TITLE: Why do we consider special relativity for electrons consider their drift velocity is really small?
QUESTION [1 upvotes]: I know that when looking at the electron and a current carrying wire in different inertial frames there will be different kind of forces acting between these two objects. And I know that it is due to length contraction which cause the charge density to change. But I was confused about something that is , we know that the drift velocity is really small compare to speed of light. Why do we still need to consider this special relativistic effect? (If I am not explaining myself clearly please forgive me, sorry)

REPLY [2 votes]: (a) If you ignore special relativity altogether, there is no length contraction and so no electrostatic force in the electron frame. Therefore the special relativistic effect, even if it is small, is larger than the (zero) non-relatvisitic effect. This is a general phenomenon: small effects can still be important if the effect that could have been there at zero-th order vanishes.
(b) Even though the force due to one non-relativist electron is small, a current carrying wire has a huge number of electrons. This huge number amplifies the effect so that it is macroscopically observable. Again, this is another general effect that can happen, where a small effect becomes large when multiplied by a large parameter, such as (in this case) the number of particles, or (in other cases) the length of time of the experiment.


TITLE: Let $L:K$ be a Galois extention, show that $L:M$ is a normal.
QUESTION [2 upvotes]: Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field.

Show that $L:M$ and $M:K$ are finite separable field extensions.

Attempt: Both $M:K$ and $L:M$ are algebraic ($M:K$ since $M \subset L$, so every $\alpha \in M$ is algebraic over $K$ and $L:M$ since any $\alpha \in L$ is algebraic over $K$, so also over $M$).
Then $M:K$ is separable again since $M \subset L$ (every element of $L$ so in particular every element of $M$ is separable over $K$).
For $L:M$, take any $\alpha \in L$, then min$_M(\alpha)$ divides min$_K(\alpha)$ in $M[X]$. Since min$_K(\alpha)$ has no multiple zeros in a splitting field, neither does min$_M(\alpha)$, i.e. $\alpha$ is separable over $M$.
I'm not quite about the proof for finite, would it involve the Tower Law?

Show that $L:M$ is a normal extension.

I'm a bit stuck here, any help for this one?

REPLY [1 votes]: Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law,
$$
[L:K] = [L:M][M:K].
$$
To show that $L/M$ is a normal extension, let $f \in M[x]$ be an irreducible polynomial which has a root $\alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g \in K[x]$ be the minimal polynomial of $\alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.


text:
\section{Introduction}
\noindent Microfluidics is the study of fluids at the microscopic scale. It is a field of study with a range of applications \cite{droplet_review}, such as: drug delivery, diagnostic chips and microreactors. Droplets in microfluidic systems have a large surface-area to volume ratio, which makes interfacial forces important; a good review is provided by \citeauthor{Rallison1984} \cite{Rallison1984}. Surfactants locally influence the surface tension of a droplet interface. This may create non-uniformity in surface tension which in turn generates a stress opposed to the flow, known as the Marangoni stress. Thus, the addition of surfactants can strongly influence the behaviour of the system \cite{anna2016,droplet_reviewTan}.

\citeauthor{Stone1990}
\cite{Stone1990} studied how surfactants change the drop deformation and breakup using a numerical boundary integral method. They found that for the same strain rate, surfactant-covered drops would reach a more deformed steady state than their clean counterparts. The effect of viscosity-ratio on deformation was investigated for surfactant-covered droplets in an extension of this work \cite{Milliken1993a}. It was shown that for cases with near-zero diffusion of surfactants along the interface of the droplet, the steady state deformation was independent of viscosity-ratio.

A microfluidic system is typically one in which viscous forces are dominant. Microfluidic multiphase flows can therefore be accurately modelled using Stokes equations. With the addition of surfactants, it is also necessary to couple Stokes equations with a convection-diffusion equation on the droplet interfaces. A discontinuity in normal stress across the droplet interface is created by the surface tension forces. This represents a major challenge for the simulation of these flow types. One way to avoid this difficulty is to ensure that the computational grid is always coincident with the interfaces. However, this becomes very expensive when droplets deform or advect as it requires the domain to be remeshed at every time-step. The most common approach for multiphase flow simulation is instead to utilise some form of interface tracking. This involves solving Stokes equations on a fixed computational grid in the entire domain. The locations of droplet interfaces are computed separately using either an explicit boundary discretisation or an implicit representation; such as a level-set function or volume-of-fluid method. Some examples of this kind of methods which also deal with surfactants include those by \citeauthor{khatri_soluble} \cite{Khatri2011,khatri_soluble} and \citeauthor{Muradoglu2014} \cite{Muradoglu2014}, who considered the Navier-Stokes equations and surfactants using a front-tracking scheme and finite differences in 2D and 3D respectively. Another approach is to use a diffuse interface method as
\citeauthor{Teigen2011} \cite{Teigen2011}.

Accurate treatment of the discontinuity in normal stress remains a significant challenge for all interface tracking methods when the interfaces do not align with the computational grid. The most common approach to handle this has been to regularise the surface tension forces, such as introduced in e.g. \citeauthor{Brackbill1992} \cite{Brackbill1992} or in the immersed boundary method by \citeauthor{Peskin1977} \cite{Peskin1977}. However, regularisation limits the accuracy of the method to first order near the interfaces. Another approach is to impose the jump conditions in the normal stress directly. \citeauthor{Leveque1994} \cite{Leveque1994} obtained second order accuracy using an immersed interface method based on finite differences for clean drops, but required drops and bulk to have the same viscosity. This constraint still remained as \citeauthor{xu2006level} \cite{xu2006level} extended the method in \cite{Leveque1994} to simulate surfactant-covered droplets in Stokes flow. Care has to be taken when imposing the jump condition to avoid  a time step constraint that is dependent on how the interfaces cut the underlying grid. This is thoroughly discussed for finite element methods in  \citeauthor{Hansbo2014} \cite{Hansbo2014}, where they manage to design a second order method and avoid such a constraint by adding suitable stabilising terms.

Reformulating Stokes equations in integral form on the droplet interfaces avoids the aforementioned issues. This reduces the dimension of the problem as only the interfaces have to be discretised and gives an explicit representation of the interface which does not have to be coupled to an underlying grid. Furthermore, it naturally handles the discontinuity in normal stress. A review of boundary integral methods for Stokes flow is given in \cite{Pozrikidis2001}. For drops, boundary integral methods including surfactants in 3D are among others \cite{Bazhlekov2006numerical,Pozrikidis2018,Sorgentone2018}. In two dimensions, boundary integral methods have been used for e.g. vesicles as by \citeauthor{Marple2016} \cite{Marple2016} who simulated vesicle suspensions in confined flows and \citeauthor{Quaife2016}  \cite{Quaife2016} who focused on developing an adaptive time-stepping scheme using an implicit spectral deferred correction method. This was extended by \citeauthor{Bystricky} \cite{Bystricky} for rigid body suspensions. \citeauthor{kropinskilushi2011} \cite{kropinskilushi2011} simulated surfactant-covered bubbles with a boundary integral method and used a spectral method to compute the surfactant concentration. \citeauthor{Booty2013} \cite{Booty2013} extended this work to include a model for soluble surfactants in the no-diffusion limit in the bulk. However, the numerical solution of the boundary integral equations has its own challenges. Primarily, the near-singular behaviour of the discretised integrals when evaluating interfaces in close proximity. In 2D, \citeauthor{Ojala2015} \cite{Ojala2015} used a special quadrature scheme to ensure high accuracy also for close to touching clean drops. Other options for closely interacting drops includes quadrature by expansion (QBX) by \citeauthor{Klockner2013} \cite{Klockner2013}, which can be extended to 3D, and was done so efficiently for the special case of solid spheroids in \cite{KlintebergQBX}.

This paper proposes an efficient and accurate boundary integral method for the simulation of deforming bubbles and droplets in two dimensional Stokes flow; for both clean and surfactant-covered droplets. The method is an extension of that proposed by \citeauthor{kropinskilushi2011} \cite{kropinskilushi2011} and uses the special quadrature method in \cite{Ojala2015} to enable accurate simulations of close drop-drop interactions. To match the spectral accuracy in space, an adaptive time-integration scheme is employed and also described herein.

This paper also aims to validate the proposed numerical method using conformal mapping theory inspired by the approaches in \cite{crowdy2005pair,Tanveer1995}. For single surfactant-covered bubbles in an extensional flow at steady state, exact solutions are computed  by \citeauthor{siegelvalid} \cite{siegelvalid}. Furthermore, \citeauthor{crowdy2005pair} \cite{crowdy2005pair} used conformal mapping theory to semi-analytically compute the deformation of a pair of clean bubbles in an extensional flow. With this approach, a conformal mapping of the interface is considered and the time-dependent problem of deforming bubbles is rewritten into a system of ODEs for the conformal mapping parameters, whilst the flowfield can be described analytically. In this paper, this approach has been extended to surfactant-covered bubbles. Together, these solutions enable a precise and quantitative validation of the numerical method proposed by this paper. A comprehensive evaluation of the method is undertaken, examining temporal development, droplets in close proximity and the distribution of surfactants.

Additionally, the errors introduced when evaluating the near-singular integrals numerically using standard Gauss-Legendre quadrature are estimated. This work extends that of \citeauthor{AfKlinteberg2017} \cite{AfKlinteberg2017,klintebergadapt} where estimates based on contour integrals where derived for Laplace's and Helmholtz equations. The estimates are in excellent agreement with the observed numerical errors. Details of the estimates and their derivation can be found in \ref{sec:app_est}.

The paper is organised as follows: in \S\ref{sec:probform} the model, governing equations and boundary integral formulation are introduced. \S\ref{sec:methodBIE} explains the numerical solution of the boundary integral equation, the special quadrature method and provides results for the error estimates. The complete numerical method for the coupled problem is explained in \S\ref{sec:nummethod} and \S\ref{sec:valid} details the methods used for validation. In \S\ref{sec:results} numerical results for drop simulation and validation is presented.

config: arxiv


TITLE: Proving the analyticity of a function of analytic function
QUESTION [1 upvotes]: Suppose $f(z)\in H(\Omega) $  and $|f(z)|<1$ for $ z\in \Omega$ .Show that $ g(z)\in H(\Omega)$ ,where
$$ g(z ) = \sum_{n=1}^\infty n \{f(z)\}^n.$$
$g(z)= \sum_{n=1}^\infty n .w^n $ where $w=f(z)$
$g(z)= \frac{d}{dw}\sum_{n=1}^\infty  w^{n+1} -\sum_{n=1}^\infty w^n $
$g(z)=   \frac{d}{dw}\{ w^2/(1-w)\} -w/(1-w) $
$g(z)= 1/(1-w)$
$g(z)= 1/(1-f(z)) $
Since $f(z) \in H(\Omega) $ and $|f(z)|<1$
Hence $g(z) \in H(\Omega)$
Can anyone confirm validity of this?

REPLY [0 votes]: You need to check the neighborhood. If the function f(z) is analytic in a neighborhood of z0 and f(z) is analytic at z0, then f(z) is analytic at z0. If it is analytic everywhere then it is an entire function. You can check by neighborhoods or also by derivative rules. Try these out.


TITLE: Unfair and fair coin Probability
QUESTION [1 upvotes]: So there are 3 coins. 2 coins are fair, and one is not (both sides are Heads). If person 1 and person 2 each pick a coin, and discard the third, (1)what is the probability that person 1 gets Tails on their first toss.
(2)What is the probability that person 1 gets Tails on the second toss.
(3)What is the probability that person 1 gets Tails on their two tosses.
I am not sure if I am doing these correctly.
For the first one I did P = (1/3)(0) + (2/3)(1/2) = 1/3
My reasoning is that if person 1 got the unfair coin, it would be 1/3, and if so there isn't any possible way to get a tais, that's for the 0. Then I add the probability that person 1 has a fair coin, 2/3, and the chance they can get tails, 1/2. Am I approaching this correctly?
If the above is correct, then the second problem should be equal to the first, no?
And for the last one, I did P = (1/3)(0)^2 + (2/3)(1/2)^2 = 1/6
My reasoning for problem 3 is that if person 1 got the unfair coin it would be 1/3 and to get tails, 0^2. You square it because each trial is dependent of each other? And if person 1 got the fair coin, 2/3, and to get tails, 1/2^2 for the two tosses.
So now how would you start doing this problem:
(4)Find the probability that person 1 gets Tails on their third toss if they got Heads, Heads on the first two.
And how does it change if the person got Tails, Tails, on the first two tosses.

REPLY [0 votes]: Consider the ways to get heads on both of the first two tosses: you can have the unfair coin, or you can have a fair coin and toss heads twice in a row. The probability of having the unfair coin is $\frac13$. The probability of having a fair coin and tossing heads twice in a row is $\frac23\cdot\frac14=\frac16$. If you know that you’ve tossed heads twice in a row, then you know that you’re in one of these two cases; the probability that you’re in the first case (have the unfair coin) is $$\frac{\frac13}{\frac13+\frac16}=\frac23\;,$$ and the probability that you’re in the second case (have a fair coin) is $1-\frac23=\frac13$.
You know the probability of getting tails with the unfair coin and the probability of getting tails with a fair coin, so you can now calculate the overall probability of getting tails if you’ve tossed heads on your first two tries; just use the same basic approach that you used in the first three questions.
If you began by tossing tails, twice, on the other hand, you know that you have a fair coin.


TITLE: Additive functors preserve projective objects?
QUESTION [1 upvotes]: Let $F:\mathscr{A}\rightarrow\mathscr{B}$ be an additive functor between abelian categories. Let $P\in\mathrm{Ob}(\mathscr{A})$ be a projective object.
I believe that $FP\in\mathrm{Ob}(\mathscr{B})$ doesn't remain projective in general. I'm looking for a concrete counterexample that I failed to construct.
Moreover, does there exist any sufficient condition which makes an additive functor preserve projective objects?
Thanks for your help!

REPLY [3 votes]: Here's a sufficient condition: if $F$ is the left adjoint of an exact functor, then $F$ preserves projective objects. This is not limited to additive functors between abelian categories: if $F$ is the left adjoint of a functor (between general categories) that preserves epimorphisms, then $F$ preserves projective objects.


TITLE: How do I compute the probability that $(X,Y)$ lies in the following set?
QUESTION [0 upvotes]: We have given two random variables $X,Y$ such that $(X,Y)\sim U([0,1]^2)$. I need to compute the probability that $(X,Y)$ is in $C:=\{(x,y)\in [0,1]^2: x^2+y^2\leq 1\}$

Now I know that since $(X,Y)$ is uniformly distributed we have $$\Bbb{P}((X,Y)\in C)=A(C)=\frac{\pi}{4}$$ where $A$ is the area.
Now I also thought about computing $\Bbb{P}((X,Y)\in C)$ really "mathematically" with integrals. I thought that $$\Bbb{P}((X,Y)\in C)=\int_0^1 \int_0^{\sqrt{1-x^2}}dydx$$ which indeed gives the same but was that just luck that the integrals give also $\frac{\pi}{4}$ or can we really compute $\Bbb{P}((X,Y)\in C)$ in this way?
Thanks for your help.

REPLY [2 votes]: Does this answer your question?

In short:  no... it wasn't "luck."


TITLE: How do I find dx/dt implicitly with the function $\tan(x)=4t^{-1}$?
QUESTION [2 upvotes]: This question is a trigonometric one from what I can tell. I understand that you need to differentiate the given function and the unknown variable on top ($dx$) will correspond to the $x$ unknowns, for example $5x^2$ will give us $10x\frac{dx}{dt}$.
With this specific question I can only get to $-\frac{\sec^2x}{4t^2}$ but the answer is $-\frac{\cos^2x}{4t^2}$. I also do not understand how $\sec^2x$ becomes $\cos^2x$.
Thanks.

REPLY [0 votes]: The easiest way to do implicit derivatives is to separate differentiation from the derivative.  First, differentiate, then solve for your derivative.
Example: $y = x^2$.  Differential of $y$ is $dy$.  Differential of $x^2$ is $2x\,dx$.  Therefore $dy = 2x\,dx$.  Solving for the derivative gives $\frac{dy}{dx} = 2x$.
So, to your problem:
$$ tan(x) = 4t^{-1} \\
d(tan(x)) = d(4t^{-1}) \\
sec^2(x)\,dx = -4t^{-2}\,dt \\
$$
Now just solve for $\frac{dx}{dt}$:
$$\frac{dx}{dt} = \frac{-4t^{-2}}{sec^2(x)}$$
Or, more simply:
$$\frac{dx}{dt} = -4\frac{cos^2(x)}{t^2} $$


TITLE: When can we replace countably valued simple functions by finitely valued simple functions
QUESTION [0 upvotes]: Suppose that $(\Omega,\mathcal{A},\mu)$ is a finite measure space and $X$ is a Banach space. Let $f:\Omega \to X$ be a function that is an a.e. pointwise limit of countable-valued functions $f_n:\Omega \to X$, i.e. $f_n(\Omega)$ is countable and $f_n^{-1}(\{x\})\in \mathcal{A}$ for all $x \in X$.

Is it then true that $f$ is an a.e. pointwise limit of finitely valued functions? What if we drop the finiteness of $\mu(\Omega)$?

REPLY [1 votes]: Define $d(f,g)=\int \frac {\|f-g\|} {1+\|f-g\|}d\mu$ for strongly measurable functions $f,g: \Omega \to X$. Then  $f_n \to f$ in measure (in the sense $\|f_n-f\| \to 0$ in measure iff $d(f_n,f) \to 0$. We can first choose $n$ such that $d(f_n,f)<\epsilon $ and then  choose a finitely valued function $g$ such that $d(f_n,g)<\epsilon $. This shows that there is a sequence of finitely valued measurable functions converging to $f$ in measure. There is a subsequence which converges almost everywhere.
If $\mu (E)=\infty$ for every non-empty set $E$ and $f:\Omega \to \mathbb R$ is a measurable functions which takes each of the values $1,2,...$ with positive (hence infinite!) measure then we cannot write this as the almost everywhere limit of a sequence of finite values measurable functions.


TITLE: Area enclosed by cardioid using Green's theorem
QUESTION [2 upvotes]: Let $$\gamma(t) = \begin{pmatrix} (1+\cos t)\cos t \\ (1+ \cos t) \sin t \end{pmatrix}, \qquad t \in [0,2\pi].$$ Find the area enclosed by $\gamma$ using Green's theorem.
So the area enclosed by $\gamma$ is a cardioid, let's denote it as $B$. By Green's theorem we have for $f=(f_1, f_2) \in C^1(\mathbb{R}^2, \mathbb{R}^2):$
$$\int_B \text{div} \begin{pmatrix} f_2 \\ -f_1 \end{pmatrix} d(x,y) = \int_{\partial B} f \cdot ds$$
So if we choose $f(x,y) = \begin{pmatrix} -y \\ 0 \end{pmatrix}$ for example, we get
$$\begin{eqnarray}
\text{Area of $B$} &=& \int_{\partial B} f \cdot ds
\\&=& \int_{\gamma} f(\gamma(t)) \cdot \gamma'(t) dt
\\&=& \int_0^{2\pi} \begin{pmatrix} -(1+ \cos t) \sin t \\ 0 \end{pmatrix} \cdot\begin{pmatrix} \sin t ( 1 - 2 \cos t) \\ \cos^2t - \sin^2t + \cos t \end{pmatrix} dt ,\end{eqnarray}$$
which I guess we can evaluate but if I keep going, this will become very nasty and tedious.
There must be a nicer way to do this. Please help me see it.

REPLY [2 votes]: As a rule of thumb, it's better to use $$\frac12\int_{\partial B}(-y\,dx+x\,dy)$$ whenever trig functions are involved. That will give a simpler expression once you use standard trig identities.
Comment: It's more straight forward in this particular case to do the area integral directly in polar coordinates: $$\int_0^{2\pi}\int_0^{1+\cos\theta} r\,drd\theta.$$


TITLE: Confusion in Euler-Bernoulli beam theory
QUESTION [1 upvotes]: Euler-Bernoulli beam equation is given by
$$
EI \frac{\mathrm d^2 u}{\mathrm d x^2} = M'(x) \\
EI \frac{\mathrm d u}{\mathrm d x} = xM'(x) + C_1
$$
Where, $E$ is modulus, $I$ is second moment of area, $u$ is displacement as function of $x$, $M'(x)$ is (moment) loading.
From classical beam bending theory,
$$
\sigma = \frac{My}{I} \\ \epsilon = \frac{\sigma}{E}=\frac{My}{EI}
$$
$\sigma$ is normal stress, $M$ is bending moment, $y$ is distance of fiber from neutral axis, $I$ is second moment of area, $\epsilon$ is normal strain, $E$ is elastic modulus.
I am confused here. With infinitesimal strain theory (http://en.wikipedia.org/wiki/Infinitesimal_strain_theory)
$$
\epsilon = \frac{du}{dx} \implies \epsilon = \frac{xM'(x) + C_1}{EI}
$$
(from second equation)
Which strain relation is correct? I am messing up somewhere, but couldn't find.

REPLY [2 votes]: The "vertical" (lateral) co-ordinate has two different meanings. In the static Euler Bernoulli equations (your first set),  $u$ is the lateral displacement of the centre of the beam from its zero strain position, i.e. its position when there are no stresses or forces on the beam. Since the beam cross section is assumed not to deform, $u$ is the lateral displacement of any point in the cross section from its no-strain position.
The second set analyses the stress in a given cross section as a function of position in the cross section. Here $y$ is not the lateral displacement but is instead the orthogonal distance of the point in question from the neutral axis for the cross section.
EB beam theory models the beam as a bundle of fibres with no friction and only normal forces between them. If there is a constant moment on the beam, the fibres bend into circular arcs, and the shape of any fibre is indeed an arc of a circle. On the "inside" edge of the beam, i.e. that nearest to the axis made up of all the centres of curvature of the fibres, the fibres are in compression; on the other edge they are in tension. There is an axis through the cross section along which all of the fibres are in a zero strain state. This is the neutral axis.
One thing you seem to be having trouble grasping is that there is no one strain for each cross section. The strain varies over the cross section, as my derivation below should show. There is zero strain on the neutral axis of each cross section. So the fifth equation would be wrong even if the second were right.
Now for your equations:
The third and fourth are correct.
The first is incorrect. It should be $M(x) =-E\,I\,\frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}$, where $u(x)$ is the lateral displacement of any point in a cross section from its zero strain position. I shouldn't worry about the difference in sign too much as long as you are consistent (although the negative sign shows the wonted convention), but the second derivative of $u$ is the moment, not its first derivative. You can derive this equation as follows: for small gradients, $\frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}\approx \kappa(x)=\frac{1}{R(x)}$ is approximately the curvature $\kappa(x)$ of the beam as measured at the neutral axis. One radian of the circular beam arc at this point therefore has a length $R(x)$. This is the length of the beam as measured at the neutral axis, and therefore this represents an unstrained length. The length of one radian of the fibre which an orthogonal distance $y$ from the neutral axis is $(R + y)$. Therefore the strain, the fractional change relative to the unstrained length, is  $((R + y)/R) - 1=y\,/\,R$. The stress normal to the cross section at this point is therefore $E\,y/R$. We then integrate the moment of this stress across the whole beam's cross section to get:
$$M(x) = -\frac{E}{R(x)}\,\int_\mathscr{S}\,y^2\,{\rm d} A = -\frac{E\,I}{R(x)}\tag{1}$$
As we have seen, $1/R(x)\approx \frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}$, whence the correct equation:
$$M(x) = -E\,I\,\frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}\tag{2}$$
Since your first equation is wrong, so is the second.
Now, the fourth equation is correct as follows. As we have seen, the strain $\epsilon(x,\,y)$ an orthogonal distance $y$ from the neutral axis is $\frac{y}{R(x)}$, where $R(x)$ is the radius of curvature of the beam (at the neutral axis). We have just seen  (eq (1)) that $\frac{1}{R(x)} = -\frac{M}{E\,I}$ therefore $\epsilon(x,\,y)=\frac{y}{R(x)} =-\frac{M(x)\,y}{E\,I}$, whence your fourth equation follows. The third simply follows by multiplying the strain $\epsilon(x,\,y)$ by the Young's modulus to get the stress.


TITLE: Motion of a falling object with air resistance
QUESTION [8 upvotes]: I was wondering how you would model the velocity of a falling object, taking into account air resistance. Bear in mind I have only studied basic calculus, and have no experience with differential equations.
Considering a general solution for mass $m$ and drag $kv^2$, where $k=\frac{1}{2} \rho AC_D$. Let's say that when $t=0$, $v=0$.
Using Newton's Second Law: $mg - kv^2 = ma$, hence $a = g - \frac{k}{m} v^2$.
From here, I tried to integrate with respect to $t$, which gave $v = gt-\frac{kt}{m}v^2$. This, when solved for v, seems to model the velocity correctly, but I have been led to believe that solutions to differential equations of this kind will involve $e^x$ in some way. Is my solution correct, or have I integrated incorrectly? How do we go from here to a function for v in terms of t?

REPLY [8 votes]: Remembering that $a=\dfrac{dv}{dt}$,
$$m\frac{dv}{dt}=mg-kv^2$$
$$\frac{dv}{mg-kv^2}=\frac{dt}{m}$$
Integrating both sides,
$$\int\frac{dv}{mg-kv^2}=\int\frac{dt}{m}$$
You can find an analytical solution from here.
Here is the full solution on my blog if you are interested:https://philosophicalmath.wordpress.com/2017/10/21/terminal-velocity-derivation/


text:
\begin{document}
\baselineskip=18pt

\title[Augmented Star Products]{Can Star Products be Augmented by Classical Physics?}

\author[M.P.G. Robbins]{Matthew P.G.  Robbins$^{1,2}$}
\author[M.A. Walton]{Mark A. Walton$^2$}

\date{\today}

\vskip1cm
\begin{abstract}
\large\baselineskip=16pt
${}$ \\ It has been suggested that star products in phase-space quantization may be  augmented to describe additional, classical effects. That proposal is examined critically here. Two known star products that  introduce classical effects are: the generalized Husimi product of coarse-grained quantization, and a non-Hermitian damped star product for the harmonic oscillator. Following these examples, we consider products related by transition differential operators to the classic Moyal star product. We restrict to Hermitian star products, avoiding problems already pointed out for the original damped product. It is shown, however, that with such star products, augmented quantization is impossible, since an appropriate classical limit does not result.

For a more complete study, we then also consider generalized, or local, transition operators, that depend on the local phase-space coordinates, as well as their derivatives. In this framework, one example of possible physical interest is constructed. Because of its limited validity and complicated form, however, it cannot be concluded that augmented quantization with local transition operators is practical.  \\
\end{abstract}
\keywords{phase-space quantum mechanics, quantization, star products, transition operators}

\maketitle
\noindent
$^1$
Department of Physics and Astronomy, University of Waterloo, Waterloo ON, Canada N2L 3G1;\\
${}\ \,$
Perimeter Institute for Theoretical Physics, Waterloo ON, Canada, N2L 2Y5 $^3$\\
$^2$
Department of Physics and Astronomy, University of Lethbridge, Lethbridge AB, Canada T1K 3M4\\

\noindent
\textit{E-mail:} m2robbins@uwaterloo.ca, walton@uleth.ca
\hfb $^3$  Current addresses.

\vfill\eject

\section{Introduction}

Observables in quantum mechanics can be described by operators or by functions (and distributions) in phase space \cite{Zachos2005, Cohen2013, Hillery1984, Balazs1983, Lee1995, Bayen1978a, Bayen1978b, Osborn1995,Takahashi1989}.  Operator and phase-space quantum mechanics are equivalent, however, as can be demonstrated through maps from one to the other. A  quantization map takes the phase-space quantum mechanics to the operator version, and its inverse is known as the dequantization map. The non-commutativity of operator observables is reflected in phase space by the non-commutative star product that must be used to multiply phase-space obervables.

Quantization of a classical system is not unique either way, however.  Different quantizations give rise to different star products \cite{Bayen1978a, Bayen1978b, Cohen2013, Lee1995, Mehta1964}.  For example, distinct operator-ordering rules determine distinct quantization maps.  The corresponding dequantizations prescribe distinct star products.  Weyl operator ordering and the Moyal star product are paired, but so are standard operator ordering and the standard star product,  Born-Jordan ordering and another star product, etc.

An advantage of phase space quantization (also known as deformation quantization) is that different quantizations and corresponding star products may be related in a simple way. For example, a transition (differential) operator \cite{Bayen1978a, Bayen1978b} can associate the phase-space observables and star products of 2 quantizations.

The same mathematical machinery can introduce physical effects distinct from quantization, however.  Coarse graining in phase space is described by the generalized Husimi star product \cite{Husimi1940, Takahashi1989}.  It is obtainable from the Moyal product, e.g., by a transition differential operator. Similarly, another transition operator produces a star product that converts the equation of motion of the harmonic oscillator into that of the damped harmonic oscillator \cite{Dito2006, Belchev2009}.

Here we study such star products modified to include additional classical effects not described by the original Hamiltonian. This scheme was suggested originally in \cite{Dito2006}. The damped star product introduced there was shown in \cite{Belchev2009} to have serious problems, however, because it is non-Hermitian. But we will see here that there exists a Hermitian damped star product similar to the original non-Hermitian one described in \cite{Dito2006}. We also note that the (generalized) Husimi star product introduces a classical coarse-graining effect, and is described by a transition differential operator in a similar way. It seems then, that the scheme might be viable.

For brevity, the construction of such augmented star products will be called augmented (phase-space) quantization.  Taking the coarse-grained and damped star products as our guides, we undertake a serious study of augmented star products related to the Moyal star product by transition operators. Our goal is to see whether or not such augmented star products can provide effective descriptions of certain quantum systems.

We find that augmented, Hermitian star products described by transition differential operators cannot. The classical limit is problematic. Only that part of the transition differential operator that is independent of Planck's constant is relevant to the classical limit. If it is 1, the  classical limit is not modified -  there is no additional  physics (no augmentation).  If the $\hbar$-independent part is non-trivial, then a multiplicative, rather than additive, modification of the equations of motion results.

Augmented quantization is therefore not possible unless the framework is changed. For a more thorough treatment, we therefore go on to consider  generalized transition operators, depending not only on derivatives of phase-space coordinates, but also on the coordinates themselves.  This generalization is quite drastic - although they have been considered before (see \cite{Crehan1989, Pinzul2008}), such ``local'' transition operators are rarely invoked.  In this framework, one example of possible physical interest is constructed: a Hermitian phase-space star product for the weak damping of a simple harmonic oscillator.  This result is of limited validity and has a rather complicated form, however.  The practical feasibility of using local transition operators in augmented quantization remains to be demonstrated.

Star products have been and continue to be very useful tools in theoretical physics.  Before we outline the organization of our paper, let us provide recent points of contact with the vast literature on star products in deformation quantization and other subjects. Reference \cite{Lizzi2014} includes a recent review in the context of this paper, non-relativistic quantum mechanics. A formal star product and associated quantization has been described for any finite-dimensional Poisson manifold as phase space in \cite{Kontsevich2003}; a generalization \cite{Dherin2015} has more recently been described for when the Lie algebra of classical observables is replaced by a Leibniz algebra.  Quantum field theory \cite{Dito1989} invokes star products with ordinary derivatives replaced by functional derivatives. Star products can encode the non-commutativity of non-commutative geometry \cite{Szabo2003}.  String theory in certain limits is described by such non-commutative geometry \cite{Seiberg1999} (see \cite{Blumenhagen2014} for a recent, pedagogical survey).  In a similar way, associativity is lost in certain regimes of string theory, motivating the very recent study of non-associative star products \cite{Kupriyanov2018, Szabo2018}.

The next section of this paper is a quick introduction to the elements of phase-space quantization that are relevant to our study, and to our notation. Section 3 describes the 2 star products we take as our guides: the coarse-grained Husimi \cite{Husimi1940, Takahashi1989} and damped \cite{Dito2006, Belchev2009} products.

Section 4 treats augmented star products abstractly and generally. In subsection 4.1 we find that transition differential operators (``global'' transition operators) do not lead to classical equations of motion augmented by additional terms; this is our main result.  Subsection 4.2 demonstrates that local transition operators can, in principle, furnish new examples of modified star products that incorporate additional classical physics.  A single, simple example is constructed: a new, Hermitian  damped star product. It is, however, rather unwieldy, and we therefore believe that local transition operators may have limited usefulness.

The final section is our conclusion.

\section{Phase-space quantum mechanics and star products}

We will restrict to 1-dimensional systems on position space $\R$ with coordinate $q$, and conjugate momentum $p\in \R$, so that the phase-space is $\R^2$. It is straightforward to generalize to several degrees of freedom.  Only time-independent Hamiltonians will be treated:  $\partial\sub{\,t}H=0$.

A brief account of quantization in phase-space will now be given.  Subsection \ref{subsec: Moyal} then outlines the canonical example, involving the Weyl map, the Moyal star product, and the Wigner transform.  It will be our reference quantization, so that all other examples treated here will be related to it by transition operators, which we discuss in Subsection \ref{subsec: transition operators}.

\subsection{Quantization in Phase-Space}

Suppose that the distribution on phase-space, $f=f(q,p)$, is a classical observable.  Then the quantization map $\cQ$ produces a quantum observable:
\begin{align}
\cQ\big( f(q,p) \big)\ =\ \hat f\ \ .\label{Qn}
\end{align}
The $\hspace{0.05cm}\hat{}\hspace{0.05cm}$ indicates that $\hat f$ is an operator, which is a function of position operator, $\hat q$, and momentum operator, $\hat p$.  The inverse (dequantization map) $\cW$ is given by
\begin{align}
\cW(\hat f )\ =\ f(q,p)\ .
\label{cW}\end{align}
Strictly, both $\cQ$ and $\cW$ should be labelled by the phase-space coordinates, so that
\begin{align}
\cQ^{(q,p)}\Big( f(q,p)  \Big)\ =\ \hat f\ ,\ \ \ \ \cW\sub{(q,p)}\Big( \hat f \Big)\ =\ f(q,p)\ .
\label{QWqp}\end{align}
To avoid overly cumbersome notation, however, we will follow convention and drop these labels when confusion is unlikely.

Dequantization maps operators into phase-space distributions (the symbols of the operators). Operator products become star products, homomorphically:
\begin{align}
\cW\big(\, \hat f\, \hat g   \,\big)\ =\ \cW(\hat f)\, *\, \cW(\hat g)\ .
\label{homostar}\end{align}
The star product $*$ is a bi-differential operator expressible in terms of the left derivatives $\Lpq, \Lpp$, and right derivatives, $\Rpq, \Rpp$, defined by
\begin{align}
f(q,p)\Lpq g(q,p)\ :=\ \I(1,2)\,\,\partial_{q_1}\, f(q_1,p_1)\, g(q_2,p_2)\ ,
\label{defLpq}\end{align}
and in a similar manner for the right derivatives. Here $\I(1,2)$ enacts the identifications ${q_1\, =\, q_2\, =q}$ and ${p_1\,=\, p_2\, =\, p\,}.$
In useful shorthand notation,  (\ref{defLpq}) is
\begin{align}
f\Lpq g\ :=\ \I(1,2)\,\partial_{q}(1)\, \Big( f(1)\, g(2) \Big)\,.
\label{shorth}\end{align}

As an illustration of the homomorphism of (\ref{homostar}), consider the Heisenberg-Weyl group relation in operator quantum mechanics:
\begin{align}
\exp\left[ i(\varphi \hat q + \xi \hat p )/\hbar \right]\exp\left[ i(\varphi' \hat q + \xi' \hat p )/\hbar \right]=\text{e}^{-\frac i{2\hbar} \big( \varphi\xi'-\xi\varphi' \big)}\,\exp\left\{ i[ (\varphi+\varphi')\hat q + (\xi+\xi')\hat p ]/\hbar\right\}\,,
\end{align}
where $\varphi, \xi, \varphi', \xi'\in\R$, a consequence of the simple Baker-Campbell-Hausdorff formula. Application of the dequantization map $\cW$  yields
\begin{align}
\cW\big(\exp\left[ i(\varphi \hat q + \xi \hat p )/\hbar \right]\big)\, *\,& \cW\big(\exp\left[ i(\varphi' \hat q + \xi' \hat p )/\hbar \right]\big)\ \nn =\ &\text{e}^{-\frac i{2\hbar} \big( \varphi\xi'-\xi\varphi' \big)}\, \cW\Big(\exp\left\{ i[ (\varphi+\varphi')\hat q + (\xi+\xi')\hat p ]/\hbar\right\}  \Big)\ .
\label{starHW}\end{align}
This demonstrates that a phase-space quantization produces a $*$-realization of the Heisenberg-Weyl group.

Let $\cD$ denote an arbitrary bi-differential operator (such as the star product $*$, or a left- or right-derivative $\Lpq, \Lpp$ or $\Rpq, \Rpp$, for examples). Transpose exchanges left- and right-derivatives:
\begin{align}
\big( \Lp \big)^{\, t}\ =\ \Rp\ ,\ \ \big( \Rp\big)^{\, t}\ =\ \Lp\ ,
\end{align}
so the transpose $\cD^{t}$ satisfies
\begin{align}
f\,\cD^t\,g\ =\ g\,\cD\,f\ ,
\label{cDtranspose}\end{align}
for arbitrary phase-space distributions (i.e. observables) $f$ and $g$. Let $\overline{\cD}$ and $\overline f$ be the complex conjugates of bi-differential operator $\cD$ and phase-space distribution $f$. The adjoint, or Hermitian conjugate, $\cD^\dagger$ of $\cD$ is the complex-conjugate transpose: $\cD^\dagger = \overline{\cD^{t}}$, so that
\begin{align}
\overline{f\,\cD\,g}\ =\ \bar g\,\cD^\dagger\,\bar f\ .
\label{cDadjoint}\end{align}
A bi-differential operator, such as a star product $*$, can be Hermitian, $\cD^\dagger = \cD$, or symmetric, $\cD^t = \cD$, or real, $\overline{\cD} = \cD$, or none of the above.

Application of $\cW$ to any relation involving operator observables yields the phase-space counterpart. With this in mind, consider the equation of motion for a quantum observable $\hat f$ in the Heisenberg picture,
\begin{align}
i\hbar\, \frac{\text{d}\hat f}{\text{d}t}\ =\ [ \hat f, \hat H ]\ ,
\label{HeisEMf}
\end{align}
assuming $\hat H^\dagger = \hat H$ is the Hamiltonian.

As a result of (\ref{homostar}), commutators of operators are mapped to $*$-commutators,
\begin{align}
\cW\big(\, [ \hat f, \hat g  ]\,\big)\ =\ [ \cW\hat f, \cW\hat g ]\sub{*}\ :=\ \cW\hat f\, *\, \cW\hat g\, -\, \cW\hat g\, *\, \cW\hat f\ .
\label{homobrack}
\end{align}
Applying the dequantization map $\cW$ to (\ref{HeisEMf}) yields
\begin{align}
\dot f\ =\ \frac{[ f, H]\sub{*}}{i\hbar}\ =\ \{ f, H \}\sub{*}\ ,
\label{HeisMoyalEqn}
\end{align}
where $H:=\cW(\hat H)$, and we have introduced the Moyal bracket $\{\cdot, \cdot\}\sub{*}$:
\begin{align}
\frac{[ f, g]\sub{*}}{i\hbar}\ =:\ \{ f, g \}\sub{*}\ =:\ f\, \cM\, g\ =\ f\left(\frac{*\, -\, *^{\, t}}{i\hbar}\right)g\ .
\label{MoyalB}\end{align}
$\cM$ denotes the associated Moyal bi-differential operator.

The formal solution to (\ref{HeisMoyalEqn}) is
\begin{align}
f(q,p;t)\
 =\ f(q,p;0)\, \exp\{\cM H t\}\ =\ \text{e}^{\frac{it}{\hbar} H*}\, f(q,p;0)\, \text{e}^{- * \frac{it}{\hbar} H}\ =\ U(-t) * f(q,p;0) * U(t)\ ,
\label{intHMEqn}
\end{align}
where
\begin{align}
U(t)\
 =\ \exp\sub{*}\Big\{- \frac{it}{\hbar} H \Big\}\ =\ \cW\big( \text{e}^{-it\hat H/\hbar} \big)\
\label{propU}
\end{align}
is the symbol of the propagator, and $\exp\sub{*}$ indicates the $*$-exponential \cite{Bayen1978a,Bayen1978b}.

In the so-called classical limit $\hbar\to 0$, the Moyal bracket reverts to a Poisson bracket,
\begin{align}
\lim_{\hbar\to 0}\, \{ f, g \}\sub{*}\ =\ \{ f, g \}\ .
\label{MoyalBPoissonB}
\end{align}
The corresponding bi-differential operators have the same relation,
\begin{align}
\lim_{\hbar\to 0}\, \cM\ =\ \cP\ .
\label{MoyalPDOPoisson}
\end{align}
Here the Poisson bi-differential operator $\cP$ is defined by the Poisson bracket:
\begin{align}
\{ f, g \}\ =:\ f\, \cP\, g\ ,\ {\text{i.e.}\,,}\ \ \cP\ =\ \Lpq\Rpp - \Lpp\Rpq\ .
\label{PoissonPDO}
\end{align}
Therefore, in the classical limit, the equation of motion (\ref{HeisMoyalEqn}) reverts to
\begin{align}
\dot f\ =\ \{ f, H \}\ ,
\label{EqMcl}\end{align}
which is the classical equation of motion, as expected.

The quantum state of the system is described by the density operator in the Schroedinger picture, $\hat\rho$.  Its equation of motion is
\begin{align}
i\hbar\, \frac{\text{d}\hat\rho}{\text{d}t}\ =\ [ \hat H, \hat\rho ]\ .
\label{EMrho}
\end{align}
The dequantization map $\cW$ transforms the density operator $\hat\rho$ into a function $W(q,p;t)=\cW(\hat\rho)$ on phase space, obeying
\begin{align}
i\hbar\, \frac{\partial W}{\partial t}\ =\ \{ H, W \}\sub{*}\ .
\label{HeisEM}
\end{align}
$W(q,p;t)$ is called a quasi-probability distribution on phase space because, although it determines expectation values as a probability distribution would:
\begin{align}
\langle A \rangle\ =\ \int {\text d}q\, {\text d}p\, W(x,p;t)\, A(x,p;t)\ ,
\label{expval}
\end{align}
it takes negative values. The first example of such a quasi-probability distribution is the Wigner function (see the next section).

Now that we have discussed the general form of phase-space quantization, we are in a position to identify the crux of our paper. We investigate the possibility that the star product may be augmented such that instead of (\ref{MoyalPDOPoisson}), we find
\begin{align}
\lim_{\hbar\to 0}\, \cM\ =\ \cP\, +\, \delta\cP\ ,
\label{limMaugP}
\end{align}
where the additional term $\delta\cP$ describes extra, classical effects. We will return to this question in Section \ref{sec: Augmented star products}.  In the original example \cite{Dito2006}, the effect produced a damping term for the harmonic oscillator.

We should point out that since
\begin{align}
\cM\ =\ \frac{*\, -\, *^t}{i\hbar}\ ,
\label{cMstar}
\end{align}
(\ref{limMaugP}) is only possible because the pointwise multiplication of classical observables is modified in the classical limit:
\begin{align}
\lim_{\hbar\to 0}\, *\ \not=\ 1\ .
\label{limstarnotid}
\end{align}
Admittedly, this is a strange feature of augmented quantization. But it is  (\ref{limMaugP}) that determines the classical equations of motion, and physical examples are not ruled out by (\ref{limstarnotid}).

\subsection{Reference phase-space quantization}
\label{subsec: Moyal}

As our reference example, we will use the Weyl quantization map $\cQ_0$, which can be defined by
\begin{align}
\cQ_0\Big( \exp\left[ i(\varphi q + \xi p )/\hbar \right] \Big)\ =\  \exp\left[ i(\varphi \hat q + \xi \hat p )/\hbar \right]\ .  \label{Qzero}
\end{align}
Expanding exponentials and equating terms proportional to $\varphi^n\, \xi^m$ produces the Weyl operator-ordering rule
\begin{equation}
\begin{aligned}
\cQ_0\Big( q^n\, p^m \Big)\ &= \ \frac{1}{(n+m)!}\left(\hat{q}^n\hspace{0.03cm}\hat{p}^m+\hat{q}^{n-1}\hspace{0.03cm}\hat{p}^m\hspace{0.03cm}\hat{q}+\hat{p}\hspace{0.05cm}\hat{q}^n\hspace{0.03cm}\hat{p}^{m-1}+\hat{q}\hspace{0.05cm}\hat{p}\hspace{0.05cm}\hat{q}^{n-1}\hspace{0.03cm}\hat{p}^{m-1}\ + \ \cdots\right) \ , \\
&=\ \frac 1{(n+m)!} \sum_{\pi\in S_{n+m}} \stackrel{\pi(1)}{\hat q}\cdots\stackrel{\pi(n)}{\hat q}\,\,\stackrel{\pi(n+1)}{\hat p}\cdots\stackrel{\pi(n+m)}{\hat p}\ .
\label{Qzeronm}\end{aligned}
\end{equation}
Here the sum is over all permutations $\pi\in S_{n+m}$ of the $n+m$ factors in $\hat q^n \hat p^m$, and the numbers above the operators indicate their place in the product. The sum is somewhat redundant: it can be restricted to permutations $\pi$ in the coset $S_{n+m}/(S_n\times S_m)$, if $1/(n+m)!$ is replaced by $n!\, m!/(n+m)!$.  $\cQ_0(q^n\, p^m )$ is the average of distinct terms obtained by permuting the factors of $\hat q^n \hat p^m$.   Alternate expressions, such as
\begin{align}
\cQ_0\Big( q^n\, p^m \Big)\ =\ \frac 1 {2^n} \sum_{\ell=0}^m \binom{m}{\ell}\, \hat q^{\, n-\ell} \hat p^{\, m} \hat q^{\,\ell} =\ \frac 1 {2^m} \sum_{\ell=0}^n \binom{n}{\ell}\, \hat p^{\, m-\ell} \hat q^{\, n} \hat p^{\,\ell} \ ,
\end{align}
for example, can be derived using the Heisenberg commutation relation \cite{Cohen2013}.

The dequantization map $\cW_0$, the inverse of the Weyl map (\ref{Qzero}),
\begin{align}
\cW_0\Big( \exp\left[ i(\varphi \hat q + \xi \hat p )/\hbar \right] \Big)\ =\  \exp\left[ i(\varphi q + \xi p )/\hbar \right]\ ,  \label{cWzero}
\end{align}
is also known as the Wigner transform. The homomorphism (\ref{homostar}) between $*$- and operator products, along with the Heisenberg-Weyl relation (\ref{starHW}), gives the famous (Groenewold-)Moyal star product
\begin{align}
*\sub{\, 0}\ =\ \exp\left\{  \frac{i\hbar}{2} \left[ \Lpq\,\Rpp - \Lpp\,\Rpq \right] \right\}\ .
\label{starz}\end{align}
The transpose of the Moyal star product equals its complex conjugate, $\bar*\sub{\, 0}$:
\begin{align}
*\sub{\, 0}^{\, t}\ =\ \exp\left\{  \frac{i\hbar}{2} \left[ \Lpq\,\Rpp - \Lpp\,\Rpq \right] \right\}\ = \bar{*}\sub{\, 0}\ ,
\end{align}
so $*\sub{\, 0} = *\sub{\, 0}^{\,\,\dagger}$  is Hermitian.

The Moyal product is associative:
\begin{align}
\Big( f\, *\sub{\, 0}\, g \Big)\, *\sub{\, 0}\, h\ =\ f\, *\sub{\, 0}\, \Big( g \, *\sub{\, 0}\, h \Big)\ ,
\label{starzassoc}\end{align}
which follows from
\begin{align}
*\sub{\, 0}(1+2,3)\,*\sub{\, 0}(1,2)\ =\ *\sub{\, 0}(1,2+3)\,*\sub{\, 0}(2,3)\ .
\label{PREstarzassoc}\end{align}
Here the notation of (\ref{shorth}) is used, with $*\sub{\, 0}(1+2,3)$, for example,  obtained from $*\sub{\, 0}$ by replacing $\Lpq$ $\rightarrow$ $\partial_{q_1}+ \partial_{q_2}$, $\Lpp$ $\rightarrow$ $\partial_{p_1}+ \partial_{p_2}$, and $\Rpq, \Rpp,$  $\rightarrow$ $\partial_{q_3}, \partial_{p_3}$, respectively.

As the Moyal product (\ref{starz}) is
\begin{align}
*\sub{\, 0}\ =\ \exp\left\{ \frac{i\hbar}{2}\, \cP\, \right\}\ ,
\end{align}
the corresponding Moyal bracket bi-differential operator  (\ref{MoyalB}) is
\begin{align}
\cM\sub{\, 0}\ =\ \frac{*\sub{\, 0}\ -\ *\sub{\, 0}^{\, t}}{i\hbar}\ =\ 2\, \sin\left(\hbar\, \cP\,/ 2 \right)/\hbar\ .
\label{cMzsin}\end{align}
In agreement with (\ref{MoyalPDOPoisson}), we find
\begin{align}
\lim_{\hbar\to 0}\, \cM\sub{\, 0}\ =\ \cP\ ,
\label{ClimPB}\end{align}
as expected. As already stated above, we will be interested in modifying this last relation to include additional classical physics.

The reference quasi-probability distribution function is the famous Wigner function
\begin{align}
W\sub{0}(q,p;t)\
 =\  \frac{1}{2\pi}\int dy\,\text{e}^{-ipy}\bigg\langle q+\frac{\hbar y}{2}\Big|\,\hat{\rho}\,\Big|q-\frac{\hbar y}{2}\bigg\rangle\ .
\end{align}
Here $\hat{\rho}$ is the density operator, and $W\sub{0}(q,p;t) = \cW\sub{0}(\hat\rho)$.

\subsection{Other phase-space quantizations: transition differential operators}
\label{subsec: transition operators}

Phase-space quantization is not unique.  In many cases, however, the different quantizations can be related by a transition differential operator $T=T(\partial_q, \partial_p)$ \cite{Bayen1978a, Bayen1978b}.

For example, an operator ordering different from the Weyl ordering of equations (\ref{Qzero}, \ref{Qzeronm}) may be used.  The Born-Jordan quantization map is
\begin{align}
\cQ_{BJ}\Big( \exp\left[ i(\varphi q + \xi p )/\hbar \right] \Big)\ =\  \int_0^1d\alpha\, \text{e}^{i\alpha\varphi\hat q/\hbar}\, \text{e}^{ i\xi \hat p/\hbar}\, \text{e}^{i(1-\alpha)\varphi\hat q/\hbar}\ ,
\label{QBJ}\end{align}
giving the ordering prescription \cite{Cohen2013}
\begin{align}
\cQ_{BJ}\Big( q^n\, p^m  \Big)\ =\  \frac 1{n+1}\, \sum_{k=0}^n \hat q^{\, n-k}\, {\hat p}^{\, m}\, \hat q^{\, k}\ .
\label{QBJnm}\end{align}
The quantization maps $\cQ\sub{\, 0}$ and $\cQ_{BJ}$ are related. Applying the simple Baker-Campbell-Hausdorff formula to (\ref{QBJ}) yields
\begin{align}
\cQ_{BJ}\Big( \exp\left[ i(\varphi q + \xi p )/\hbar \right] \Big)\ =\ \frac{\sin(\varphi\xi/2\hbar)}{\varphi\xi/2\hbar}\, \exp\left[ i(\varphi \hat q + \xi \hat p )/\hbar \right]\ ,
\end{align}
so that
\begin{align}
\cQ_{BJ}\Big( \exp\left[ i(\varphi q + \xi p )/\hbar \right] \Big)\ = \frac{\sin(\varphi\xi/2\hbar)}{\varphi\xi/2\hbar}\, \cQ\sub{\, 0}\Big( \exp\left[ i(\varphi q + \xi p )/\hbar \right] \Big)\ .
\end{align}

Both Weyl and Born-Jordan operator orderings are Hermitian.  A non-Hermitian example is the so-called standard operator ordering, with rule \cite{Lee1995}
\begin{align}
\cQ_{S}\Big( q^n\, p^m  \Big)\ =\   \hat q^{\, n}\, {\hat p}^{\, m}\ .
\label{QSnm}\end{align}
The relation to Weyl ordering is encoded in
\begin{align}
\cQ_{S}\Big( \exp\left[ i(\varphi q + \xi p )/\hbar \right] \Big)\ =\ \exp\left( i\varphi \hat q/\hbar \right)\, \exp\left( i\xi \hat p/\hbar \right) \ =\ \text{e}^{\frac{-i\varphi\xi}{2\hbar}}\, \cQ\sub{\, 0}\Big( \exp\left[ i(\varphi q + \xi p )/\hbar \right] \Big)\ .
\label{QSQz}\end{align}
Notice that this relation, for a non-Hermitian ordering, involves a complex multiplicative function.

Suppose $\cQ, \cW$ are the quantization and dequantization maps of a phase-space quantization. We can connect these maps to our reference quantization of $\cQ\sub{\, 0}, \cW\sub{\, 0}$ using an invertible differential operator $T$ such that
\begin{align}
\cW\ =\ T\, \cW\sub{\, 0}\ ,\ \ \cQ\ =\ \cQ\sub{\, 0}\, T^{-1}\ .
\end{align}
We show pictorially how to relate different maps in Figure \ref{figure:maps}.

\begin{figure}[t!]
\begin{equation*}
\xymatrixcolsep{6pc}
\xymatrixrowsep{4pc}
\xymatrix{
   {\hat\cA_0} \ar@<-4pt>[r]_{\hat T} \ar@<-4pt>[d]_{{\cal W}_0} & {\hat\cA} \ar@<-4pt>[d]_{{\cal W}} \ar@<-4pt>[l]_{\hat T^{-1}}\\
   {\cA(*\sub{0})}\ar@<-4pt>[u]_{{\cal Q}_0} \ar@<-4pt>[r]_{T} & {\cA(*)} \ar@<-4pt>[l]_{T^{-1}} \ar@<-4pt>[u]_{{\cal Q}}
}
\end{equation*}
\caption{For a fixed classical system,  consider a reference operator quantization map $\cQ_0$ and its related dequantization $\cW_0$. The algebra of quantum observables is the operator algebra $\hat\cA_0$ of operator quantum mechanics, or the $*\sub{0}$-algebra $\cA(*\sub{0})$ in phase-space. Suppose a different quantization map $\cQ$ and corresponding dequantization $\cW$ (with their accompanying algebras $\hat\cA$ and $\cA(*)$, respectively) are related to the reference quantizations through the transition operator $T$. The quantization and dequantization maps, and the  maps  between the observable algebras are shown. The arrows labelled by $\hat T$ and $\hat T^{-1}$ are included for completeness. $\hat T$ indicates what might be called the transition superoperator.  }
\label{figure:maps}
\end{figure}

As both $\cW$ and $\cW\sub{\, 0}$ are homomorphisms from  operator products to $*$-products (see (\ref{homostar})), we find
\begin{align}
T\left(\, f\, *\sub{\, 0}\, g \,\right)\ =\ Tf\, *\sub{\, T}\, Tg\ ,
\label{starT}\end{align}
for any two phase-space distributions $f, g$.  This equation determines the product $*\sub{\, T}$. With $F=Tf$ and $G=Tg$, we have
\begin{align}
T\left(\, T^{-1}F\, *\sub{\, 0}\, T^{-1}G \,\right)\ =\ F\, *\sub{\, T}\, G\ .
\label{starTFG}\end{align}
Consider transition differential operators $T=T(\pq, \pp)$, i.e. those without any dependence on $q, p$. Since $F, G$ are arbitrary, (\ref{starT}) yields
\begin{align}
*\sub{\, T}(1,2)\ =\ T(1+2)\, *\sub{\, 0}\hskip-2pt(1,2)\,\, T^{-1}(1)\, T^{-1}(2)\ ,
\label{starTstar}\end{align}
using the notation introduced in (\ref{defLpq}), so that, because of the Liebniz rule,
\begin{align}
T(1+2)\ =\ T(\partial_{q_1}+\partial_{q_2}, \partial_{p_1}+\partial_{p_2})\ .
\end{align}
We therefore obtain
\begin{align}
*\sub{\, T}\ =\ *\sub{\, 0}\,\, \odot\sub{\, T}\ ,
\label{starodot}\end{align}
with
\begin{align}
\odot\sub{\, T}\ =\ T^{-1}(\Lpq,\Lpp)\, T(\Lpq+\Rpq, \Lpp+\Rpp)\, T^{-1}(\Rpq, \Rpp)\; ,\  {\text{when}}\ T=T(\pq, \pp)\ .
\label{Tnoqp}\end{align}
This formula appeared in \cite{Belchev2009}.

The Born-Jordan quantization discussed above provides an example:  the transition operator and star product
\begin{align}
T\sub{BJ}\ =\ \text{sinc}\big(\,  \hbar\, \partial_p\partial_q /2 \,\big)\ \ ,\ \ \ *\sub{BJ}\ =\ *\sub{0}\  \, \frac{\, \text{sinc}\big[\, \hbar\, (\Lpp\, +\, \Rpp )\, (\Lpq\, +\, \Rpq) /2  \,\big] }{\text{sinc}( \hbar \Lpp\Lpq /2 )\, \text{sinc}( \hbar \Rpp\Rpq /2 ) }
\label{BJTstar}\end{align}
are related by (\ref{starodot}, \ref{Tnoqp}). In this case,  we have a real transition operator, $T\sub{BJ} = \overline{T\sub{BJ}}$ and so a Hermitian star product $*\sub{BJ} = *\sub{BJ}^\dagger$.

As an illustration of a non-Hermitian star product, consider the standard operator ordering of (\ref{QSnm}, \ref{QSQz}). Then the relevant transition operator and star product are
\begin{align}
T\sub{S}\ =\ \exp\big[  i\hbar\, \partial_p \partial_q /2 \big]\ \ , \ \ \ *\sub{S}\ =\ \exp\big[ i\hbar\, \Lpq \Rpp \big]\ .
\label{sTstar}\end{align}
Similarly, (\ref{starodot}, \ref{Tnoqp}) relate this non-real transition operator and non-Hermitian star product.

\section{Augmented star products: guiding examples}

Intriguingly, attempts have been made to use transition operators to introduce classical, physical effects, and not just to relate alternative quantizations.  We wish to examine the feasibility of this technique.

Two guiding examples are discussed in the following subsections. The first classical effect is Gaussian coarse-graining in phase-space, which gives rise to a (generalized) Husimi star product and phase-space quantization \cite{Husimi1940, Takahashi1989}. The second case  introduced damping into the simple harmonic oscillator equations of motion, and produced a modified star product depending on the damping coefficient \cite{Dito2006, Belchev2009}.

\subsection{Coarse-grained Husimi quantization}

Consider a distribution in phase-space, $f(q,p)$, coarse grained as follows:
\begin{align}
\frac 1{\pi\eta}\, \int dq'\, dp'\, f(q', p') \exp\left\{ - \frac 1 \eta\,\left[ \frac{(q-q')^2}{\sigma^2}  +  \sigma^2(p-p')^2  \right]  \right\}\, =\, \exp\left[ \frac \eta 4 \left( \sigma^2\pq^2 + \frac 1{\sigma^2} \pp^2  \right) \right] f(q,p)\ .
\label{coarse}\end{align}
Here $\eta$ is a classical coarse-graining scale, independent of $\hbar$, and $\sigma$ is a squeezing parameter. When $\eta=\hbar$, and $f$ is the Wigner function $W$, the expressions in (\ref{coarse}) equal the original Husimi quasi-probability distribution \cite{Husimi1940, Takahashi1989}.

By (\ref{coarse}),
\begin{align}
T\sub{\,\eta}\ :=\ \exp\left[ \frac \eta 4 \left( \sigma^2\pq^2 + \frac 1{\sigma^2} \pp^2  \right) \right]
\label{Teta}\end{align}
can be interpreted as a transition differential operator. As (\ref{coarse}) is a coarse-grained Wigner function, it would be improper to say that the Husimi distribution is the result of an alternative quantization.  An additional classical physical effect (coarse-graining) is introduced with the transition operator.  In other words, $T_{\eta}$ converts Weyl quantization to an augmented quantization.

From  (\ref{starodot}, \ref{Tnoqp}), we find the (generalized) Husimi star product\footnote{\,We mention that the Husimi star product (with $\eta=\hbar$) and the Moyal star product can be unified into a family of one-parameter star products, the so-called $s$-ordered products \cite{Cahill1969a,Lizzi2014}. For $s=0$, the Moyal product is recovered, while $s=1$ yields the Husimi star product.}
\begin{align}
*_{T_\eta}\ =:\ *\sub{\,\eta}\ =\ \exp\left[\frac{i\hbar}2\left( \Lpq\Rpp - \Lpp\Rpq \right) + \frac{\eta}2\left( \sigma^2\Lpq\Rpq + \Lpp\Rpp/\sigma^2 \right)  \right]\ =\ *\sub{0}\,\odot\sub{\, \eta}\ .
\end{align}
Notice that $\odot\sub{\,\eta}^t = \odot\sub{\,\eta}$, so that, from (\ref{MoyalB}),
\begin{align}
\cM\sub{\,\eta}\ =\ \frac{*\sub{\,\eta}\, -\, *\sub{\,\eta}^{\, t}}{i\hbar}\ =\ \left( \frac{*\sub{0}\, -\, *\sub{0}^{\, t}}{i\hbar} \right)\,\odot\sub{\, \eta}\ =\ \cM\sub{0}\, \odot\sub{\, \eta}\  .
\label{MMzOeta}
\end{align}
Therefore the classical limit produces a multiplicative modification of the Poisson bi-differential operator:
\begin{align}
\lim_{\hbar\to 0}\, \cM\sub{\eta}\ =\ \lim_{\hbar\to 0}\, \frac{(*\sub{0} - *\sub{0}^t )}{i\hbar}\, \odot\sub{\eta}\ =\  \cP\, \exp\left[\frac{\eta}2\left( \sigma^2\Lpq\Rpq + \Lpp\Rpp/\sigma^2 \right) \right]\  ,
\label{ClassStarEta}
\end{align}
rather than an augmentation of the form (\ref{limMaugP}).

\subsection{Damped quantization}

Consider the simple harmonic oscillator, with Hamiltonian
\begin{align}
H\ =\ \frac{1}{2m}p^2\ +\ \frac1 2 m\omega^2 q^2\ .
\label{Hsho}\end{align}
Replace the Poisson bi-differential operator $\cP$ with \cite{Dito2006}
\begin{align}
\cP\sub{\gamma}\ :=\ \cP\ -\ 2\gamma m \Lpp\,\Rpp \ .
\label{Pgamma}\end{align}
The canonical equations of motion are changed to
\begin{align}
\dot q\ =\ q\cP\sub{\gamma}H\ =\ p/m\ \ , \qq\dot p\ =\ p\cP\sub{\gamma}H\ =\ -m\omega^2 q \,-\, 2\gamma p\ .
\label{DHO}\end{align}
The equation of motion of a damped harmonic oscillator results:
\begin{align}
\ddot q\ =\ -\
\omega^2 q\ -\ 2\gamma\dot q\ ,
\label{qDHO}\end{align}
with $\gamma$ as the damping parameter.

The same replacement in the Moyal star product produces a damped star product $*\sub{\gamma}$:
\begin{align}
*\sub{\gamma} = \exp\left\{\frac{i\hbar}2 \cP\sub{\gamma}\right\}\ =\  *\sub{0}\, \exp\left\{-i\hbar\gamma m\, \Lpp\,\Rpp\right\}\ =:\ *\sub{0}\, \odot\sub{\gamma}\ .
\label{StarG}\end{align}
The transition operator
\begin{align}
T\sub{\gamma}\ =\ \exp\left\{ -\frac{i\hbar\gamma m}2 \, \partial\sub{p}^2  \right\}
\label{Tgamma}\end{align}
reproduces the star product $*\sub{\gamma}=*_{T_\gamma}$ when used in  (\ref{starodot}, \ref{Tnoqp})  \cite{Dito2006}. $T\sub{\gamma}$ therefore describes an augmented quantization of the simple harmonic oscillator, with classical damping introduced.

Notice, however, that the transition operator $T\sub{\gamma}$ is not real.  This causes significant problems when the time evolution of quasi-probability distributions and observables is  considered \cite{Belchev2009}.  With $T\sub{\gamma}\not=\overline{T\sub{\gamma}}$, the damped $*$-product is non-Hermitian as a result:
\begin{align}
*\sub{\gamma}^\dagger = \exp\left\{\frac{i\hbar}2 \cP\sub{\gamma}\right\}^\dagger\ =\   *\sub{0}\,\, \odot\sub{\gamma}^\dagger\ =\ *\sub{0}\, \exp\left\{+i\hbar\gamma m\, \Lpp\,\Rpp\right\}\ \not=\ *\sub{\gamma}\ .
\label{stardnonH}\end{align}
The dynamics is governed by the Moyal bi-differential operator (see (\ref{HeisMoyalEqn}) above), but with $*\sub{\gamma}^\dagger\not= *\sub{\gamma}$,
\begin{align}
\cM\sub{\gamma}\ =\ \frac{*\sub{\gamma}-*\sub{\gamma}^t}{i\hbar}\ \not=\ \overline{\cM\sub{\gamma}}\ .
\label{Mgnotreal}\end{align}
Therefore, the reality of an observable, such as $f$ in (\ref{HeisMoyalEqn}), is not preserved in evolution because
\begin{align}
f\cM_{\gamma}H\ =\ f\cM_0H+i\gamma\hbar\pq\pp f\ .
\end{align}

Furthermore, by (\ref{cMzsin}),
\begin{align}
\lim_{\hbar\to 0}\, \cM\sub{\gamma}\ =\ \lim_{\hbar\to 0}\, \frac{(*\sub{0} - *\sub{0}^t )}{i\hbar}\, \odot\sub{\gamma}\ =\ \lim_{\hbar\to 0}\, \frac{\sin\left(\hbar\, \cP\sub{\, 0}\,/ 2 \right)}{\hbar/2}\, \text{e}^{-i\hbar\gamma m \Lpp\Rpp}\ =\ \cP\ \not=\ \cP\sub{\gamma}\ .
\label{Mgclass}\end{align}
That is, the damping disappears in the classical limit.  A quantization of the classical damped harmonic oscillator is not described after all \cite{Belchev2009}!

The hopeful substitution \cite{Belchev2009}
\begin{align}
\cM\sub{\gamma}\ \rightarrow\ \frac{*\sub{\gamma} - \overline{*\sub{\gamma}}}{i\hbar}\ =\ \frac{\sin\left(\hbar\cP\sub{\gamma}/2\right)}{\hbar/2}\ =\ \frac{*\sub{\gamma} - *\sub{-\gamma}^t}{i\hbar}
\label{MgtoIg}\end{align}
would fix both problems. But then the dynamical equation would become
\begin{align}
\dot f\ \rightarrow\ \frac{f\,*\sub{\gamma}\, H\ -\ H\, *\sub{-\gamma}\, f}{i\hbar}
\label{WtIt}\end{align}
with formal solution
\begin{align}
f(q,p;t)\ \rightarrow\ \exp\left[  \frac{it}{\hbar} H *\sub{\gamma}  \right]\,f(q,p;0)\,
\exp\left[   *\sub{-\gamma}\frac{it}{\hbar} H  \right] \ .
\label{egWemg}\end{align}
The last expression is problematic, however.  Although $*\sub{\gamma}$ is associative when $\gamma$ is fixed, we find
\begin{align}
\left( a\,*\sub{\gamma}\, b \right)\, *\sub{-\gamma}\, c\ \not=\ a\, *\sub{\gamma}\, \left( b\, *\sub{-\gamma}\, c \right)
\label{gmgNoAssoc}\end{align}
for phase-space obervables $a,b$ and $c$ \cite{Belchev2009}.  This non-associativity result follows from
\begin{align}
*\sub{-\gamma}(1+2, 3)\, *\sub{\gamma}(1, 2)\ \not=\ *\sub{\gamma}(1, 2+3)\, *\sub{-\gamma}(2, 3)
\label{StarGnum}\end{align}
in the notation of (\ref{starzassoc}, \ref{PREstarzassoc}). In turn, by (\ref{StarG}), (\ref{StarGnum}) is a consequence of
\begin{align}
\cP\sub{-\gamma}(1+2, 3)\ +\ \cP\sub{\gamma}(1, 2)\ \not=\ \cP\sub{\gamma}(1, 2+3)\ +\ \cP\sub{-\gamma}(2, 3)\ .
\label{Pgnum}\end{align}

Can something similar work better? Consider a generalization of the damped transition operator $T\sub{\gamma}$ of (\ref{Tgamma}):
\begin{align}
T\sub{\gamma, \eta}\ =\ \exp\left\{ -\eta\gamma  m\, \partial\sub{p}^2  \right\}\ .
\label{Tgeta}\end{align}
As in the generalized Husimi quantization, the parameter $\eta$ has dimensions of action. It replaces $i\hbar/2$, so that it does not vanish in the $\hbar\to 0$ limit. Using $\eta\in \R$, equation (\ref{Tnoqp}) yields a Hermitian star product:
\begin{align}
*\sub{\gamma, \eta}\ =\ *\sub{0}\,\odot\sub{\gamma, \eta}\ =\ *\sub{0}\,\exp\big( -2\eta\gamma m\Lpp\Rpp  \big)\ =\ *\sub{\gamma, \eta}^\dagger\ \ \ (\eta\in \R)\ .
\label{StarGE}\end{align}
However, the classical limit still fails, as
\begin{align}
\lim_{\hbar\to 0}\, \cM\sub{\gamma, \eta}\ =\ \lim_{\hbar\to 0}\, \frac{(*\sub{0} - *\sub{0}^t )}{i\hbar}\, \odot\sub{\gamma, \eta}\ =\  \cP\, \text{e}^{-2\eta\gamma m \Lpp\Rpp} \ .
\label{Mgeclass}\end{align}
The damped Poisson bracket $\cP\sub{\gamma}$ of (\ref{Pgamma}) is not recovered. Notice this is true even if small $\gamma$ is considered,
\begin{align}
\cP\, e^{-2\eta\gamma m \Lpp\Rpp} \ =\  \cP\,\Big[ 1 - 2\eta\gamma m\Lpp\Rpp + O(\gamma^2) \Big]\ \ .
\label{MgeclassSmall}\end{align}
While $\cP\sub{\gamma}$ differs additively from $\cP$, equation (\ref{Mgeclass}) describes instead a {\it multiplicative} modification of $\cP$. A multiplicative modification of the classical limit was also found when using the Husimi transition differential operator in the previous subsection.

To progress, we need to understand what is and isn't possible in augmented quantization.  For that reason, we discuss the possibilities described by an arbitrary transition differential operator $T=T(\partial_q, \partial_p)$ in the next section.

\section{Augmented star products: generalities}
\label{sec: Augmented star products}

In this section, we consider modified star products in more general terms. The goal is to see if what has been learned from the examples leads us to results that pertain to any possible Hermitian, augmented star product.

\subsection{Transition differential operators $T=T(\partial_q, \partial_p)$}
\hfb
First, consider star products $*\sub{T}$ modified by a transition differential operator $T=T(\partial_q, \partial_p)$. $T$ describes the relation between $*\sub{T}$ and the reference Moyal star product $*_0$, see (\ref{starodot}, \ref{Tnoqp}).  We will restrict to real $T$, so that Hermitian star products $*\sub{T}$ result.

In the classical limit $\hbar\to 0$, the dependence on $\hbar$ is crucial.  Consider an arbitrary transition differential operator  $T$; if a non-singular classical limit is to be found, we can write
\begin{align}
T\ =\ T_0\ +\ \hbar\, T_1\ +\ \hbar^2\, T_2\ +\ \ldots\, ,\ \ {\text{ with}}\ \ \frac{\partial T_i}{\partial\hbar}\, =\, 0\ ,
\label{Thbarseries}
\end{align}
with no negative powers of $\hbar$.  As a consequence,\footnote{\,Notice that (\ref{Ohbar}) implies that the classical limit of the star product is $\lim_{\hbar\to 0} *\sub{T} = \odot\sub{T_0}$, as in (\ref{limstarnotid}).}
\begin{align}
*\sub{\, T}\ =\ *\sub{0}\, \odot\sub{\, T}\ =\ *\sub{0}\, \odot\sub{\, T_0}\ +\ \cO(\hbar^1)\ ,
\label{Ohbar}
\end{align}
with
\begin{align}
\odot\sub{\, T_0}\ =\ T_0^{-1}(\Lpq,\Lpp)\, T_0(\Lpq+\Rpq, \Lpp+\Rpp)\, T_0^{-1}(\Rpq, \Rpp)\ .
\label{Tznoqp}\end{align}
Another critical observation is that
\begin{align}\odot\sub{\, T}^{\, t}\ =\ \odot\sub{\, T}\ ,
\label{odotSymm}
\end{align}
by (\ref{Tnoqp}). Similarly, $\odot\sub{\, T_0}$ is symmetric, by (\ref{Tznoqp}).

The classical limit therefore yields
\begin{align}
\lim_{\hbar\to 0}\, \cM\sub{\, T}\ =\ \lim_{\hbar\to 0}\, \frac{(*\sub{\, T} - *\sub{\, T}^t )}{i\hbar}\ =\  \left(\lim_{\hbar\to 0}\, \frac{\left(*\sub{0} - *\sub{0}^t \right)}{i\hbar}\, \odot\sub{\, T} \right)\ =\ \left(\lim_{\hbar\to 0}\, \frac{\left(*\sub{0} - *\sub{0}^t \right)}{i\hbar} \right)\, \odot\sub{\, T_0} \  ,
\label{ClassMTnoh}
\end{align}
so that
\begin{align}
\lim_{\hbar\to 0}\, \cM\sub{\, T}\ =\ \cP\,\odot\sub{\, T_0}\  .
\label{ClassMTPodot}
\end{align}

The classical limit is indeed modified, by that part of the transition differential operator that is independent of $\hbar$.  However, an augmentation (\ref{limMaugP}) of the desired additive form does not result - instead we find a multiplicative modification, (\ref{ClassMTPodot}).

This is the kind of multiplicative modification described by the generalized Husimi product in (\ref{ClassStarEta}).  This is the only kind of modification that a real transition differential operator can describe.

Of course, if we can write $\odot\sub{T_0}\approx 1 + \delta\odot\sub{T_0}$, we recover an approximate additive augmentation of the form  (\ref{limMaugP}), with $\delta\cP\approx \cP\,\delta\odot\sub{T_0}$.  But this is a very restrictive form.  For example, Hamilton's equations of motion for $q$ and $p$ can only be modified by the extra terms $q\cP\,\delta\hskip-2pt\odot\sub{T} H = \delta\hskip-2pt\odot\sub{T} \partial_pH$ and $p\cP\, \delta\hskip-2pt\odot\sub{T_0} H = -\delta\hskip-2pt\odot\sub{T_0} \partial_qH$. They vanish unless $\delta\odot\sub{T}$ contains terms  of the form $\Rpq^{\, m}$ or $\Rpp^{\, n}$, for some integer powers $m,n\in\N$. It never does, however, since
\begin{align}
T_0\, \approx\, 1 + \theta_0\ \ \ \Rightarrow\ \ \ \delta\,\odot\sub{T_0}\, \approx\, -\,\theta_0(\Lpq, \Lpp)\, +\, \theta_0(\Lpq+\Rpq, \Lpp+\Rpp)\, -\,  \theta_0(\Rpq, \Rpp)\ ,
\label{dOdottheta}
\end{align}
by (\ref{Tznoqp}). We will therefore not consider this possibility further.

Recall that  the Dito-Turrubiates transition operator (\ref{Tgamma})  resulted in a classical limit (\ref{Mgclass}) with no augmentation.
To understand this, consider the transition differential operators on which (\ref{Tgamma}) was modelled: those that describe operator orderings that differ from the Weyl ordering, such as (\ref{BJTstar}).  They are necessarily $\hbar$-dependent, since different orderings can be related by the Heisenberg commutation relations. For them, however, a non-augmented classical limit is necessary if they are only to describe different quantizations of the same classical system. In other words, when these transition operators are applied, additional physical effects should not appear in the classical limit.

We see now that the feature that the Dito-Turrubiates transition operator (\ref{Tgamma}) shares with transition differential operators describing changes in operator-ordering rules is
\begin{align}
T_0\ =\ 1\ \ \ \Rightarrow\ \ \ \odot\sub{T_0}\ =\ 1\ .
\label{Tzone}
\end{align}

To summarize, only the $\hbar$-independent part $T_0$ of a transition differential operator $T(\partial_q, \partial_p)$ is relevant to the classical limit of the Moyal bi-differential operator $\cM\sub{T}$. If $T_0=1$, there is no augmentation, since the classical limit is unchanged by the transition operator. If $T_0\not=1$, however, there is an augmentation produced, but the classical Poisson bi-differential operator is not changed in the desired additive way, as in (\ref{limMaugP}). A multiplicative modification,
(\ref{ClassMTPodot}), instead results.

\subsection{Local transition operators $T=T(q,p,\partial_q, \partial_p)$}
\hfb
Since transition differential operators do not yield augmented star products, we will now consider generalized transition operators that depend on the phase-space coordinates: $T=T(q, p; \pq, \pp)$.

As already noted, this is a radical step, but there are precedents. Generalizations in a similar spirit were previously discussed in \cite{Crehan1989, Pinzul2008}.\footnote{Although ref.~[14] does not  make explicit use of transition operators or star products, it does work in a mathematically equivalent formalism.  It should be noted, however, that only (``local'') quantizations dependent on $\hbar$ were considered. Ref. [15] focused on using transition operators to gauge the star product by expanding in powers of $\hbar$.}

\subsubsection{Star product}
\hfb
In the more general case, we conjecture that (\ref{starTFG}) is solved by
\begin{align}
*\sub{\, T}(1,2)\ =\ T(1,2)\, *\sub{\, 0}\hskip-2pt(1,2)\,\, T^{-1}(1)\, T^{-1}(2)\ ,
\label{starTTstar}\end{align}
where now
\begin{align}
T(1,2)\ :=\ T\Big( (q_1+q_2)/2, (p_1+p_2)/2; \partial_{q_1}+\partial_{q_2}, \partial_{p_1}+\partial_{p_2} \Big)\, ,\  {\text{when}}\  T=T(q, p; \pq, \pp)\ .
\label{Tqp}\end{align}
The bi-differential operator $*\sub{\, T}$ is obtained from $*\sub{\, T}(1,2)$ by identifying $(q_1, p_1) = (q_2, p_2) = (q, p)$:
\begin{align}
*\sub{\, T}\,  =\, \I(1,2)\,\, *\sub{\, T}(1,2)\ \ .
\end{align}
Notice that when $T=T(\partial_q, \partial_p)$ is a differential operator, the simpler result (\ref{Tnoqp}) is recovered. An explicit, general expression for $*\sub{\, T}$, however,  in terms of $q, p, \Lpq, \Rpq, \Lpp$ and $\Rpp$, is out of reach.

The result  (\ref{starTTstar}, \ref{Tqp}) appears to be new.  The formulas can be given some justification as follows. We must show that
\begin{align}
\I(1,2)\, T(1,2)\, K(1,2)\ =\ T\, \I(1,2)\, K(1,2)\ ,
\end{align}
for arbitrary $K(1,2):= K(q_1,p_1; q_2, p_2)$.

Using the commutation relations
\begin{align}
[\partial_q, q]\ =\ [\partial_p, p]\ =\ 1\ ,
\end{align}
we can rewrite the transition operator in the form
\begin{align}
T\ =\ T(q,p,\partial_q, \partial_p)\ =\ \sum\, t_{m,n}(q,p)\, \partial_q^{\, m}\, \partial_p^{\, n}\ .
\end{align}
It follows that $T(1,2)$ obeys
\begin{align}
T(1,2) \ =\ \sum\, t_{m,n}\left(\frac{q_1+q_2}{2}, \frac{p_1+p_2}{2}\right)\, (\partial_{q_1}+\partial_{q_2})^{\, m}\,\, (\partial_{p_1}+\partial_{p_2})^{\, n}\ ,
\end{align}
since
\begin{align}
[\, \partial_{q_1}+\partial_{q_2}, (q_1+q_2)/2\, ]\ &=\ [ \partial_q, q]\ =\ 1\ \ \&\nn
[\, \partial_{p_1}+\partial_{p_2}, (p_1+p_2)/2\, ]\ &=\ [ \partial_p, p]\ =\ 1\ .
\end{align}
A similar result holds if $(q_1+q_2)/2$ and $(p_1+p_2)/2$ are replaced by $\alpha q_1 + (1-\alpha)q_2$ and $\beta p_1 + (1-\beta) p_2$, for any $0\leq \alpha, \beta \leq 1$, since the necessary commutation  relations are obeyed. We choose the most symmetrical solution here, however, for simplicity.

Since $\I(1,2) t_{m,n}\big( (q_1+q_2)/2, (p_1+p_2)/2 \big) = t_{m,n}(q,p)$, what remains to be shown is
\begin{align}
\I(1,2)\, (\partial_{q_1}+\partial_{q_2})^{\, m}\,\, (\partial_{p_1}+\partial_{p_2})^{\, n}\, K(1,2)\ =\ \partial_q^{\, m}\, \partial_p^{\, n}\, \I(1,2)\, K(1,2)\ .
\end{align}
Assuming we can expand
\begin{align}
K(1,2)\ =\ K(q_1, p_1; q_2, p_2)\ =\ \sum\, k_{a_1,b_1; a_2,b_2}\, q_1^{a_1} p_1^{b_1} q_2^{a_2} p_2^{b_2} \ ,
\end{align}
we need only show that
\begin{align}
\I(1,2)\, (\partial_{q_1}+\partial_{q_2})^{\, m}\,\, (\partial_{p_1}+\partial_{p_2})^{\, n}\,\, q_1^{a_1} p_1^{b_1} q_2^{a_2} p_2^{b_2}\ =\ \partial_q^{\, m}\, \partial_p^{\, n}\, q^{a_1+a_2}\, p^{b_1+b_2}\ ,
\end{align}
which reduces to showing
\begin{align}
\I(1,2)\, (\partial_{q_1}+\partial_{q_2})^{\, m}\,\, q_1^{a_1}  q_2^{a_2}\ =\ \partial_q^{\,\, m}\, q^{a_1+a_2}\ .
\label{reduced}\end{align}
But this last equation is satisfied.

It is important to note that the associativity of $*\sub{\, T}$ follows from that of the Moyal star product $*\sub{\, 0}$, for any invertible $T$, whether it depends on $q$ and $p$ or not. One obtains
\begin{align}
\Big( Tf\, *\sub{\, T}\, Tg \Big)\, *\sub{\, T}\, Th\ =\ Tf\, *\sub{\, T}\, \Big( Tg \, *\sub{\, T}\, Th \Big)\ ,
\label{starTassoc}\end{align}
by applying (\ref{starT}) twice to (\ref{starzassoc}).

\subsubsection{Augmented equations of motion}
\label{sec: Augmented equations of motion}
\hfb
In this section, we will use  local transition operators $T=T(q,p; \partial_q, \partial_p)$ to try to find star products that yield classical limits augmented by additional physics. We will focus on the equations of motion for the phase-space coordinates $q, p$.

Let $x$ denote either $q$ or $p$. Precisely, we will ask that
\begin{align}
\dot x\ =\ \lim_{\hbar\to 0}\, \{ x, H \}\sub{*_T}\
\label{LimEqMx}
\end{align}
describes the augmented quantization in the classical limit.

With $T=T(q,p; \partial_q, \partial_p )$ a local transition operator, we have no general formula for $*\sub{\, T}$ written directly in terms of left- and right-derivatives. It is therefore easiest to work with $T$ directly and use (\ref{starTFG}) to rewrite (\ref{LimEqMx}) as
\begin{align}
\dot x\ =\ \lim_{\hbar\to 0}\, T\Big(\,  \{ T^{-1}x, T^{-1}H \}\sub{\, *_0} \,\Big)\ =\
T\Big(\,  \{ T^{-1}x, T^{-1}H \}\,\Big)\ ,
\label{LimEqMxT}
\end{align}
where we have used $\partial T/\partial\hbar = 0$ and (\ref{ClimPB}).

We will consider augmentations that are weak, by writing
\begin{align}
T\ =\  \approx\ 1\,+\, \theta\ ,\ \ \ \ T^{-1}\ \approx\ 1 \,-\, \theta\ ,
\label{weakTtheta}
\end{align}
with $\theta$ a bi-differential operator. Then (\ref{LimEqMxT}) becomes
\begin{align}
\dot x\ -\ \{ x, H \}\ \approx\ \theta\big(\, \{ x, H \} \,\big)  \ -\ \{\theta(x), H\}\ -\ \{ x, \theta(H) \}\ ,
\label{ThetaAug}
\end{align}
where the augmenting terms are all on the right-hand side.  It is helpful, perhaps, to rewrite them in the notation of (\ref{shorth}).  Denoting the  terms augmenting the equations of motion by $\cA\sub{\theta}(x)$, we have
\begin{align}
\cA\sub{\theta}(x)\ =\ \I(1,2)\,\Big\{  \theta(1+2)\,\cP(1,2) \, -\, \cP(1,2)\, \big[ \theta(1) + \theta(2)   \big] \Big\}\, x(1)\,H(2)\ .
\label{AugShort}
\end{align}

From these expressions, one sees that $\theta = \beta \partial_{x'}$ ($\beta$ a constant, and $x'=q$ or $p$) produces no augmentation.  Furthermore, no term contributes that is higher order in derivatives $\partial_q, \partial_p$, if it is multiplied by a constant. A multiplicative function of $q$ and $p$ in $\theta$ is necessary - this confirms that a local transition operator is required.

Consider then the ansatz
\begin{align}
\theta\ =\ \sum_{{m,n}\atop{1\leq m+n}}\, \theta\sub{m,n}\, \partial_{q}^{\, m}\, \partial_{p}^{\, n}\ ,
\label{ThetaQuad}
\end{align}
producing
\begin{align}
\cA\sub{\theta}(q)\ =&\  -\{\theta\sub{1,0}, H\}\ -\sum_{m,n} \left(\partial_p\theta\sub{m,n}\right) \left( \partial_q^{\, m} \partial_p^{\, n}H\right)\ =\  -\{\theta\sub{1,0}, H\}\ -\ \big[\, (\partial_p\theta) H \,\big] , \nn
\cA\sub{\theta}(p)\ =&\  -\{\theta\sub{0,1}, H\}\ +\sum_{m,n} \left(\partial_q\theta\sub{m,n}\right) \left( \partial_q^{\, m} \partial_p^{\, n}H\right)\ =\  -\{\theta\sub{0,1}, H\}\ -\ \big[\, (\partial_q\theta) H \,\big] \ .
\label{AugThetaqp}
\end{align}
\subsubsection{Example: weakly damped harmonic oscillator}
\hfb
As an example, consider augmenting the simple harmonic oscillator by introducing weak damping, with damping coefficient $\gamma$.  To produce the equations of motion (\ref{DHO}) for the damped oscillator, we need
\begin{align}
\cA\sub\theta(q)\ =\ 0\ ,\ \ \ \cA\sub\theta(p)\ =\ -2\gamma p\ ,
\label{DHOaug}
\end{align}
when the Hamiltonian (\ref{Hsho}) is used in (\ref{AugThetaqp}).
We find that
\begin{align}
\theta\ =&\ \theta_{0,1}\,\partial_p\ +\ \theta_{0,2}\,\partial_p^{\, 2}\ , \nn
\theta_{0,1}\ =&\ 2\frac\gamma\omega\, \int \arctan \left( \frac{m\omega q}p \right)\, dp\ ,\nn
\theta_{0,2}\ =&\ -\frac\gamma\omega\, \left\{ 2mH\,\arctan \left( \frac{m\omega q}p \right)\ +\ m\omega q p \right\}\ ,
\label{TgaLoc}
\end{align}
yields (\ref{DHOaug}). Here $\int\arctan({m\omega q}/p)\, dp$ indicates the indefinite integral.\footnote{${}\,$For simplicity, we have not written the somewhat more general solution we obtained for  $\theta$ of (\ref{AugThetaqp}, \ref{DHOaug}). It is recorded in Appendix C of \cite{Robbins2017}. Also written there is a $\theta$ producing quadratic damping.}

This example demonstrates that local transition operators may be capable of  producing star products describing systems with additional classical effects. A Hermitian star product incorporating weak damping, at least in the equations of motion for phase-space coordinates $q$ and $p$, is described by the local transition operator of (\ref{weakTtheta}, \ref{TgaLoc}).

However, we have not worked out the modified Moyal bi-differential operator $\cM$  (\,see (\ref{MoyalB}, \ref{cMstar})\,) in general, and we have no explicit formula for the star product in terms of left- and right-derivatives. The result (\ref{TgaLoc}) is quite complicated, while only producing the desired $q$ and $p$ equations of motion, and being only valid for weak coupling. Furthermore, the damping it describes applies only to the simple harmonic oscillator - a different form would be required for the damping of a different system.

All this suggests that even if local transition operators may produce quantizations augmented by classical effects, their use may turn out to be too unweildy to be practical in what is an effective description of certain physical systems.

\section{Conclusion}

Our goal was to investigate the possibility, suggested in \cite{Dito2006}, of introducing additional classical physics during phase-space quantization. To do this, we analyzed transition operators describing Hermitian quantizations related to the Weyl-Wigner quantization, with its Moyal star product. Can transition operators yield such augmented quantizations?

By examining the classical limit, we showed that transition differential operators $T(\pq, \pp)$ cannot. Only the $\hbar$-independent part $T_0$ of the transition operator is relevant for the classical limit. If $T_0=1$, then there is no augmentation: the classical limit recovers the usual Poisson brackets, with no extra classical physics. If $T_0\not=1$, the classical physics is indeed modified, but in a multiplicative, rather than additive way.

So, augmented quantization using  transition differential operators $T=T(\pq, \pp)$ does not produce a classical limit of the desired form (\ref{limMaugP}).

In retrospect, we see that the Dito-Turrubiates \cite{Dito2006, Belchev2009} transition operator (\ref{Tgeta}) is an example for which $T_0=1$.  Also, the (generalized) Husimi \cite{Husimi1940, Takahashi1989} transition operator  (\ref{Teta}) illustrates the other case, with $T_0\not=1$, but describing a multiplicative modification of the classical physics.

For a more thorough treatment, we also considered a significant generalization. Local transition operators $T=T(q, p, \pq, \pp)$, dependent on the phase-space point, were also examined. We managed to construct a single example, a real (local) transition operator (\ref{weakTtheta}, \ref{TgaLoc}) for a Hermitian star product that introduces a weak damping into the $q, p$ equations of motion of the simple harmonic oscillator.

However, the analysis is significantly more difficult when using local transition operators. Only the equations of motion for the phase-space coordinates were examined; the modified Moyal bi-differential operator was not worked out.  Furthermore, the transition operator (\ref{weakTtheta}, \ref{TgaLoc}) is quite complicated.  At some point, an effective description is not practical if it is too involved.

The weak-coupling result we found is also of limited validity, and the transition operator has the undesirable feature of being specific to the harmonic oscillator. It would have to take a different form to introduce damping into a different quantum system.  This contrasts sharply with  the standard procedure: to incorporate an additional physical effect into different systems, an identical term is added to the different Hamiltonians.

We conclude, therefore, that the significant generalization to local transition operators yields a rather unwieldy machinery.  In its original form, augmented quantization with a Hermitian star product does not work, and with local transition operators, it is not clear that it is practical.

\vskip1cm
\noindent{\bf Acknowledgements}\hfb
This research was supported in part by a Postgraduate Scholarship (MR) held at the University of Lethbridge and a Discovery Grant (MW, grant number RGPIN 5809/2015) from the Natural Sciences and Engineering Research Council (NSERC) of Canada. Some of the results of this paper were already reported in \cite{Robbins2017}, available at https://www.uleth.ca/dspace/handle/10133/4923.

\vfill\eject

\bibliographystyle{unsrt}
\bibliography{RobbinsWalton2018References}

\end{document}


TITLE: Prove that $|G:A|=|G:B||B:A|$ for $A\subseteq B\subseteq G$
QUESTION [1 upvotes]: Given $G$ a group and $A,B$ subgroups of $G$ such that $A\subseteq B\subseteq G$, I want to prove $|G:A|=|G:B||B:A|$ where these are the indexes of the respective groups.
I have been presented with a proof as follows but get stuck near to the end;
Let $Bx_i$ and $Ab_j$ be the set of distinct right cosets in $G$, $B$ respectively for some $i\in I, j\in J$.
Then $|G:B|=|I|$ and $|B:A|=|J|$.
With this set up, the proof claims that $Ab_jx_i$ are all the distinct right cosets of $A$ in $G$.
I understand how the proof has shown that these are the distinct right cosets of $A$, but do not understand the deduction that these are distinct.
This is the given argument;
Suppose $Ab_jx_i=Ab_kx_l$. As $Ab_j,Ab_k \subseteq B$, this implies $Bx_i =Bx_l$ and hence $i=l$. A similar argument is used to show $Ab_j=Ab_k$ implies $j=k$.
Surely we can have two cosets being equal to each other with distinct elements representing each coset and so why would $i=l$ here?
eg. if $G=(\mathbb Q^*, \times)$ and $H=\{-1,1\}$ then the right cosets $H(1)=H(-1)=H$.

REPLY [1 votes]: Essential is: if two cosets have a non-empty intersection then they are equal.
$Bx_i\supseteq Ab_jx_i=Ab_kx_l\subseteq Bx_l$ tells us that the cosets $Bx_i$ and $Bx_l$ have a non-empty intersection hence are equal.
Then $x_i$ and $x_l$ both represent the same coset, but among all representatives $x_k$ there is only one that represents the coset $Bx_i=Bx_l$.
We could rephrase that by saying that among all indices $k$ there is only one such that $x_k$ represents coset $Bx_i=Bx_l$.
This justifies the conclusion that $i=l$
Then $i=l$ and setting $x_i=x=x_j$ now from $Ab_jx=Ab_kx$ we can conclude that also $Ab_j=Ab_k$.
Then cosets $Ab_j$ and $Ab_k$ have non-empty intersection hence are equal, so that also $j=k$.


text:
\section{Experiments}
\label{sec:simulations}

We consider \emph{Infection Spread} described in Section \ref{sec:formulation}. State space is $\mathcal{S}=\{0,\ldots, 24\}$ with $0$ being the lowest health level. Action space is $\mathcal{A}=\{0, \ldots, 4\}$ with $4$ being the strongest preventive action, and $c_f$ is the infection intensity constant. Details of the parameters and more experiments are presented in the supplementary materials.  We evaluate the performance of our algorithms, TMFQ-Learning (TMFQ),  GMBL, and Online TMFQ-Learning (O-TMFQ) on this model, along with comparisons with Independent Q-Learning (IQL) and Mean Field Q-learning (MFQ) \citep{yang2018mean}.  In IQL,  each agent ignores other agents, maintains an individual Q-function and performs TQ-learning independently. We implement a variant of MFQ, where each agent maintains a Q-function parameterized by the average states of a subset of the population and performs TQ-Learning. We average over 20 runs in each experiment, and the dashed line and band in figures show the average and standard deviation, respectively.

Figure \ref{fig:CDF_Algos} shows the final mean field distribution obtained by each of our algorithms, simulated with $1000$ agents. Note that all of them converge to the same T-MFE, indicating the accuracy of O-TMFQ.  We next compare the performance O-TMFQ with  IQL and MFQ.  Figure \ref{fig:Algos_CDF} shows that the final equilibrium distributions of IQL and MFQ are inaccurate (not the true T-MFE), and that O-TMFQ results in higher states (health levels).  Figure \ref{fig:Algos_mean_state} shows the evolution of the mean health of the population, $\sum_{s} s z_{k}(s)$ with iteration number $k.$  The  mean health of the population quickly converges to a higher value under O-TMFQ while the corresponding value is lower under MFQ and IQL.

Figure \ref{fig:z_converge} shows rate of convergence of the mean field under different numbers of agents. Here, $z^*$ is the final mean field obtained by O-TMFQ.  We see that the asymptotically accurate mean field approximation becomes increasingly correct even with a relatively small number of agents of 500 or 1000.  We show the evolution of average health level of the population for O-TMFQ in Figure~\ref{fig:mean_state} for different values of $c_f$.  This plot indicates that the convergence of the algorithm is fairly fast.  Finally, Figure \ref{fig:CDF_OL} explores the impact of the mean field on the model and its associated equilibrium via $c_f$. As expected, more agents are in lower health states for larger $c_{f}$.

\section{Conclusions}

We introduced the notion of trembling-hand perfection to MFG as a means of providing strategically consistent exploration.  We showed existence of T-MFE in MFG with strategic complementarities, and developed an algorithm for computation.  Based on this algorithm, we developed model-free, model-based and fully online learning algorithms, and provided PAC bounds on their performance.  Experiments illustrated the accuracy and good convergence properties of our algorithms.

\clearpage


text:
\begin{definition}[Definition:Apotome of Medial/Terms]
The '''terms''' of $a - b$ are the elements $a$ and $b$.
=== [[Definition:Apotome of Medial/Whole|Whole]] ===
{{Definition:Apotome of Medial/Whole}}
=== [[Definition:Apotome of Medial/Annex|Annex]] ===
{{Definition:Apotome of Medial/Annex}}
\end{definition}

config: wiki


TITLE: Decreasing water pressure toward top of tank walls
QUESTION [1 upvotes]: A contractor who is building a freestanding water storage tank of 6m (length) x 2m (width) x 1.5m (depth) suggests stepping out halfway up the walls (fibreglass sat in an iron frame) by 2cm to increase structural integrity of walls (i.e. measurements of lower section of tank 5m x 2m and upper half 5.04m x 2.04m so they do not bow outward.  This makes sense in terms of putting this profile in to strengthen the fibreglass but they also state the water pressure at the top will be less in doing so.  Is this correct and what is the scientific reasoning?

REPLY [1 votes]: There is more pressure as you go deeper in the tank - so making the wall thicker as you go deeper makes sense.
The reason for the higher pressure is the weight of the liquid column above you. Imagine a thin tube (1 cm$^2$ cross sectional area) filled with water, and with a piston at the bottom. If you fill the tube to a height of 1 m, the mass of the water (at a density of 1000 kg/m$^3$) will be 100 g, and the weight about 1 N. This means there will be a force of 1 N needed to hold the piston in place.
If you now had a second hole at the bottom of the tube, but at right angles, and it had the same dimensions, then the piston would again need a force of 1 N to keep it in place, as pressure is isotropic (because water is a liquid).
So pressure is linearly related to depth (for an incompressible liquid). You need a thicker wall near the bottom of your tank to hold in the greater pressure. Of course you can make the whole thing of equal thickness - but then you are using much more material than needed near the top. Building up the lower part of the wall only makes sense.
On the other hand, the statement

the water pressure at the top will be less in doing so

does not make sense. The pressure of the water is not a function of the area (unless you say - the area is larger, I have the same amount of water, so I have a lower height). And for a circular container (yours is rectangular if I understand the description) the hoop stress (the stress on the wall) is greater (for a given wall thickness) when the area is larger - it scales with the radius of the tank. So that's backwards.


TITLE: Prove for x real and p integer that $\begin{pmatrix}x\\ p\end{pmatrix}=\left(-1\right)^p\:\cdot \begin{pmatrix}p-x-1\\ \:p\end{pmatrix}$
QUESTION [0 upvotes]: I have to prove for x real and p integer that
$\begin{pmatrix}x\\ p\end{pmatrix}=\left(-1\right)^p\:\cdot \begin{pmatrix}p-x-1\\ \:p\end{pmatrix}$
I have tried to remember all the properties of falling and rising factorials before starting working, i didn't find a relation of $\left(-1\right)^p\:\cdot \begin{pmatrix}p-x-1\\ \:p\end{pmatrix}$. So what i did was to use the definition of combinatorics.

$\frac{x!}{p!\left(x-p\right)!}=\left(-1\right)^p\:\cdot
    \:\left(\frac{p-x-1}{p!\left(-x-1\right)!}\right)$

$\frac{x!}{\frac{p!\left(x-p\right)!}{\frac{\left(p-x-1\right)!}{p!\left(-x-1\right)!}}}=\left(-1\right)^p$

$\frac{\left(-x-1\right)!x!}{\left(x-p\right)!\left(p-x-1\right)!}=\left(-1\right)^p$

$x!\left(-x-1\right)!=\left(-1\right)^p\:\cdot \:\left(x-p\right)!\left(p-x-1\right)!$

REPLY [1 votes]: This is an exercise
in the definition of generalized binomial coefficients,
use of product notation,
and manipulation of indices.
$\begin{array}\\
\binom{x}{p}
&=\dfrac{\prod_{k=0}^{p-1}(x-k)}{p!}
\qquad\text{(Definition of generalized binomial coefficient)}\\
&=\dfrac{(-1)^p\prod_{k=0}^{p-1}(k-x)}{p!}
\qquad\text{(Reverse the sign of each term)}\\
&=(-1)^p\dfrac{\prod_{k=0}^{p-1}((p-1-k)-x)}{p!}
\qquad\text{(Reverse the order of the product so } k \to p-1-k)\\
&=(-1)^p\dfrac{\prod_{k=0}^{p-1}(p-x-1-k)}{p!}
\qquad\text{(Rearrange the expression)}\\
&=(-1)^p \binom{p-x-1}{p}
\qquad\text{(Definition, again)}\\
\end{array}
$