spectroscopy - Bleaching groundstate

I'm reading an article about two-dimensional infrared spectroscopy and I don't understand the following sentence.

Bleach or stimulated emission contributions yield negative signals.

What are "bleach contributions"? I have never heard of it and cannot find a suitable explanation on the Internet. Can somebody explain that to me?

The article I'm referring to is J. Chem. Phys. 121, 5935 (2004)

Answer

I feel obliged to post another answer since the first one contains a mistake.

As you deplete the ground state, you also populate the excited state, and when the photon interacts with the excited molecule, then instead of being absorbed it generates another photon through stimulated emission. The paper you cite treats this contribution as a negative signal.

Eventually, at 50/50 population you reach an equilibrium, where the amount of absorbed light is equal to the amount of light produced by stimulated emission, and it looks like your medium is not absorbing light at all, which is why it is called bleaching (i.e. loosing its color).

classical mechanics - Water bottle moment of inertia

I've noticed that I can make a full water bottle spin about its short axis easier than I can make it spin when it is 1/4 or 1/2 full. Also, when it is spun and is not full, the geometric center of the water bottle moves in a bizarre way (kind of circular).

In general the moment of inertia is directly proportional to the mass. So why is this the case?

I can see that the water in the bottle has more freedom to move when it is not full than when it is full, and this affects the spin. My idea is that it maybe has something to do with the fact that the center of mass of the system is not fixed to a certain point relative to the bottle when this bottle is not completely full.

But I'd like to see a mathematical explanation of this phenomenon.

Answer

This is a late answer; a recent question was marked as a duplicate of this.

The phenomenon discussed in the question goes under the general concept of the dynamics of sloshing liquid. Sloshing liquids can overturn tank trucks, derail railroad tanker cars, capsize ships at sea, crash aircraft, and cause spacecraft to lose controllability. This makes this a very important concept for economic and safety reasons and hence is the subject of many journal articles and entire technical books.

The dynamics of slosh are nonlinear, rather complex (particularly so if the sloshing is extreme and creates bubbles), and are highly dependent on container geometry. As a general rule, a container that is nearly full or nearly empty of fluid doesn't slosh much, and sloshing is at its worst when the container is close to half full.

Excitations from vehicle suspension, from vehicle acceleration and braking, from ocean waves, and from the control systems of aircraft and spacecraft can turn low amplitude sloshing into high amplitude sloshing, something that is best avoided.

But I'd like to see a mathematical explanation of this phenomenon.

You are inadvertently asking me to write a lot. Go to scholar.google.com and books.google.com and search for "slosh dynamics" and you'll see how much has been written on understanding and mitigating slosh. I'll instead provide an overview.

Most slosh models are a bit ad hoc. A simple approach is to model the fluid as being partitioned into a fixed part (one that moves with the container) and a sloshing part, with the sloshing part modeled as a spring/mass/damper system or as a damped pendulum system. The slosh wave slams into the container wall, and this has to be modeled as well. This works well for low amplitude slosh, not so well for high amplitude slosh. These low amplitude slosh models yield a natural slosh frequency. These models predate modern computing.

More recently, slosh has been studied using computational fluid dynamics, and even more recently, with smoothed-particle hydrodynamics originally developed by astrophysicists to model galaxy formation, star formation, supernovae, etc. The same techniques work quite nicely to model more mundane fluids such as sloshing in a container.

quantum mechanics - How can one derive Schrödinger equation?

The Schrödinger equation is the basis to understanding quantum mechanics, but how can one derive it? I asked my instructor but he told me that it came from the experience of Schrödinger and his experiments. My question is, can one derive the Schrödinger equation mathematically?

Answer

Be aware that a "mathematical derivation" of a physical principle is, in general, not possible. Mathematics does not concern the real world, we always need empirical input to decide which mathematical frameworks correspond to the real world.

However, the Schrödinger equation can be seen arising naturally from classical mechanics through the process of quantization. More precisely, we can motivate quantum mechanics from classical mechanics purely through Lie theory, as is discussed here, yielding the quantization prescription

$$ \{\dot{},\dot{}\} \mapsto \frac{1}{\mathrm{i}\hbar}[\dot{},\dot{}]$$

for the classical Poisson bracket. Now, the classical evolution of observables on the phase space is

$$ \frac{\mathrm{d}}{\mathrm{d}t} f = \{f,H\} + \partial_t f$$

and so its quantization is the operator equation

$$ \frac{\mathrm{d}}{\mathrm{d}t} f = \frac{\mathrm{i}}{\hbar}[H,f] + \partial_t f$$

which is the equation of motion in the Heisenberg picture. Since the Heisenberg and Schrödinger picture are unitarily equivalent, this is a "derivation" of the Schrödinger equation from classical phase space mechanics.

forces - Newton's second law of motion in terms of momentum

I am reading a document and in answer to the question "State Newton’s second law of motion", the candidate answers that

The force acting on an object equals the rate of change of momentum of the object.

While this is not a complete answer, the examiner picks on on a word equal and says that it is proportional instead.

Now I understand that $F=\frac{dp}{dt}$ where both $F$ and $p$ are vectors — what is "proportional" about it?

I checked my Good Old Ohanian (2nd edition) and it says explicitly "The rate of change of momentum equals force" in section 5.5 The Momentum of a Particle.

What is this examiner talking about?

Answer

This actually comes down to a question about the interpretation of units, and it's kind of a tricky issue.

On one hand, you can view any equation in physics as nothing more than a mathematical relationship between some numbers. This was the view taken in the early days of quantitative physics,^* back in the 17th and 18th centuries, when the concepts of force and momentum were just being quantified. Since there was very little in the way of collaboration, the idea of a standardized unit system hadn't really taken off, so if you were doing an experiment to establish the relationship between force, momentum, and time, the units you used would have been determined by your equipment. In other words, all you would know is that you have a force meter (scale) which will give you a number proportional to the force, a "momentum meter" which will give you a number proportional to the momentum change, and a clock of some sort (perhaps a pendulum) which will give you a number proportional to the time.

Let's say it's been established that the relationship is linear. You would probably run an experiment in which you apply a certain quantity of force for a given number of ticks of the clock, and change in momentum off your measuring device. Then you would plug those numbers - it's important to notice that you're only dealing with numbers, since there are really no meaningful units to speak of - into the discrete approximation of Newton's second law: $$F^{(N)} = K_F^{(N)}\frac{\Delta p^{(N)}}{\Delta t^{(N)}}\tag{1}$$ Here I've used the superscript $^{(N)}$ to indicate pure numerical value, i.e. the number you read off the scale/clock/meter. Plugging in these numbers will allow you to determine the value of the constant $K_F^{(N)}$.

Hopefully it's obvious that the value of this constant will depend on how your equipment is calibrated, or in other words, which unit system you're using. For example, suppose your equipment is calibrated so that:

• Your force meter measures in what we now call pounds. In other words, a reading of $3$ on the force meter corresponds to the force we would now call $3\ \mathrm{lb_F}$, and similarly for other numbers - but of course, when you take that reading, you just think of it as "$3$".


• Your momentum meter measures in what we would currently call pound-feet per second, $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}$.


• Your time meter measures in units equivalent to modern seconds.

Then you would find that $$K_F^{(N)} = 32.174$$ Back in those days, before people started really thinking about units, all multiplicative equations in physics were thought of as simple relationships between numbers. Accordingly, they included constants of proportionality which were customized to each lab's equipment.

Eventually, as more people started doing physics, there arose a need for standardized unit system so you could compare data from different labs. The units were first thought of as reference values, and your own measurements would represent how many times the reference value you were measuring. If you were lucky enough to have equipment calibrated to match the reference value, you could just read the number right off it, but otherwise you would have to convert your measurements using an equation like $$F^{(\text{standard})} = \biggl(\frac{\text{your unit}}{\text{standard reference value}}\biggr)F^{(N)}$$ and so on for other quantities.You would find some way to determine the conversion factor $\frac{\text{your unit}}{\text{standard reference value}}$ for each of your measurement devices, and then you'd have to do this conversion each time you made a measurement.

Or, you could be smart, and put the reference values right into the equation. This is the big advance in unit systems, because your measurements are not just numbers anymore. Whereas before you would work with, say, $F^{(N)}$ or $F^{(\text{standard})}$, now you can work with the quantity $F$, where $$F = F^{(\text{standard})}(\text{standard reference value}) = F^{(N)}(\text{your unit})$$ So, again assuming that your units for force, momentum, and time correspond to the modern $\mathrm{lb_F}$, $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}$, and $s$, respectively, you can write the above equation (1) as $$\frac{F}{\mathrm{lb_F}} = \frac{K_F}{\mathrm{lb_F}/(\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2})}\frac{\bigl(\frac{\Delta p}{\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}}\bigr)}{\bigl(\frac{\Delta t}{\mathrm{s}}\bigr)}$$ Then you can algebraically rearrange this to $$F = K_F\frac{\Delta p}{\Delta t} \quad\text{where}\quad K_F = K_F^{(N)}\frac{\mathrm{lb_F}}{\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2}}$$

So Newton's second law has shifted from being a simple relationship between numbers, to being a relationship between physical quantities which are expressed as multiples of a reference value. However, you still have a conversion constant in the equation. Symbolically, it's independent of units, but you still do have to plug in a different number depending on which combination of units you want to work with. This is the sense in which $F$ is only proportional to $\frac{\mathrm{d}p}{\mathrm{d}t}$.

In the modern scientific community, on the other hand, I think most people would agree that units are a human invention, and that physical quantities should exist in some sense independently of the units we choose to use for them. Taking that view, there should be some "natural" way to express the equations of physics that doesn't incorporate any "unit system artifacts" like these proportionality constants.

The way we do this is to define the units as abstract objects and develop a set of rules for manipulating them (kind of like unit vectors). We can then incorporate the conversion constants into those rules. For example, let's again consider the discrete approximation of Newton's second law, but this time without the conversion constant written into it:

$$F = \frac{\Delta p}{\Delta t}$$

You can still use seconds for time and $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}$ for momentum in this equation. When you read the numbers off your measuring equipment and plug them into the formula, you'll do it like this:

$$F = \frac{\Delta p^{(N)}\ \mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}}{\Delta t^{(N)}\ \mathrm{s}} = \frac{\Delta p^{(N)}}{\Delta t^{(N)}}\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2}$$

Suppose you want your answer in pounds of force. You would look up the multiplication rule for $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2} \to \mathrm{lb_F}$, which in this case can be found on Wikipedia:

$$\mathrm{lb_F} = 32.174\ \mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2}$$

(in general you might have to chain a few rules together to get the right conversion). So you wind up with

$$F = \frac{1}{32.174}\frac{\Delta p^{(N)}}{\Delta t^{(N)}}\mathrm{lb_F}$$

It works out to the same thing as before, but this time the conversion constant $K_F$ is part of the unit system, not the equation. This means that if you're not plugging actual values into the equation, you don't have to think about units or proportionality constants at all. And if you look at it this way, $F$ is actually equal to $\frac{\mathrm{d}p}{\mathrm{d}t}$.

So what's the verdict? Unfortunately, there's no unassailable answer to this question of whether Newton's second law is a proportionality or an equality. Depending on how you think about it, either answer could be valid. But I would say the "equality" answer, which corresponds to the modern view of units, is conceptually cleaner. It's accepted by all competent modern physicists, as far as I know (for mechanics, at least; electromagnetism is a whole different story), and it's certainly the interpretation we try (however unsuccessfully) to instill into introductory physics students' minds. I'd definitely agree that the examiner was being unreasonably picky (though to be fair, he did give credit for it).

---

^{*I don't have an explicit source, so I'm not entirely sure this is the way things were really developed; I'm basing my description on some fuzzy memories. That being said, the backstory does help clarify the various ways in which we treat units, so consider it historical fiction if you must.}

perturbation theory - Linear response correlation functions

I'm working my way through Methods of Molecular Quantum Mechanics by R. McWeeny and have run into a derivation I can't seem to figure out.

So in chapter 12, he obtains an expression for the first order coefficients of the perturbed wavefunction with respect to a perturbation $H'(t)=F(t)\mathbf{A}$. $\mathbf{A}$ is a hermitian operator and $F(t)$ is a time dependent strength factor and the system was assumed to have started in the state $|0\rangle$ with the perturbation being weak so that these coefficients vary slowly.

$$c_n^{(1)}=(i\hbar)^{-1}\int_{-\infty}^t\langle n|\mathbf{A}|0 \rangle F(t')\exp(i\omega_{n0}t')dt'$$

I'm fine with this expression. Where I get confused is when we try to use this expression to determine the response of some operator $\mathbf{B}$ to the perturbation described by $\mathbf{A}$. He writes that $$\langle \mathbf{B} \rangle-\langle \mathbf{B} \rangle_0=\delta\langle \mathbf{B} \rangle=$$

$$(i\hbar)^{-1}\int_{-\infty}^t\sum_{n\neq0}\bigr[ \langle 0|\mathbf{B}|n \rangle \langle n|\mathbf{A}|0 \rangle \exp(-i\omega_{n0}(t-t'))-\langle 0|\mathbf{A}|n \rangle \langle n|\mathbf{B}|0 \rangle \exp(i\omega_{n0}(t-t'))\bigr]F(t')dt'$$

I can't seem to figure out he gets this expression. My thought is to expand $$\langle \Psi'|\delta\mathbf{B}|\Psi' \rangle$$ where $$|\Psi' \rangle=\sum_{n=0} c_n(t)e^{-i\omega_{n0}t}|n\rangle$$ I would hope this would lead to terms like $\langle0|\mathbf{B}|\Psi'\rangle$, but I'm getting extra terms that I can't figure out how to remove.

Answer

Try rewriting $\delta\langle \mathbf{B} \rangle= \langle \mathbf{B} \rangle-\langle \mathbf{B} \rangle_0$ more explicitly.

$$\begin{align} \langle \mathbf{B} \rangle-\langle \mathbf{B} \rangle_0 &= \text{Tr}[\mathbf{P}(t)\cdot \mathbf{B}] - \text{Tr}[\mathbf{P}(0)\cdot \mathbf{B}] \\ &= \langle 0 | \mathbf{P}(t) \mathbf{B} | 0 \rangle - \langle 0 | \mathbf{P}(0) \mathbf{B} | 0 \rangle \\ \end{align}$$

Recall that to first order, the density matrix $\mathbf{P}(t)$ can be written

$$\mathbf{P}(t) = |0\rangle\langle0| + |\Psi^{(1)}(t)\rangle\langle0| + |0\rangle \langle \Psi^{(1)}(t)| + \cdots $$

You can plug in your expression

$$|\Psi' \rangle = |\Psi^{(1)}(t)\rangle =\sum_{n=0} c^{(1)}_n(t)e^{-i\omega_{n0}t}|n\rangle$$

This should give you (treating $\mathbf{B}$ as Hermitian operator)

$$\delta\langle \mathbf{B} \rangle = \langle 0 | \mathbf{B}\sum_{n=0} c^{(1)}_n(t)e^{-i\omega_{n0}t}|n\rangle\langle0|0\rangle + \langle 0 | 0 \rangle \sum_{n=0} \langle n | c^{(1)*}_n(t)e^{+i\omega_{n0}t}\mathbf{B} |0\rangle $$

From here it is a matter of plugging in your first expression

$$c_n^{(1)}=(i\hbar)^{-1}\int_{-\infty}^t\langle n|\mathbf{A}|0 \rangle F(t')\exp(i\omega_{n0}t')dt'$$

And cleaning up until you get

$$(i\hbar)^{-1}\int_{-\infty}^t\sum_{n\neq0}\bigr[ \langle 0|\mathbf{B}|n \rangle \langle n|\mathbf{A}|0 \rangle \exp(-i\omega_{n0}(t-t'))-\langle 0|\mathbf{A}|n \rangle \langle n|\mathbf{B}|0 \rangle \exp(i\omega_{n0}(t-t'))\bigr]F(t')dt'$$

quantization - What are the reasons to expect that gravity should be quantized?

What I am interested to see are specific examples/reasons why gravity should be quantized. Something more than "well, everything else is, so why not gravity too". For example, isn't it possible that a quantum field theory on curved space-time would be the way treat QFT and gravity in questions where the effects of neither can be ignored?

electricity - Can energy be extracted from clouds?

Can cloud charge imbalance be used as an energy source?

First off quite some energy must be present in clouds: a lightning path is quite long, and electrical breakdown of air requires about 1MV/m. Most (many, smaller) electric discharges are not seen because they occur within clouds. The electric energy built up can also be lost in other ways (i.e. Lorentz force).

Some of it comes to ground and when it happens it has such large peak power (on average about a trillion watts) that it can not efficiently be harvested.

I assume that charges build up quite continuously during the lifetime of a cloud: if we knew what cloud to select for discharge we could control (by selecting the right cloud or part of a cloud) the power flux, though not easy its hard to rule out as impossible.

Now how exactly do I propose to do this? Air can be made conducting through bruteforce electrical breakdown (not efficient), or by heating it (not efficient) or by an energy form encouraging with higher specificity the ionization of specific molecules present in air (be it oxygen or nitrogen or ...).

Apparently non-lethal weapons were designed which do not use projectiles: a UV laser with 2 optical, parallel paths, each connected to an electrode at the weapon side can tetanize (stimulate at such a pulse rate as to keep the muscle contracted) with specificity any of the 3 broad muscle types (voluntary skeletal muscle, involuntary smooth muscle, and cardiac muscle all respond to different pulse patterns/frequencies). A weapon on same principle was designed to stop cars up to 2km by frying the control chip in the car.

According to hstvi (not sure how credible this source is) ionizing wavelengths with lasers can make a path of air conductive to 2km:

"With an ultraviolet wavelength of 193 nanometers, the maximum effective range is about 100 meters. The effective range increases to two kilometers when a wavelength of 248 nanometers and a more intense beam are used."

Now for our clouds we would not need 2 optical paths to the laser.

Problems I envision: clouds heights are on kilometer scales, so possibly stages of such lasers would need to be mounted on weatherballoons, preferably locked by cable to a certain height/location.

Each laser would could be surrounded by a thick piece of transparent insulator (glass?) so as to allow UV to pass but not current to enter the setup. On the glass electrodes are placed which are connected to a similar electrode on say a retroreflector of this balloon (a lower one targets this balloon).

If forming a conductor with the lasers consumes more energy than would be got from the lesser charged clouds (or parts of them), we may need to allow electrical breakdown to occur at which point switching of the laser beam would not stop the lightning.

Problem lightning?