the indiscrete topology and $16-n$ copies of $\mathbb{R}$ with the discrete topology, and it would still define a topology on $\mathbb{O}(1,3)$.

[EDIT: my original answer discussed whether the supremum norm might imbue the resulting topological space with non-physical structure; but as Valter Moretti pointed out in a comment, all norm topologies on a finite-dimensional vector space such as $\mathbb{R}^{16} \sim M_4(\mathbb{R})$ are equivalent, so this point is somewhat moot.]

symmetry - Noether's theorem under arbitrary coordinate transformation

Noether's theorem states that every differentiable symmetry of the action of a physical system has a corresponding conservation law.

Suppose our action is of the form $S = \int d^4x\, \mathcal{L}(\phi,\partial_\mu\phi).\tag{1}$

if $x \rightarrow x'$ then if $S \rightarrow S'$ where
$S' = \int d^4x'\, \mathcal{L'}(\phi',\partial_\mu\phi').\tag{2}$

But from calculus we know that $S=S'$ so does that mean that every change of variable correspond to a conserved quantity? why the quantities conserved under Poincare transformation, for example, is more especial?

Answer

The action shown in the question is a functional of $\phi$, not of $x$. A change of the integration variable $x$ is just a relabeling of the index set. It does not transform the dynamic variables $\phi$ at all, so no: a change of variable does not correspond to a conserved quantity.

More explicitly, if $y(x)$ is a monotonic smooth function of $x$, then $$ \int d^4y\ {\cal L}\left(\phi\big(y(x)\big),\, \frac{\partial}{\partial y_\mu}\phi\big(y(x)\big)\right) = \int d^4x\ {\cal L}\left(\phi(x),\frac{\partial}{\partial x_\mu}\phi(x)\right) \tag{1} $$ identically, for any ${\cal L}$ whatsoever (as long as it depends on $x$ only via $\phi$). This is just a change of variable (a relabeling of the index-set), and there is no associated conserved quantity.

In contrast, suppose that the action has this property: $$ \int d^4x\ {\cal L}\left(\phi\big(y(x)\big),\, \frac{\partial}{\partial x_\mu}\phi\big(y(x)\big)\right) = \int d^4x\ {\cal L}\left(\phi(x),\frac{\partial}{\partial x_\mu}\phi(x)\right). \tag{2} $$ Unlike equation (1), equation (2) is not identically true for any ${\cal L}$ and any $y(x)$, though it may be true for some choices of ${\cal L}$ and $y(x)$. The transformation represented in equation (2) replaces the original function $x$, namely $\phi(x)$, with a new function of $x$, namely $\phi\big(y(x)\big)$. This is the kind of transformation we have in mind when we talk about Poincaré invariance and its associated conserved quantities: it is a change of the function $\phi$ which we then insert into the original action, not a change of the integration variable.

Gaps between adjacent light rays from a light source

I can not seem to grasp light rays FILLING space like air and liquid in an enclosure. Two rays adjacent to each other coming from Sun, - do they not diverge as they go into space? If so is there not dark space gap between them, constantly increasing? How come when the sun is up in the sky, we do not perceive these dark spaces as night, as we travel around the sun while facing it, like an interference pattern? I hope my Q is clear.

quantum mechanics - Natural units of information

In physics entropy is usually measured in nats. I wonder is there a possible model of a physical system which has entropy of discrete number of nats?

How particles and degrees of freedom should be arranged so this to happen?

I would be interested in both

• 
  

  example of an analog (non-discrete) system

  
  


• 
  

  example of a quantum system

  
  

gravity - General relativity theory

As I understand general relativity theory (please correct me if I'm wrong), time becomes dilated and space becomes compressed around mass, and this is responsible for gravity. I'm struggling with precisely how that results in gravitational force between masses. Can someone explain that to me (please keep any required math understandable to a lay person).

radioactivity - How does a half-life work?

Carbon-14 has a half-life of 5,730 years. That means that after 5,730 years, half of that sample decays. After another 5,730 years, a quarter of the original sample decays (and the cycle goes on and on, and one could use virtually any radioactive isotope). Why is this so? Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?

Answer

The right way to think about this is that, over 5,730 years, each single carbon-14 atom has a 50% chance of decaying. Since a typical sample has a huge number of atoms¹, and since they decay more or less independently², we can statistically say, with a very high accuracy, that after 5,730 years half of all the original carbon-14 atoms will have decayed, while the rest still remain.

To answer your next natural question, no, this does not mean that the remaining carbon-14 atoms would be "just about to decay". Generally speaking, atomic nuclei do not have a memory³: as long as it has not decayed, a carbon-14 nucleus created yesterday is exactly identical to one created a year ago or 10,000 years ago or even a million years ago. All those nuclei, if they're still around today, have the same 50% probability of decaying within the next 5,730 years.

If you like, you could imagine each carbon-14 nucleus repeatedly tossing a very biased imaginary coin very fast (faster than we could possibly measure): on each toss, with a very, very tiny chance, the coin comes up heads and the nucleus decays; otherwise, it comes up tails, and the nucleus stays together for now. Over a period of, say, a second or a day, the odds of any of the coin tosses coming up heads are still tiny — but, over 5,730 years, the many, many tiny odds gradually add up to a cumulative decay probability of about 50%.

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_{¹ A gram of carbon contains about 0.08 moles, or about 5 × 10²² atoms. In a typical natural sample, about one in a trillion (1 / 10¹²) of these will be carbon-14, giving us about 50 billion (5 × 10¹⁰) carbon-14 atoms in each gram of carbon.}

_{² Induced radioactive decay does occur, most notably in fission chain reactions. Carbon-14, however, undergoes spontaneous β⁻ decay, whose rate is not normally affected by external influences to any significant degree.}

_{³ Nuclear isomers and other excited nuclear states do exist, so it's not quite right to say that all nuclei of a given isotope are always identical. Still, even these can, in practice, be effectively modeled as discrete states, with spontaneous transitions between different states occurring randomly with a fixed rate over time, just as nuclear decay events do.}

general relativity - Is the gravitational effect of distant galaxies lost forever?

Hubble's law is usually expressed by the equation $$v = H_0D$$

According to this equation, the space between us and very distant galaxies, is expanding with a speed greater than the speed of light $c$.

As a result the light from these galaxies can no longer be detected.

Can we also assume that the 'gravitational effect' that these galaxies exert can no longer influence our visible universe?

Since these galaxies can no longer interact in any way with other galaxies, does this means that in a way they form their own 'universe'?

How the theory of the 'big crunch' deals with this?

Answer

The influence of gravity and gravitational waves are thought to travel at the speed of light. So what goes for light also goes for gravity.

Galaxies that we see now can already be receding at greater than the speed of light. As Thriveth says in his comments, this is the case for galaxies at redshift more than 1.4. We see them because the light we see was emitted in the past.

The edge of the observable universe and therefore the most distant that objects can be to influence us now, either through light or gravity, is some 46 billion light years away. This called the particle horizon.

There is another horizon at about 16 billion light years which refers to how far away an object can be now such that its light and gravitational waves never reach us in the future. This is called the event horizon.

The exact values of these numbers depend on the cosmological parameters and, in the case of the vacuum energy density, their time dependence.

In an expanding, accelerating universe, these horizon distances do increase, but all galaxies will eventually reach a point where they lie beyond the event horizon and their influence will no longer be felt in the future.

Of course, a big crunch does not happen in an expanding, accelerating universe.

electricity - Graph of Electrical Resistivity of Air vs Air Pressure

I've search many place ( Google, forums etc ) but can't seem to find anything that explains the Relationship between:

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Electrical Resistivity of Air
vs
Air Pressure

Constant Variable: Temperature ( Room Temperature, not Ionized Gas ), Humidity

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So the question is, what is the Graph of 'Electrical Resistivity of Air' vs 'Air Pressure'?
- Or in other words,
Is Dry normal air better electrical insulator than Vacuum?

Important Note:
This Question is not about Electric Arc vs Air Pressure.

( As what most search engine gave me as answer previously... )

Answer

Related question on EE: Does perfect insulation exist? (especially the part about vacuum)

Insulators and conductors

The property of a material to carry charges from one point to another is what electric current is.

The difference between insulators and conductors lies in the electron band structure they posses. In conductors the Fermi-level (thermodynamically probable energy of an electron) is in a conducting(valence) band. This means most electrons will be ready to move and electricity will flow easily.

In an insulator the Fermi level is far away from the next conduction band, this means that very few electrons will be ready to move(because of a low probability being at a specific state). Other electrons can only be excited by very strong fields or other circumstances.

Therefore the conduction in an insulator can be described as by chance. As air is an insulator this is the case.

How does pressure change this behavior?

Very low, and very high pressure will reduce the number of charged particles that get transported. The first case is because there are only few particles, the second is because it is too stuffed and they collide and loose energy all the time. Actually you shouldn't look at pressure, but the product of pressure and electrode distance, as their product is the thing that is important. That means low and high pressure reduce conductivity.

Except for probabilities, in real life you have cosmic and other rays that can ionize air molecules. This is actually a very important part of the conduction mechanism for gasses, and can't be neglected.

Breakdown laws (but for voltages lower than the breakdown voltage)

As you were only interested in conductivity I will just tell you a few keywords regarding the other aspects of conduction during breakdown: Pashen's law, Townsend Mechanism, Streamers.

These mechanisms can be fed parameters which won't result in breakdown. In these cases they will describe the conduction mechanism (to some degree).

newtonian mechanics - Why does frictional force cause a car to move? Also, is friction a reaction force?

My teacher told me that this was because frictional force resists relative motion between the engine and the car. However, in this case it seems to allow relative motion between the ground and the car.

Answer

[...] it seems to allow relative motion between the ground and the car.

While there is relative motion between car and ground, there is not relative motion between wheel and ground. And that's what matters, because that's where the friction is.

Think of walking. Your body moves relative to the ground. But at each step, your foot on the ground is stationary. Your foot pushes on the ground backwards. Static friction holds back forwards to avoid that your foot slides.

On a car's wheel, the same happens, just at each new point that continuously comes in contact with the ground. For that short moment that it is in contact, that point is stationary and there is no sliding. That point pushes backwards, and so a static friction pushes forwards to avoid sliding (to avoid wheel spin).

Also, is friction a reaction force?

You can think of static friction as a reaction force, if you will. It only exists because your leg - or the car's wheel - applies a force backwards on the ground.

quantum mechanics - What does a light wave look like (3d model)

What does a light wave look like?

The only models I can seem to find online are 2D waves, they just look like sin() graphs.

I have seen the models of the two components of "light waves" (electric field and magnetic field) and they are represented on a 3D Cartesian coordinate system, but they are still just two 2D waves.

Surely light isn't really FLAT like this is it?

I guess I have always assumed it pules out in all directions greater then lesser as it travels, giving off the shape that you see during a sonic boom:

I have drawn what I picture to be a 3D model of the electric field of a light wave as it travels from left to right:

(cone image from: http://www.presentation-process.com/images/3d-powerpoint-cone.jpg)

Is this an accurate representation of what the 3D radiation emitting from a light beam looks like, like a 3D wave? (Obviously it would be more wavey, using a 3D cone graphic to create this diagram caused the edges to look spiked and sharp, a better object to use would have been something like a bullet (3D parabola) but I'm not the best with photoshop).

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Also, if this is a somewhat accurate model of light wave pulsing in 3D, what does the magnetic wave look like in model form? Do they just overlap with possibly a slightly larger or smaller amplitude but the same maxima and minima locations along t (x-axis in my model)

Answer

The first 2-D image you posted is a typical simplification for teaching purposes. In it, they use the height of the sine wave to represent magnitude, and the directions of the sine waves to show how the fields point relative to each other. The light itself however is not itself at all cone-like. You have to imagine this sine wave existing at multiple points in space, not localized in this cone-like volume. This can be difficult to visualize.

A common method for visualizing these kinds of fields is a 3-D vector field plot.

Length or colour will typically correspond to amplitude, and the arrows show direction of the electric field. The magnetic field is always perpendicular, and the amplitude is proportional, so there's little sense in plotting both together. This one I've included shows how the field actually permeates a volume.

Out of interest, here's a gif of dipole radiation. This is just a 2-D slice of the field, and doesn't show vector direction, but is a very good visualization of a more real-life kind of radiation. Colour corresponds to magnitude in this case.