What makes an electron flip to spin-up?

The normal mode of spin for an electron is spin down:

with angular momentum L pointing in the opposite direction of motion V (right)

but, when a free electron approaches , say, a Helium ion it must , according to Pauli's, flip its spin in the the direction of motion (left: spin up)

What makes the electron flip over? who provides the energy for the change? Does the energy required vary form case to case, and if so, what is the formula?

Answer

The spin of an electron is an intrinsic and unambiguous property. The spin is related to the electrons magnetic dipole moment and to the direction of the deflection in a magnetic field (Lorentz force). If - by convention - the electrons magnetic dipole moment and the spin are showing in the same direction, then for a positron these two parameters are anti-aligned (and in a magnetic field the moving positron gets deflected in the opposite direction).

In an atom two electrons can occupy the same three quantum numbers if and only if their magnetic dipole moments are anti-aligned. Pauli realized...

... that the complicated numbers of electrons in closed shells can be reduced to the simple rule of one electron per state, if the electron states are defined using four quantum numbers. For this purpose he introduced a new two-valued quantum number, identified by Samuel Goudsmit and George Uhlenbeck as electron spin.

That is the origin of why there are two understandings of what spin is. But the flip or alignment of the magnetic dipole moments of two electrons in opposite directions has nothing to do with the spin of an electron and the opposite spin in positrons (emanated in Lorentz force and in the direction of induced magnetic fields).

In the comments you ask

... the question is a bit different: whatever changes (helicity/chirality), what makes a free electron adapt it to Pauli's (principle).

Simply the electrons orienting each other in such a way that north and south poles near the opposite poles of the other electron. This is the lowest possible energetic and most stable level.

---

user157860 in a comment talks about the 21cm hydrogen line. The most detailed information I found on this Wikipedia page (Google translation):

The forbidden line of neutral hydrogen is caused by the interaction of the magnetic moments of an electron and a proton in a hydrogen atom. The energy of the hydrogen atom with parallel arrangement of the magnetic moments of the electron and the proton is somewhat larger than in the case of an antiparallel one, so - when a spontaneous change in the orientation of the magnetic moment of the electron in the opposite orientation - atom emits a quantum of electromagnetic radiation with a wavelength of 21.1 cm (frequency 1420.40575 MHz).

In parallel with the emission of the EM radiation, the reverse process also occurs - the excitation of hydrogen atoms by electromagnetic quanta with high energies or in collision between atoms. Therefore, in interstellar atomic hydrogen, a dynamic equilibrium is established between the radiation events of radio quanta and the excitation of atoms by EM quanta and collisions.

general relativity - Perturbation of a Schwarzschild Black Hole

If we have a perfect Schwarzschild black hole (uncharged and stationary), and we "perturb" the black hole by dropping in a some small object. For simplicity "dropping" means sending the object on straight inward trajectory near the speed of light.

Clearly the falling object will cause some small (time dependent) curvature of space due to its mass and trajectory, and in particular, once it passes the even horizon, the object will cause some perturbation to the null surface (horizon) surrounding the singularity (intuitively I would think they would resemble waves or ripples). Analogously to how a pebble dropped in a pond causes ripples along the surface.

Is there any way to calculate (i.e. approximate numerically) the effect of such a perturbation of the metric surrounding the black hole?, and specifically to calculate the "wobbling" of the null surface as a result of the perturbation,maybe something analogous to quantum perturbation theory?

Or more broadly, does anyone know of any papers or relevant articles about a problem such as this?

Answer

Your intuitive picture is basically correct. If you perturb a black hole it will respond by "ringing". However, due to the emission of gravitational waves and because you have to impose ingoing boundary conditions at the black hole horizon, the black hole will not ring with normal-modes, but with quasi-normal modes (QNMs), i.e., with damped oscillations. These oscillations depend on the black hole parameters (mass, charge, angular momentum), and are therefore a characteristic feature for a given black hole.

Historically, the field of black hole perturbations was pioneered by Regge and Wheeler in the 1950ies.

For a review article see gr-qc/9909058

For the specific case of the Schwarzschild black hole there is a very nice analytical calculation of the asymptotic QNM spectrum in the limit of high damping by Lubos Motl, see here. See also his paper with Andy Neitzke for a generalization.

Otherwise usually you have to rely on numerical calculations to extract the QNMs.

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?

mathematics - A walk of 3000 meters, but one foot has moved more, how so?

My math teacher has struck again.

Here's his newest riddle:

Today I went for a normal walk of 3000 meters.
One of my feet had to move exactly 3000 meters.
However, the second foot moved 3100 meters.

Can you justify how did that happen?

Edit:

I am not sure how this question has been selected as too broad.
It clearly only has a mathematics tag and the correct answer has been given and accepted in a fast way. If people decided to answer it using lateral-thinking it doesn't mean that the question is too broad, it means that posters just want to get the "funny" comments and a "+1".

Answer

Based on @TwoBitOperation's answer

and assuming his feet are 25 centimeters apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one foot walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 meters.

Whether these

63.6 turns around e.g. a fountain or pond with a 15m diameter constitute a "normal walk" remains an open discussion.

special relativity - How to get the accurate relativistic momentum form for photons?

I have studied from Griffiths, the relativistic form of momentum is $$p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} m_0v$$

Now when I evaluate the momentum for photon, I just insert $v=c$ and $m_0=0$ and I get $p= 0/0$. How does it make sense?

Can you tell me that where I am wrong?

Answer

You should consider a particle with some finite energy $E$ and use that constraint to take the $v\rightarrow c$ limit.

With Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$, the relativistic total energy is $E = \gamma mc^2$. Therefore, $p/E = v/c^2$. With the particular case of $v = c$, it follows that $E = pc$.

Although really, you should simply consider $E = pc$ for massless particles to be more fundamental. The general relation is $(mc^2)^2 = E^2 - (pc)^2$, which corresponds to the the norm-squared of the four-momentum vector in relativity.

photoelectric effect - Why are photoelectrons emitted in the direction of incident photons?

In the experiments for photoelectric emission, the light is incident on one face of the emitting plate, for example the anode, when determining the stopping potential. The electrons are emitted by this face of the plate. Why are the electrons emitted in the direction of the incident light, and not opposite to it?

I have read a little about photons (the key words being, a little), and from what i gather, they have an intrinsic momentum $p=h/\lambda$, which is transferred entirely to the particle that absorbs it (along with the energy $h\nu$). This should imply that the electrons try to move further into the cathode, and not escape it! What is the fault in the reasoning?

Thanks!

Answer

Your reasoning is quite correct, and you can see exactly this phenomenon in a photomultiplier tube. The photomutiplier tube uses very thin metal sheets, and when a photon strikes the sheet the primary photoelectron is emitted in the same direction as the incident photon and escapes from the far side of the sheet:

The quantum yield for this process is close to 100% i.e. almost every photon ejects a photoelectron. However the quantum yield in a typical photoelectric experiment is about $10^{-5}$ to $10^{-6}$ i.e. up to a factor of a million times poorer. The reason for this is because the initial photoelectron is emitted travelled down into the body of the metal. For the photoelectron to escape it has to backscatter off other electrons in the metal and ricochet back to the surface without losing so much energy it can no longer escape. This is a very low probability process so the overall quantum yield is very low.

quantum mechanics - How does an electron absorb or emit light?

It's a common understanding that atoms emit or absorb light when the energy of the photons is equal to the the difference in the energy levels in the atom. What I don't understand is how does an electron absorb light inside an atom? Is it that the atom as a whole absorbs light or the individual electrons absorb the light packets? Also I would like to know, what does one mean when you say that an electron absorbs photon. How does an electron absorb a photon??

Answer

An atom is nothing but a bounded state of electrons and a positively charged core called nucleus. The electrons in the atom are in bound state and so their energy levels are quantized. Also, it is possible to have quantized rotational and vibrational energy levels of the molecules. The way in which they differ is in the difference in the energy characterizing the transition from one state to another.

Possible ways in which a photon is absorbed by an atom or a molecule

If the energy level of the incoming photon is such that the electrons can have a transition from a state to some higher permissible state, then the photon energy level will be in the visible or ultraviolet range and we make use of this principle in electronic spectroscopy.

Suppose, a particular electron is in the energy state with energy eigenvalue $E_i$. There exists a higher energy level $E_f$. If the energy levels of the electron bound states are such that it precisely matches with the energy of the photon: $h\nu=E_f-E_i$, then the electron will get excited to the energy state $E_f$.

Now, if the incident photon energy matches the difference in the vibrational energy levels of any pair of states of the molecule, then it can cause transition from that vibrational energy state to the higher energy state. This energy usually lies in the infrared region and the technique is used in infrared spectroscopy.

For example, in the case of diatomic molecules, the vibrational energy levels are quantized and in a good sense they can be approximated to that of a harmonic oscillator: $E_n=\left(n+\frac{1}{2}\right)\bar{h}\omega$. So, if the photon energy is such that $h\nu=E_f-E_i$, the electron transits from the state $E_i$ to $E_f$, where $E_i$ and $E_f$ are given by the above equation of the harmonic oscillator and the states are defined by the quantum number $n=i$ and $n=f$.

Now, if the absorption of a photon can only affect the rotational energy levels of the molecule, then the absorbed photon will be in the microwave region. The spectroscopic technique making use of this principle is the microwave spectroscopy.

For example, the rotational energy levels of a diatomic molecule are given by: $\displaystyle{E_j=\frac{j(j+1){\bar{h}}^2}{2I}}$, where $I$ is the moment of inertia and $j$ is the angular momentum quantum number. In such a case, we can write: $h\nu=E_f-E_i$ and the bound state absorbs the photon and will get excited to the state with energy $E_f$, with $E_f$ and $E_i$ determined by the quantum number $j=f$ and $j=i$.

Now, the energy can be absorbed by the nuclei also. It can be elastic nuclei scattering (analog to very low energy Compton scattering by an electron. In this process, a photon interacts with a nucleon in such a manner that a photon is re-emitted with the same energy), inelastic nuclei scattering (the nucleus is raised to an excited level by absorbing a photon. The excited nucleus subsequently de-excites by emitting a photon of equal or lower energy) and Delbruck scattering (the phenomenon of photon scattering by the Coulomb field of a nucleus, also called nuclear potential scattering, which can be thought of as virtual pair production in the field of the nucleus. i.e., pair production followed by annihilation of the created pair). However, these processes are negligible in photon interactions.

Conclusion:

Absorption of a photon will occur only when the quantum energy of the photon precisely matches the energy gap between the initial and final states of the system. (the atom or a molecule as a whole) i.e., by the absorption of a photon, the system could access to some higher permissible quantum mechanical energy state. If there is no pair of energy states such that the photon energy can elevate the system from the lower to the upper energy state, then the matter will be transparent to that radiation.

So, if any of the above types of energy transition take place, that will affect the quantum state of the system as a whole (transits the system from one state to another). So one could say, as @annav pointed out, it is the atom (or the molecule) that absorbs the radiation and changes the energy levels of its constituent particles, depending on the energy absorbed. Anyway, a change in energy level of the electron, or rotational or vibrational energy levels of the molecules can be seen as changing the quantum state of the molecule. So, it's better to stick with the concept of the molecule as a whole absorbs the energy and changes its state to some higher energy state by changing the quantum state of its constituent particles.

logical deduction - How did the police know?

One night, a man receives a call from the police. The police tell the man that his wife was robbed, murdered, and her body dumped in a remote location, and that he should reach the crime scene as soon as possible.

The man drops his phone in shock, and drives 20 minutes to the crime scene. Though the man does not have any evidence of committing the murder on his personal being (no blood or other signs of foul play on the car or any possessions) as soon as he reaches the crime scene, and before he can say or do anything, the police arrest him on suspicion of committing the murder (he will, of course be judged by a jury of his peers, etc.).

Without any sign of typical evidence, why do the police think that he committed the crime?

_{It should be noted that this takes place in the typical world, without aliens, super powers, or unlimited quantities of any material.}

Answer

I think

They didn't tell him where the murder scene was but he could find it.

What exactly is light?

What is light in terms of physics? Whenever looking for definition on the internet getting explanation or translation of my language, a common definition/explanation for light was

Light is light. We get it from the Sun, when we flip on the switch, or turn on the flashlight. We see it! But that explanation doesn't really explain anything

This was explanation or it just was an example of light. Can anyone define what exactly light is? Is it motion? Is it energy? Or vice versa?

Answer

Light is a phenomenon which transfers energy and momentum through many substances, including vacuum, with characteristic velocity $c$ through the vacuum and characteristic energy-to-momentum ratio $E/c.$ It differs from other similar phenomena in many other characteristic ways: it can bend through glass, it has a 2D polarization transverse to its propagation (so there exist polaroid filters which work on it), it can be emitted and absorbed by antennas, it is absorbed (at least at visible wavelengths) by our eyes, and it tends to cause electrons and protons to vibrate, but not isolated neutrons: its effect is proportional to a strongly-related physical phenomenon called "electric charge." We like to think of it as a wave in the "electromagnetic field" but ultimately it comes in lumps of energy where the energy per lump depends on the frequency via Planck's constant, $E = h~f,$ and these diffract due to the rules of quantum mechanics.

special relativity - Why are usually 4x4 gamma matrices used?

As far as I understand gamma matrices are a representation of the Dirac algebra and there is a representation of the Lorentz group that can be expressed as

$$S^{\mu \nu} = \frac{1}{4} \left[ \gamma^\mu, \gamma^\nu \right]$$

Usually the representations used for them are the Dirac representation, the Chiral representation or the Majorana representation.

All of these are 4x4 matrices. I would like to know what the physical reason is that we always use 4x4, since surely higher dimensional representations exist.

My guess is that these are the smallest possible representation and give spin half fermions as the physical particles, which are common in nature. Would higher dimensional representations give higher spin particles?

Answer

You have no other choice than to use $4\times 4$ matrices. All these "representations" are different realizations (related by similarity transformations) of the only possible irreducible representation of the Clifford algebra that is spanned by the abstract $\gamma^\mu$. This representation, in a way, is the definition of what a "Dirac spinor" is, and it is usually a representation of the covering group of the rotation group, but only a projective representation of the rotation group itself. Also, it is not always irreducible as a representation of the rotation group (e.g. the 4D Diac spinor decomposes into the two Weyl spinors and also into two Majorana spinors).

You can show in general that the Clifford algebra in $(1,d-1)$ dimensions has its only irreducible representations given by a vector space of dimension $2^{\lfloor {d/2}\rfloor}$, which is $2^2 = 4$, by considering the "raising/lowering operators" ${\gamma^\pm}^k = \gamma^{2k}\pm\gamma^{2k+1}$ in close analogy to the usual ladder operator method for $\mathfrak{su}(2)$. It turns out that the space spanned by $\lvert s_1,\dots,s_k\rangle $ for $s_i=\pm 1/2$ (the $s_i$ are the eigenvalues of $S^k = [\gamma^{+k},\gamma^{-k}]$) is the only consistent non-trivial irreducible representation you can construct. In odd dimensions, there are two different ones of these that differ by chirality.

Another way uses the group of the $\Gamma^M$ constructed by taking products $\gamma^{\mu_1}\dots\gamma^{\mu_k}$ for $k \leq d$ and $\mu_1 < \mu_2 < \dots \mu_k$. The $M$ runs from $1$ to $2^d$ (another thing one must show...). Any irreducible representation of the Clifford algebra is an irreducible group representation of this group.

Now consider $S = \sum_M \rho(\Gamma^M) N\sigma(\Gamma^M)^{-1}$ for two irreducible representations $\rho$ of dimension $n$ and $\sigma$ of dimension $n'$ and any $n\times n'$-matrix $N$. You can show that $S\rho(\gamma^M) = \sigma(\gamma^M)S$, so $S$ is an intertwiner, and by Schur's lemma either $S$ is invertible, so $n=n'$, or $S=0$. So if there are two different irreducible representations, this says that $\sum_M\rho(\Gamma^M)N\sigma(\Gamma^M)^{-1} = 0$ for any choice of $N$. Therefore $$ \sum_M \rho(\Gamma^M)_{kl}\sigma(\Gamma^M)_{ij} = 0$$ for all $k,l,i,j$. Choosing $k=l$ and $i=j$ summing (i.e. taking the trace of the two matrices independently) and thinking about which gamma matrices contribute to these traces, one can conclude both for the even and the odd case that $n=n'$ must hold, and that there is one irreducible representation for even $d$ and two of them for the odd case.

The interpretation of mass in quantum field theories

Consider a free theory with one real scalar field: $$ \mathcal{L}:=-\frac{1}{2}\partial _\mu \phi \partial ^\mu \phi -\frac{1}{2}m^2\phi ^2. $$ We write this positive coefficient in front of $\phi ^2$ as $\frac{1}{2}m^2$, and then start calling $m$ the mass (of who knows what at this point) and even interpret it as such. But pretend for a moment that you've never seen any sort of field theory before: if someone were to just write this Lagrangian down, it's not immediately apparent why this should be the mass of anything.

So then, first of all, what is it precisely that we mean when we say the word "mass", and how is our constant related to this physical notion in a way that justifies the interpretation of $m$ as mass?

If it helps to clarify, this is how I think about it. There are two notions of mass involved: the mathematical one that is part of our model, and the physical one which we are trying to model. The physical mass needs to be defined by an idealized thought experiment, and then, if our model is to be any good, we should be able to come up with a 'proof' that our mathematical definition agrees with the physical one.

(Of course, none of this at all has anything to do with this particular field theory; it was just the simplest Lagrangian to write down.)

quantum mechanics - Problem in Hamiltonian

I need to elaborate the equation ,and need to know what is the physical significance and how matrices will manipulate in the equation $$ \hat{H} = (\hat{\tau_3}+i\hat{\tau_2})\frac{\hat{p}^2}{2m_0}+ \hat{\tau_3}m_0 c^2 = \left| \begin{array}{ccc} 1 & 1 \\ -1 & -1 \\ \end{array}\right| \frac{\hat{p}^2}{2m_0} + \left| \begin{array}{ccc} 1 & 0 \\ 0 & -1 \\ \end{array}\right| m_0 c^2 $$

Where $$\tau_1 , \tau_2,\tau_3 $$ are Pauli matrices and Hamiltonian comes from "Schrodinger form of the free Klein_Gordon equation And also why did we added Pauli matrices in the free Hamiltonian ?

Answer

Your second equation is somehow mistyped: it should have only one $\frac{p^2}{2m}$ factor in the first term.

Also, the notation $m_0$ devoid of physical meaning. There is not "rest mass" (different from any other "mass") - what used to be called by some authors (e.g. early in the 20th century) the "rest mass" is now understood to be the only mass that make sense to assign to a (massive) particle. Therefore the only correct notation is $m$, not $m_0$.

I am not sure what is your question. Perhaps showing that that Hamiltonian is somehow equivalent to the Klein-Gordon's Hamiltonian? Or something else? If that is what you are asking, then the derivation is straight-forward. Let's have a slightly more general starting expression: $H= a \frac{p^2}{2m} + b m c^2$, where $a$ and $b$ are $2\times2$ matrices whose properties are to be determined. Square $H$ to get: $H^2 = a^2 \left(\frac{p^2}{2m}\right)^2 + \{a,b\} \frac{p^2}{2m} mc^2 + b^2 m^2c^4$. If we can find $a$ and $b$ such to have $a^2=0$ $\{a,b\}=2\cdot{\bf 1}$ (here $\{A,B\}:\equiv A B + B A$) and $b^2={\bf 1}$ (and ${\bf 1}$ is the unit matrix), then we would get the original dispersion equation for a free relativistic particle: $H^2 = p^2c^2 + m^2c^4$. One such a choice is $a= \tau_3+i\tau_2$ and $b=\tau_3$, but there are many other choices, too.

Working with the components:

$H \pmatrix{\chi \cr \xi} = \left[\left(\frac{p^2}{2m}+mc^2\right) \pmatrix{1 &0 \\ 0 & -1} + \frac{p^2}{2m} \pmatrix{0 & 1 \\ -1 & 0}\right]\pmatrix{\chi \cr \xi}$

gives:

$H \chi = \left(\frac{p^2}{2m}+mc^2\right) \chi + \frac{p^2}{2m} \xi$

and

$H \xi = - \left(\frac{p^2}{2m}+mc^2\right) \xi - \frac{p^2}{2m} \chi$

Notice that $[H, p^2] =0$ so one can apply $H$ one more time to the left side of any of these two equations:

$H^2 \chi = \left(\frac{p^2}{2m}+mc^2\right) H \chi + \frac{p^2}{2m} H \xi = \left(\frac{p^2}{2m}+mc^2\right)^2 \chi + \left(\frac{p^2}{2m}+mc^2\right)\frac{p^2}{2m}\xi - \frac{p^2}{2m}\left(\frac{p^2}{2m}+mc^2\right)\xi - \left(\frac{p^2}{2m}\right)^2\chi = \left[\left(\frac{p^2}{2m}+mc^2\right)^2- \left(\frac{p^2}{2m}\right)^2\right]\chi = \left[p^2 c^2 + m^2c^4\right]\chi$.

Etc.

homework and exercises - Find electric potential due to line charge distribution?

I need help how to set up this integral $$V(\mathbf r)=\frac{1}{4\pi\epsilon_0} \int_L \frac{\rho'_l}{\lvert \mathbf r - \mathbf{r'} \rvert}\mathrm{d}l'. $$

I have a uniform line charge along the $z$-axis and want to calculate the electric potential between two points $A=(r_A,\phi_A,0)$ and $B=(r_B,\phi_B,0)$ (cylindrical coordinates).

Solution: The line charge is aligned along the $z$-axis and the the source vector is $\mathbf{r}=z\hat{\mathbf z}$ and the field vector is $\mathbf{r'}=r'\hat{\mathbf r}+z'\hat{\mathbf z}$ so $$\lvert \mathbf{r}-\mathbf{r'} \rvert=\sqrt{(r')^2+(z-z')^2}.$$ I integrate along $z'$ from $-\infty$ to $\infty$ \begin{align} V(\mathbf r) &=\frac{1}{4\pi\epsilon_0} \int_L \frac{\rho'_l}{\lvert \mathbf r - \mathbf{r'} \rvert}\mathrm{d}l' \\ &=\frac{\rho_l}{4\pi\epsilon_0} \int_{-\infty}^{\infty}\frac{1}{\sqrt{(r')^2+(z-z')^2}}\mathrm{d}z'\\ &=\frac{1}{4\pi\epsilon_0} \big [ \ln(z-z'+\sqrt{(r')^2+(z-z')^2})\big ]^{\infty}_{-\infty}\\ &= -\infty +\infty \end{align} The integral is indeterminate and I'm stuck here. Are the vectors wrong, the limits? What have I missed?

Thanks!

If I calculate the line integral of the electric field I find the correct potential (however, I want to calculate with the integral above).

I integrate in the radial direction $\mathrm{d}\mathbf{r}$ from $r_A$ to $r_B$. The electric field is $\mathbf{E}(\mathbf{r})=\frac{\rho_l}{2\pi\epsilon_0 r}\hat{\mathbf{r}}$ so \begin{align}V(\mathbf{r})&=-\int_L \mathbf{E}(\mathbf r) \cdot \mathrm{d}\mathbf{l}\\&=\frac{\rho_l}{2\pi\epsilon_0}\int_{r_A}^{r_B}\frac{1}{r}\mathrm{d}r=\frac{\rho_l}{2\pi\epsilon_0}\ln{\frac{\rho_B}{\rho_A}}\end{align}

homework and exercises - Why must identical lightbulbs in series have identical voltage drops?

When I connect two identical lightbulbs in series, how come they have equal brightness? Why can't one lightbulb have a larger voltage drop than the other? (i.e. the first lightbulb "uses up all the energy" and the second lightbulb is barely lit because "there isn't much energy left"?

earth - What is the deepest man-made underground tunnel?

What is the deepest tunnel in the Earth's surface?

I have meet numbers, around 300km.

What is the main problem to make tunnel more deeper? What is the problem with 1000km?

newtonian gravity - Newton's Law of Graviation: Why $G$ and not e.g. $dfrac{1}{4pi G_0}$?

I've been wondering, in Coulomb's Law, $k_e = \dfrac{1}{4\pi\epsilon_0}$. Therefore, why do we use $G$ in Newton's Law of Gravitation? What if the constant is more like Coulomb's Law, e.g. $G = \dfrac{1}{4\pi G_0}$ where $G_0$ is some constant.

This would make Newton's Law of Gravitation look like the following: $$\bf{F}_{12} = -\dfrac{m_1 m_2}{4\pi G_0 |\bf{r}_{12}|^2} \bf{\hat{r}}_{12}$$

$GM = \mu$ is used for calculations but so could $\dfrac{M}{4\pi G_0} = \mu$.

If this is not the case, what is the significance of this definition?

quantum field theory - Is there a standard convention for whether the term "handedness" refers to helicity or chirality?

I was under the impression that the "handedness" of a massive spin-1/2 particle refers to its chirality rather than its helicity. This answer, this one and Srednicki's QFT textbook seem to use the term in that way. But this answer and the web pages here, here, here, and here use the term "handedness" to mean helicity.

I often see the statement "only left-handed neutrinos interact (non-gravitationally) with the other Standard Model particles", but (incorporating neutrino masses) this statement clearly only makes sense if "handedness" refers to chirality, because whether or not two particle species interact is clearly Lorentz invariant, like chirality but unlike helicity.

Is the usage of the term "-handed" consistent enough that we can identify one of these sets of sources as "wrong"? Or is the usage inconsistent enough that the word is fundamentally ambiguous and should never be used in technical discussions involving massive spin-1/2 particles without specifying its definition?

Answer

The uses are absolutely inconsistent. The problem is that "handedness" is a wonderfully intuitive term for both "chirality" and "helicity". That means that courses and textbooks will use "handedness" to describe whichever word is used more. There is no standard.

Particle physics books that avoid QFT have no need for chirality, so they use handedness to refer to helicity. For example, Griffiths' standard book states on p.138

Neutrinos are left-handed; antineutrinos are right-handed.

in all caps for emphasis. He means there's a left-chiral neutrino field that makes left-helicity neutrino particles and right-helicity antineutrino particles, but he has no need for fields at all.

On the other hand, books that focus on QFT will likely use handedness to refer to chirality, and this is doubly true if the book spends a lot of time on classical field theory. Less careful books such as Zee's will freely switch back and forth between the two definitions. Personally I try to completely avoid using the word "handedness" at all.

Pop science typically focuses on particles because they're less abstract, so they'll use handedness to refer to helicity. However, you should take caution trying to fit popsci statements you "often see" into a cohesive framework, because half of these statements have been known to be wrong for decades. Almost every popular source still believes that virtual particles can "borrow energy from the vacuum" by the uncertainty principle, that the mass of a particle increases with its speed, that the Higgs gives mass to the proton, that antimatter falls up, and so on. The fact is that there are two Standard Models, a real one made of mathematics and a fake one patched together from dubious analogies, and in the public sphere the fake one has completely won.

special relativity - According to Einstein & Brian Greene, does the photon remain stationary in the fourth dimension?

According to Einstein and Brian Greene, does it logically follow that the photon remains stationary in the fourth dimension?

In An Elegant Universe, Brian Greene writes:

“Einstein found that precisely this idea—the sharing of motion between different dimensions—underlies all of the remarkable physics of special relativity, so long as we realize that not only can spatial dimensions share an object’s motion, but the time dimension can share this motion as well. In fact, in the majority of circumstances, most of an object’s motion is through time, not space. Let’s see what this means.” Space, Time, and the Eye of the Beholder, An Elegant Universe, Brian Greene, p. 49

Brian Greene and Albert Einstein also state that there is one and only one velocity for all entities through the four dimensions--the velocity of light or c.

A photon travels at c through the three spatial dimensions. All of its velocity is directed through the three spatial dimensions. Thus Brian and Einstein are stating that a photon must be stationary in the fourth dimension. For if the photon had any velocity component in the the fourth dimension, its velocity would be different from c, which is not the case.

On the other hand, if an object is stationary in the three spatial dimensions, it must be moving at c through the fourth dimension.

We can summarize this as:

Axiom: All entities have one velocity through the four dimensions--c. (Einstein & Brian Greene).

Axiom: The velocity of light (photons) is c through the three spatial dimensions. (Einstein)

Theorem: The photon remains stationary in the fourth dimension, as all of its velocity c is through the three spatial dimensions.

Does this logic make sense?

Also, do you prefer using the word "Axiom" or "Postulate"?

An Axiom or a Postulate is a Truth.

A Theorem is that which follows logically from Axioms.

Answer

Photon experience infinite time dilation and hence, time is stationary for it. Does photon experience time

 photon travels at c through the three spatial dimensions. All of its velocity is directed through the three spatial dimensions. Thus Brian and Einstein are stating that a photon must be stationary in the fourth dimension. For if the photon had any velocity component in the the fourth dimension, its velocity would be different from c, which is not the case.

All this is non sense because there is nothing like speed of time in actual world. It's just a minkowski space concept. You cannot compare speed of object in time axis and 3 dimensional world. Calculating speed in 4D

Axioms and postulate serve as a basis for deducing other truths. The ancient Greeks recognized the difference between these two concepts. Axioms are self-evident assumptions, which are common to all branches of science, whilepostulates are related to the particular science.

The two axioms are postulates and not axioms , since they are relevant to only one branch, physics.

Theorem seems to correct.

Is there a Quantum-Mechanics analog to QFT's quartic $lambda phi^{4}$ interaction?

In QFT; for the quartic-interacting real scalar field $\phi$ we have the Lagrangian density: $$ \mathcal{L} \ = \ \frac{1}{2} \left( \partial^{\mu} \phi \right)\left( \partial_{\mu} \phi \right) + \frac{1}{2} m^2 \phi^2 + \lambda \phi^{4} $$ For $\lambda =0$ we have a free bosonic field - when $\lambda \neq 0$ one way to interpret the interaction term is as the quanta of the field, the bosons, having a 2-body repulsive interaction.

My question is; what is the analog for the above interaction in ordinary Quantum Mechanics?

Suppose I have a particle $A$ with free Hamiltonian $H_{A} = \frac{\mathbf{p}_{A}^2}{2m}$, whose states span some Hilbert Space $\mathcal{H}$. Take another identical particle $B$ with free Hamiltonian $H_{B} = \frac{\mathbf{p}_{B}^2}{2m}$. The non-interacting free Hamiltonian for the combined system of the two particles is something like: $$ H \ = \ \frac{\mathbf{p}^{2}_{A}}{2m} \otimes \mathbb{I}_{\mathcal{H}} + \mathbb{I}_{\mathcal{H}}\otimes \frac{\mathbf{p}^{2}_{B}}{2m} + V(\mathbf{x}_{A}, \mathbf{x}_{B}) $$

Is there a potential $V(\mathbf{x}_{A}, \mathbf{x}_{B})$ which emulates the $\lambda \phi^{4}$ interaction from QFT? My naive guess is something along the lines of; $$ V(\mathbf{x}_{A}, \mathbf{x}_{B}) = \lambda \big| \mathbf{x}_{A} \otimes \mathbb{I}_{\mathcal{H}} - \mathbb{I}_{\mathcal{H}} \otimes \mathbf{x}_{B} \big|^{4} $$

Is this correct?

Answer

The transition back to ordinary quantum mechanics is accomplished by working out the way probability amplitudes in the $N$-particle states evolve with time. When that is done, the particles interact with "contact terms", potentials that are proportional to $\delta(\mathbf{x}_i-\mathbf{x}_j)$. For details, see the development done in the first chapter of Brian Hatfield's "Quantum Field Theory of Point Particles and Strings".

That said, it is often repeated that ordinary quantum mechanics is just quantum field theory with zero spatial dimensions. Under the mapping implied the $\phi^4$ theory would become $$H = \frac{p^2}{2m} + \frac{m\omega^2}{2}x^2 + \lambda x^4.$$ With a countable number of particles being just a sum by adding indices to $p$ and $x$.

Transitioning back to continuum mechanics is tricky, and I don't know all of the details for certain (textbooks usually get a little hand-wavy about this path). I can say with confidence, though, that any term that mixes $x_i \rightarrow f_i \sqrt{h^3}$ with different indices, where $h$ is the spacing between adjacent sites, will either become a derivative or vanish on the continuum transition. The $\lambda$ of the discrete theory would have to be remapped by $\lambda\rightarrow \lambda h^{-3}$, for instance, in order to keep the $\lambda \phi^4$ term from vanishing in the limit $h \rightarrow 0$. All of the $m_i$ would have to be the same (say, $1$), by Lorentz invariance, and the springs that connect the $x_i$ would have to have spring constants that diverge like $h^{-1}$ in order to produce the spatial derivatives between needed sites. So, the intermediate Hamiltonian is given by \begin{align} H &= \sum_{i,j,k=-N}^N \left[\frac{1}{2}p_{i,j,k}^2 h^3 + \frac{1}{2} (f_{i+1,j,k}-f_{i,j,k})^2 h^2 + \frac{1}{2} (f_{i,j+1,k}-f_{i,j,k})^2 h^2 + \frac{1}{2} (f_{i,j,k+1}-f_{i,j,k})^2 h^2 \right. \\ & \hphantom{= \sum_{i,j,k=-N}^N}\ \ \left. + \frac{m^2}{2}f_{i,j,k}^2 h^3 + \lambda f_{i,j,k}^4 h^3\right] \end{align} and the continuum mechanics are recovered by applying $\lim_{h\rightarrow 0}\lim_{N\rightarrow \infty}$.

Note that the transformation where $x_i\rightarrow f_i\sqrt{h^3}$ and $p_i\rightarrow p_i\sqrt{h^3}$ is not a canonical change of variables. Starting from the Lagrangian it is straightforward to show from canonically conjugate momentum's definition that $p_i$ would transform as $p_i \rightarrow p_i h^{-3/2}$ if it were. What's going on here is that, in order to get a clean Hamiltonian that becomes an integral of a density, we need to modify the canonical commutation relation to read $$[f_{i,j,k},\, p_{n,\ell,m}] = i\hbar \delta_{in}\delta_{j\ell}\delta_{km} h^{-3},$$ where the right hand side now actually becomes a Dirac delta function in the continuum limit, instead of just saying that Kronecker deltas become Dirac deltas.

All of that said, there's a reason that the usual transition back to ordinary QM mentioned in the first paragraph is so convoluted. In ordinary QM $x$ is the observable position of some particle. In QFT the positions of particles are still the fundamental observables, it's just that now the particles are described as excitations in a field. The field strength is not directly observable, though, it can just be inferred on average when it is strong enough to not be significantly disturbed by the injection of multiple small test particles that we measure the positions of.

Ideas on puzzles for a scientific event

I hope this kind of question is welcome here.

A festival of science will take place in a few months in my university, and I would like to set up a simple but exciting escape room for people to discover scientific notions in a fun way.

Ideas on puzzles?

The main theme would be the light; do you know manual puzzles which are simple to setup and would fit (loosely) in this subject? Ideally the puzzles should be solved using some basic knowledge about the physics of light.

The puzzles should be safe for children, repeatable many times, understandable by the general public (after scientific explanation if necessary). Note that I have some budget to build some sort of large cubicle, or even decorate a whole room.

Some well-known tools for puzzles related to light:

• UV flashlight to reveal hidden messages


• (safe) laser pointers


• objects' shadows

Ideas on the scenario?

EDIT: let me add that I am also very bad at finding a scenario that would:

• make sense of the need for the group to escape


• make sense of the limited time


• make sense of the puzzles sitting around, waiting for a solution

... so any ideas on crafting a scenario around the puzzles would be great.

Technical issues

What size should the escape room take? I am worried about a room too small for having sufficient space for the people and the puzzles.

What time should it take and for how many people? Ten minutes would be the maximum time in order to allow many groups to try the escape room. But is this enough to solve puzzles?

Do you have experience in this kind of event? I welcome any general suggestions (organization, budget, technical problems).

Answer

I have no experience making escape rooms, so I can't help too much with the logistics or know if these ideas are practical.

Here are some puzzle ideas:

• 
  

  Use the science of color perception to hide a message in plain sight. Have a picture featuring lots of colored dots in a seemingly nonsense pattern. People to find a magenta gel, and when they put it over the light source, a code is revealed. The idea is that under a magenta light, the color pairs green/black, cyan/blue, yellow/red, and white/magenta become indistinguishable. If you're careful, you can have three distinct codes revealed using magenta, cyan and yellow filters, which would be cool if you had three combination locks whose colors corresponded to the filter you needed to see its code.

  
  


• 
  
  

  A laser maze, where they have to rotate mirrors to get a laser to shine on a certain place. Having a cheap fog machine nearby would reveal the full path of the laser, and be much more fun. Solving the maze causes the light to shine on a particular place on a wall, perhaps a wall with many lock combinations where only the one which is pointed at is correct.

  
  


• 
  

  Using refraction: a laser is shining through a glass club, barely missing a mirror. Filling the cup with water causes the beam to bend, so the laser hits a mirror and points out an interesting message.

  
  


• 
  

  Using polarized light: Make a grid of squares, where each square is a piece of polarizing film. Some of the squares are turned 90 degrees relative to the others, spelling out a message. This message is invisible under plain light, but when looked at through polarized sunglasses (which you can make yourself by attaching two pieces of the polarized film to empty frames), the rotated squares will appear black while the unrotated ones are still transparent. I got the idea from this how-to page.

  
  

None of these puzzles are that hard from a puzzling point of view, so they should be able to be done in 10 min time, especially if you have someone offer hints at the 5 min mark. Their main point is how they have surprising reveals and only work because of the science of light.

enigmatic puzzle - The Madman's Speech

You are walking through the prairie when you find a madman wandering around talking to himself. The following is what you manage to hear of his speech:

"How? I - I'll ask her. I owe her much, again. I'd a home on town, a tax as florid as out the coat, a virgin a year. Oh, yodel - aware you take all or I do. Never the road: I'll land in the Anna-Marie land. Main can's a sore gone; tennis is out t'car. Oh, line a canned turkey!"

What is the man really talking about?

Answer

I guess he's talking about

States in America.

"How? - I" = Hawaii
"I'll ask her." = Alaska
"I owe her" = Iowa
"much again" = Michigan
"I'd a ho..." = Idaho
"...me on town, a" = Montana

"tax as" = Texas
"florid a..." = Florida
"...s out the coat, a" = South Dakota
"virgin a year." = Virginia
"Oh yo..." = Ohio
"...del - aware" = Delaware
"You ta..." = Utah
"...ke all or i do" = Colorado
"Never the" = Nevada
"road: I'll land" = Rhode Island

"in the Anna-..." = Indiana
"...Marie land" = Maryland
"Main" = Maine
"can's a s..." = Kansas
"...ore gone;" = Oregon
"tennis i..." = Tennessee
"...s out t'car. Oh line a" = South Carolina
"canned Turkey" = Kentucky

logical deduction - The counterfeit coin

You have 9 coins. One of the nine is counterfeit. The counterfeit coin can be distinguished by weight - it is heavier than the rest.

Using a balance scale only twice, find the counterfeit coin.

Answer

Ternary search.

Split into groups of 3. Call them first3group, second3group, third3group.

Now weigh first3group vs second3group. If first test shows equal weights, the heavy coin is in third3group. Split third3group and weigh 2. Where one is heavier, QED, otherwise, it is the non-weighed one, QED.

Else take the heavier group of 3 and do a similar exercise as described above when left with 3, 1 of which is heavier.

QED

optics - Why does a circle of light appear when I shine a laser pointer at a wire?

When I point a laser at a wire , a circle of light appears on a wall behind it . why this happens.

What is the circle of light and why does it appear?

Could anyone give some tips on what I should search for or any references?

wire have diameter 3.17 mm

PS.I don't good at English,,, sorry

Wire perpendicular to the projection screen

Laser under on the wire Then born circle of light

special relativity - Why don't electromagnetic waves require a medium?

As I understand it, electromagnetic waves have two components which are the result of each other, i.e., when a moving electric charge creates a changing magnetic field at point X then a changing electric field is created at point Y and this repeating process is what creates EM waves, so therefore, it requires no medium. Is my understanding correct?

One thing that I'm surprised to know is that light is also called an electromagnetic wave.

Does this include light of any kind, for example: light from a bulb, a tube, and also from the Sun? How do they contain electric and magnetic fields?

mathematics - Pouring problem

You are in the possession of two bottles, one with a capacity of 7 litres and one with a capacity of 11 litres. Next to you is an infinitely large tub of water.

You need to measure exactly 2 litres in one of the bottles.

You are only allowed to entirely empty or fill the bottles. You can't fill them partially since there is no indication on the bottles saying how much liquid is in them.

How do you measure exactly 2 litres?

Answer

Fill the 7 litre bottle completely and pour the whole thing to the 11 litre bottle.
Fill the 7 litre bottle completety again and pour the water to the 11 litre bottle until it becomes full.
Now the 7 litre bottle contains 3 litres.
Empty the 11 litre bottle and then pour the 3 litres into the 11 litre bottle.
Fill the 7 litre bottle completely and pour the whole thing to the 11 litre bottle.
Fill the 7 litre bottle completely again and pour the water to the 11 litre bottle until it becomes full.
Now the 7 litre bottle contains 6 litres.
Empty the 11 litre bottle and then pour the 6 litres into the 11 litre bottle.
Fill the 7 litre bottle completely and pour the water to the 11 litre bottle until it becomes full.
Now, the 7 litre bottle contains 2 litres.

---

An other possible solution would be to start by filling the 11 litre bottle as shown below.

Einstein equation $E=mc^2$: Does it mean an object without mass does not have energy?

Einstein equation $E=mc^2$ where $E$ is energy, $m$ mass, and $c$ the speed of light in vacuum. So does it mean objects without any mass does not posses energy for eg lights photons does not have mass but how can they posses energy

electromagnetism - Does existence of magnetic monopole break covariant form of Maxwell’s equations for potentials?

Absence of magnetic charges is reflected in one of Maxwell's fundamental equations: $$\operatorname{div} \vec B = 0. \tag1$$ This equation allows us to introducte concept of vector potential: $$\vec B = \operatorname{rot} \vec A.$$ Using this concept, it is possible to express Maxwell's equations in a graceful symmetric form: $$\nabla^2 \vec A - \frac{1}{c^2}\frac{\partial^2 \vec A}{\partial t^2} = - \frac{\vec j}{\epsilon_0 c^2}, \tag2$$ $$ \nabla^2 \phi -\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = - \frac{\rho}{\epsilon_0}. \tag3 $$

Noticing, that vector $\vec A$ and scalar $\phi$ potentials, as well as electric current density $\vec j$ and charge density $\rho$, form a 4-vector in Minkovsky space-time. Therefore, Maxwell's equations can be expressed in a covariant form, using d'Alembertian: $$\nabla_{\mu}\nabla^{\mu} A_{\nu} = \frac{j_{\nu}}{\epsilon_0}. \tag4$$

If magnetic monopols exist, Maxwell's equation $(1)$ will look as: $$\operatorname{div} \vec B = \mu_0 c \rho_{\mathrm{magnet}}.$$

As the divergence of $\vec{B}$ isn't equal to zero, it impossible to introduce concept of vector potential. Thus, the equation in the form of $(4)$ will not be possible to express.

Answer

Another option, besides modifying the potential $A_\mu = (A_i, \phi)$ in some way, is to introduce another 4-potential $C_\mu = (C_i, \psi)$. Then the electric and magnetic field are given by $$E = - \nabla \times C - \frac{\partial A}{\partial t} - \nabla \phi$$ $$B = \nabla \times A - \frac{\partial C}{\partial t} - \nabla \psi$$

More on this 2-potential approach can be found here: http://arxiv.org/abs/math-ph/0203043

lagrangian formalism - Do an action and its Euler-Lagrange equations have the same symmetries?

Assume a certain action $S$ with certain symmetries, from which according to the Lagrangian formalism, the equations of motion (EOM) of the system are the corresponding Euler-Lagrange equations.

Can it happen that the equations of motion derived by this procedure have different kinds and/or numbers of symmetries than the action one has started with? And if yes, are there underlying principles that state why which kind of symmetries the action does not have can emerge in corresponding EOMs or which kind of symmetries of the action can potentially disappear in the EOMs derived from the Euler-Lagrange equations?

Answer

Setting. We are considering a transformation, which may transform the field variables $\phi^{\alpha}(x)$ and which may transform the space-time points $x^{\mu}$. The transformation in turn apply to

1. 
   

   The action $S_V[\phi]=\int_V \! d^nx~{\cal L} $.

   
   


2. 
   

   The Euler-Lagrange equations = the equations of motion (EOM).

   
   


3. 
   

   A solution $\phi$ of EOM.

   
   

Definition. If any of the items 1-3 are invariant under the transformation, we speak of a symmetry of the corresponding item 1-3 .

Definition. If a solution (3) doesn't have a symmetry that the EOM (2) have, we speak of a spontaneously broken symmetry.

Definition. Next let us recall the definition of an (off-shell$^1$) quasi-symmetry of the action. It means that the action changes by a boundary integral $$\tag{0.1} S_{V^{\prime}}[\phi^{\prime}] +\int_{\partial V^{\prime}} \!d^{n-1}x~(\ldots) ~=~S_V[\phi]+ \int_{\partial V} \!d^{n-1}x~(\ldots) $$ under the transformation.

Proposition. In general, if an action (1) has a quasi-symmetry, then the EOM (2) must have a symmetry (wrt. the same transformation), cf. e.g. this Phys.SE post.

Examples:

1. 
   

   One example is the Maxwell Lagrangian density (in vacuum without the $J^{\mu}A_{\mu}$ source term) $$\tag{1.1}{\cal L} ~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\frac{1}{2}(\vec{E}^2-\vec{B}^2), $$ which doesn't have electromagnetic $SO(2,\mathbb{R})$ duality symmetry $$\tag{1.2}(\vec{E}, \vec{B})\quad \longrightarrow \quad(\vec{E}\cos\theta - \vec{B}\sin\theta, \vec{B}\cos\theta + \vec{E}\sin\theta),$$ while the Euler-Lagrange equations (the Maxwell's equations in vacuum) are symmetric under electromagnetic duality.

   
   


2. 
   
   

   Another example is a non-relativistic free point particle where the Lagrangian $$\tag{2.1}L~=~\frac{1}{2}m\dot{q}^2$$ is not invariant under the Galilean symmetry $$\tag{2.2}\dot{q}\quad \longrightarrow \quad\dot{q}+v,$$ nor the dilation/scale symmetry $$\tag{2.3} q \quad \longrightarrow \quad \lambda q,$$ but the EOM $$\tag{2.4}\ddot{q}~=~0$$ is invariant. In the case of the Galilean symmetry (2.2), the Lagrangian changes by a total time derivative $$\tag{2.5} L \quad \longrightarrow \quad L +mv\frac{d}{dt}\left( q +\frac{vt}{2}\right).$$ See also this Phys.SE post. Thus (2.2) is actually an example of a quasi-symmetry of the action. [It is an instructive exercise to derive the corresponding Noether charge $Q$. At the infinitesimal level, the Galilean transformation (2.2) reads $$ \tag{2.6}\delta \dot{q}~=~\delta v~=~\varepsilon, \qquad \delta q~=~\varepsilon t,\qquad \delta L ~=~ \varepsilon\frac{df}{dt}, \qquad f ~:=~mq. $$ The bare Noether charge is $$ \tag{2.7} Q^0~=~t \frac{\partial L}{\partial \dot{q}}~=~t m\dot{q}, $$ while the full Noether charge is $$ \tag{2.8}Q~=~Q^0-f~=~m(\dot{q}t-q),$$ which is conserved on-shell, cf. Noether's Theorem. The (non-relativistic) Galilean boosts generator (2.8) should be compared to the (relativistic) Lorentz boosts generators $tP-xE$ in relativistic theories, cf. e.g. this Phys.SE post.]

   
   


3. 
   

   The dilation/scale transformation $$\tag{3.1} q \quad \longrightarrow \quad \lambda q, $$ is not a quasi-symmetry of the Lagrangian action $$ \tag{3.2} S[q]~= ~\int\! dt ~L, \qquad L ~=~\frac{m}{2}\dot{q}^2- \frac{k}{2}q^2, $$ for the simple harmonic oscillator (SHO), but it is a symmetry of the EOM $$\tag{3.3} m\ddot{q}~=~-kq. $$

   
   


4. 
   

   The dilation/scale transformation $$\tag{4.1} q \quad \longrightarrow \quad \lambda q, \qquad p \quad \longrightarrow \quad \lambda p, $$ is not a quasi-symmetry of the Hamiltonian action $$ \tag{4.2} S_H[q,p]~= ~\int\! dt ~L_H, \qquad L_H ~=~p\dot{q}-H, \qquad H ~=~\frac{p^2}{2m}+ \frac{k}{2}q^2, $$ for the SHO, but it is a symmetry of Hamilton's EOM $$\tag{4.3} p~=~m\dot{q} , \qquad \dot{p}~=~-kq. $$

   
   


5. 
   

   The EOM of the SHO $$\tag{5.1} m\ddot{q}~=~-kq $$ is not invariant under the temporal symmetry $$\tag{5.2} t \quad \longrightarrow \quad \lambda t,\qquad \lambda~\neq~\pm 1,$$ but the trivial solution $q=0$ is.

   
   
   

--

$^1$ Here the word off-shell indicates that the EOM are not assumed to hold under the specific transformation. In case of continuous transformations, if we assume the EOM to hold, then any infinitesimal variation of the action is trivially a boundary integral.

quantum field theory - Gupta-Bleuler and Lorenz Gauge: I don't understand the principle behind Gupta-Bleuler

I would like to make the link between the Gupta-Bleuler Lagrangian and the Lorenz Gauge for Electromagnetism because everything is not clear to me.

I am looking for a simple explanation without too many references if possible.

---

So, from what I have understood :

We have $$\mathcal{L}_0=-\frac 14 F_{\mu \nu} F^{\mu \nu}$$

The problem with this Lagrangian is that we have $\Pi_0=0$: the momentum associated to $A_0$ is $0$.

Also, the field equations are:

$$ \Box A_{\mu}-\partial_\mu \partial^{\nu} A_\nu=0$$

And there is no propagator associated with this equation (no Green function).

Two problems here. The momentum associated to $A_0$ is $0$ and there is no propagator associated with the equation of motion.

The "trick" we do is to add a term $\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2$ to the Lagrangian.

When we do it, the Lagrangian is now $$\mathcal{L}=-\frac 14 F_{\mu \nu} F^{\mu \nu}+\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2$$

---

Then, we can show if we take $\xi=1$, and do a few integrations by parts, that this new Lagrangian is equivalent to $\mathcal{L}=\partial_\mu A_\nu \partial^\mu A^\nu$ and thus, will give the equations of motion :

$$ \Box A_{\mu} = 0$$

Then, we can quantize everything as the Klein Gordon field.

$$A_\mu=\int d\widetilde{k}\left(A_\mu(k)^+ e^{ikx} + A_\mu(k)^- e^{-ikx}\right) $$

And we implement the Lorenz Gauge by saying $\partial_\mu A^{\mu, +} |\psi_\text{Phys} \rangle=0$ where $|\psi_\text{Phys} \rangle$ are "physical" states of our Hilbert space. (we define them like this).

---

My questions:

I don't understand the global spirit of this. For me we modified the dynamics because we modified the Lagrangian. I think everything relies upon the fact that we work with an equivalent Lagrangian as the one we started with but I don't find where this is true? Why can we modify our Lagrangian like this?

I would like the most simple answer even if it is not very rigorous, I just would like to get the average Idea of the thing.

Answer

The sketch of the philosophy:

1. 
   
   

   You modified the Lagrangian by introducing the $\xi$ term. This shouldn't really bother you because the Lagrangian is not measurable, so there is, in principle, nothing wrong about modifying it. But the Lagrangian does tell you about the underlying theory, so it is understandable that you don't like the gauge-fixing term.

   
   


2. 
   

   What is measurable is the $S$-matrix. To define this matrix you have to specify what states we are allowed to scatter off each other. This is where the Gupta-Bleuler condition comes in. The in- and out-states are restricted to be those that satisfy $\langle a|\partial\cdot A|b\rangle\equiv 0$. You set up the $S$-matrix by evolving these states in time, $$ S_{1\cdots n;1'\cdots n'}\equiv \langle a_1\dots a_n;\text{in}|b_{1'}\dots b_{n'};\text{out}\rangle $$

   
   


3. 
   

   Using the Ward-Takahashi identity, you can prove that $S$ is actually independent of $\xi$. This is a non-trivial statement because $S$ is calculated using $\xi$-dependent objects. This way, we conclude that the gauge-fixing term has not really affected the theory, and its introduction has been nothing but a formal tool. It would be nice to be able to set up the $S$ matrix directly, without the need to use non-physical auxiliary objects. This is not really possible, as one can learn from several no-go theorem (that we shall not discuss here).

   
   


4. 
   

   Formally, you can eliminate the unwanted term $\frac{1}{2\xi}(\partial\cdot A)^2$ by taking the so-called unitary limit, $\xi\to\infty$. As $S$ is $\xi$ independent, you may want to argue that you have taken the unitary limit, thus getting rid of the unwanted term, while at the same time taking $\xi=1$, thus simplifying the evaluation of $S$. The 't Hooft-Feynman gauge is more convenient from the practical point of view, and the unitary one is more convincing from the Lagrangian point of view. Gauge invariance is essential if these different POVs are to coexist.

   
   
   


5. 
   

   This formalism is admittedly ad-hoc. Its ingredients can be motivated but not in a completely convincing manner. Moreover, it is not really possible to extend it to other gauge theories. In this sense, Gupta-Bleuler is best thought of as a formalism that is proved to work a posteriori, but that should not be pursued beyond what an historical analysis is concerned. Modern methods -- being more general and systematic -- are preferred (cf. the BV-BRST formalism; for the equivalence to Gupta-Bleuler in the abelian case, see this PSE post).

   
   

---

Note: the recent paper 1801.06311 argues that there is an equivalent formalism that rests on a different philosophy: one may take the gauge-fixed Lagrangian as fundamental, and forget that it comes from a gauge-invariant theory. In this scenario, the Hamiltonian is in principle not positive-definite. By insisting that $H\ge0$ we essentially get the Gupta-Bleuler condition, so the latter may be understood as the requirement for the theory either to contain only physical states, or for it to contain only positive-energy states. The two points of view are equivalent.

thermodynamics - Why a mono-atomic crystal layer (2D) can't be stable?

According to Peierls and Landau, 2D crystals were thermodynamically unstable. They can't exist! Of course, this theory was disapproved in 2004 (example: graphene).

What is the general definition of stability of a general system?

What is the thermodynamics' stability?

classical mechanics - The quantum analogue of Liouville's theorem

In classical mechanics, we have the Liouville theorem stating that the Hamiltonian dynamics is volume-preserving.

What is the quantum analogue of this theorem?

Answer

It's subtle. The theorem is not there: quantum flows are compressible (Moyal, 1949).

I'll follow Ch. 0.12 of our book, Concise Treatise of Quantum Mechanics in Phase Space, 2014.

The analog of the Liouville density of classical mechanics is the Wigner function in phase space quantum mechanics. Its evolution equation (generalizing Liouville's) is $$ {\partial f \over \partial t} =\{\!\!\{H ,f\}\!\!\}~, $$ where the double (Moyal) brackets indicate a celebrated quantum modification of Poisson brackets by terms of $O(\hbar^2)$, and serve to prove Ehrenfest's theorem for the evolution of expectation values.

For any phase-space function $k(x,p)$ with no explicit time-dependence, $$ \begin{eqnarray} \frac{d\langle k \rangle }{dt} &=& \int\! dx dp~ {\partial f \over \partial t} k \nonumber \\ &=& {1\over i\hbar} \int\! dx dp~ ( H\star f- f\star H)\star k \nonumber\\ &=& \int\! dx dp~ f \{\!\!\{k,H\}\!\!\} = \left \langle \{\!\!\{ k,H\}\!\!\} \right \rangle , \end{eqnarray} $$ where the star product and its manipulations are detailed in said text.

Moyal stressed (discovered?) that his eponymous quantum evolution equation above contrasts to Liouville's theorem (collisionless Boltzmann equation) for classical phase-space densities, $$ {d f_{cl}\over dt}= {\partial f_{cl} \over \partial t} + \dot{x} ~\partial_x f_{cl} + \dot{p}~ \partial_p f_{cl} =0~. $$

Specifically, unlike its classical counterpart, in general, $f$ does not flow like an incompressible fluid in phase space, thus depriving physical phase-space trajectories of meaning, in this context. (Only the harmonic oscillator evolution is trajectoral, exceptionally.)

For an arbitrary region $\Omega$ about some representative point in phase space, the efflux fails to vanish, $$ \begin{eqnarray} {d \over dt}\! \int_{\Omega}\! \! dx dp ~f&=& \int_{\Omega}\!\! dx dp \left ({\partial f \over \partial t}+ \partial_x (\dot{x} f) + \partial_p (\dot{p} f ) \right)\\ &= & \int_{\Omega}\! \!\! dx dp~ (\{\!\!\{ H,f\}\!\!\}-\{H,f\})\neq 0 ~.\nonumber \end{eqnarray} $$

That is, the phase-space region does not conserve in time the number of points swarming about the representative point: points diffuse away, in general, at a rate of O($\hbar^2$), without maintaining the density of the quantum quasi-probability fluid; and, conversely, they are not prevented from coming together, in contrast to deterministic (incompressible flow) behavior.

Still, for infinite $\Omega$ encompassing the entire phase space, both surface terms above vanish to yield a time-invariant normalization for the WF.

The $O(\hbar^2)$ higher momentum derivatives of the WF present in the MB (but absent in the PB---higher space derivatives probing nonlinearity in the potential) modify the Liouville flow into characteristic quantum configurations. So, negative probability regions moving to the left amount to probability flows to the right, etc... Wigner flows are a recondite field, cf. Steuernagel et al, 2013.

For a Hamiltonian $H=p^2/(2m)+V(x)$, the above evolution equation amounts to an Eulerian probability transport continuity equation, $$\frac{\partial f}{\partial t} +\partial_x J_x + \partial_p J_p=0~,$$ where, for $\mathrm{sinc}(z)\equiv \sin z/~ z$ , the phase-space flux is $$J_x=pf/m~ ,\\ J_p= -f \mathrm{sinc} \left( {\hbar \over 2} \overleftarrow {\partial _p} \overrightarrow {\partial _x} \right)~~ \partial_x V(x). $$
Note added. For a recent discussion/proof of the zeros, singularities,and negative probability density features, hence the ineluctable violations of Liouville's theorem in anharmonic quantum systems see Kakofengitis et al, 2017.

homework and exercises - Shouldn't the work function of a metal increase when photo electrons are ejected from it?

"When a metal is irradiated with radiation of suitable frequency, photo electrons are ejected from it, making the metal (slightly) positively charged. So more energy is required to eject more electrons from the same metal. Hence the the work function should increase"

Please mention any flaw in the above paragraph.

Answer

Well, if the metal is connected to something else (as it is in most experiments, often to ground) then the induced charge will pull electrons from ground, which will restore the neutral state.

If the metal is completely isolated, then yes, you will induce some charge, but it's important to keep an eye on the significance: that charge is diluted over a macroscopically-sized sample, and in most cases it will result in a negligible change in the work function. In less-realistic cases, if you just keep going and going, then yes, the work function will start some meaningful increase at some point.

newtonian mechanics - Why is it that a force is not required for a body to move at constant velocity?

A body continues in its state of motion unless a force is applied to it. But how does an object stay in motion in the first place? A force must have caused it to move right?

Answer

While you know the statement of Newton's first law, Newton's second law can be used to answer your question. Newton's second law is stated mathematically as $$\vec{F} = m\vec{a}$$

This statement tells us that, for a given object of mass $m$, the acceleration of that object (whether it speeds up, slows down, or travels at a constant velocity), is directly proportional to the net force applied to it. Thus, if a net force is applied to an object initially at rest, the object will accelerate in the direction of the force (so yes, a force is required for an object to begin moving initially). However, if the force is then no longer applied to the object, then $\vec{a} = \frac{\vec{F}}{m} = \frac{0\text{N}}{m} = 0 \frac{\text{m}}{\text{s}^2}$. Thus, the object's speed will not change. So, in short, "staying in motion" (traveling at a constant velocity and following a straight line) is the "default" action of a given object. If the object was speeding up or slowing down, $\vec{F} \neq 0$.

What justifies dimensional analysis?

Dimensional analysis, and the notion that quantities with different units cannot be equal, is often used to justify very specific arguments, for example, you might use it to argue that a particular formula cannot possibly be the correct expression for a particular quantity. The usual approach to teaching this is to go "well kids, you can't add apples and oranges!" and then assume that the student will just find it obvious that you can't add meters and seconds.

I'm sorry, but... I don't. I'm not convinced. $5$ meters plus $10$ seconds is $15$! Screw your rules! What are the units? I don't know, I actually don't understand what that question means.

I'm specifically not convinced when this sort of thing is used to prove that certain formulae can't possibly be right. Maybe the speed of a comet is given by its period multiplied by its mass. Why not? It's a perfectly meaningful operation - just measure the quantities, multiply them, and I claim that the number you get will always equal the current speed of the comet. I don't see how "but it doesn't make sense to say mass times time is equal to distance divided by time" can be a valid counterargument, particularly because I don't really know what "mass times time" is, but that's a different issue.

If it's relevant, I'm a math student and know extremely little about physics.

Answer

Physics is independent of our choice of units

And for something like a length plus a time, there is no way to uniquely specify a result that does not depend on the units you choose for the length or for the time.

Any measurable quantity belongs to some set $\mathcal{M}$. Often, this measurable quantity comes with some notion of "addition" or "concatenation". For example, the length of a rod $L \in \mathcal{L}$ is a measurable quantity. You can define an addition operation $+$ on $\mathcal{L}$ by saying that $L_1 + L_2$ is the length of the rod formed by sticking rods 1 and 2 end-to-end.

The fact that we attach a real number to it means that we have an isomorphism $$ u_{\mathcal{M}} \colon \mathcal{M} \to \mathbb{R}, $$ in which $$ u_{\mathcal{M}}(L_1 + L_2) = u_{\mathcal{M}}(L_1) + u_{\mathcal{M}}(L_2). $$ A choice of units is essentially a choice of this isomorphism. Recall that an isomorphism is invertible, so for any real number $x$ you have a possible measurement $u_{\mathcal{M}}^{-1}(x)$. I'm being fuzzy about whether $\mathbb{R}$ is the set of real numbers or just the positive numbers; i.e. whether these are groups, monoids, or something else. I don't think it matters a lot for this post and, more importantly, I haven't figured it all out.

Now, since physics should be independent of our choice of units, it should be independent of the particular isomorphisms $u_Q$, $u_R$, $u_S$, etc. that we use for our measurables $Q$, $R$, $S$, etc. A change of units is an automorphism of the real numbers; given two units $u_Q$ and $u'_Q$, the change of units is $$ \omega_{u,u'} \equiv u'_Q \circ u_Q^{-1}$$ or, equivalently, $$ \omega_{u,u'} \colon \mathbb{R} \to \mathbb{R} \ni \omega(x) = u'_Q(u_Q^{-1}(x)). $$ Therefore, \begin{align} \omega(x+y) &= u'_Q(u_Q^{-1}(x+y)) \\ &= u'_Q(u_Q^{-1}(x)+u_Q^{-1}(y)) \\ &= u'_Q(u_Q^{-1}(x)) + u'_Q(u_Q^{-1}(y)) \\ &= \omega(x) + \omega(y). \end{align}

So, since $\omega$ is an automorphism of the reals, it must be a rescaling $\omega(x) = \lambda x$ with some relative scale $\lambda$ (As pointed out by @WetSavannaAnimalakaRodVance, this requires the weak assumption that $\omega$ is a continuous function -- there are everywhere discontinuous solutions as well. Obviously units aren't useful if they are everywhere discontinuous; in particular, so that instrumental measurement error maps an allowed space of $u_\mathcal{M}$ into an interval of $\mathbb{R}$. If we allow the existence of an order operation on $\mathcal{M}$, or perhaps a unit-independent topology, this could be made more precise.).

Consider a typical physical formula, e.g., $$ F \colon Q \times R \to S \ni F(q,r) = s, $$ where $Q$, $R$, and $S$ are all additive measurable in the sense defined above. Give all three of these measurables units. Then there is a function $$ f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R} $$ defined by $$ f(x,y) = u_S(F(u_Q^{-1}(x),u_R^{-1}(y)). $$

The requirement that physics must be independent of units means that if the units for $Q$ and $R$ are scaled by some amounts $\lambda_Q$ and $\lambda_R$, then there must be a rescaling of $S$, $\lambda_S$, such that

$$ f(\lambda_Q x, \lambda_R y) = \lambda_S f(x,y). $$

For example, imagine the momentum function taking a mass $m \in M$ and a velocity $v \in V$ to give a momentum $p \in P$. Choosing $\text{kg}$ for mass, $\text{m/s}$ for velocity, and $\text{kg}\,\text{m/s}$ for momentum, this equation is $$ p(m,v) = m*v. $$ Now, if the mass unit is changed to $\text{g}$, it is scaled by $1000$, and if the velocity is changed to $\text{cm/s}$, it is scaled by $100$. Unit dependence requires that there be a rescaling of momentum such that $$ p(1000m,100v) = \lambda p(m,v). $$ This is simple -- $10^5 mv = \lambda mv$ and so $\lambda = 10^5$. In other words, $$ p[\text{g} \, \text{cm/s}] = 10^5 p[\text{kg} \, \text{m/s}]. $$

Now, let's consider a hypothetical situation where we have a quantity called "length plus time", defined that when length is measured in meters and time in seconds, and "length plus time" in some hypothetical unit called "meter+second", the equation for "length plus time" is $$ f(l,t) = l + t. $$ This is what you've said - $10 \text{ m} + 5 \text{ s} = 15 \text{ “m+s”}$. Now, is this equation invariant under a change of units? Change the length scale by $\lambda_L$ and the time scale by $\lambda_T$. Is there a number $\Lambda$ such that $$ f(\lambda_L l, \lambda_T t) = \lambda_L l + \lambda_T t $$ is equal to $$ \Lambda f(l,t) = \Lambda(l+t) $$ for all lengths and times $l$ and $t$? No! Therefore, this equation $f = l + t$ cannot be a valid representation in real numbers of a physical formula.

newtonian mechanics - Why is the work done by a rocket engine greater at higher speeds?

From this comment by orlp:

If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?

Answer

The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?

To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).

Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.

So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?

Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.

At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.

What characteristics of a ciphertext can be indicators of a particular cipher?

If I have a ciphertext, what information about the probable ciphers used can I infer from the ciphertext itself? For example, information from:

• the length


• the spacing / grouping of characters


• the range of characters used


• frequency counts

and any other information that is intrinsic to the ciphertext itself.

I'm talking about pencil-and-paper ciphers of the kind often used in cryptogram puzzles, rather than real, 'strong', modern forms of crypto.

mathematics - Why are radians more natural than any other angle unit?

I'm convinced that radians are, at the very least, the most convenient unit for angles in mathematics and physics. In addition to this I suspect that they are the most fundamentally natural unit for angles. What I want to know is why this is so (or why not).

I understand that using radians is useful in calculus involving trigonometric functions because there are no messy factors like $\pi/180$. I also understand that this is because $\sin(x) / x \rightarrow 1$ as $x \rightarrow 0$ when $x$ is in radians. But why does this mean radians are fundamentally more natural? What is mathematically wrong with these messy factors?

So maybe it's nice and clean to pick a unit which makes $\frac{d}{dx} \sin x = \cos x$. But why not choose to swap it around, by putting the 'nice and clean' bit at the unit of angle measurement itself? Why not define 1 Angle as a full turn, then measure angles as a fraction of this full turn (in a similar way to measuring velocities as a fraction of the speed of light $c = 1$). Sure, you would have messy factors of $2 \pi$ in calculus but what's wrong with this mathematically?

I think part of what I'm looking for is an explanation why the radius is the most important part of a circle. Could you not define another angle unit in a similar way to the radian, but with using the diameter instead of the radius?

Also, if radians are the fundamentally natural unit, does this mean that not only $\pi \,\textrm{rad} = 180 ^\circ$, but also $\pi = 180 ^\circ$, that is $1\,\textrm{rad}=1$?

Answer

Angles are defined as the ratio of arc-length to radius multiplied by some constant $k$ which equals one in the case of radians, $360/2\pi$ for degrees. What you're effectively asking is what's natural about setting $k$ = 1? Again it's tidyness as pointed out in dmckee's alternative answer.

rubiks cube - Simple 4x4x4 OLL parity algs?

I am learning to solve a 4x4x4 Rubik's cube and I think that the beginner OLL parity alg:

(r U2 x r U2 r' U2 l U2 r' U2 r U2 r' U2 r')

is a little too complex.

Does anyone know a simpler alg for it?

fluid dynamics - Pseudo Force Direction

A cork and a metal Bob are connected by a string as shown in the figure. In case the beaker is given an acceleration towards left, the cork moves towards ..?

I answered that the cork moves towards right because of a pseudo force but the answer key says that it moves towards left due to a pseudo force. But if you look at it, it seems intuitive that the pseudo force gets applied to the right, isn't it?

riddle - And now I'm sitting in my Frozen Chamber

Many have seen my crown,
but none have seen my throne.
I have hundreds in my harem,

but none of them is my own.

My people work from sunrise to sunset,
just as I commanded.
And sometime I just sit on the roof
and watch where the dandelions landed.

Answer

I think it is

A rooster

Many has seen my crown, but none have seen my throne.

A rooster has a crown on it's head, but no throne

I have hundreds in my harem, but none of them is my own.

All the hens in the "harem" belong to the farmer, not the rooster.

My people work from sunrise to sunset, just as I commanded.

The rooster crowing at sunrise used to be the alarm for the people to wake up and start working.

And sometime I just sit on the roof and watch where the dandelions landed.

Weather vanes on the top of barn roofs are sometimes shaped like a rooster. They point in direction of wind, and the dandelions will land in the direction of wind. So they see the dandelions land.

And now I'm sitting in my Frozen Chamber

The rooster is culled and put in a freezer.

experimental physics - Photomultiplier/ voltage divider troubleshooting

I have a Photonis XP5301 PMT to use for fast neutron detection (detector optically glued to the PMT window). I can't quite figure out why its not producing a spectrum. First of all, I have multiple different values of operating voltage that I found on the internet, but I think it's either -800 V or - 1000V. I used the same setup with a NaITl scintillation detector + attached PMT, and it gave me proper results, but not mine. What could I do to locate the issue? The Voltage divider I'm using is a Photonis VD202K/40. ALso, while the NaITl detector's PMT has a power cable that connects to the amplifier, my PMT doesnt have a power cable. What could be the issue here? What am I not addressing? Thank you!

Answer

You say you are using the setup for a neutron detector.

What is the material you are using? What is the expected light output (in photons / event) and how fast would they appear? When you use the NaI based detector, what is your source? What is the activity?

Here are the troubleshooting steps I would take:

1. To make sure the PMT is working, I would hook the output (straight from the PMT anode) up to a scope, into a low-ish impedance (1 k to ground). Turn the gain of the scope to maximum (or 10 mV / cm). Slowly increase the operating voltage (while PMT is in darkness); eventually, you should see "grass" on the baseline of the scope: this is the signal from individual (thermally induced) photoelectrons.



2. Turn the voltage down about 100 V from the point where noise started appearing, and introduce a very weak flashing LED into the enclosure of the PMT (if you can see the flash with the naked eye it's too bright for the PMT). You should now see a massive spike of signal for every flash of the LED. Turn the HV off and demonstrate to yourself that the signal is gone (so there is no cross talk)


3. Now connect the PMT output to a shaping amplifier (it's not clear whether you have one - if you don't, then that might be your problem). The shaping amplifier should give a "slow" output at every pulse coming in - this is usually needed as the input to a spectrum analyzer. Check the output of the amplifier on your scope (you will need a much less sensitive setting). Again, using the LED could be very helpful here. If you don't get a signal from the amplifier, check its power etc.


4. Now that you have a PMT with roughly appropriate voltage and a working amplifier, connect to the spectrum analyzer. Introduce a robust source (so you get good signal) and slowly sweep the HV in the range you found above, until you see triggers on the spectrum

When you have done all the above, and you found a place where things don't work as expected, you've likely narrowed your problem significantly. If you still have trouble, leave a comment.

homework and exercises - QFT Propagator across spacelike separation

I have this general formula for the propagator.

$$ D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\omega_k}[e^{-i[\omega_kt-\vec k\cdot\vec x]}\Theta(t)+e^{i[\omega_kt-\vec k\cdot\vec x]}\Theta(-t)] $$

I am supposed to verify that it decays exponentially for spacelike separation. To impose this condition I use $t=0$ and $x>0$. The symbols $\vec k$ and $\vec x$ are 3-vectors and $\omega_k=+\sqrt{\vec k^2+m^2}$. Also, for the Heaviside step function $\Theta(0)=0.5$.

This is problem 1.3.1 in Zee's QFT in a Nutshell book, and the solution to this problem is in the back of the book. He writes that the first step should look like

$$ D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}e^{-i\vec k\cdot\vec x} $$

but when I plug it in the values for $x$ and $\Theta$, I get

$$ D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}\dfrac{1}{2}[e^{i\vec k\cdot\vec x}+e^{-i\vec k\cdot\vec x}]=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}\cos(\vec k\cdot\vec x) $$

What is the reasoning behind throwing out one of the exponentials instead of using both of them? I haven't done my integral yet, so I don't know if it still produces exponential decay, but it seems like won't, so I want to know what the discrepancy is before I start.

classical mechanics - On Superposition In Diffraction

I am a high school student learning Physics. I am currently learning about diffraction of waves, be it water waves, light waves each, in systems where there are 2 slits for it to diffract through.

I have 2 questions about this diffraction:

1. 
   

   Where are the peaks and where are the troughs of the waves in the diagrams? Are the blank spaces the troughs and the red lines the peaks?

   
   


2. 
   
   

   I understand that constructive interference occurs when the peaks of 2 waves align, and destructive interference when the peak of the wave hits the trough of another wave. But why in the diagram above does constructive interference occur in the center (represented by bright)? And why does destructive interference occur on the 2 sides of the center (represented by dark), followed by another 2 zones of constructive interference (represented by bright)?

   
   

Can all this be explained at the level of a high school student, and without Calculus based mechanics?

Answer

1. Yes, the red lines represent the peaks, called wavefronts. The spaces between them are troughs.


2. 
   

   Constructive interference occurs when two wavefronts coincide (their amplitudes add up). Destructive interference occurs when a wavefront coincides with the trough (their amplitudes cancel out). This is established by calculating the path difference. The path difference is the difference in distance travelled by the two waves to the screen. Assuming that the two sources are coherent with each other, meaning that they are in phase, the path difference then determines the difference in phase when the light arrives at a point on the screen.

   
   

   We assume the screen is far away compared to the slit separation (which is the standard assumption). Let the slit separation be $d$, and look at a point on the screen A (it might be helpful to draw it out). Let the line connecting point A to the center of the slits make an angle $\theta$ with the line connecting the center of the screen to the center of the slits. Connecting each of the slits to point A and using similar triangles, for constructive interference, We have $$d\, sin \theta=n\lambda,\;\; n \in \mathbb Z^+$$ where $\lambda$ is the wavelength. Let me explain this formula. $d\, sin \theta$ is the difference in distance traveled by light from the 2 slits to point A. $n \in \mathbb Z^+$ because the two coherent beams must differ by whole number of wavelengths for the wavefronts to meet again. Similarly, for destructive interference, we have $$d\, sin \theta=(n + \frac{1}{2})\lambda,\;\; n \in \mathbb Z^+$$ because their distances must differ by half a wavelength for their amplitudes to cancel out. This can be used to further derive the fringe widths.

   
   
   

I hope I have explained it simply enough.