atmospheric science - Is there any way to anticipate exceptional Aurora Borealis viewing conditions?

I've always wanted to see the Aurora Borealis, but never had the opportunity, since it is so rarely visible in my area (Pennsylvania).

However, last week was apparently a rare opportunity of exceptional visibility.

Unfortunately, I missed my opportunity, as I did not find out about it until the next day.

Are there any good resources or strategies for anticipating opportunities like this? Are coronal mass ejections like the one that precipitated this event predictable? If so, are such predictions available to the amateur public?

Answer

I use Twitter user Aurora_Alerts and its corresponding RSS feed to make aurora alerts show up in my RSS reader (which happens to be Opera).

Aurora_Alerts posts messages like In 53 minutes the Aurora should be Quiet 2.33 Kp and 31 minutes the Aurora should be at 'STORM' LEVEL! It's On!! 7.33 Kp (K_p is related to the K-index which quantifies disturbances in the horizontal component of earth's magnetic field with an integer in the range 0-9 with 1 being calm and 5 or more indicating a geomagnetic storm.).

SpaceWeather is a source for following the general state of the Sun and alerts of coronal mass ejections (CMEs), but it may not be updated frequently enough to get alerts in time (and it is quite tedious to keep checking it).

newtonian mechanics - Why won't a tight cable ever be fully straight?

I posted this picture of someone on a zipline on Facebook.

One of my friends saw it and asked this question, so he could try to calculate the speed at which someone on the zipline would be going when they hit the water.

The only answer, which was accepted, includes the disclaimer, "Assuming the pulley being used to slide to be friction less.Though not possible.Also the rope is assumed to be in-extensible and straight."

I used to have a zipline of about the same length in my back yard as a kid and even when I was young, I noticed that we could never straighten the line completely, even when it was slack, we could not make it completely straight. And, naturally, once weight was added, there was a curve where the weight pulled the line down.

One of the comments from the member providing the answer is "Well i can show you why the string cannot be ever straight." I know that from experience. We could never make it completely straight with no sagging. I asked the reason for this and was directed to a book on Amazon. Having just spend $50 on a number of books for summer reading, my book budget is gone for a while.

So can someone answer that? Why will the line never be straight when it's set up (and when there is no load on it)?

Answer

Imagine a heavy chord raised off the ground between two blocks. Rather than consider all of the mass pieces of the rope, and the forces on them, we can simplify the problem a little bit by considering a slightly different one.

The chord can be represented by a heavy ball (in the middle of the chord) connected by two massless strings to the blocks. From experience, we know that this mass/string combination forms a triangle with two sides that are the same length and one other side. Each of the slanted sides forms some angle with respect to the ground.

When you pull tighter on the strings (in the picture of the tightrope walker below, for example) the ball (or tightrope walker) goes up a little. The angle between the angled sides and the ground gets smaller. But the tighter you make the strings, and the higher the ball goes, the more the tension in the string (which always points along a string) is going into pulling the ball sideways.

So consider this. If the ball were hanging at its highest point, that is, with the strings forming a straight line instead of a triangle, then the tension force of the rope would be pulling totally horizontally on the ball. But this doesn't cancel the force of gravity, which pulls the ball downward -- regardless of the tension in the rope. So the ball will sink a little.

Therefore there can never be a configuration where a ball hangs on a straight rope. The rope must have a kink in it. This is the same reason why if you have a "straight" rope and a tightrope walker walks on it, the rope must sag a little. See the diagram below. In a static problem, all of the components of the arrows (left-right, up-down) have to add up to zero. Notice how the tension arrows point a tiny bit upward.

For a heavy rope (a rope with mass but without a ball) the hanging doesn't form a kink, but a slightly slopey curve. The principle is the same.

faster than light - How would you connect a destination to a wormhole from your starting point to travel through it?

Lets say it was possible to create as massive traversable wormholes as we want. Say we started from earth and wanted to cross to alpha centauri. How would I create the wormhole so that it leads directly to alpha centauri? So we start from earth and want the wormhole to start from their and end at alpha centauri, how would you extend the wormhole there from your starting point?

quantum mechanics - Pegg-Barnett phase implementation does not seem to work

I attempt to monitor the phase of a wavevector $|\psi\rangle$.

As I found (e.g. here ), a matrix representation for the Pegg-Barnett phase operator in Fock base can be obtained as

$$\Phi=\sum_{m,n,n'=0}^s\frac{2\pi m}{(s+1)^2}\,\exp\left[\frac{2\pi i m(n-n')}{s+1}\right]|n\rangle\langle n'|$$

for a particle number cutoff $s$. Making the indices start at one instead of zero (for implementation in matlab), this becomes

$$\Phi=\sum_{\tilde{n},\tilde{n'}=1}^\tilde{s}\sum_{m=0}^{\tilde{s}-1}\frac{2\pi m}{\tilde{s}^2}\,\exp\left[\frac{2\pi i m(\tilde{n}-\tilde{n'})}{\tilde{s}}\right]|\tilde{n}\rangle\langle \tilde{n'}|$$ such that $|\widetilde{n+1}\rangle=|n\rangle$.

When I construct the wavefunction though of e.g. a coherent state with $\alpha=\sqrt{30}$; $\langle\alpha|\Phi|\alpha\rangle$ with $\tilde{s}=200$ gives me a phase ~2.7, whereas this coherent state is located in phase-space on the positive X-axis (so that phase zero would be expected.)

What went wrong?

NOTE ADDED: by trial and error, I found that we can obtain a phase prediction of zero if the summation over m also goes up to $\tilde{s}$ and in addition the reference phase $\theta_0$ is chosen $-\pi$ instead of zero. I don't understand why though, and if this is to be trusted as a measure of phase. At first sight, it does not seem to fullfil the small angle approximation $\phi\approx \frac{P}{X}$. Any insight is still welcome.

CODE USED with the original parameters

%% Constructing the initial state
alpha=sqrt(30);Nmax=200;
a=sparse(Nmax,Nmax);for it=1:Nmax-1,  a(it,it+1)=sqrt(it); end;adag=a';  
%annihilation and creation operators for use in the displacement operator.
vacuumpsi=zeros(Nmax,1); vacuumpsi(1,:)=1;
largepsiinit=exp(-0.5*abs(alpha)^2)*(expm(alpha*adag)*vacuumpsi);


%% The PB operator (tildes omitted)
s=Nmax;
PBphaseop=zeros(s);
thetazero=0;
for n=1:s
 for nprime=1:s
  for m=0:s-1
  PBphaseop(n,nprime)=PBphaseop(n,nprime)+(-
  thetazero+2*pi*m/(s))/Nmax*exp(1i*(n-nprime)*2*pi*m/(s));
   end

 end
end

%% Calculate expectation value of phase
phaseexp=largepsiinit'*PBphaseop*largepsiinit

Answer

What works quite nicely is defining the operator from eq.33 in https://arxiv.org/pdf/hep-th/9304036.pdf .

Also, it was to be expected that the PB-phase of a coherent state on the positive real axis is not zero for reference phase $\theta_0=0$, namely the eigenvalue spectrum is bounded by 0 and $2\pi$ and as a coherent state has a finite phase uncertainty, it crosses the cut in phase.

Aside from these issues, this PB-operator used has proven useful in monitoring the phase of a numerical wavevector and gives quite intuitive results. For further improvement, one can also iteratively update the used $\theta_0$ to avoid crossing of the phase-cut.

electromagnetic radiation - Difference between an electron wiggler and an undulator?

Both wigglers and undulators use periodic magnetic fields applied to stored relativistic electron beams to produce intense beams of UV or X-rays that can be used in a wide range of condensed matter physics and materials science experiments.

This helpful answer to a different question states:

As for the difference between undulators and wigglers: synchrotron radiation has a characteristic opening angle (that goes as $1/\gamma$). In an undulator, the electron motion in the transverse direction is set to be on the order of the opening angle. In a wiggler, however, the motion is made to be larger than the opening angle and therefore a wider beam results.

That answer also links to a google books webpage for Synchrotron Radiation Sources: A Primer which says in section 14.2.2 (undulator radiation):

In an undulator, radiation from the various periods interfere coherently. Sharp peaks are produced at harmonics of the resonant frequency, which depends on the electron energy, the undulation period and field strength, and the observation position. The optical wavelength is a Lorentz transformation of the undulation period into the beam frame followed by a relativistic Doppler shift back into the laboratory frame. The velocity used in the Lorentz transformation and the Doppler shift is the longitudinal electron velocity, which is less than the full electron velocity because of the electron’s curved path through the undulator.

Functionally speaking, it sounds like a wiggler could be characterized as an incoherent, underperforming undulator.

An undulator used with the wrong electron beam energy would be in effect a wiggler.

The narrow energy spectrum from a properly operated undulator, the result of the coherent addition in the forward direction would spread out to a broadband spectrum several orders of magnitude wider in energy and lower in brightness (energy per unit energy and solid angle) when operated with the wrong electron energy or the wrong alternating magnetic field strength.

Am I missing something fundamental, or does this pretty much describe the difference between the two devices?

note: I'm not asking for the difference between the two beams, I'm asking about the devices themselves, so if there are specific issues with their magnetic field designs that would be more interesting than differences in beam characteristics.

Answer

Short answer: Yes, you pretty much summed it up. However, I wouldn't go as far as describing a wiggler as an "underperforming" undulator.

Both these devices convert kinetic energy into radiation. A wiggler can produce magnitudes more radiation than an undulator, precisely because it radiates in a wider spectrum. Its overall output power is therefore much higher. (it's essentially a row of bending-magnets.)

An undulator provides a higher power per frequency, since its output is generally confined to a single frequency (and its harmonics). However, the output of a wiggler at that specific frequency can be comparable or higher (with the same electron beam / the same number of pair magnets). The undulator setup acts mainly as a filter, so unwanted frequencies are absorbed by the electrons and returned to kinetic energy of the electron beam. It does however produce a much narrower beam and thereby its brilliance can be much higher. The main advantage, if a specific frequency is desired, is in reducing heat load to monochromators after the insertion device.

Both undulator and wiggler are so similar, they are usually built as the same device. By increasing or decreasing magnetic field strength or moving permanent magnets closer or farther apart, the device can be set up as wiggler or undulator. The device can be classified as one or the other by its strength parameter:

$$K=\frac{e B \lambda_U}{2 \pi m_e c}$$

• K << 1 : classic undulator, waves can interfere and radiation will be a harmonic of the spatial magnet separation $\lambda_U$.


• K >> 1 : classic wiggler, electrons radiate independently and a wide high-power spectrum is the result.


• K ~ 1 : in between, generally results in a spectrum similar to that of a single bending magnet, though undulator/wiggler specific polarization.

classical mechanics - Floating Objects and Weight

The Situation:

A ball is placed in a beaker filled with water and floats. It is also attached to the bottom of the beaker via a string.

The Question:

The ball is attached to the beaker, thus making the ball and beaker (and string included) a system. However, it is floating. Does the ball add to the weight of the beaker?

Answer

Yes, it does. When an object floats, its mass is not affected. It only affect the force experienced by it, as the water exerts a "buoyant force" on the object:

Basically, there is a pressure difference between the top and bottom surfaces, and this corresponds to a force difference, leading to a net upwards force:

However, remember that the force exerted by the water on the object leads to an equal and opposite force exerted by the object on the water. Since at equilibrium, $m_{obj}g=\text{buoyant force}$, a force equal to the weight of the body is exerted on the water. When the beaker is weighed, this extra force is balanced by the normal force.

Here are some free body diagrams. BF is the buoyant force, T is the string tension. Note that N is the weight that the weighing platform measures.

quantum mechanics - Are virtual pions necessary to mediate the residual strong interaction?

In relation to another question I asked recently, I was wondering if the forces between the up and down quarks of the protons and neutrons in nuclei would be sufficient to keep the nuclei bound together without invoking virtual pions as pseudo-Goldstone bosons for the residual strong interaction.

Also, a quick note on the detection of these pions via proton-proton collisions, it seems like pions would be produced anyway and that this says little about the residual strong interaction. That is, it doesn't prove that these virtual pions do indeed mediate that interaction.

Any help is much appreciated.

Answer

You seem to be misunderstanding what the effective theory with the residual nuclear force between hadrons mediated by pions is: It is an effective theory which is a good approximation of the underlying nuclear force between quarks mediated by gluons. It is not that one of these is correct and the other false, or that they represent two different forces - they are both descriptions of fundamentally the same phenomenon at different levels of detail.

You cannot detect virtual particles mediating an interaction anyway - since a virtual particle is just a mathematical fiction arising from the diagrammatic description of our perturbation series used to compute QFT amplitudes, it does not "exist" in the sense that it could be detected.