quantum field theory - Calculating the commutator of Pauli-Lubanski operator and generators of Lorentz group

The Pauli-Lubanski operator is defined as $${W^\alpha } = \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta}{M_{\mu \nu }},\qquad ({\varepsilon ^{0123}} = + 1,\;{\varepsilon _{0123}} = - 1)$$ where $M_{\mu\nu}$ is the generators of Lorentz group.

The commutation relation between generators of Poincare group is know as $$i[{M^{\mu \nu }},{M^{\rho \sigma }}] = {\eta ^{\nu \rho }}{M^{\mu \sigma }} - {\eta ^{\mu \rho }}{M^{\nu \sigma }} - {\eta ^{\mu \sigma }}{M^{\rho \nu }} + {\eta ^{\nu \sigma }}{M^{\rho \mu }},$$ $$i[{P^\mu },{M^{\rho \sigma }}] = {\eta ^{\mu \rho }}{P^\sigma } - {\eta ^{\mu \sigma }}{P^\rho }.$$

I try to derive the commutator between Pauli-Lubanski operator and a generator of Lorentz group, which is also given in our lecture $$i[{W^\alpha },{M^{\rho \sigma }}] = {\eta ^{\alpha \rho }}{W^\sigma } - {\eta ^{\alpha \sigma }}{W^\rho }.\tag{*}$$

But I only get $$\begin{align} i[{W^\alpha },{M^{\rho \sigma }}] =& i[\frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }{M_{\mu \nu }},{M^{\rho \sigma }}] \\ =& \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }\left( {\delta _\nu ^\rho {M_\mu }^\sigma - \delta _\mu ^\rho {M_\nu }^\sigma - \delta _\mu ^\sigma {M^\rho }_\nu + \delta _\nu ^\sigma {M^\rho }_\mu } \right) \\ & + \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}\left( {\delta _\beta ^\rho {P^\sigma } - \delta _\beta ^\sigma {P^\rho }} \right)M_{\mu\nu}. \end{align}$$

Obviously I can contract out the delta's, but that does not get me any closer to the simpler result of (*). Can anyone point out what to do next?

How did LIGO scientists determine the source of the gravitational waves they detected?

I read the newspaper today and I found that LIGO had detected gravitational waves. This is from their official website:

For the first time, scientists have observed ripples in the fabric of spacetime called gravitational waves, arriving at the earth from a cataclysmic event in the distant universe. This confirms a major prediction of Albert Einstein’s 1915 general theory of relativity and opens an unprecedented new window onto the cosmos.

How did they know the source of the gravitational wave was a binary black-hole merging event? Wikipedia says:

Measurable emissions of gravitational waves are expected from binary systems (collisions and coalescences of neutron stars or black holes), supernova explosions of massive stars (which form neutron stars and black holes), accreting neutron stars, rotations of neutron stars with deformed crusts, and the remnants of gravitational radiation created by the birth of the universe.

There are more possible sources. Can the LIGO detect black holes as well?

Answer

Firstly, they were able to calculate the masses of the two objects involved and their sizes. This was an indicator that they were compact objects since the masses were two objects roughly a few hundred kilometres across and weighing about 30 times the mass of the sun. This is a strong indication that such an object is a black hole.

Secondly, like Gabi Gonzalez mentioned in the announcement, the waves detected actually tell a lot more than you can actually read off. In particular, the model that fits the observed data the best is the model of a binary black hole merger.

There is no direct proof that can confirm that these were black holes because a good estimate of the location is not known. If it were known, then the telescopes would be turned to view that patch of the sky and most likely will see nothing, confirming that the gravitational waves were from black holes.

photons - How does the frequency of a particle manifest itself?

In terms of wave-particle duality for, let's say a photon; how would the frequency practically manifest/demonstrate itself? Like, i understand that the frequency is related to the energy a particle has, but frequency in my mind suggests oscillation about a point. Is the photon physically oscillating through space as it travels? I wouldnt imagine so. Which periodic occurrence is referred to when one talks about the frequency of a particle?

Answer

Is the photon physically oscillating through space as it travels? I wouldnt imagine so. Which periodic occurrence is referred to when one talks about the frequency of a particle?

No the photon is not oscillating through space. It is an elementary particle of the standard model which is the quantum mechanical description of most of our experimental knowledge on elementary particle to date. Elementary particles are point particles.

The classical wave is built up from an enormous number of photons and as physics theories have to be consistent when the parameters and variables change from microscopic to macroscopic the frequency entering the E=h*nu for the individual photon is the frequency built up in a coherent classical beam emerging from a large ensemble of photons.

Conversely, when one starts with a classical coherent beam of frequency nu, as discussed in the other answer, and one goes to the microscopic level of individual quanta that compose the beam, the photons, the frequency identifies the quantum of energy the photon carries.

For example, a large number of excited atoms at the same energy level ( lasers for example) de exciting to photons will have the energy h*nu that builds up the classical wave.

For what is a particle in the quantum mechanical microcosm see my answer here.

electromagnetism - Simple explanation to the induction from the slowly changing $vec B$ of a solenoid in the region of $0$ magnetic field

I would like to get some elementary intuition into the problem a solenoid fed with a time-dependent current, and the resulting current that such the solenoid field would induce in a loop completely outside the solenoid.

The classic situation is to imagine a perfect solenoid of $n_1$ turns per meter, having radius $a$ and fed by a current $I(t)=I_0e^{-t/\tau}$. One places $N_2$ coils (all having radius $R>a$) around the solenoid as illustrated by this picture from Haliday&Resnick 10th Edition.

Using the usual Ampère's law argument shows the $\vec B$ field of the solenoid would be completely contained inside the solenoid, and homogenous across the cross section of the solenoid.

If there is a slowly varying current in the solenoid, will there be an induced current?

Presumably the answer is yes (at least according the HR solution manual): if we take the $\vec B$ field of the solenoid to be changing with time as a result of the current changing in time, but $\vec B$ still homogenous across its cross-section and $0$ outside, there is a change in the flux of this field through the surface bounded by the coils, and even though there is no magnetic field outside the solenoid where the coils are located. The resulting change in flux induces an EMF independent of $R$, the radius of the coils, or for that matter the shape of the coil, provided it completely contains the solenoid.

If you are suspicious about the use of Ampère's law for slowly varying currents, the same solenoid field is obtained explicitly by Das Gupta in "Magnetic field due to a solenoid." American Journal of Physics 52 (1984): 258-258, starting from Biot-Savart, which certainly holds for quasi-static currents.

But how can one intuitively grasp that no field outside the solenoid induces a current in a stack of coils located completely outside this solenoid, in the region where $\vec B=0$?

A "classic" explanation might that the field might be $0$ but the vector potential is $\ne 0$, yet it seems this is invoking a lot of heavy machinery for a 1st year physics problem. Moreover, Aharonov-Bohm-like explanations are really quantum in their nature and show that in quantum mechanics the potentials are the essential quantities.

Nota: a possible path of solution would be to invoke hidden momentum, along the argument in this file by K.T. McDonald. Is there a simpler explanation?

double slit experiment - Delayed Choice Quantum Eraser?

Note that I'm looking for a layman explanations for these layman ideas.

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Introduction

Can someone help me understand why my idea wouldn't work?

Explain the the Delayed Choice Quantum Eraser?

First, I saw this interpretation of the experiment: Video I (at $4:30$)

Then, an idea of how to "cheat on lottery", by sending information to the past: Video II

Then, explained why it would fail here: Video III (at $4:00$)

But I have another idea of how to try something like that.

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Theoretical Experiment - predict future

Leave the board on the earth, and the detectors A & B on the moon. Let's say you can send the first pairs to the earth board and their pairs to the moon detectors.

If you turn the machine, you see a scattered pattern, meaning that there are detectors on moon that will detect it few seconds in the future. Same case as the initial experiment, other than the distance to detectors is greater. So far, so good?

But what if there is the second person on the moon?

He then decides to replace the detectors with the eraser, before the photons reach them (You can synchronize watches to know when the machine was started, so he can replace them in that one second, assuming you can do that or just assume greater distance instead). Meaning if you start the machine and see a wave pattern, it means someone will replace the detectors with erasers in a couple of seconds. If you see a scattered pattern again, it means he will decide not to replace detectors with erasers, or fail to do so for some reason. You know the future?

But you can't change it since you can't send him a message to change his mind faster than the second pairs hit his detectors/erasers. (Faster than the speed of light)

Thus you can predict the future, if it involved a choice of whether to detect the path or to erase it?

Based on operating under premise that this is true, if you assume you have a theoretical way to trap light indefinitely, you could send messages to the past?

Example - alter future

Instead of using one such setup, why not use one for each bit of information? Each setup will end to be either $0$ or $1$, depending on if it has a wave pattern or just scattered dots on its board.

Lets say you play lottery and want to send results back.

Then, lets say you start the machine and trap the moon photons until you see the results. Then, you encode those results in binary and travel to the moon, and replace the detectors/erasers accordingly. Then you release the trapped light.

Now, look at when you first started the machine. The boards on earth will, after turning the machine on, yield wave patterns and you can read that as $0$s, or will be scattered and read as $1$s. After reading each bit of information, you got the results because you know you will send them back in the future, after the results really get announced.

But what if you decide not to send yourself lottery numbers after you get them ? That means you shouldn't have received them in the first place.

But what if you do not get them, and still decide to send them back. That means you should've got them in the first place.

This paradox makes me thing something is wrong.

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Is the first idea valid?

What is wrong with the second idea?

Why wouldn't a similar idea ever work?

Help me understand of what the things I got wrong? Because I feel like I did not grasp a proper idea of these quantum properties.

Answer

It is not going to work. This is because regardless of what the person on the Moon does, the person on Earth will never see an interference pattern. He can only see an interference pattern if the person on the Moon sends him a message telling him: 1) At what time he used a "detector", and if he detected a signal or not. 2) At what time he used an "eraser", and if he detected a signal or not in that case.

Using that information, the person on Earth will see an interference pattern only if he plots the photons that were detected using the "eraser" or not detected if using the "detector".

electromagnetism - Time varying magnetic field, yielding two effects of induction?

$B_s$ is nonuniform, and it's generated from a movable source(e.g magnet or electromagnet).

A rectangular loop of area $A$ is stationary.

The variation of flux for this case is caused from the following:

1. Spatial movement of $B_s$.


2. Strength variation $B_s$ of due to (1).

How can I quantitatively formulate $\varepsilon_{induced}$ here?

I could assume that:

$$\varepsilon_{induced} =\oint \vec{E} \cdot \vec{dl} = -\frac{\delta \Phi}{\delta t}$$

to:

$$\varepsilon_{induced} =\oint \vec{E} \cdot \vec{dl} = -\int \frac{\delta B_s}{\delta t} \cdot \vec{da}$$

But how can I factor in the effects of 1&2 to $\delta B_s$?

From my textbook,the examples brought up had the effects(1&2) mutually exclusive, but for this case both add to $\varepsilon_{induced}$

newtonian mechanics - Why does a rocket spin during launching

Why does a rocket spin during launching?

https://www.youtube.com/watch?v=BuPY5l45UC8

harmonic oscillator - Metronome synchronisation applied to swings

The movement of several metronomes can be synchronised when a movable floor is utilised which couples the movement of the different metronomes.

Is it possible to apply this sort of synchronisation to several swings occupied by people (the ones you have on children playgrounds)?

I visualise this as several swings connected to a rigid platform that is free to move parallel to the ground. I assume that the length of each swing line ($L$) is the same and that the length of the swing line is the only factor that influences the swing period ($T$) according to:

$$T = 2 \cdot \pi \cdot \sqrt\frac{L}{g}$$

I understand that the metronome is driven by a wound spring and that this part is not present in a swing. People can move their c.o.g. back and forth during the swing, emulating the effect of the spring.

Can the principles present in the metronome sync be carried over to swing sync such that swings start swaying together after a certain amount of time?
Are my assumptions correct? Which additional assumptions do I need to make this work?

soft question - Is there a historical name for an inductor that starts with the letter $L$? (RL circuits)

Why is a circuit with a resistor and inductor called an RL circuit and not an RI circuit?

Was there a now-obsolete name for an inductor that started with the letter $L$?

Answer

Quoted from Wikipedia: Inductance:

It is customary to use the symbol $L$ for inductance, in honour of the physicist Heinrich Lenz.

hilbert space - Hamiltonian of quantum harmonic oscillator with $psi(x)=delta(x)$: comparison to classical mechanics

I was just reading the question Why can't $\psi(x)=\delta(x)$ in the case of a harmonic oscillator? The accepted answer says that $\psi(x)=\delta(x)$ is a mathematically valid state, though it's not physically possible because delta distributions aren't normalizable. If it's possible, I'd like to ignore the fact that it's un-physical, and instead try to fill in the holes in my understanding of the mathematical model.

The answer concludes that the expectation value of the Hamiltonian of the system is infinity, which makes sense to me because it follows from the proposition that $\left$ is infinity.

However, I see this as a contradiction of my impression that quantum mechanical systems at high energies are easily approximated to classical behavior. For instance, with a particle in a square well, the position probability distribution at high energies is approximately constant across the whole allowed region, which is exactly what classical physics predicts for the same energy.

Going back to the case of harmonic oscillators, I think that a classical model with zero energy would have a position distribution of $\delta (x)$, because the particle's localized in one exact place and it isn't oscillating or moving in any way. So this tells me that a quantum harmonic oscillator with an extremely high energy behaves like a classical zero-energy system, which just generally sounds wrong.

Is it wrong/meaningless to make such comparisons of a zero-energy classical particle to a quantum harmonic oscillator with $\psi(x)=\delta(x)$? Alternatively, is this apparent contradiction actually logically justifiable?

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I think that there's something fishy in my comparison when I say that a classical system shows a position distribution of $\delta(x)$, since the momentum distribution is also a delta function (with a spike at $p=0$). However, the QM handling of momentum would predict a completely different state. But I can't build more out of that reasoning to see if it's relevant.

Furthermore, the lowest energy state of a QM oscillator is nonzero, so there is no valid QM counterpart of a classical oscillator with $\psi(x)=\delta(x)$. However, this holds true for the particle in an infinite well too, so I would expect a symmetry whereby in both cases, high-energy QM descriptions are similar to classical mechanics.

Answer

The most classical-like states of harmonic oscillator are Coherent states of harmonic oscillator, not $\delta(x)$. The reason is that coherent state has balanced uncertainty of coordinate and momentum: $$\Delta x =\sqrt{\frac{\hbar}{2m\omega}}; \Delta p = \sqrt{\frac{m\hbar\omega}{2}},$$ and in the classical limit ($\hbar\rightarrow 0$) $\Delta x \approx 0$, $\Delta p \approx 0$.

In a state with $\phi(x)=\delta(x)$ the momentum uncertainty $\Delta p=\infty$, it does not tend to zero.

UPDATE:

It is not correct to assume that a particle with high energy can always be described by classical mechanics. Consider for example two states $|\psi_1\rangle$ and $|\psi_2\rangle$, in which particle has high energy. Any superposition of them is a valid state in quantum mechanics but does not make sense in classical mechanics.

Why photons are emitted because of changes to electron behavior

Explanations I have read of why photons are emitted from atoms mention electrons being 'excited' to another energy level, and then returning to their base level, releasing a photon. I have also seen the occasional mention of 'fields' and I vaguely expect from what I've read that interaction of the electrons with these fields has some effect. I've also seen, but not used, Feynmann diagrams that involve additional particles.

Please can you (ideally) clarify what are the different theories and experimental evidence relating to electrons and photons because of changes to electron behavior?

optics - Optical Drive Physics

I have been recently wondering how is data stored on and retrieved from optical devices like CDs, DVDs, and Bluray. What makes these different storage types different from each other?

particle physics - Why is stringless supergravity not considered by many to be a candidate theory of quantum gravity?

This paper seems to show that $d=4, N=8$ supergravity is finite. Yet the paper only has three citations in spires, and I certainly haven't heard talk of a new candidate theory of gravity.

Why isn't perturbative supergravity with some supersymmetry breaking principle, coupled with the standard model considered a possible theory of the universe? Has someone checked the coupling to matter? Is that the problem?

Answer

In order to not be entirely negative, I'll answer your question first and then provide a reason or two why research on the subject is interesting for other reasons.

1. 
   

   The finiteness conjecture has to do with perturbation theory. Even if true, it is still believed that in order for the theory to be non-perturbatively finite and otherwise consistent (e.g unitary), it has to include more degrees of freedom. The most plausible scenario is that the completion is the full string theory, for which N=8 SUGRA is very closely related.

   
   


2. 
   

   The theory does not have enough structure to be a realistic description of nature by itself. For example it does not have chiral fermions, or a mechanism to break the extended SUSY spontaneously. All of these things become possible when you embed the theory within string theory.

   
   


3. 
   
   

   If you keep the theory as a QFT, but add some ingredients by hand to get more a realistic theory, the extended SUSY is broken and the magic is gone. Any finiteness conjecture is only valid for the pure (super)gravity case, and goes away as soon as you make the situation slightly less symmetric.

   
   

On the other hand, purely as a theoretical laboratory it is fascinating that a quantum field theory which contains gravity is better behaved than expected, and the theory has a close relationship to other highly symmetric quantum field theories (for example N=4 SYM). There is great excitement and hope recently that by understanding precisely why it is finite (or at least well-behaved) we'd be able to understand the structure of quantum field theories better. It is in some sense the "simplest quantum field theory" (arxiv.org/abs/0808.1446).

research level - Boundary currents for Asymptotic Symmetry Group (ASG)

In the context of asymptotic symmetry groups, what is a boundary current? Why is it called a "current"?

Context: I'm reading Strominger's recent paper on Asymptotic symmetry group of Yang-Mills (link here) and he has a section on the boundary current (section 2.3). I can follow the math completely fine, but some of the words are confusing to me.

Answer

In this context, a "current" is an object obeying an affine Lie algebra, also called current algebra and a special case of a Kac-Moody algebra. It is an algebra formed by unit weight operators: take for example a current $J^a(z)$, where $a$ is a label and $z$ is a complex coordinate. The algebra is given by

$$[J^a_n,J^b_m]=i{f^{ab}}_cJ^c_{n+m}+mkd^{ab}\delta_{n+m},$$

where

$$J^a_n=\frac{1}{2\pi i}\oint dz \, z^{-(n+1)}J^a(z).$$

The integer $n$ denotes the mode number, the integer $k$ is the level and $d^{ab}=(t^a,t^b)$ defines the inner product between generators.

The word "boundary" refers to the fact that the symmetry group underlying the algebra preserves a certain structure at the boundary of the geometry at infinity. In the case of the paper you are reading, the symmetry group is $U(1)$ and the boundary is given by $\mathcal{I}^+$.

Additional information:

Affine Lie algebras play a role in string theory/conformal field theory, where they can be used to generate states in certain representations of a group. For example, the state

$$J^a_{-1}\tilde{\alpha}^{\mu}_{-1}|0\rangle$$

corresponds to a massless vector $A^{\mu a}$ in the adjoint representation of the underlying group ($\tilde{\alpha}^{\mu}_{-1}$ is a creation operator).

density - What kind of matter are black holes made of?

Imagine a black hole originally formed from, for example, Rubidium atoms. On the other hand, one made from, for example, Helium atoms. Will it be there any difference between the two? Or perhaps once formed it all turns into the same type of matter?

Can the "composition" of the matter-energy in the black hole be inferred somehow?

Could one say that "black hole substance" is a distinct form or state of matter different from other forms or states of matter?

Answer

A not-entirely-silly answer: Black holes seem to act like a very, very heavy fundamental particle, described only by its mass and quantum numbers like spin and charge. Those numbers just happen to be very large compared to typical fundamental particles.

If you made them out of different kinds of atoms, they'd have the sum of the quantum properties of those atoms. If you're using neutral atoms, the black hole will be neutral, but if some of the atoms are ionized, you'll end up with a positively charged black hole. If you use entirely atoms that are prepared in a way that makes them mostly spin-up, the black hole will be spin-up. Most of the information about what you put in disappears, though.

particle physics - Asymptotic Freedom - Qualitative Explanation

I am doing a (mostly qualitative) course on Particle Physics, and am confused about the concept of asymptotic freedom. The lecture notes basically say that a quark may experience no force/be "unbound" temporarily as a result of a collision. (due to properties of the strong force) Is there all there is to it?

Later, it mentioned that asymptotic freedom is important in electron-positron annihilation into hadrons. Were there no asymptotic freedom, the cross section of the process would be different. I can't see how this follows on from what was said above.

So I am seeking an qualitative explanation of this concept, and perhaps something about its consequences as well.

Answer

In order to understand asymptotic freedom, you need to be aware of the concept of renormalization. Since you want a qualitative description, just think of renormalization a modification of the coupling strengths and masses of particles at high energies. This is roughly like pushing a ball through the water; the harder you push, the more the water sticks around it and the harder it is to move. This can be modeled with Newton's 2nd law $F=ma$ by replacing the mass with a slightly larger mass $m+\delta m$, and this $\delta m$ depends on the velocity of the ball in the water.

(that discussion can be found in section 3.2 of Connes and Marcolli, "Noncommutative Geometry, Quantum Fields and Motives")

Once you have the concept of renormalization, asymptotic freedom is a property the strong force has as you scale the coupling constant to high energy. Rather then the coupling getting stronger, it gets weaker. This has major consequences for confinement - that is, bound quarks. At low energies, quarks in bound states are forever bound - it becomes harder and harder to pull them apart the further apart you pull them. At high enough energies (say, colliding two protons at 7 TeV like the LHC) the quark coupling gets small and quarks are essentially free and unbound. It should be easy to see how this would change the cross section.

As a sidenote, only the strong force is asymptotically free. The E/M and weak force become stronger as the energy gets higher. In addition, it is important to realize that we cannot solve problems involving the strong force at low energies (if you could, the Clay Mathematics Institute would give you $1 million!). Once they are at high energies, the strong coupling is weak so QCD acts quite a bit like QED.

Why quantising gravity necessarily give us gravitons?

Gravitons are supposed to be the quanta of gravitational field

My question is, if we do not know how to quantize gravity yet, how do we know that quantizing it in principle should give us gravitons, what guarantees that? Why it is always mentioned that gravitons are the gravity force mediators, although we have not quantized gravity yet.

newtonian mechanics - Defy gravity torques with gyroscopes?

Context

On the following drawing, a platform is hung from the ceiling not exactly from its centre of gravity. Because of this it can't sustain an arbitrary orientation for long; I want to increase its stiffness with the use of 3 orthogonal fast spinning flywheels, but before building those flywheels I need to know how big they need to be by making a dynamic model...

... And I have been unable to model the gyroscopic effect [Edit: end removed for clarity].

Question

How can I model the 3 flywheels reaction to (themselves and) the gravity torque, be it through a rate increment or a torque (to inject in the plant model)?

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Implementing Lelesquiz's answer

Top level equations: Euler-Lagrange

The goal is to solve for each timestep for the angles $(\phi, \theta, \psi)$ using the equations, in inertial frame: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right)-\frac{\partial L}{\partial \phi}= (\boldsymbol{M_g}-\sum_{k=1}^3 \boldsymbol{M_{wk}})\cdot \frac{\partial \boldsymbol{\omega_b}}{\partial \dot{\phi}}$$ $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)-\frac{\partial L}{\partial \theta}= (\boldsymbol{M_g}-\sum_{k=1}^3 \boldsymbol{M_{wk}})\cdot \frac{\partial \boldsymbol{\omega_b}}{\partial \dot{\theta}}$$ $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\psi}}\right)-\frac{\partial L}{\partial \psi}= (\boldsymbol{M_g}-\sum_{k=1}^3 \boldsymbol{M_{wk}})\cdot \frac{\partial \boldsymbol{\omega_b}}{\partial \dot{\psi}}$$ Where $M_g$ is the gravity torque, $M_{wk}$ is the motor torque applied on the flywheel k (opposite of what's applied on the platform), and $\omega_b$ is the platform rate vector in inertial frame. 3 other equations based on $\alpha_k$, the angles of the flywheels, should have been added to completely characterize the system but the rates of the wheels are inputs, measured directly. The torques on the wheels are inputs as well (and the command of the closed loop system).

Knowing that the lagrangian $L=T-U$ in the inertial frame.

Kinetic energy T

$$ T(\phi, \theta, \psi, \dot{\phi}, \dot{\theta}, \dot{\psi}) = \frac{1}{2} \sum_{k=1}^3 \left(\boldsymbol{N_{k,Ri}^T}\boldsymbol{[I_{wO,Ri}]}\boldsymbol{N_{k,Ri}} \right) + \frac{1}{2} \left( m_{b} \boldsymbol{v_{b,Ri}}^T\boldsymbol{v_{b,Ri}}+\boldsymbol{\omega_{b,Ri}^T} \boldsymbol{[I_{bG,Ri}]} \boldsymbol{\omega_{b,Ri}} \right)$$ Where $m_b$ is the mass of the system (with wheels on), $\boldsymbol{v_{b}}$ the velocity vector of the center of gravity of the system, $\boldsymbol{\omega_{b}}$ the system's rate vector (wheels not spinning) and $\boldsymbol{[I_{bG}]}$ the inertia matrix of the system about its center of gravity (which needs to be transported from the guimbal centre 0), $[I_{wO}]$ the inertia matrix of the flywheels about their axis, and $\boldsymbol{N_{k}}$ the rate vector of the flywheels in the inertial frame. $$\begin{array}{lc}\boldsymbol{[I_{wO,Ri}]}=\boldsymbol{[R_{ib}]}\left[\begin{matrix}J_w & 0 & 0\\ 0 & J_w &0\\0 & 0& J_w\end{matrix}\right] \boldsymbol{[R_{bi}]} \end{array}$$ $$\boldsymbol{v_{b}}=\boldsymbol{\omega_{b,Ri}}\times\boldsymbol{\Delta_{b,Ri}}$$ $$\begin{array}{lc} \boldsymbol{[I_{bG,Ri}]}= \boldsymbol{[R_{ib}]}\boldsymbol{[I_{bO,Rb}]}\boldsymbol{[R_{bi}]}+m_b& \left(\begin{matrix} \Delta_{by,Ri}^2+\Delta_{bz,Ri}^2 & -\Delta_{bx,Ri}\Delta_{by,Ri} & -\Delta_{bx,Ri}\Delta_{bz,Ri} \\ -\Delta_{bx,Ri}\Delta_{by,Ri} & \Delta_{bx,Ri}^2+\Delta_{bz,Ri}^2 & -\Delta_{by,Ri}\Delta_{bz,Ri} \\ -\Delta_{bx,Ri}\Delta_{bz,Ri} & -\Delta_{by,Ri}\Delta_{bz,Ri} & \Delta_{bx,Ri}^2+\Delta_{by,Ri}^2 \end{matrix}\right) \end{array}$$ $\boldsymbol{\Delta_{b,Ri}}$ being the guimbal centre - center of gravity vector of the system expressed in the inertial frame. See below "Euler angle expressions" for the transformation body-inertial $\boldsymbol{_i}=\boldsymbol{[R_{ib}]}\boldsymbol{_b}$.

Potential energy U

$$U(\phi, \theta, \psi) = -m\boldsymbol{g_{Ri}}.\boldsymbol{\Delta_{b,Ri}}$$ The minus sign comes from the fact U is minimum with the guimbal-CG vector pointing downward (dot product positive).

Euler angles expressions

The Euler angles 313 have been selected as a coordinate system, which gives the rotation matrix from inertial frame $(0 \boldsymbol{X_i Y_i Z_i})$ to body frame $(0 \boldsymbol{X_b Y_b Z_b})$: $$ \begin{array}{lc} \boldsymbol{R_{RbRi}}(\phi, \theta, \psi)= & \left[\begin{matrix} cos \psi & sin \psi & 0 \\ -sin \psi & cos \psi & 0 \\ 0 & 0 & 1 \end{matrix}\right] \left[\begin{matrix} 1 & 0 & 0 \\ 0 & cos \theta & sin \theta \\ 0 & -sin \theta & cos \theta \end{matrix}\right] \left[\begin{matrix} cos \phi & sin \phi & 0 \\ -sin \phi & cos \phi & 0 \\ 0 & 0 & 1 \end{matrix}\right] \end{array}$$ The rate vector is obtained by summing the projections on each axis of the inertial frame the $\boldsymbol{\dot{\phi},\dot{\theta},\dot{\psi}}$ vectors): $$\begin{array}{lc} \boldsymbol{w_{b,Ri}}= & \left[\begin{matrix} \dot{\theta}cos\phi+\dot{\psi} sin\theta sin\phi \\ \dot{\theta}sin\phi+\dot{\psi} sin\theta cos\phi \\ \dot{\phi}+\dot{\psi}cos\theta \end{matrix}\right]\end{array}$$

Problem

The kinetic energy of the wheels is independent on the Euler angles, so the rates of the wheels have no effect, which is wrong! Since it's absolutely normal that no matter the rotation matrix the norm of the wheels rates vector does not change, the kinetic energy expression or even the top level equations must be wrong.

nuclear physics - What is the "penetrability factor"?

I have read/heard this term a few times in nuclear physics papers. I'm guessing it has something to do with the Coloumb barrier of a nucleus. Could you maybe explain what this "penetrability factor" is?

[See page 63 of this paper]

Answer

The penetrability factor is a carefully defined factor that separates the coulomb-force and the interesting nuclear-forces in the relationship between the partial decay width and the reduced width (for R-Matrix analysis) of a particular resonance.

$$\Gamma = 2\gamma^2P(l,\rho,\eta)$$

Here $\Gamma$ is the partial decay width, $\gamma$ is the reduced width, $l$ is the orbital angular momentum of the state, $\rho$ is defined as $k$ (wavenumber) multiplied by $r$ the target particle radius, $\eta$ is the sommerfeld [nuclear] parameter, and $P(l,\eta,\rho)$ is the penetrability factor. The penetrability factor is defined as,

$$P(l,\rho,\eta)=\frac{\rho}{\left(F_l (\eta,\rho)\right)^2+\left(G_l (\eta,\rho)\right)^2}$$

where $F$ and $G$ are the regular and irregular coulomb wave functions, respectively.

Source: http://www.scholarpedia.org/article/The_R-matrix_theory_in_nuclear_and_atomic_physics (equation 10 and 13)

energy - How do momentum get transferred?

Simple Question ,

Consider two objects namely $A$ and $B$ where $B$ is stationary and $A$ is moving towards $B$ with velocity $v$.

When the two objects touch each other what does actually happen between them that $B$ moves.

• 
  

  Does the momentum get transferred.

  
  


• 
  

  Also is momentum a physically property that can be transferred.

  
  

(If you think somehow really somehow a non physical property like energy is transferred please answer question with regard to it).

• If so How?

I mean what is the process of transfer of momentum.

If the momentum is transferred physically or in very simple words somehow flows .And if It can .Will there be a change in fundamental idea of the process of transfer of it.

For sake of simplicity lets take both objects rigid particles.

I really need Help .

Thank You .

newtonian mechanics - Energy dissipation due to frictional force in rotational motion

Let me have a ramp with length l and height h. What I want is to roll that ball on the plane without slipping .

When I face this types of problem, I try to consider the rotational energy and kinetic energy in total. Therefore total energy $E= E_k + E_w(rotational)$ I see that a lot of problems have been solved by using the above formulae. But what about the dissipation of energy by frictional force? Don't we have to consider it?

Answer

Where you write 'roll that ball without slipping', that is a crucial point. If it wasn't for the friction force you've drawn in your diagram the ball would indeed be simply slipping (sliding) down the slope without any rotation going on.

It's that friction force that prevents slippage and that provides the torque around the centre of the ball: torque = friction force times radius ($r$) of the ball. This way the friction force provides the work that is converted to rotational energy.

What you indicated as $P_x$ provides the force for acceleration along the slope and the work it does is converted to translational energy.

Without slippage, the ball will rotate with frequency $f$ and is related to the translational speed $v$ via $v=2\pi rf$.

The total kinetic energy is $E_k=\frac{mv^2}{2}+\frac{I\omega^2}{2}$. In this simple case, $E_k=mgh$ with $h$ the height of the slope. All other things being equal a rolling ball will reach a lower final speed $v$ than one purely sliding (without any friction) because the same amount of energy ($mgh$) is now partitioned into rotational and translational energy.

That rotation has a reducing effect on the acceleration along the slope $a$ can also be shown as follows.

The equation of motion along the slope is given by:

$ma=mg\sin\alpha-F_f$.

The equation of motion for the rotation is:

$F_fr=I\dot{\omega}$.

With $v=2\pi fr=\omega r$, differentiate to $t$ and get $a=\dot{\omega}r$ or $\dot{\omega}=\frac{a}{r}$.

So that $F_fr^2=Ia$.

So we have two linear equations in $a$ and $F_f$ and we get:

$F_f=\frac{Ig\sin\alpha}{R^2+\frac{I}{m}}$.

But this is an ideal case: smooth ball on a smooth surface with just the right amount of friction to avoid slippage. In 'real world' cases there will be extra friction and this will do work that is irreversibly lost: for example a golf ball rolling down a grassy slope. Resistance by the grassy surface will cause some potential energy to be lost to friction work.

optics - Why does the sky change color? Why is the sky blue during the day, red during sunrise/set and black during the night?

Why does the sky change color? Why is the sky blue during the day, red during sunrise/set and black during the night?

Answer

The keywords here are Rayleigh scattering. See also diffuse sky radiation.

But much more simply, it has to do with the way that sunlight interacts with air molecules. Blue light is scattered more than red light, so during the day when we look at parts of the sky that are away from the sun, we see more blue than red. During sunset or sunrise, most of the light from the sun comes towards the earth at a sharp angle, so now the blue light is mostly scattered away, and we see mostly red light.

quantum mechanics - Electron shells in atoms: What causes them to exist as they do?

I have seen similar posts, but I haven't seen what seems to be a clear and direct answer.

Why do only a certain number of electrons occupy each shell? Why are the shells arranged in certain distances from the nucleus? Why don't electrons just collapse into the nucleus or fly away?

It seems there are lots of equations and theories that describe HOW electrons behave (pauli exclusion principle), predictions about WHERE they may be located (Schrödinger equation, uncertainty principle), etc. But hard to find the WHY and/or causality behind these descriptive properties. What is it about the nucleus and the electrons that causes them to attract/repel in the form of these shells at regular intervals and numbers of electrons per shell?

Please be patient with me, new to this forum and just an amateur fan of physics.

Answer

Any answer based on analogies rather than mathematics is going to be misleading, so please bear this in mind when you read this.

Most of us will have discovered that if you tie one end of a rope to a wall and wave the other you can get standing waves on it like this:

Depending on how fast you wave the end of the rope you can get half a wave (A), one wave (B), one and a half waves (C), and so on. But you can't have 3/5 of a wave or 4.4328425 waves. You can only have a half integral number of waves. The number of waves is quantised.

This is basically why electron energies in an atom are quantised. You've probably heard that electrons behave as waves as well as particles. Well if you're trying to cram an electron into a confined space you'll only be able to do so if the electron wavelength fits neatly into the space. This is a lot more complicated than just waving a rope because an atom is a 3D object so you have 3D waves. However take for example the first three $s$ wavefunctions, which are spherically symmetric, and look how they vary with distance - you get (these are for a hydrogen atom) $^1$:

Unlike the rope the waves aren't all the same size and length because the potential around a hydrogen atom varies with distance, however you can see a general similarity with the first three modes of the rope.

And that's basically it. Energy increases with decreasing wavelength, so the "half wave" $1s$ level has a lower energy than the "one wave" $2s$ level, and the $2s$ has a lower energy than the "one and a half wave" $3s$ level.

---

_{$^1$ the graphs are actually the electron probability distribution $P(r) = \psi\psi^*4\pi r^2$. I did try plotting the wavefunction, but it was less visually effective.}

newtonian mechanics - What is terminal velocity?

What is terminal velocity? I've heard the term especially when the Discovery Channel is covering something about sky diving. Also, it is commonly known that HALO (Hi-Altitude, Lo-Opening) infantry reaches terminal velocity before their chutes open.

Can the terminal velocity be different for one individual weighing 180 pounds versus an individual weighing 250 pounds?

Answer

Terminal velocity is the (asymptotic) maximum velocity that you can reach during free-fall. If you imagine yourself falling in gravity, and ignore air resistance, you would fall with acceleration $g$, and your velocity would grow unbounded (well, until special relativity takes over). This effect is independent of your mass, since

$F = ma = mg \Rightarrow a = g$

Where terminal velocity arises is that air resistance is a velocity-dependent force acting against your free fall. If we had, for example, a drag force of $F_D=KAv^2$ ($K$ is just a constant to make all the units work out and depends on the properties of the fluid you're falling through, and $A$ is your cross-sectional area perpendicular to the direction of motion) then the terminal velocity is the velocity at which the forces cancel (i.e., no more acceleration, so the velocity becomes constant):

$F = 0 = mg - KAv_t^2 \Rightarrow v_t=\sqrt{mg/KA}$

So we see that a more massive object can in fact have a larger terminal velocity.

statistical mechanics - Can a system entirely of photons be a Bose-Einsten condensate?

Background:

In Bose-Einstein stats the quantum concentration $N_q$ (particles per volume) is proportional to the total mass M of the system:

$$ N_q = (M k T/2 \pi \hbar^2)^{3/2} $$ where k Boltzmann constant, T temperature

Questions:

A) For a B-E system "entirely of Photons" - what is the total mass of the system? (answered, see below)

B) Does an ensemble of photons have a temperature? (answered, see below)

C) Is this a Bose-Einstein condensate?

I've found a paper here (like the paper put forward by Chris Gerig below) which finds a BEC, but it is within a chamber filled with dye, and the interaction of the photons with the dye molecules makes it a dual system, as to one purely of photons. I think in this case there is a coupling between the dye molecules and the photons that is responsible for the chemical potential in the partition equation
$$N_q = \frac{g_i}{e^{\left.\left(\epsilon _i- \mu \right)\right/\text{kT} - 1}}$$

where $g_i$ is the degeneracy of state i, $\mu$ is the chemical potential, $\epsilon_i$ is the energy of the ith state.

I suspect an Ansatz along the lines of $\mu$ = 0, and $\epsilon_i$ = $\hbar \nu_i$, where $\nu_i$ is the frequency of the i photon.

another edit:

After going for a walk, I've realized the Ansatz is almost identical to Planck's Radiation Law but the degeneracy = 1 and chemical potential = 0.

So, in answer to my own questions:

A) is a nonsensical question, as photons have no mass, noting from wiki on Quantum Concentration: "Quantum effects become appreciable when the particle concentration is greater than or equal to the quantum concentration", but this shouldn't apply to non-coupling bosons.

B) yes the ensemble has a temperature, but I was too stupid to remember photons are subject to Planck's Law.

C) Is this a Bose-Einstein condensate? No, as photons have no coupling or chemical potential required for a BEC.

So, for an exotic star composed entirely photons, all the photons should sit in their lowest energy levels and the star will do nothing more than disperse.

Is this right?

statistical mechanics - What are the differences between indistinguishable and identical?

What is the difference between indistinguishable particles and identical particles?

Answer

If I create an electron on earth and someone else creates an electron on Andromeda, they're identical particles. They have the same quantum numbers, they're both excitations of the electron field. However they're distinguishable by means of their spatial separation. Their wavefunctions don't overlap.

Edit: perhaps I should add that not everyone uses the two words in this strict sense. Sometimes they're used interchangeably, but blurring them carries with it the danger of taking seriously the entanglement implied by antisymmetrizing across all existing electrons.

quantum mechanics - Quantization of energy in semi-infinite well

Consider an electron with total energy $E>V_2$ in a potential with $$V(x)= \begin{cases} \infty & x< 0 \\ V_1 & 0< x< L \\ V_2 & x>L \end{cases} $$ where $V_2>V_1>0$.

We can determine that $$\phi_E(x)= \begin{cases} 0 & x< 0 \\ A\sin(kx)+B\cos(kx) & 0< x< L \\ Ce^{qx}+De^{-qx} & x>L \end{cases} $$ where $k^2=\frac{2m_e E}{\hbar}$ and $q=k\sqrt{V_1-E}$.

We can also apply the boundary condition at $x=0$ to determine that $$\phi_e(x)=A\sin(kx)$$ for $x\in[0,L]$.

We can also apply boundary conditions at $x=L$ to find that $$A\sin(kL)=De^{-qL}$$ $$Ak\cos(kL)=-Dqe^{-qL}$$ (since $C=0$ due to the corresponding positive exponent), and $$k\cot(kL)=-q$$

I'm stuck with the question: are the energy states with $E>V_2$ quantized?

I can see that, because the boundary condition at $x=L$ is not homogeneous, we cannot determine the eigenvalues in discrete form. Does this mean that the energy states are not quantized in this case?

Would appreciate some help.

Answer

$$A\sin(k_iL)=De^{-qL}$$ $$Ak_i\cos(k_iL)=-Dqe^{-qL}$$ $$k_i\cot(k_iL)=-q$$ Insert the values for $k_i$ and $q$: $$[2m(V_1-E_i)/\hbar^2]^{1/2}\cot[2m(V_1-E_i)/\hbar]^{1/2}L=-[2mE_i/\hbar^2]^{1/2}$$ The allowed energy levels ($E_i

But particles with energy $E>V_2$ can not be bound (contained in the well). Such a particle, coming in from the right e.g., would simply bounce off the infinite potential wall.

newtonian mechanics - Does it take more energy to open a door when applying force close to the hinge?

Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is?

"Opening the door" should be interpreted as accelerating the door to a certain rotational speed.

My own answer is no, since the change in force would be proportional to the distance required to open the door and therefore the total energy would remain the same.

Answer

You are right.

To open the door during the same time interval (for pushing at $a$ and $b$), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives $$ \frac{F_a}{F_b} = \frac{r_b}{r_a} $$ where $r$ is the distance from the point of contact to the axis.

However the distance that the force should be applied at $a$ and at $b$ are related because they are arcs of the same angle $$ \frac{r_b}{r_a}=\frac{l_b}{l_a} $$ This implies that $F_a l_a = F_b l_b$ and equal work needs to be done.

metric tensor - Trace of generators of Lie group

In most textbooks (Georgi, for example) a scalar product on the generators of a Lie Algebra is introduced (the Cartan-Killing form) as $$tr[T^{a}T^{b}]$$ which is promptly diagonalised (for compact algebras) and the generators scaled such that $$tr[T^{a}T^{b}] = \delta^{ab}.$$ In this basis we get that, for example, $$f_{abc} = -i\, tr ([T^{a}, T^{b}]T^{c})$$ that are fully antisymmetric.

Yet I have seen the these relations used for arbitrary (it particular the fundamental) representation as matter of course (maybe up to some normalisation). Is this because $tr[T^{a}T^{b}]$ defines a symmetric matrix in any rep that can thus be diagonalised? Is it a general truth? Or does the diagonalisation in the adjoint imply a diagonal for in any other rep?

I know that the structure constants are essentially fixed for all reps by smoothness and the group product -- is this why fixing the form in one basis for one rep fixes it for that basis in all reps?

For a concrete example, let's suppose I look at SU(2). The adjoint rep is 3 dimensional and I can linearly transform and scale my generators (i.e. the structure constants) so that I get the trace to be diagonal and normalised. This fixes once and for all that the structure constants of SU(2) are $f_{ijk} = \epsilon_{ijk}$, say.

Now I ask someone to construct the fundamental rep; they look for 2x2 matrices satisfying the Lie algebra with these structure constants. They find the Pauli matrices. Why do these come out such that the trace $tr [\sigma^{a} \sigma^{b}] \propto \delta_{ab}$ automatically? It's a different rep...why is it guaranteed?

homework and exercises - How do I find work done by friction over a curve represented by a polynomial?

I am facing a problem in Physics.

Problem: What will be the work done by the frictional force over a polynomial curve if a body is sliding on this polynomial($a+bx+cx^2+dx^3+\ldots$) curve from rest from the height $h_1$ to height $h_2$ (where $h_1 > h_2$).

I tried to solve this as follows:

frictional force $F = k mg \cos\theta$, where $mg \cos\theta$ is normal force at that point. $k$ is coefficient of friction

Total work done=Line Integration over the polynomial(dot product of F and displacement).

But to go ahead from this point,i do not know.

special relativity - Simultaneity (working backwards) and absolute time

If two events happen at different times and at different places in two different reference frames (observers) couldn't the observers work backwards (considering the finite speed of light) to find out when and where the events took place ? Is this God's-eye perspective correct ?

homework and exercises - Top angular speed of electric motor

I recently came across a question asking the following:

If a motor is switched on, it quickly reaches a top speed. Why does it not just go faster and faster and faster?

I thought it might be related to the fact that the electrical power is constant, so I tried approaching with $P=\tau \cdot \omega$. However I still do not have any clue to reaching the conclusion.

Is it somehow related to the energy loss (due to friction)? Like the terminal speed of a parachute or something? But I was thinking that friction should be constant ($f=\mu \cdot N$) while air drag is related to the speed, so this may not be the answer.

Thank you.

Answer

Surpringingly the top speed is not necessarily anything to do with friction, though friction will of course have some effect.

A motor acts as a generator, i.e. if you turn a motor it will generate a potential difference just like a generator, and this potential difference (usually called the back EMF) is proportional to the motor speed. So if you connect a X volt battery to an unloaded motor the armature will accelerate until the back EMF rises to the same value as the applied voltage. At this point the motor will settle to a steady angular velocity.

The equations describing this are given in the Wikipedia article on the DC motor, though this is far from the best written article I've seen. Basically the back EMF, $E_b$, is given by:

$$ E_b = \omega \Phi k_b $$

where $\phi$ and $k_b$ are dependant on the motor design and for the sake of this discussion can be treated as constants. So ignoring friction, the internal resistance of the motor and with no load on the motor the top speed will be when $E_b$ is the same as the applied voltage $V$ so:

$$ \omega = \frac{V}{\Phi k_b} $$

So it predicts angular velocity is proportional to applied voltage.

The effect of internal resistance, $R_i$, is easily included since it just creates a potential drop of $IR_i$ and the equation changes to:

$$ \omega = \frac{V - IR_i}{\Phi k_b} $$

condensed matter - How to construct the matrix of Hamiltonians for a hexagonal lattice

For part of a project I need to solve the time-independent Schrödinger equation, $\mathbf H\Psi = E\Psi$ (where $\mathbf H$ is the matrix with elements $\langle\Psi_i|H|\Psi_j\rangle$, and $\mathbf S$ is a matrix with elements $\langle \Psi_i|\Psi_j\rangle$) for different lattices. I've done this for a square lattice but I'm having trouble with the hexagonal lattice. The difficulty I'm having with this is that the matrix is different depending on which sub lattice, A or B (see image below), you start on since they different possible directions (therefore different corresponding energies) to travel in.

Edit: I forgot to mention that I am using the tight binding approximation so unless $\Psi_i$ and $\Psi_j$ are next to each other then $\langle\Psi_i|H|\Psi_j\rangle = 0$. In terms of the image above, only $\delta$ movement is allowed. $\delta_1$, $\delta_2$ and $\delta_3$ are all different in this case.

Anyone know how to make the matrix of Hamiltonians for this lattice? Any help would be greatly appreciated.

special relativity - What if photons are not the fastest particles?

Einstein originally thought that special relativity was about light and how it always travelled at the same speed. Nowadays, we think that special relativity is about the idea that there is some universal speed limit on the transfer of information (and experiments tell us that photons, the quanta of light, move with the largest speed, $c$).

But what if tomorrow we happen to observe a particle $X$ that travels with a speed $v>c$? What changes would have to be made to special relativity?

Answer

If (and that's a big if) tomorrow we had a $70\sigma$ detection in a repeatable experiment of a particle that travelled faster than $c$, then one of several things would be true.

1) We would be forced to conclude that $c$ is not, in fact, the limiting speed of information transfer; everything based on this assumption would have to be scrapped (pretty much all of research-level physics); and we would have to start over in developing even the mathematics that allows us to start re-describing the universe.

2) We would be forced to conclude that $c$ is not the limiting speed of information transfer; we would assume that special relativity and everything based on it is the special-case effective theory of much broader physical laws and behaviours; and we would have to find a way of modifying relativity (and basically everything that relies on it) so that it can causally allow for this particle to exist and yet have everything else we see still basically operate under the idea that $c$ is the max speed.

3) We find a way to use this particle to communicate with the past and future, travel faster than light, and then we go home every night and laugh at Einstein.

4) We perform the experiment thousands of times in different laboratories, find the same result, then go back and discover that there was a fundamental flaw with the theory. Once the flaw is corrected, we see that we are not actually observing a superluminal particle.

5) We also discover flying pigs, perpetual motion, and that we really can believe it's not butter. Then I wake up from my nightmare.

My money is on (4) with (2) being a close second (although (5) has happened before).

---

Note: This answer assumes that the speed $c$ referenced is the assumed maximum speed; the speed of a massless particle in a vacuum. This is why I did not include a "we determine the photon is not massless and then have to change EM" option.

homework and exercises - Why isn't the relative y-velocity in this problem $.8c$?

I am trying to solve the following problem:

A radioactive atom is moving in the $x$ direction in the laboratory at speed $0.3c$. It emits an electron having speed $0.8c$ in the rest frame of the atom. What will be the velocity of the electron in the laboratory when it is ejected in the $y$ direction in the atom's rest frame? (Find $u_y$, $u_x$, $|u|$, and $\theta$.)

I thought that since the atom is moving in the $x$ direction relative to the laboratory, then any motion in the $y$ direction relative to it must be the same for the laboratory. However, when I put $u_y = 0.8c$, the teacher marked it wrong; he refuses to explain why. Why is $u_y \neq 0.8c$?

Answer

To see why it is wrong it is easier to consider the case where the x direction of the initial particle is $w^{'}_{x} = 0.8c$ (velocity of radioactive particle) in the lab frame (indicated by prime). Now since the second particle was traveling at $u_{y} = 0.8c$ in the rest frame, if that was also how fast it was traveling in the lab frame then $u^{'}_{x} = u^{'}_{y} = 0.8c$ using pythagorus' theorem this means the velocity of the ejected particle is greater then the speed of light which indicates that something is wrong.

The source of the error comes about from the fact that there is time dilation between the reference frames. This is why you can never get above the speed of light. To work out the solution it is better to think about the world line that the ejected particle has, in the rest frame we get $\vec{x}(t) = (ct,0,u_{y} t)$. Here I'm giving time in units of length because that feels nicer to me and the transformation becomes nice and symmetric and I'll give that below, for a change in velocity $v$ the transformations are:

$y^{'} = y$

$x^{'} = \gamma \left(x-\frac{v}{c}(ct)\right)$

$(ct^{'}) = \gamma \left((ct)-\frac{v}{c}x\right)$

which means in the lab coordinates (where $v = -u_{x}$) this is $\vec{x}^{'}(t) = (\gamma ct,\gamma u_{x}t,u_{y} t)$.

At first glance this looks like the velocity in the y direction is still $u_{y}$ but t is the time coordinate for the original frame, not the lab frame. Instead we can see that it is $t^{'} = \gamma t$ by looking at the first term in $\vec{x}^{'}$ and we get:

$\vec{x}^{'}(t) = \left(ct^{'},u_{x}t^{'},\frac{u_{y}}{\gamma} t^{'}\right)$

So the correct speed is $\frac{u_{y}}{\gamma}$ which when we sub in the numbers comes to be $\frac{0.8c}{\sqrt{1-(0.3)^{2}}} = \frac{0.8}{\sqrt{0.91}}c$, so its slightly slower then what you would think, and this comes down to time dilation effects.

Main thing that might trip you up here is the sign for the coordinate change but that can be easily checked by working out what direction a stationary particle would be moving and check it would be moving in the correct direction)

(edit: Corrected subscript)

units - Why are significant figure rules in Multiplication/Division different than in Addition/Subtraction?

I've never understood specifically why this is.

Here's what I mean.

---

In Addition/Subtraction, what matters are the digits after the decimal point. So for example:

1.689 + 4.3 =

  1.629
+ 4.3XX

---------
  5.929
---------
  5.9

This makes sense to me. I filled in uncertain values with X, and it makes sense why I can't use the 0.029 in the answer - because I added it to an uncertain value.

---

However, I don't understand the rules when it comes to Multiplication/Division. The same little trick with X's doesn't help me here.

I know that what matters in Multiplication/Division are the significant figures. So for example:

12.3 * 4.6 =

  12.3
*  4.6
-------
   738
  492X
-------
  56.58
-------
  57

The answer is 57 according to significant figure rules of Multiplication/Division, but I just can't make sense of those rules like the way I did with Addition/Subtraction.

Does anyone have an intuitive explanation for the significant figure rules of Multiplication/Division?

Answer

The number of significant figures is a representation of the uncertainty of a number. $123.4$ has an uncertainty of $0.1$ since the first uncertain digit is usually included. So, your multiplication of $$12.3 \times 4.6$$ is better represented as $$(12.3 \pm 0.1) (4.6 \pm 0.1).$$ I'm going to use @barrycarter's trick of using scientific notation to better represent the values $$(1.23 \pm 0.01) \cdot 10^1 \times (4.6 \pm 0.1) \cdot 10^0.$$ $$(1.23 \pm 0.01) (4.6 \pm 0.1) \times 10^1.$$ From this we can see that the number with more significant digits has less relative uncertainty. Relative uncertainty is approximately the measured uncertainty (also called absolute uncertainty) divided by the value.

Multiplying out: $$((1.23)(4.6) + (1.23)(\pm 0.1) + (4.6)(\pm 0.01) + (\pm 0.01)(\pm 0.1)) \times 10^1.$$ $$(5.658 + (\pm 0.123) + (\pm 0.046) + (\pm 0.001)) \times 10^1.$$ The number of digits we can write down is determined by the total uncertainty, which here is dominated by $\pm 0.123$. So, our result is $$(5.658 \pm 0.123) \times 10^1.$$ $$(5.7 \pm 0.1) \times 10^1.$$ $$57 \pm 1$$ which results in $57$.

To summarize, in multiplication, the largest relative uncertainty dominates the uncertainty of the final answer. This translates into the the number with fewer significant digits determining the significant digits of the final answer. This contrasts with addition, where the largest absolute uncertainty determines the uncertainty of the answer.

quantum gravity - Is the 125 GeV Higgs boson some kind of a "almost-commutative graviton" at the electroweak scale?

The clumsy "almost-commutative graviton" is provocative. I use it on purpose, to ask two questions in one :

• 
  

  Is the observation of only one Higgs and no supersymmetric particle below 8 TeV (up to now) a sufficient evidence to substantiate the almost commutative spectral model?

  
  


• 
  
  

  Can physicists consider now this kind of models pioneered by Connes and Chamseddine to be an effective (physical) and not only formal tentative unification of gravitation and Yang-Mill-Higgs interactions?

  
  

Recent developments of the almost-commutative spectral model regarding the Higgs boson and its mass:

Last comments:

(motives for "graviton" as a metaphore and "almost commutative" as a pedagogical reminder)

• I know that graviton is a spin 2 gauge boson associated to the gravitational field in a tentative quantification of general relativity. In the framework of Quantum Field Theory it is thus an object independant a priori from the Higgs that is a scalar boson responsible for masses of elementary particles from the Standard Model. Nevertheless I remind that Higgs interaction can be considered as derivated from gravitation in the noncommutative geometric setting (following Thomas Schücker).


• The adjective almost-commutative has a precise technical meaning but I use it also in my question to underline the fact that in any theoretical framework non-commutativity is a necessary but not sufficient tool to describe quantum phenomena, therefore it is clear that gravitation has not been quantized yet!

mass - How can you weigh your own head in an accurate way?

I read some methods but they're not accurate. They use the Archimedes principle and they assume uniform body density which of course is far from true. Others are silly like this one:

Take a knife then remove your head.
Place it on some scale
Take the reading
re-attach your head.

I'm looking for some ingenious way of doing this accurately without having to lie on your back and put your head on a scale which isn't a good idea.

fluid dynamics - Speed of sound in astrophysics

Why is the speed of sound given so much importance in Astrophysics? For example in gas outflow (and accretion) problems, we often calculate the sonic point (the point at which the outflow speed becomes equal to the speed of sound). Is it simply because speed of sound is a convenient number that can be used to scale down the variables, or is something deep involved?

Answer

For one thing the speed of sound determines the rate at which disturbances can propagate in the medium or the "cone of causality". And this characteristic scale in turn determines at what point the system becomes unstable and heads towards gravitational collapse. In cosmology this is known as the Jeans instability. The Jeans length - i.e. the characteristic scale which demarcates perturbations which will induce gravitational collapse from stable oscillations of the medium is given by:

$$ \lambda_J = c_s \left( \frac{\pi}{G\rho} \right)^{1/2} $$

where $c_s$ is the speed of sound in the gas and $\rho$ is the mean density.

homework and exercises - Magnitude of magnetic field at the center of circular wire

I'm preparing for an exam by solving the sample questions , here is the one I'm having difficulty with :

Following is the given circuit.
Which contains two resistance $R_1$ and $R_2$ in form of circle of radius $r = 1 m$ with a battery having e.m.f. $V = 10π $ volt.
Upper resistance is having resistivity $p_1 = 4$ & lower resistance having resistivity $p_2 = 2$.
Angle between two points $A$ and $B$ is $60°$. (wires have same cross section $A_1 = A_2 = 2 cm^2$)

Diagram :

Here is the question based on this passage that I'm having difficulty with :
Find magnitude of magnetic field at center.

(A) 0T (B) 1T (C) 2T (D) πT

The problem is I don't know any formula to calculate the magnetic field at a point nor am I expected to know (probably , at least it is not in syllabus) .

Also what is 'T' ? Is it unit of magnetic field?
The answer is 0 so I guess I should be able to reason why it should be 0 but I don't know how. Please help.

special relativity - Is it really possible to create some mass only from equivalent energy?

From Special theory of relativity we know that $E=m_0 c^2$, which says about mass energy equivalency. But my question : **Is there any real experiment where some mass is created purely from energy? **

electrostatics - Why doesn't a gaussian surface pass through discrete charges?

I have read that Gaussian surface cannot pass through discrete charges. Why is it so? I have even seen in application of Gauss' Law when we imagine a Gaussian Surface passing through a charge distribution, e.g. in case of infinite plane charge carrying sheet .

If it cannot pass through discrete charges how do we use it in continuous charge distributions as same 'objection' must be there for it also.

Please explain the reason.

Here $E \rightarrow \infty$ as, $r\rightarrow 0$

If this is ambiguity then this must be same in continuous charge distribution , otherwise please state it more clearly because we can define charge to be a spherical ball and half charge can be considered inside surface (as in pic and even agreed by @JoshuaBarr).

homework and exercises - Finding the normal ordered momentum operator for free theory

I am asked to show that the normal ordered momentum operator for free theory is

$$\hat{p^\mu} = \int \frac{d^3 p}{(2 \pi)^3} p^\mu \: a_p^\dagger \:a_p.$$

The free theory Lagrangian is given by

$$\mathcal{L}=\frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi.$$

I have shown that for this Lagrangian the energy-momentum tensor is

$$T^{\mu \nu} = \partial^\mu \phi \partial^\nu \phi - \eta^{\mu \nu}\mathcal{L}.$$

From here we can show that the energy is

$$p_0 = \int d^3x T^{00} = \int d^3x \left( \frac{1}{2} \phi^2 + \frac{1}{2} (\nabla \phi) ^2 + \frac{1}{2}m^2 \phi^2\right),$$

and the three-momentum is

$$p_i = \int d^3x T^{0i} = \int d^3x \:\dot{\phi} \partial^i \phi.$$

When quantizing these expressions, I am trying to work in the Schrodinger picture. In order to show the first expression, I will work with the spatial and temporal parts of the four-momentum separately due to the difference in their expressions in the classical field theory. At this point I'm only trying to show that the spatial part of the normal ordered momentum is consistent with the first equation of this post. Quantizing gives us

$$\hat{\phi}(x) = \int \frac{d^3 p}{(2 \pi)^3 \sqrt{2E_p}} \left[ a_p e^{i \vec{p} \cdot \vec{x}} + a_p^\dagger e^{-i \vec{p} \cdot \vec{x}} \right].$$

I am told that this expression actually contains the time part in the exponentials, but since this is the Schrodinger picture we choose to define this operator at some fixed time which we conveniently choose to be $t=0.$ Given that, the time derivative of the above expression is

$$\dot{\phi}(x)= -i \int \frac{d^3p}{(2 \pi)^3} \sqrt{\frac{E_p}{2}} \left[ a_p e^{i \vec{p} \cdot \vec{x}} - a_p^\dagger e^{-i \vec{p} \cdot \vec{x}} \right],$$

and

$$\partial^i \phi = i \int \frac{d^3p}{(2 \pi)^3 \sqrt{2 E_p}} p^i \left[ a_p e^{i \vec{p} \cdot \vec{x}} - a_p^\dagger e^{-i \vec{p} \cdot \vec{x}} \right].$$

Per my rough calculation this means that

$$\dot{\phi} \partial^i \phi=\int \frac{d^3p d^3k}{(2 \pi)^6} \sqrt{\frac{E_p}{4 E_k}} k_i \left[ a_p a_k e^{i(p+k)x} - a_p a_k^\dagger e^{i(p-k)x} - a_p^\dagger a_k e^{i(k-p)x} + a_p^\dagger a_k^\dagger e^{-i(p+k)x} \right].$$

Then, using the exponential definition of the delta function and integrating through x and p I found

$$p^i = \int d^3x \dot{\phi} \partial^i \phi= \int \frac{d^3k}{(2 \pi)^3} \frac{1}{2} k_i \left[a_k a_{-k} - a_k a_k^\dagger - a_k^\dagger a_k + a_k^\dagger a_{-k}^\dagger \right].$$

I appear to have made a mistake in sign somewhere but I'm not too worried about that -- I can go back and find it on my own time. That said,, this is something like what I want except for the first and last terms in square brackets. If I do this in the Heisenberg picture I will get an exponentially decaying factor attached to the first and last terms which I think will make those two terms disappear but I feel like I should be able to get the same result in the Schrodinger picture, and I can't quite understand why I'm not.

New physics at high energies, cosmic rays, particle-detectors in space

New physics is expected at high energies and cosmic rays have high energies, so have there been or are there any plans to put particle detectors in space to study cosmic rays for new physics ?

Answer

JEM-EUSO is designed to look for air showers caused by extremely high energy cosmic rays (>$10^{19}$ eV). According to the website, it will be attached to the International Space Station in 2016.

newtonian mechanics - Physics of Carving on a Skateboard

How does carving on a skateboard work, why is it easier to carve on a longboard as opposed to a shortboard? I feel like I'm more prone to hurt myself trying to carve or throw my weight out on a shortboard, why is this? Is it just because you have more control over your weight distribution on a longboard? Can anyone give me an indepth analysis of this?

Answer

Can anyone give me an indepth analysis of this?

I might be able to. It would have been easier if you had told people what carving means, fortunately, as a skater (and skier), I know what you mean, so for the benefit of the non-skaters reading, carving is used in skiing, snowboarding, skateboarding and probably some other sports. It is a technique used to maintain a controllable speed when faced with a gradient that would otherwise provide too much of an acceleration.

This is done by moving in S paths down a slope. This video will give you an idea about what I mean. It's not a perfect technique (correct carving shouldn't skid) and it is of skiing, but the same thing applies for skateboarding too.
https://www.youtube.com/watch?v=qhZmnHtgwec

How does carving on a skateboard work

The same way that carving in general works, it slows you down (or reduces your overall acceleration down a hill) by making you travel partly across the slope. This picture demonstrates what I mean:

It shows two paths down a slope. The areas in red are where the skater will accelerate due to being headed directly down the slope, and the areas in blue mark where the skater will be decelerating due to moving across the slope and the friction slowing him/her down. Obviously the path on the left will cause the skater to travel faster and the path on the right is longer, slower and more controllable.

why is it easier to carve on a longboard as opposed to a shortboard?

Longboards turn better, it's part of their design. Better trucks for turning, better wheels for grip, you can lean into the turn further to ballance the centripetal force. Longboards were designed for carving and smooth turning because they were trying to mimic surfing on land.

I feel like I'm more prone to hurt myself trying to carve or throw my weight out on a shortboard, why is this? Is it just because you have more control over your weight distribution on a longboard?

Yes, the same point as above, longboards are built for carving, skateboards are not.

statistical mechanics - Calculation of the partition function for a classical 2D gas lying on the surface of a sphere of constant radius $R$

I'm kind of confused with this system. My first question is. Is the Hamiltonian of one particle of this gas the following? $$H(x,y,z,p_{x},p_{y},p_z)=\frac{1}{2m}\left(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\right)\to \frac{1}{2m}\left(\frac{p_{\theta}^{2}}{R^{2}}+\frac{p_{\phi}^{2}}{R^{2}\sin^{2}\theta}\right)$$ with $p_{\phi}=$constant.

If so, then my second question is: in order to obtain the classical partition function of one particle, should I integrate over the 6-dimensional phase space (Considering in spherical coordinates $r=$constant, $p_r=0$ and $p_{\phi}=$constant)? i.e. $$Z=\int \exp\left[-\frac{\beta}{2m}\left(\frac{p_{\theta}^{2}}{R^{2}}+\frac{p_{\phi}^{2}}{R^{2}\sin^{2}\theta}\right)\right] J\, drd\theta d\phi dp_rdp_{\theta}dp_{\phi} \qquad (1)$$

or should I just integrate over the 4-D phase space, since $r=$constant, $p_r=0$?

$$Z=\int \exp\left[-\frac{\beta}{2m}\left(\frac{p_{\theta}^{2}}{R^{2}}+\frac{p_{\phi}^{2}}{R^{2}\sin^{2}\theta}\right)\right] J\, d\theta d\phi dp_{\theta}dp_{\phi} \qquad (2)$$

And my final question is: in the expressions (1) and (2), is $J=1$ or $J=R^2\sin \theta $ ?

Answer

1. 
   

   Yes. The Hamiltonian is your $H$. $p_r=0$ because $\frac{\partial L}{\partial \dot r}=0$ with $L:=\frac{m}{2}(R^2\dot\theta^2+R^2\sin^2\theta\dot\phi^2)$.

   
   


2. 
   

   The integral is over the 4-dimensional phase space: $(\theta,\phi,p_{\theta},p_{\phi})$, because the particles just move over the 2D surface of the sphere.

   
   
   


3. 
   

   $J=1$ regardless of the coordinate system you are expressing the Hamiltonian. This is because of the Liouville's theorem that states that the phase volume is conserved under canonical transformations. In particular a coordinate transformation is included. So, in 3 dimensions:

   
   

$$Z=\frac{1}{h^3}\int d^3q d^3p \, \, \exp[-\beta H(\bar q,\bar p)] \qquad (*)$$

being $\bar q = (x,y,z)$ or $q=(r,\theta,\phi)$, etc.

*Note. The constant $h$ introduced in the integral (*) in order to maintain $Z$ dimensionless. So $h$ is just a constant (yet unknown), with units of action, i.e. units of angular momentum.

thermodynamics - Why does the gas get cold when I spray it?

When you spray gas from a compressed spray, the gas gets very cold, even though, the compressed spray is in the room temperature.

I think, when it goes from high pressure to lower one, it gets cold, right? but what is the reason behind that literally?

Answer

The temperature of the gas that is sprayed goes down because it adiabatically expands. This is simply because there is no heat transferred to or from the gas as it is sprayed, for the process is too fast. (See this Wikipedia article for more details on adiabatic processes.)

The mathematical explanation goes as follows: let the volume of the gas in the container be $V_i$, and its temperature $T_i$. After the gas is sprayed it occupies volume $V_f$ and has temperature $T_f$. In an adiabatic process $TV^{\,\gamma-1}=\text{constant}$ ($\gamma$ is a number bigger than one), and so $$ T_iV_i^{\,\gamma-1}=T_fV_f^{\,\gamma-1}, $$ or $$ T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1}. $$ Since $\gamma>1$ and, clearly, $V_f>V_i$ (the volume available to the gas after it's sprayed is much bigger than the one in the container), we get that $T_f

By the way, adiabatic expansion is the reason why you are able to blow both hot and cold air from your mouth. When you want to blow hot air you open your mouth wide, but when you want to blow cold air you tighten your lips and force the air through a small hole. That way the air goes from a small volume to the big volume around you, and cools down according to the equations above.

quantum mechanics - What would be likely to completely stop a subatomic particle assuming it was possible?

Suppose that completely stopping a subatomic particle, such as an electron, could happen under certain conditions. What would be likely ways to get an electron to be perfectly still, or even just stop rotating the nucleus and collapse into it by electromagnetic forces? What would likely be required, below absolute-zero temperatures? Negative energy? Or could a 0 energy rest state not exist in any form, of any possible universe imaginable?

Let's say there was a magnetic field of a certain shape that we could postulate that is so intensely strong that if we put an electron in the center of it, it could not move at all in any direction. Would the energy requirement of the field be infinite? What would be the particle's recourse under this condition?

Further, suppose it were possible and one could trap an electron and stop all motion completely. What would this do to Heisenberg's Uncertainty Principle and/or Quantum Mechanics, because its position and momentum (0) would both be known? If it can be done, is Quantum Mechanics no longer an accurate model of reality under these conditions? Could we say QM is an accurate model under most conditions, except where it is possible to measure both position and momentum of a particle with zero uncertainty?

Please assume, confined in a thought experiment, that it IS possible to stop a particle so that is has 0 fixed energy. This may mean Quantum Mechanics is false, and it may also mean that under certain conditions the uncertainty commutation is 0. ASSUMING that it could physically be done, what would be likely to do it, and what would be the implications on the rest of physics?

Now here's the step I'm really after - can anyone tell me why a model in which particles can be stopped is so obviously not the reality we live in? Consider the 'corrected' model is QM everywhere else (so all it's predictions hold in the 'normal' regions of the universe), but particles can be COMPLETELY stopped {{inside black holes, between supermagnetic fields, or insert other extremely difficult/rare conditions here}}. How do we know it's the case that because the uncertainty principle has lived up to testing on earth-accessable conditions, that it holds up under ALL conditions, everywhere, for all times?

Answer

Your question is interesting, and gets specifically to the kinds of questions that quantum mechanics was intended to answer in the first place. It helps to understand the motivation behind the original Bohr model of the atom, and how that led to QM in the first place.

The problem Bohr was trying to address can be paraphrased as, "If an electron orbits a nucleus like a planet, why doesn't it gradually lose energy and spiral into the nucleus?" The answer came when Bohr realized that the orbital momentum was quantized, effectively meaning that since the electron had mass then by the relationship $p=mv$, the velocity was also quantized (note: this simple expression is more complicated when relativity is included, but the discussion can continue without including it). These quantized values of momentum/velocity are what one would call eigenvalues, or observables in quantum mechanics. Since the electron can only change orbits by given off specific quantities of energy instead of giving off energy continuously, it remained in a stable orbit relative to the nucleus, thus preventing it from spiraling in.

What is important to understand is the idea of the potential well. In the Bohr model, the electrons closest to the nucleus have higher velocity than the electrons further away. In other words, they have greater kinetic energy ($K.E. = \frac{1}{2} mass \times velocity^2$), but it had less potential energy since it was closer to the nucleus (obeying the relationship $P.E. = mass \times distance \times gravity$). However, the total energy ($K.E. + P.E.$) associated with orbits closer to the nucleus is less than those further away. So in order for the electron to move closer to the nucleus, energy must be given up. This is accomplished by the emission of a photon. Alternatively, if one wants to cause an electron to move into a more distant orbit, then one must add energy through use of a photon.

It is in contemplating how to determine the orbit of the electron that the uncertainty principle first became apparent. The only probe that we have available to determine the position of an electron in its orbit is a photon, and the photon must be of sufficient energy in order for it to be small enough to give a meaningful result, however if we use a photon small enough (in terms of wavelength), it will have enough energy to shift the electron into a different orbit, and then we would have to start the process all over again.

A free electron has sufficient energy to escape the nucleus. In other words, it has acquired sufficient energy to fill its potential energy deficit. If there were only one nucleus in the universe, the potential deficit would only be eliminated when the electron was at infinite distance from the nucleus. In that situation, the implication is that the electron would also have zero velocity. This situation is obviously unrealistic, first there are more than one nuclei in the universe, and second, to verify that a particle has zero velocity at infinite distance is clearly an impossible task.

If we move past the Bohr model and into more modern quantum mechanics, the question then is whether there are eigenstates that have eigenvalues for momentum of a particle that are equal to zero? It is important to review some basic facts about matrix operations and linear algebra

If 0 is an eigenvalue of a matrix A, then the equation A x = λ x = 0 x = 0 must have nonzero solutions, which are the eigenvectors associated with λ = 0. But if A is square and A x = 0 has nonzero solutions, then A must be singular, that is, det A must be 0. This observation establishes the following fact: Zero is an eigenvalue of a matrix if and only if the matrix is singular.

This means that the matrix in question is not of full rank. In QFT this has a very specific interpretation. The annihilation operator has the power to destroy the vacuum state and map it to zero. This situation is understood to be associated with the free field vacuum state with no particles. This state is necessary because it allows us to find vacuum state solutions for the associated quantum mechanical system.

By means of analogy, we can see that the solution to ground state problem is the solution to the homogeneous part of a differential equation.

The Schrodinger equation is a linear, homogeneous equation which governs evolution of the wave function of a particle. The solutions of the Schrodinger equation can be used to understand particle motion. The exact position and momentum of a particle can only be known if h (planck's constant) approaches zero, however, in quantum mechanics, planck's constant is fundamental to the theory, so this cannot occur in the single particle case. Because of this, momentum and position uncertainty establish an inverse relation to each other, and if the uncertainty of momentum is zero, then the uncertainty in position is infinite.

For these reason's it is not possible to talk about a particle "stopping" or being "stopped" in any meaningful or non-contrived sense.

quantum gravity - Would a directional "graviton" emitter violate any known laws of physics?

Setting aside that we don't known what the mediating particle in quantum gravity looks like and have no way to manipulate it, what would the implications of a directional graviton source be? Would it allow a "reaction-less" drive without creating other problems (e.g. that a violation of conservation of momentum has corresponding implications with respect to translational invariance)?

Also, presuming that such a device (unlike mass) can be made to turn on and off, would that imply (via conservation of momentum and the finite propagation rate of gravity), that a graviton must have (negative?) momentum?

---

By "directional", I'm referring to a gravitational effect that, on average and over an arbitrarily long time, will attract particles more strongly in one hemisphere than the other.

optics - Could a human beat light in a footrace?

Is there anything preventing the following experiment from being done right now?

Imagine that a human ran from point 'a' to point 'b' while light particles that reflected off a clock moved through a special medium from point 'a' to point 'b' as well. Could a human arrive at point 'b' before light? (As for the special medium, I'm imagining something like a complex maze of mirrors submerged in a very dense material.)

If so, if the human waited at the finish line to view light arriving in second place, would they see the time that the race began on the clock?

Answer

No physical laws are being broken in this thought experiment. If you are concerned with the relativistic requirement "nothing can go faster than the speed of light", that only applies to the speed light goes in a vacuum: $c = 3 \times 10^8$ m/s. The reference to light in that relativity postulate makes it sound like if you could only find a situation where you slowed light down, you could break the laws of physics; not so. A better statement of the postulate would be "nothing can go faster than $3 \times 10^8$ m/s, which happens to also be the speed light travels at in a vacuum." I don't see anyone going faster than $3 \times 10^8$ m/s in this thought experiment, so no physics violations.

As for what the human at the end of the race sees:

He sees a blinding blue light from the all the Cherenkov Radiation from even the slightest charged particle passing through the medium. And perhaps the time at the start of the race. It's exactly what you would imagine since we are talking non-relativistic speeds. What an anti-climactic answer, eh?