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# Parameter estimation using MCMC

I am working on Parameter estimation. I've been told to use MCMC. I've chosen the Metropolis-Hasting algorithm. I think I understand the algorithm.

I have around 8 input parameters to estimate. How do I go about choosing a multivariate proposal distribution and acceptance ratio? what factors should be considered? Also, the values for parameters need to be varied only by +\- 50% of their default values, How can I specify this?

I have seen the code examples of MCMC in Python and R, I don't understand them that well, most of them are for 2-3 variables.

Any help is appreciated. It would also be sufficient if you could just explain me the intuition behind choosing the proposal distribution and the acceptance ratio function.

• you don't choose the acceptance ratio. You calculate it at every iteration using your choice of proposal and model. – Taylor Jul 6 '17 at 19:27
• @Taylor: the average acceptance ratio is governed by the proposal. Proposal that usually depends on parameters like scale. Hence by choosing the scale you can modify the average acceptance ratio. This is the essence of adaptive MCMC algorithms. – Xi'an Jul 12 '17 at 20:07
• @Xi'an, I assumed OP was under the impression that both the proposal distribution and the acceptance ratio determined by it (and the model) were left up to choice. I was just trying to be helpful by reducing the number of things that have to be selected. Regarding adaptive MCMC: it's on my reading list, and I am not too familiar with it at the moment. – Taylor Jul 13 '17 at 2:40

## 1 Answer

Here's a quote from page 306 of the second edition of Bayesian Data Analysis that has helped me personally with the same question:

It is hard to give general advice on efficient jumping rules, but some results have been obtained for normal random walk jumping distribtutions that seem to be useful in many problems. Suppose there are $d$ parameters, and the posterior distribution $\theta = (\theta_1, \ldots, \theta_d)$, after appropriate transformation, is multivariate normal with known variance matrix $\Sigma$. Further suppose that we will take draws using the Metropolis algorithm with a normal jumping kernel centered on the current point and with the same shape as the target distribution: that is $J(\theta^*|\theta^{t-1}) = N(\theta^*|\theta^{t-1},c^2\Sigma)$. Among this class of jumping rules, the most efficient has scale $c \approx 2.4/\sqrt{d}$, where efficiency is defined relative to independent sampling from the posterior distributuon.


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## An obvious remark

### Aside

An obvious remark:

If a sequence of independent random variables $X_n$ converge almost surely to some limit $X$, this limit must be a constant (almost surely).

I’ve been thinking about the Central Limit Theorem about related Large Deviations results this afternoon, and wasted almost an hour worrying about situations which were effectively well-disguised special cases of the above.

Why is it true? Well, suppose each $X_n$ is $\mathcal{F}_n$-measurable. But by independence, we might as well take $\mathcal{F}_n=\sigma(X_n)$. Then the limit variable $X$ is independent of $\mathcal{F}_n$ for all $n$, and thus independent of $\cup F_n=\mathcal{F}\supset \sigma(X)$. If $X$ is independent of itself, it must be almost surely constant.

# Recurrence and Transience of BM

In this post, we consider Brownian motion as a Markov process, and consider the recurrence and transience properties in several dimensions. As motivation, observe from Question 5 of this exam paper that it is a very much non-trivial operation to show that Brownian motion in two-dimensions almost surely has zero Lebesgue measure. We would expect this to be true by default, as we visualise BM as a curve. So it is interesting to see how much we can deduce without significant technical analysis. We will make use of Ito’s formula. This material comes from the Part III course Advanced Probability, which doesn’t explicitly mention Ito’s result, and instead proves the result required separately, making use of the nature of solutions to the diffusion equation. In this context we assume that for $f\in C_b^{1,2}$:

$M_t:=f(t,B_t)-f(0,B_0)-\int_0^t(\frac{\partial}{\partial t}+\frac12\Delta)f(s,B_s)ds$

is a martingale. Of course, precisely, from Ito’s formula, this can be expressed as the stochastic integral of a bounded function with respect to Brownian motion, which is therefore a (continuous local, but bounded) martingale.

d=1: In one dimension, BM is point-recurrent. This means that almost surely, BM returns to zero infinitely many times. This is easiest shown by using the time-reversal equivalence to deduce that $\lim\sup B_t=-\lim\inf B_t=\infty$.

d=2: BM is two dimensions is point-transient. That means that the probability of returning to a given point is <1. In fact it is 0, as one might suspect from the fact that BM is space-invariant and, intuitively at least, has measure 0. However, it is neighbourhood-recurrent, meaning that it almost surely returns to a ball around a given point infinitely often. We discuss small balls around 0, but obviously the conclusions apply equally well elsewhere.

The aim is to choose a function so that the expression in Ito’s formula as above is as simple as possible. Taking f a function of space alone and harmonic causes the integral term to vanish. In this case, $f(y)=\log|y|$ will suffice. Obviously we have to restrict attention to $\epsilon\leq |y|\leq R$. We stop M at $T_\epsilon\wedge T_R$, that is the first time that the BM hits the boundary of the annulus on which f is defined, and apply OST, since $\log|B_t|$ is bounded here and the stopping time is a.s. finite. We obviously have to assume the BM starts in this annulus, but then we obtain:

$\mathbb{E}_x\log|B_{T_\epsilon\wedge T_R}|=\log|x|$

and so we can consider the two possibilities for $B_{T_\epsilon\wedge T_R}$ to deduce:

$\mathbb{P}_x(T_\epsilon

Now let $\epsilon\downarrow 0$ to see that $\mathbb{P}_x(B_t=0,\text{ some }t>0)=0$. Now apply the (weak) Markov property at a small fixed time a, to deduce, with a mild abuse of notation:

$\mathbb{P}_0(B_t=0,\text{ some }t>a)=\int \mathbb{P}_x(B_t=0,t>0)\mathbb{P}_0(B_a=dx)=0$

as the first term in the integral we have shown to be 0 $B_a$-a.e. Then let $a\downarrow 0$ to obtain the result about point-transience.

For neighbourhood recurrence, instead let $R\uparrow\infty$, so $\mathbb{P}_x(T_\epsilon<\infty)=1$. As before, can integrate over law of $B_n$ to obtain

$\mathbb{P}_0(|B_t|\leq \epsilon,\text{ some }t\geq n)=1$

which is precisely what we require for transience.

d=>3: BM is transient. That is, $|B_t|\rightarrow\infty$ a.s. Note that for d>3, the first three components have the same distribution as BM in three dimensions, and so it suffices to consider the case d=3.

Here, the correct choice of harmonic function is $f(y)=\frac{1}{|y|}$, so conclude as before that

$\mathbb{P}_x(T_\epsilon

From this, we can take a limit to see that

$\mathbb{P}_x(T_\epsilon<\infty)\leq \frac{\epsilon}{|x|}$

We deploy a neat trick to lift this result to a global statement about transience. Define the events that the modulus never returns to $n$ after hitting $n^3$

$A_n:=\{|B_t|>n\quad \forall t\geq T_{n^3}\}$

Calculate

$\mathbb{P}_0(A_n^c)\stackrel{\text{SMP}}{=}\mathbb{E}_0[\mathbb{P}_{B_{T_{n^3}}}(T_n<\infty)]=\mathbb{E}_0[\frac{1}{n^2}]=\frac{1}{n^2}$

Applying Borel-Cantelli 1, $A_n$ eventually holds almost surely, which certainly implies the desired result.

# Martingale Convergence

I continue the theme of explaining important bookwork from the Part III course, Advanced Probability, as succinctly as possible. In this post, we consider the convergence properties of discrete time martingales.

1) Theorem: Assume $X$ is a martingale bounded in $L^1$. Then

$X_n\rightarrow X_\infty\in L^1(\mathcal{F}_\infty)$ a.s.

Remark: The theorem and proof works equally well for $X$ a supermartingale.

Proof: Essentially, we want to reduce convergence of the random variables which make up the martingale to a countable collection of events. We do this by considering upcrossings, which counts the number of times the process alternates from less than a given interval to greater than a given interval. The formal definition will be too wide for this format, so we summarise as

$N_n([a,b],X)=$ the number of disjoint time intervals up to time n in which X goes from $[-\infty a)$ to $(b,\infty]$. Define $N([a,b],X)$ to be the limit as n increases to infinity.

It is a genuinely easy check that a sequence converges (possibly to $\infty$) iff this number of upcrossings of any interval with rational bounds is finite. We will show that the martingale almost surely has this property. The key lemma is a bound due to Doob:

Lemma: $(b-a)\mathbb{E}[N_n([a,b],X)]\leq \mathbb{E}[(X_n-a)^-]$

Proof: Say $S_1 are the successive hitting times of $[-\infty,a),(b,\infty]$ respectively. So $N_n=\inf\{k:T_k\leq n\}$. We decompose, abbreviating the number of upcrossings as N.

$\sum_{k=1}^n(X_{T_k\wedge n}-X_{S_k\wedge n})=\sum_{k=1}^N(X_{T_k}-X_{S_k})+(X_n-X_{S_{N+1}})1_{\{S_{N+1}\leq n\}}$

Now take an expectation of both sides, applying the Optional Stopping Theorem to the bounded stopping times on the LHS. (If we are working with a supermartingale, formally we need to take $\mathbb{E}[\cdot|\mathcal{F}_{S_k}]$ of each summand on LHS to show that they are non-negative, and so taking a further expectation over $\mathcal{F}_{S_k}$ gives the required result.) We obtain:

$0\geq (b-a)\mathbb{E}N-\mathbb{E}(X_n-X_{S_{N+1}})1_{\{S_{N+1}\leq n\}}$

If $S_{N+1}>n$ then both $1_{\{S_{N+1}\leq n\}}=(X_n-a)^-=0$. Otherwise $(X_n-a)^-\geq X_{S_{N+1}}-X_n$. This complete the proof of the lemma.

Since $\mathbb{E}[(X_n-a)^-]\leq \mathbb{E}|X_n|+a<\infty$, where this last bound is uniform in by assumption, applying monotone convergence, we get that $N([a,b],X)$ is almost surely finite for every pair $a. Because this set is countable, we can deduce that this holds almost surely for every pair simultaneously. We therefore define $X_\infty(w)=\lim X_n(w)$ when this limit exists, and 0 otherwise. With probability one the limit exists. Fatou’s lemma confirms that $X_\infty\in L^1(\mathcal{F}_\infty)$.

2) We often want to have convergence in $L^1$ as well. Recall for Part II Probability and Measure (or elsewhere) that

UI + Convergence almost surely is necessary and sufficient for convergence in $L^1$.

This applies equally well to this situation. Note that for a martingale, this condition is often convenient, because, for example, we know that the set $\{\mathbb{E}[X_\infty|\mathcal{F}_n],n\}$ is UI for any integrable $X_\infty$.

3) Convergence in $L^p$ is easier to guarantee.

Theorem: i) X a martingale bounded in $L^p$ iff ii) $X_n\rightarrow X_\infty$ in $L^p$ and almost surely iff iii) $\exists Z\in L^p$ s.t. $X_n=\mathbb{E}[Z|\mathcal{F}_n]$ a.s.

Remark: As part of the proof, we will show, as expected, that $X_\infty,Z$ are the same.

Proof: i)->ii) Almost sure convergence follows from the above result applied to the p-th power process. We apply Doob’s inequality about running maxima in a martingale process:

$||X_n^*||_p:=||\sup_{m\leq n}X_m||_p\leq \frac{p}{p-1}||X_n||_p$

Using this, we see that $X_n^*\uparrow X_\infty^*:=\sup|X_k|$. Now consider $|X_n-X_\infty|\leq 2X_\infty^*\in L^p$ and use Dominated Convergence to confirm convergence in $L^p$.

Note that Doob’s $L^p$ inequality can be proven using the same author’s Maximal inequality and Holder.

ii)->iii) As we suspected, we show $Z=X_\infty$ is suitable.

$||X_n-\mathbb{E}[X_\infty|\mathcal{F}_n]||_p\stackrel{\text{large }m}{=}||\mathbb{E}[X_m-X_\infty|\mathcal{F}_n]||_p\stackrel{\text{Jensen}}{\leq}||X_m-X_\infty||_p\rightarrow 0$

iii)->i) is easy. Z bounded in $L^p$ implies X bounded by a simple application of the triangle inequality in the definition of conditional expectation.

# Remarkable fact about Brownian Motion #2: Blumenthal’s 0-1 Law and its Consequences

Brownian motion is a martingale, so it is known that it has the Markovian property. That is, $(B_{t+s}-B_s,t\geq 0)$ is a Brownian motion independent of $\mathcal{F}_s$. But in fact we can show that this is independent of $\mathcal{F}_s^+=\cap_{t>s}\mathcal{F}_t$, the sigma algebra that (informally) deals with events which are determined by the process up to time s and in the infinitesimal period after time s.

Is this surprising? Perhaps. This larger sigma algebra contains, for example, the existence and value of the right-derivative of the process at time s. But the events determined by the local behaviour after time s are clearly also determined by the whole process after time s, so the independence property looks likely to give a 0-1 type law, like in the proof of Kolmogorov’s 0-1 law.

Theorem: $(B_{t+s}-B_s,t\geq 0)$ is independent of $\mathcal{F}_s^+$.

Proof: Take a sequence of times $s and $A\in\mathcal{F}_s^+$. It will suffice to show that the joint law of the new process at these times is independent of event A. So take F a bounded continuous function. The plan is to approximate $B_s$ from above, as then we will definitely have independence from $\mathcal{F}_s$, and hope that we have enough machinery to carry through the statements through the limit down to s. So take $s_n\downarrow s$, and then: $\mathbb{E}[F(B_{t_1+s}-B_s,\ldots,B_{t_k+s}-B_s)1_A]=\lim_n \mathbb{E}[F(B_{t_1+s_n}-B_{s_n},\ldots,B_{t_k+s_n}-B_{s_n})1_A]$ as continuity gives a.s. pointwise convergence, and we can lift to expectations by Dominated Convergence. The function in the limit on the right separates by independence as $\mathbb{E}[F(_{t_1+s_n}-B_{s_n},\ldots,B_{t_k+s_n}-B_{s_n})]\mathbb{P}(A)$. Applying the previous argument in reserve gives that the limit of this is $\mathbb{E}[F(B_{t_1+s}-B_s,\ldots,B_{t_k+s}-B_s)]\mathbb{P}(A)$ as desired.

In particular, this gives Blumenthal’s 0-1 Law, which states that $\mathcal{F}_0^+$ is trivial. This is apparent by setting s=0 in the above result, because then the process under discussion is the original process, and so $\mathcal{F}_0^+$ is independent of $\mathcal{F}\supset \mathcal{F}_0^+$.

A consequence is the following. For a BM in one dimension, let $\tau=\inf\{t>0:B_t>0\}, \sigma=\inf\{t>0:B_t<0\}$. By the fact that BM is almost surely non-constant, for any sample path, at least one of these is 0, and by symmetry $\mathbb{P}(\tau=0)=\mathbb{P}(\sigma=0)$ so these are greater than or equal to 1/2. But it is easy to see that the event $\{\tau=0\}\in\mathcal{F}_0^+$, and so by the triviality of the sigma-field, this probability must be 1. With continuity, this means that every interval $(0,\epsilon)$ contains a zero of the Brownian motion almost surely. Patching together on rational intervals (so we can use countable additivity) gives that BM is almost surely monotonic on no interval. A similar argument can be used to show that BM is almost surely not differentiable at t=0. For example, the existence and (conditional on existence) value of the derivative at t=0 is a trivial event, so by symmetry, either the derivative is a.s. =0, or a.s. doesn’t exist. Ruling out the former option can be done in a few ways. Predictably, in fact it is almost surely differentiable nowhere, but that is probably something to save for another post.

# Remarkable fact about Brownian Motion #1: It exists.

A Brownian Motion in one dimension is a stochastic process $B_t$ adapted to $(\mathcal{F}_t)$, defined on some probability space, which is almost surely continuous, and has the properties that $B_0=0$ a.s. and for every $0\leq s\leq t, B_t-B_s\sim N(0,t-s)$ and is independent of $\mathcal{F}_s$. So it has independent increments.

It can be shown that Brownian motion is also surely differentiable nowhere, and is invariant under suitable time-space rescaling. It is therefore not obvious that a probability space with rich enough structure exists to construct such a process.

Theorem (Wiener): There exists a Brownian motion on some probability space.

Proof: We will construct Brownian motion on D, the dyadic rationals in [0,1], then develop the machinery which will enable us to conclude that taking a limit onto the reals retains all the properties we need. The Strong Markov property allows us to extend this to the real line by taking countably many copies, and d independent copies will give BM on $\mathbb{R}^d$.

We make the following observation. Given independent $X_1,X_2\sim N(0,1)$, it is a simple check that conditional on $(X_1|X_1+X_2=a)\sim N(\frac{a}{2},\frac{1}{2})$. From this, it is clear how our construction on D will proceed. Take a family of independent N(0,1) RVs, one for each dyadic rational. Rescale these appropriately to construct the BM on $D_{n+1}$ from the values on $D_n$. Can check the covariance of the new finer increments, and it is clear that these are independent, provided the original increments were independent.

We need continuity. The precise result needed will be dealt with at the end, but it is easy to check that $B_d$ has dyadic increments bounded as $\mathbb{E}|B_t-B_s|^p=|t-s|^{p/2}\mathbb{E}|N|^p$. The Kolmogorov criterion thus gives that $D\ni t\mapsto B_t(\omega)$ is Holder continuous for any exponent in (0,1/2). Then define: $B_t=\lim_{D\ni s\downarrow t} B_s, t\in[0,1]$, which will also have the a.s. Holder property.

Need to check increments property. Given some increments $t_0, approximate from above by dyadic $t_i^n\rightarrow t_i$. Then $(B_{t_0}^n,\ldots, B_{t_k}^n)\stackrel{\text{a.s.}}{\rightarrow} (B_{t_0},\ldots, B_{t_k})$, so the joint distributions converge also. Using Levy then gives both the Gaussian property and the independence in one go.

Theorem (Kolmogorov’s Criterion): If there exist $p,\epsilon>0$ such that: $\mathbb{E}|X_t-X_s|^p\leq C|t-s|^{1+\epsilon}\; \forall s,t\in D$ then for every $\alpha\in (0,\frac{\epsilon}{p})$, X is $\alpha$-Holder continuous almost surely.

Proof: By Markov and BC, can deduce from the condition the existence of a RV M such that almost surely: $\sup_{n}\max_k 2^{-n\alpha}|X_{k2^{-n}}-X_{(k+1)2^{-n}}|\leq M<\infty$. Now given dyadic s,t, there is a unique dyadic rational with smallest denominator, say $2^{r}$, between them. Can then express the difference t-s as a sum of dyadic reciprocals with denominators greater than $2^{(r+1)}$, where each denominator occurs at most twice. So can write: $|X_t-X_s|\leq 2\sum_{n\geq r+1}M2^{-n\alpha}=2M\cdot\frac{2^{-(r+1)\alpha}}{1-2^{-alpha}}$. Then $M<\infty$ a.s. gives the Holder criterion.

# Convergence of Random Variables

The relationship between the different modes of convergence of random variables is one of the more important topics in any introduction to probability theory. For some reason, many of the textbooks leave the proofs as exercises, so it seems worthwhile to present a sketched but comprehensive summary.

Almost sure convergence: $X_n\rightarrow X\;\mathbb{P}$-a.s. if $\mathbb{P}(X_n\rightarrow X)=1$.

Convergence in Probability: $X_n\rightarrow X$ in $\mathbb{P}$-probability if $\mathbb{P}(|X_n-X|>\epsilon)\rightarrow 0$ for any $\epsilon>0$.

Convergence in Distribution: $X_n\stackrel{d}{\rightarrow} X$ if $\mathbb{E}f(X_n)\rightarrow \mathbb{E}f(X)$ for any bounded, continuous function f. Note that this definition is valid for RVs defined on any metric space. When they are real-valued, this is equivalent to the condition that $F_{X_n}(x)\rightarrow F_X(x)$ for every point $x\in \mathbb{R}$ where $F_X$ is continuous. It is further equivalent (by Levy’s Convergence Theorem) to its own special case, convergence of characteristic functions: $\phi_{X_n}(u)\rightarrow \phi_X(U)$ for all $u\in\mathbb{R}$.

Note: In contrast to the other conditions for convergence, convergence in distribution (also known as weak convergence) doesn’t require the RVs to be defined on the same probability space. This thought can be useful when constructing counterexamples.

$L^p$-convergence: $X_n\rightarrow X$ in $L^p$ if $||X_n-X||_p\rightarrow 0$; that is, $\mathbb{E}|X_n-X|^p\rightarrow 0$.

Uniform Integrability: Informally, a set of RVs is UI if the integrals over small sets tend to zero uniformly. Formally: $(X_n)$ is UI if $\sup_{n,A\in\mathcal{F}}\{\mathbb{E}[|X_n|1(A)]|\mathbb{P}(A)\leq \delta\}\rightarrow 0$ as $\delta\rightarrow 0$.

Note: In particular, a single RV, and a collection of independent RVs are UI. If X~U[0,1] and $X_n=n1(X\leq \frac{1}{n})$, then the collection is not UI.

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## Functional Completeness and Optimization

### Background

A logical connective is a function which takes boolean values and returns a boolean value. AND, OR, and NAND are all examples of logical connectives. Sometimes, one logical connective can be written in terms of another. For example, OR can be written with NAND: $$\text{OR}(p,q) = \text{NAND}(\text{NAND}(p,p), \text{NAND}(q,q))$$ and XOR can be written with NOR: $$\text{XOR}(p,q) = \text{NOR}(\text{NOR}(\text{NOR}(p,p), \text{NOR}(q,q)), \text{NOR}(p,q))$$ Sometimes, however, this isn't possible. For example, there is no way to write NAND using only OR.

A connective $$f$$ is functionally complete if all other connectives can be written with $$f$$. Of the familiar 2-input connectives, only two are functionally complete: NAND and NOR. The functional completeness of NAND is the basis of virtually all semiconductors, as the behavior of a chip can be written in terms of logical connectives and it is easier to manufacture a chip using only a single logical connective.

### 3-Input Connectives

So far, I have only been talking about 2-input connectives like AND and XOR. However, these concepts can be generalized to 3-input connectives.

The question is, if you wanted to build a CPU with 3-input connectives, and you want to minimize the total number of gates, what connective should you choose? (This is not as theoretical as it might sound. The Apollo Guidance Computer used 3-input logic gates, and the Soviet Setun computer used 3-valued logic.) In order to answer this question, we need to know which 3-input connective can be used to write all other connectives using the fewest gates. That is, which connective $$f$$ minimizes the total expression size of all the expressions of other connectives in terms of $$f$$?

This problem is simple in the 2-input case because the numbers are small enough that everything can be solved by brute force. For example, it is easy to find the minimal expression for writing OR in terms of NAND just by trying all possible expressions until you find one that works. However, the number of expressions that need to be considered grows so quickly that brute force is impossible for the 3-input case.

I designed and implemented a dynamic program in C++ which takes a 3-input connective $$f$$ and finds the minimal expressions for writing all other 3-input connectives in terms of $$f$$, as long as it is possible. I'm not going to write too much about the details of the algorithm, but feel free to contact me if you are interested. Using my dynamic program, I found that the connective with the following truth table was the most efficient: $$\begin{array}{c c c | c} p & q & r & f(p,q,r) \\ \hline T & T & T & F \\ T & T & F & F \\ T & F & T & F \\ T & F & F & T \\ F & T & T & T \\ F & T & F & F \\ F & F & T & T \\ F & F & F & T \end{array}$$ Here are the expressions for representing $$g(p,q,r)$$ in terms of this connective $$f$$, for every one of the 256, 3-input connectives $$g$$:

This connective minimizes the total leaf count of those expressions. (I should also mention that there are 6 connectives tied for first place, but they are all equivalent by symmetry.) The animation at the top of the page is a representation of these expressions. My program can also minimize the total depth, but I will focus on leaf count here.

As mentioned above, the Apollo Guidance Computer used 3-input logic gates. In particular, it used a generalized version of NOR. As you can see looking at the truth table above, NOR is not the most efficient choice, and in fact, it is actually tied for the least efficient choice! The total leaf count of the most efficient connective is 647, while the total leaf count with NOR is 1643.

### A Note About the Number of Functionally Complete Connectives

While working on this project, I also wondered about the number of functionally-complete connectives. It turns out that this can be found relatively easily, and the authors of this paper computed that the number of functionally complete $$n$$-input connectives is $$2^{2^n-2}-2^{2^{n-1}-1}$$. That might seem like a lot, but there are $$2^{2^n}$$ connectives, so the fraction of $$n$$-input connectives which are functionally complete is $$\frac{2^{2^n-2}-2^{2^{n-1}-1}}{2^{2^n}} = \frac{1}{4} - \frac{1}{2^{1+2^{n-1}}}$$ which very quickly limits to $$\frac{1}{4}$$ as $$n$$ goes to $$\infty$$ (all of which I computed with Mathematica).


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## Alternative Fuels

The use of Alternative Fuels, also known as “Co-Processing”, is the use of combustible waste materials as a source of energy in pyro-processing of cement manufacturing plants, lime plants, and in the calcining of other minerals.

Some forms of alternative fuels are:

• Used tires
• Plastic waste (bags, bottles etc.)
• Waste Oils
• Solvents
• Saw dust
• Paint
• Hazardous Wastes
• RDF (Refuse Derived Fuels)

## Benefits

Conversion to Alternative Fuels benefits the economy of operation by lowering fuel costs. In some cases, a facility that uses Alternative Fuels receives economic compensation for waste elimination (e.g. use of Hazardous Wastes). In addition to energy recovery, there are considerable environmental/green benefits, such as the reduction of emissions and greenhouse gases into the atmosphere by not using natural non-renewable raw materials and fuels such as coal. In certain countries, these savings can be utilized for Carbon Trading credits.

However, because of their inherent characteristics, the implementation of Alternative Fuels present a challenge to use which, if not handled properly, may impact the manufacturing process, product quality, and environmental exposure.

## What We Do

At PEC Consulting we provide innovative and thoughtful solutions to help the cement industry and lime industry implement co-processing at their plants and become more cost efficient, which improves their economic competitiveness.

An Alternative Fuels Project will evaluate the characteristics of co-processing and recommend ways to handle and utilize fuels in the pyro-process in the most effective and secure way.

Our scope of work includes:

• Studies for handling and firing Solid Waste Fuels
• Studies for handling and firing Liquid Waste Fuels
• Studies for handling and firing Hazardous Waste Fuels
• Conceptual studies to use whole or shredded tires
• Studies to use gas or petroleum coke (petcoke) as replacement fuel
• Alternative Fuels / Co-Processing Studies for Cement Plants
• Alternative Fuels / Co-Processing Studies for Lime Plants


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xtb

GFN xTB is a method of self-consistent, third-order, density functional tight-binding (DFTB3), suitable for the calculation of structures, vibrational frequencies, and non-covalent interactions in large molecular systems. A key feature of this model is that it almost entirely avoids element-pair-wise parameterisation. The GFN1 parameterisation consequently covers all spd-block elements and the lanthanides, up to Z = 86, using reference data at the hybrid density functional theory level.

The original model was developed by Stefan Grimme, Christoph Bannwarth and Philip Shushkov, and can found in J. Chem. Theory Comput., 2017, 13(5), pp 1989–2009

A GFN xTB calculation can be run using the following example:

xtb(
structure(formula = HLi bond_length = 1.0 angstrom)
)

This command can appear in the global context.

Options

broyden_damping

Factor of new shell-resolved charges to be used in Broyden mixing: [ q_{n+1} = f * q_{n} + ... ] where $$f$$ is the factor.

• The type is real
• The default is 0.4
broyden_small_gap

Value of HOMO-LUMO gap (in hartree) to switch to using broyden_small_gap_damping.

• The type is real
• The default is 0.001
broyden_small_gap_damping

Damping factor of new shell-resolved charges used in Broyden mixing when the HOMO-LUMO gap is less than broyden_small_gap.

• The type is real
• The default is 0.05
charge

Total molecular charge. No constraint is imposed, but clearly the value has to be such that for a given molecule the number of electrons is nonnegative, and that the number of electrons does not exceed the number that can be described with the given atomic-orbital basis. Note that non-integer values are possible as well.

This option is deprecated. Please use the charge option inside structure instead.

This option is deprecated.

• The type is real
• There is no default value.
charge_rms_threshold

Convergence threshold for RMS change of shell charges.

• The type is real
• The default is 1e-5
check_fock_norm

Check the norm of the Fock matrix against the numerical derivative of the energy before starting SCF. Only for debugging purposes.

• The type is bool
• The default is false
check_molecule

Check the molecule doesn't have any known problematic atoms.

• The type is bool
• The default is true
diis

Method of DIIS.

diis_max

Maximum number of vectors for DIIS.

• The type is int
• The default is 10
• The value must be nonnegative
diis_shift

Diagonal shift for regularization of DIIS.

• The type is real
• The default is 0
• The value must be nonnegative
diis_switch_threshold_grad

In DIIS modes ADIIS+CDIIS and EDIIS+CDIIS the algorithm is switched to CDIIS if the gradient falls below this threshold.

• The type is real
• The default is 1e-1
• The value must be nonnegative
dispersion_cutoff

Cut-off for summation over unit cells in dispersion correction

• The type is quantity
• The default is 40.0 bohr
energy_threshold

SCF convergence threshold using the absolute value of the energy difference between iterations.

• The type is real
• The default is 1e-6
• The value must be nonnegative
ewald_alpha

Broadening parameter for Gaussian screening of point charges in Ewald treatment of second and third order terms of xTB Hamiltonian. This effectively determines the relative contributions of real-space and reciprocal-space terms in the Ewald sum. As $$\alpha$$ is increased, ewald_real_cutoff should be decreased, while ewald_reciprocal_cutoff should be increased. By default, entos determines this from ewald_real_cutoff and ewald_relative_error, using the expression: $$\alpha = \frac{1}{r_c} \bigg[ W \bigg( \frac{Q}{E}\sqrt{\frac{r_c}{2V}} \bigg) \bigg]^{1/2},$$ where $$r_c$$ is the ewald_real_cutoff, $$E$$ is the ewald_relative_error, $$Q = \sum^{N_{atoms}}_{i=1} q_i^2$$, $$\{q_i\}$$ are atomic partial charges of neutral atoms, $$W$$ is the Lambert W function, and $$V$$ is the system volume. This approximation is not guaranteed to give ideal results, but it works fairly well for most systems. Further details can be found in A comparison of the Spectral Ewald and Smooth Particle Mesh Ewald methods in GROMACS.

• The type is real
• There is no default value.
ewald_real_cutoff

Real-space cut-off over unit cells in Ewald sum.

• The type is quantity
• The default is 10.0 bohr
ewald_reciprocal_cutoff

Reciprocal-space cut-off over k-vectors in Ewald sum. By default, entos determines this from ewald_real_cutoff and ewald_relative_error, using the expression: $$k_c = \frac{\sqrt{3} L \alpha}{2 \pi} \bigg[ W \bigg( \frac{4}{3L^2} \bigg(\frac{Q^2}{\pi \alpha E^2}\bigg)^{2/3} \bigg) \bigg]^{1/2}$$ where $$\alpha$$ is the ewald_alpha, $$E$$ is the ewald_relative_error, $$Q = \sum^{N_{atoms}}_{i=1} q_i^2$$, $$\{q_i\}$$ are atomic partial charges of neutral atoms, $$W$$ is the Lambert W function, and $$L$$ is the average system length. This approximation is not guaranteed to give ideal results, but it works fairly well for most systems. Further details can be found in A comparison of the Spectral Ewald and Smooth Particle Mesh Ewald methods in GROMACS.

• The type is real
• There is no default value.
ewald_relative_error

Approximate relative error in both the real-space and reciprocal-space terms of the Ewald sum. This should not be set if the user wishes to manually specify ewald_reciprocal_cutoff and ewald_alpha.

• The type is real
• The default is 1.e-8
fixed_shell_charges

Specify a list of fixed, shell-resolved partial charges. This results in a non-self-consistent calculation, for which the SCF driver will perform two iterations confirming no change in the charge density.

Using this option with the fixed shell charges set to zero corresponds to doing a non-self-consistent calculation with only the core Hamiltonian.

• The type is [real]
• There is no default value.
fock_damping

Amount of previous Fock matrix to use in Fock damping. The new Fock matrix for iteration $$n$$: $$F_{n} = (1-\alpha) F_{n} + \alpha F_{n-1}$$ where $$\alpha$$ is the fock_damping_factor.

• The type is real
• The default is 0.8
fock_damping_gradient_threshold

XTB Fock matrix extrapolation is initially performed using a damping (see fock_damping). When the largest element of the absolute orbital gradient drops below fock_damping_gradient_threshold, DIIS will be used.

• The type is real
• The default is 0.2
guess

Initial guess for the xTB wavefunction.

• The type is string
• The value must be one of:
• SAD - Superposition of Atomic Densities (SAD).
• H0 - Using eigenvectors of the core Hamiltonian as the initial guess. This is equivalent to setting the initial density to zero.
• eeq - Use EEQ charges as the initial guess.
guess

Initial guess for the SCF wavefunction.

• The type is string
• The value must be one of:
• SAD - Superposition of Atomic Densities (SAD).
• H0 - Using eigenvectors of the core Hamiltonian as the initial guess. This is equivalent to setting the initial density to zero.
h0_cutoff

Cut-off for summation over unit cells in H0

• The type is quantity
• The default is 10.0 bohr
initial_charges

Specify the list of atom resolved partial charges.

• The type is [real]
• There is no default value.
initial_shell_charges

Specify a list of initial shell-resolved partial charges.

• The type is [real]
• There is no default value.
k_point

Specify a point in the reciprocal lattice.

• The type is (real, real, real)
• There is no default value.
level_shift

Turn on level shifting to improve SCF convergence. Raises the energy of virtual orbitals by level shifting value. Whether to remove the level shift at the end is controlled by option remove_level_shift.

• The type is real
• The default is 0
• The value must be nonnegative
load

Name of assigned result (containing orbitals) to start the SCF from.

• The type is string
• There is no default value.
max_iter

Maximum number of SCF iterations.

• The type is int
• The default is 128
• The value must be nonnegative
model

Specify the xTB model. When gfn0 is used, setting solver = non_iterative is recommended.

• The type is string-lowered
• The default is gfn1
• The value must be one of:
• gfn0
• gfn1
monkhorst_pack

Size of Monkhorst-Pack grid for sampling reciprocal space. The Monkhorst-Pack grid is implemented in accordance with https://doi.org/10.1103/PhysRevB.13.5188. All grids are centred on the Gamma point. An odd choice of integers includes the Gamma point in the grid, whereas an even choice of integers does not.

• The type is (int, int, int)
• There is no default value.
n_primitives

Number of primitive Gaussians per shell. This setting is only used if use_reference_basis is false

• The type is int
• The default is 6
• The value must be one of:
• 4
• 5
• 6
name

Specify the name of a set of results.

This option is deprecated.

• The type is string
• There is no default value.
no_fail

If set to true, the program will continue its execution even if the SCF does not converge. This is useful for frozen-orbital calculations.

• The type is bool
• The default is false
number_type

Number type for orbitals and operators, whether real or complex.

• The type is string
• The default is real
• The value must be one of:
• real - Use real numbers for orbital coefficents and fock matrices.
• complex - Use complex numbers for orbitals coefficents and fock matrices.
orbital_grad_threshold

SCF convergence threshold using the infinity norm of the orbital gradient.

• The type is real
• The default is 1e-5
• The value must be nonnegative
overlap_cutoff

Cut-off for summation over unit cells in overlap matrix

• The type is quantity
• The default is 10.0 bohr
overlap_screening_threshold

Value below which overlap matrix elements are considered negligible and are therefore set to zero

• The type is real
• The default is 1.e-10
potential_type

Treatment of the second-order potential in periodic GFN1-xTB

• The type is string-lowered
• There is no default value.
• The value must be one of:
• truncated - Achieve convergence in the real-space summation by smoothly forcing the potential: $$\delta \phi^{xTB}{Al, Bl’}(|\mathbf{R}|) = \sum_\mathbf{T} f(|\mathbf{R}{AB}+\mathbf{T}|) \left( \left(\frac{1}{|\mathbf{R}+\mathbf{T}|^2 + \eta^{-2}{Al,Bl'}} \right)^{1/2} -\frac{1}{ |\mathbf{R} + \mathbf{T}|} \right),$$ to zero at a pairwise-dependent cut-off, where $$(A,B)$$ are atomic indices, $$(l,'l)$$ are shell indices, $$\mathbf{T}$$ is a translation vector and $$f(|\mathbf{R}_{AB}+\mathbf{T}|)$$ is a polynomial used to send $$\delta \phi^{xTB}$$ to zero. The polynomial is defined as: $$f(|\mathbf{R}|) = -6R^5 + 15R^4 - 10R^3 + 1,$$ which follows the treatment of CP2K (see their xtb_coulomb.F module). One notes that truncating the real-space sum will affect the final energies.
• full - Define the damped xTB real-space potential as: $$\delta \phi^{xTB}{Al, Bl’}(|\mathbf{R}|) = \sum_\mathbf{T} \left(\frac{1}{|\mathbf{R}{AB}+\mathbf{T}|^2 + \eta^{-2}} \right)^{1/2} -\frac{1}{ |\mathbf{R}{AB} + \mathbf{T}|} +\frac{1}{2 \eta^2 |\mathbf{R}_{AB} + \mathbf{T}|^3},$$ and treat the $$+1/r$$ and $$-1/r^3$$ terms with standard Ewald summations.
print_basis

Print basis information.

• The type is bool
• The default is false
print_level

Print level.

• The type is int
• There is no default value.
• The value must be one of:
• -2 - No output
• -1 - Minimum output
• 0 - Output that doesn't scale with system size
• 1 - Output that scales linearly with system size
• 2 - (Debugging) output that scales quadratically with system size
• 3 - (Debugging) output that scales cubically with system size
pseudo_diagonalization

Whether to use pseudo-diagonalization instead of full diagonalization of the Fock matrix. Recommended for calculations on very large systems or with a very large basis.

• The type is bool
• The default is false
remove_level_shift

Remove level shift from final orbitals.

• The type is bool
• The default is true
repulsive_cutoff

Cut-off for summation over unit cells in repulsive energy

• The type is quantity
• The default is 10.0 bohr
scramble_complex

Add noise to the imaginary part of the guess density while keeping it hermitian. This option is only used if number_type = complex.

• The type is bool
• The default is true
scramble_complex_density_only

Scramble only the initial density (leaving the orbitals unchanged) in complex orbital calculations. Use scramble_complex to scramble the density and orbitals.

The change in the guess density implies a change in the guess orbitals and/or occupation numbers which are needed for orbital-dependent potentials. To obtain the adapted orbitals and occupations, the density must be diagonalized, which may be expensive. Usually, most parts of the potential are calculated directly from the density and scrambling in these parts is sufficient to break the symmetry. Thus, one may be able to achieve symmetry-breaking even when the orbitals and occupations are left unchanged.

• The type is bool
• The default is false
scramble_parameter

The amount of noise that is added to the imaginary part of the guess density. This option is only relevant if scramble_complex is true.

• The type is real
• The default is 1e-5
smoothing_range

If potential_type is set to truncated, define the range over which the potential is smoothly forced to zero. The default of 1 bohr is consistent with CP2K.

• The type is quantity
• The default is 1 bohr
solver

Type of solver to use for determining the density and orbitals. The default is automatically set depending on the xTB model.

• The type is string-lowered
• There is no default value.
• The value must be one of:
• scf - Perform a standard self-consistent field calculation
• scc - Perform a self-consistent charge calculation with Broyden mixing
• non_iterative - Perform a non-iterative (single diagonalization) calculation
symmetry_reduction

Use symmetry to reduce the number of k-grid sampling points via point group operations. This reduces the number of xtb calculations to perform.

• The type is bool
• The default is true
temperature

Specify the temperature for the electrons and perform the SCF calculation in the canonical-ensemble. Convergence is often poor with the default convergence settings, and using the original Pulay DIIS method is recommended (see diis option).

• The type is quantity
• The default is 300 kelvin
use_atomization_correction

Include atomization energy correction in total energy and gradient

• The type is bool
• The default is false
use_reference_basis

This utilises 4 primitives for H 1s shell, 7 primitives for H 2s shell, 4 primitives for He 1s. For Z > 2, 6 primitives are used per s or p shell and 4 primitives for d shells. The number of primitives is consistent with the basis used by the GFN1 binary (available upon request from Stefan Grimme's group).

• The type is bool
• The default is true
version

• gfn0
• gfn1


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# Tag Info

3

(BIG) HINT: $$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=\frac{2}{\sin(\phi)}\Longleftrightarrow$$ $$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=2\csc(\phi)\Longleftrightarrow$$ $$\sqrt{\cos^2(\phi)+15}=2\csc(\phi)-\cos(\phi)\Longleftrightarrow$$ $$\cos^2(\phi)+15=\left(2\csc(\phi)-\cos(\phi)\right)^2\Longleftrightarrow$$ ...

3

You can do linear algebra in terms of $n$-tuples $(x_1,x_2,\ldots, x_n)$ and matrices $\bigl[a_{ik}\bigr]$, or in terms of "abstract" vectors ${\bf x}$ and linear maps $A$. Similarly with derivatives in a multivariate setting: You can consider component functions $f_i$ $(1\leq i\leq m)$ and their partial derivatives $f_{i.k}$, as defined in $(25)$, or you ...

3

The function isn't bounded above since $$f(x,0)=2x^4-3x^2+2\rightarrow\infty$$ Since the function is even in both variables we can work only with $\;x,y\ge0\;$ , and $$f(x,y)\xrightarrow[(x,y)\to(0,0)]{} 1+1=2$$ But for any point on the circle $\;x^2+y^2=1\;$ or on the ellipse $\;2x^2+y^2=1\;$ the value of the function is $\;1\;$ : $\;f(x,y)= 1$ Yet ...

3

Let $g(x,y)=(x^3,y)$ and $h(x,y)=(\sin x\cosh y,\cos x\sinh y)$, so $f=h\circ g$. Then $g(0,0)=(0,0)$ and $g$ is invertible, and $h$ is locally invertible at $(0,0)$ since $\displaystyle\frac{\partial(h_1,h_2)}{\partial(x,y)}=\cos^2x\cosh^2y+\sin^2x\sinh^2y=\cos^2x(\sinh^2y+1)+\sin^2x\sinh^2y=\sinh^2y+\cos^2x$, so ...

3

Along $y = -x$ for example: $$\lim_{(x,y)\rightarrow(0,0)}\frac{ e^{x+y} - x - y}{\sqrt{x^2 + y^2}} = \frac{1}{\sqrt{2}}\cdot{\lim_{x\rightarrow 0}\frac{1}{|x|}} \rightarrow \infty$$

2

Consider the function $f(x, y) = x^2 + y^2$. The gradient is $\langle 2x, 2y\rangle$, so if you depart from the origin at any angle and continue straight, you will be parallel to the gradient at any point. So the trajectory isn't unique; you'll have to choose an initial velocity. For any function $f$, I believe we can identify one of these trajectories as a ...

2

You are correct until partial difference of $f(\mathbf{x})$. Next, the vector $\mathbf{x} - \nabla f(\mathbf{x})$ is replaced for $\mathbf{y}$. Therefore, \begin{align} \mathbf{y} &=\mathbf{x}-\nabla f(\mathbf{x}) \\ &=\mathbf{x}-(Q\mathbf{x}-\mathbf{b}) \\ &=(E-Q)\mathbf{x}+\mathbf{b} \end{align} Then, $E$ is an identical matrix. So, ...

2

As the whole situation has a rotational symmetry w.r.t. to $z$ we can simplify the problem by looking at a plane which goes throu the z-axis. In this image you can see that $R=4$ is the radius of the sphere, $r=2$ is the radius of the cylinder. You can also see that the intersectins are circles. And In this case you can easily find that those circles have ...

2

The designation is important. For example, we can write the scalar $I_1$ $$I_1=\int_S (\hat n\cdot \vec F) \,dS$$ the vector $I_2$ $$I_2=\int_S (\hat n\times \vec F) \,dS$$ and the dyadic (tensor, rank 2) $I_3$ $$I_3=\int_S(\hat n\,\vec F) \,dS$$ The notation for $I$, $$I=\int_S \vec F\,$$ is ambiguous without explicit designation.

2

If we write $r:=\vert\vert(x,y)-(x_0,y_0)\vert\vert$, differentiability of $f$ is equivalent to the existence of $(a,b)$ such that $f(x,y)=f(x_0,y_0)+a(x-x_0)+b(y-y_0)+o(r)$. Squaring this gives $f(x,y)^2=f(x_0,y_0)^2+2(a(x-x_0)+b(y-y_0))f(x_0,y_0)+(a(x-x_0)+b(y-y_0))^2+o(r)$ (noting that $(f(x_0,y_0)+a(x-x_0)+b(y-y_0))\times o(r)$ is $o(r)$. By switching ...

2

$$3u_x+2u_t=\cos(x)$$ HINT : The equations characteristics are : $$\frac{dx}{3}=\frac{dt}{2}=\frac{du}{\cos(x)}$$ First characteristic : $\frac{dx}{3}=\frac{du}{\cos(x)} \quad\to\quad 3du-\cos(x)dx=0 \quad\to\quad u-\frac{1}{3}\sin(x)=c_1$ Second characteristic : $\frac{dx}{3}-\frac{dt}{2}=0 \quad\to\quad 2x-3t=c_2$ General solution on implicite form, ...

1

The best way to view this in my opinion is by looking at the actual ODE you get when you solve the characteristic equation. This ODE is what you get along the path $x=x_0+\frac{3}{2}t$. For each $x_0$ you have $v(t)=u(t,x_0+\frac{3}{2}t)$ with $v(0)=f(x_0)$. It satisfies $\frac{dv}{dt}=\frac{1}{2} \cos(x(t))=\frac{1}{2} \cos(x_0+\frac{3}{2}t)$. The solution ...

1

It happens because $f(ax,ay) = af(x,y)$

1

It's true that . However, you might note that the same can be said of your ordinary integral on a line: one can easily see if $f$ is a function $[a,b]\rightarrow \mathbb R$, then it's already clear what $$\int_a^bf$$ means, and the $dx$ is just going to be tagging along. However, the advantage of writing $$\int_a^b f(x)\,dx$$ becomes more clear when one has ...

1

I would write it as $$\sqrt{\cos^2\phi+15} = \frac{2}{\sin \phi} - \cos\phi$$ Both left and right hand side are positive (where defined), so you can square both sides. You will get to $$\cos^2\phi + 15 = \frac{4}{\sin^2\phi} + \cos^2\phi - 4 \frac{\cos\phi}{\sin\phi}.$$ Cancel what you can, multiply by $\sin^2\phi$, and use double angle formulas to ...

1

Here is a simple proof using index notation and BAC-CAB identity. \begin{align} \nabla \times \left( {{\bf{A}} \times {\bf{B}}} \right) &= {{\bf{e}}_i}{\partial _i} \times \left( {{A_j}{{\bf{e}}_j} \times {B_k}{{\bf{e}}_k}} \right)\\ &= {\partial _i}\left( {{A_j}{B_k}} \right){{\bf{e}}_i} \times \left( {{{\bf{e}}_j} \times {{\bf{e}}_k}} \right)\\ ... 1 One way is to recognize that the intersection must satisfy the equation for both planes, and must therefore satisfy their sum: (3x-y+z)+(y+z) = 4+2  3x+2z = 6 $$You can then let x = t, and then 3t+2z = 6, whence we get z = 3-\frac{3}{2}t. You can then rewrite your first equation as$$ y = 3x+z-4 $$to obtain an expression for y in ... 1 The notes that you are following should be corrected by erasing the initial fraction:$$ \frac{dE}{dt}(t) = \int_\Omega (u_tu_{tt} + c(x)^2 \nabla u \cdot \nabla u_t + q(x)uu_t)\, dx. $$In particular, the derivative of (u_t)^2 is obtained exactly as you wrote. As for the middle term, you have instead$$ \frac{\partial}{\partial t}|\nabla u|^2 = ...

1

A first integration with respect to $x$ is clearly easier. I suggest you to apply Fubini theorem and reverse the order of integrals. $$\int_{0}^{1}\int_{x}^{1} y^2 \sin(2\pi \frac{x}{y})dydx = \int_{0}^{1} y^2\left( \int_{0}^{y} \sin(2\pi \frac{x}{y})dx \right)dy.$$ And $$\int_{0}^{y} \sin(2\pi \frac{x}{y}) dx = \frac{y}{2\pi} [-\cos(2\pi \frac{x}{y})]_0^y ... 1 Let's think about how this works for a curve in \Bbb R^3. We show that a curve \gamma has torsion identically 0 if and only if \gamma lies in an affine plane. Namely: If there is a constant unit vector \mathbf C so that \mathbf C\cdot(\gamma-\gamma_0) = 0, then we differentiate a few times to find that \mathbf e_1\cdot\mathbf C = \mathbf ... 1 Integrate wrt to x to get that$$\frac{\partial u}{\partial y} = z(y)$$Integrate wrt to x$$u = \int z(y)dy + g(x) = f(y) + g(x)$$for two functions f,g. Now add the boundary conditions to find$$u(x, x^3) = f(x^3 ) + g(x) = \sin x^6$$\frac{\partial u}{\partial x} (x,x^3) = g'(x) = 0 \implies g(x) = B \in \mathbb R hence f(x^3) = \sin x^6 ... 1 I think I know how to make this clear. Consider for example the function of one variable g(x)=\begin{cases}1&x>0\\-1&x<0\end{cases} How would you show that \lim\limits_ {x\to 0}g(x) does not exist? Well, you say that \underbrace{\lim\limits_{x\downarrow 0}g(x)=1}_{\text{the limit from the right}} and ... 1 Either you note that f(z)=e^z and know (or show) that e^z takes all complex values except zero, or you solve explicitly the equations e^x\cos y=a and e^x\sin y=b: These equations give e^x=\sqrt{a^2+b^2} and since the real exponential e^x never vanishes you get a\ne0 or b\ne0. In particular, x=\log\sqrt{a^2+b^2}. If a=0, then you can ... 1 The mistake you seem to be making is that you misunderstood what \Sigma is. It is the cilinder without the two 'caps' at the ends (it is what you would get is you took a piece of A4 paper and folded two sides together). This means that the boundary of \partial \Omega are the two circles at the ends of the cilinder. While checking the rest of your ... 1 HINT: multiplying the first plane by -2 we get the same plane, these two planes are the same 1 I hope I have a possible solution. Le us assume that, for all the components J_i of \boldsymbol{J}, there is a function f_i\in C^4(\mathring{A}) such that \nabla^2 f_i=J_i, with \bar{V}\subset \mathring{A} and \boldsymbol{x}\in \mathring{V} and there exists a \delta such that, for all \epsilon\le \delta, \epsilon >0, the region ... 1 Sketch/Hint: Let me show you one "follow your nose" way to tinker until you reach a conclusion. We want to show that |F(x_k,t_k) - F(x,t)| \to 0 as (x_k,t_k) \to (x,t) in \mathbb R^n \times \mathbb R. Since we know that |f(x_k,s)-f(x,s)| \to 0 as k \to \infty (why?) and f(x,s) is the integrand inside the integral definition F(x,t), we can try ... 1 A permutation matrix is a 0, 1 square matrix with a single 1 in each row and in each column. Clearly, all the row sums and all the column sums of a permutation matrix are 1. Now if P_1,P_2,\dots,P_r are permutation matrices, and c_1,c_2,\dots,c_r are any real numbers, then$$c_1P_1+c_2P_2+\cdots+c_rP_r$$has the property that every row sum and every ... 1 For the first equation, complete the square in z... that is, convert the equation to the form x^2 + y^2 + (z - q)^2 = r^2. That will give you a sphere centered at \{0,0,q\} of radius r. You must express q and r in terms of your a. For the second equation, note that the region is rotationally symmetric about the z axis because the x-y ... 1 A change to polar coordinates will suffice:$$ \lim_{r\to 0}\frac{ e^{r(\sin\theta+\cos\theta)} - r(\sin\theta+\cos\theta)}{r}=\lim_{r\to 0}[\frac{ e^{r(\sin\theta+\cos\theta)}}{r} - (\sin\theta+\cos\theta)]  Which does not exist since $\lim_{r\to 0}\frac{ e^{r(\sin\theta+\cos\theta)}}{r}$ does not exist for any $\theta$.

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text:
# Gan Air M Review

#### Cubingcubecuber

##### Member

Basic Info:
Cost: $46.99 on Gan's Amazon Store Weight: 66 grams Size: 56 mm Released: 6-22-2020 Packaging: The Air M came in a hard white box with a picture of the cube on top, and GAN356 Air M printed on the side. There are also some cool bubble designs on it, which is nice. The cube itself came in a clear plastic box, and was wrapped in plastic wrap. The box is the same type as the box from the 354 and X, not the type from the XS. I prefer this type of box, as it is much easier to open. Accessories: In the accessories box, there was a pamphlet on how to use the GES Pro, a CFOP pamphlet with algs for all of the cases, an blue Gan bag, and the GES adjustment tool with other tension nuts. First Impressions: Out of the box, the cube felt a bit dry and scratchy, and was on the light blue nuts on setting 1 with the elasticity was set to 0. Plastic: It has frosted plastic, which is annoying at first, but after a couple hundred solves goes away and becomes very grippy. Tensioning and Elasticity System: The Gan Air M uses the GES Pro system, first used in the Gan XS, and also used in the Xv2. It is adjusted by hand, and has 4 tensions: 1, 2, 3, and 4. 1 and 3 are on the light blue nut, while 2 and 4 are on the dark blue one. 1 is the tightest, and 4 is the loosest. I think that setting 1, 2, and 3, are viable, but setting 4 is too loose for me. However, some people like really loose cubes, and an extra setting doesn't hurt. The clear part is the elasticity system. It has 6 settings and is adjusted with the tool that sort of looks like a trident. It is turned counterclockwise to increase the spring strength. When it is flush with the GES, it is at setting 0, and when you have gone all the way around, it will make a loud click and jump to setting 0 again. Cornercutting: The cornercutting on this cube depends on the tensions, but on reasonable tensions it will cut easily over 45˚ forward and past line-to-line reverse. Edgecutting: The edgecutting on this cube depends on the tensions, but on reasonable tensions it will cut line-to-line forward and 30˚ reverse. Magnets: The magnet strength on this cube is comparable to that of the strong Gan XS and Gan X magnets, though it might be a bit stronger. My Setup: I have my GES on setting 2, with the elasticity on setting 4. I lubed it with Lubicle Silk, Lunar, Compound X, and Cubicle Weight 5 Comparison with Other Gan Cubes: 354 M: Compared to the 354, it feels very different. The Air M is much rounder feeling, and is 2 millimeters bigger. It is much lighter, and the tensioning system is much better. It is faster, but not as smooth. I don't really see any reason to get the 354 instead of the Air M, as I think that the Air M is superior in every category. 356 X: The Gan X is much smoother than the Air M, but sacrifices speed. The X has adjustable magnets, which is a nice customization feature. The X's tensioning system is a bit easier to understand, but requires carrying around a case on 21 extra GES. The Air M is lighter than the X, and is also not as clicky. I would say to get the X if smoother and more controllable cubes are your thing. 356 XS: This is really interesting, as when I heard about the Air M I thought it would just be the same as the XS, but just without the adjustable magnets. I was wrong. The XS feels much more swishy, but is slightly less stable. It has adjustable magnets, which might be a deciding factor if you like weak magnets. The mechanism looks exactly the same to me, so the feel could have to do with setup and break-in. Final thoughts: The Air M is a good release from Gan, though I think it is too similar to previous releases, and I hope Gan tries something different next time. It is my favorite Gan cube because of the stability, and the edge cutting is good. This cube will probably be my BLD main. Do I Recommend This Cube: It depends on who you are. If you are sure you want a Gan cube, I would recommend the Air M as it is cheaper than the XS, but is probably better for most cubers because it is more stable. If you already have a Gan cube, you don't need to get this cube, as all modern Gan cubes are essentially the same thing but with different customization options. I think the Air M is a bit overpriced, and would fit better in the market at the$35 range. If you have the money, and want a good cube, then you should consider get this, but also look at other cubes, as cubes in any price range can be good depending on you turning style and personal preferences.

Thank you to Gan for giving this to me for free to review

#### Zubin Park

##### Member
Did you lube the core of your cube, and if so, what lube did you use?

#### CyoobietheCuber

##### Member
Nice review. BTW, how do you do that pattern?


text:
# Gamma spectroscopy, Compton edge

• B
artis
Hello, I have a few uncertainties about this, Forgive the long questions, I appreciate your time.

1) Why is the maximum energy photon loses during Compton scattering about 0.238MeV (if maximum angle 180 achieved aka backscatter) irrespective of the incoming photon total energy? It seems counter-intuitive that the incoming photon irrespective of it's energy can lose no more than this amount which puts the "Compton edge" at a maximum of 0.238 MeV from the Photopeak.

2) Why is the electron in Compton scattering considered "free" yet the ones in the photoelectric effect are bound regular electrons , in both cases the material is the same , for example a HPGe detector material. Especially given the energies for the Photoelectric effect and Compton effect overlay and either one can happen.

3) Looking at the Compton edge spectrum , is it true that most photons tend to be scattered at or near their maximum scattering angle/energy, as it seems that the amplitude rises as one get's further away from the photopeak and hits it's highest value at the Compton edge?

4) Before I ask this I want to acknowledge that I know that a detector counts gamma energies as electron current and electron current arises in the detector when a gamma interacts with one which then interacts with others causing a cascade in the presence of an applied E field.
This being said I cannot fully understand why for example the Photoelectric effect causes a photopeak but the Compton scattered electrons cause their peak/edge at a energy which is close to the photopeak instead of the energy maximum that a single photon can transfer during Compton to a electron.
I would suspect there are a distribution of photons namely , those that scatter and then leave the detector VS those that scatter within the detector and then further interact within the detector.
Now the first ones , those that leave after scatter simply produce a recoil electron with an energy up to 0.238MeV , while those that scatter and further interact in the detector like create a Photoelectron get counted with the energy they had after the Compton scatter event.
So I would suspect that the Compton edge is created by the second kind of photons - those that scattered first off an electron and then further interacted within the detector and the energy registered is corresponding to that which they had after the Compton event?
If this is so then why there isn't also a peak for the photons that Compton scattered but then left the detector? Or is the increased amplitude continuum at the very beginning from 0 eV up to about 240KeV corresponding to these very electron whose photons escaped during scatter?

5) Is the reason why in pair production the "single escape" peak is higher than the value of a 0.511 MeV photon because the detector counts the value of the 511 KeV photon-electron interaction + the energy the original gamma imparted to the produced pair before pair annihilation?

## Answers and Replies

1) Why is the maximum energy photon loses during Compton scattering about 0.238MeV (if maximum angle 180 achieved aka backscatter) irrespective of the incoming photon total energy? It seems counter-intuitive that the incoming photon irrespective of it's energy can lose no more than this amount which puts the "Compton edge" at a maximum of 0.238 MeV from the Photopeak.

The kinematics of Compton Scattering gives the difference in the scattered wavelength and the incident wavelength Δλ = h/mec(1 - cosθ ) where θ is the scattering angle, me is the rest mass of the electron, h is Planck's constant and c is the velocity of light.

As you can see the loss of energy as determined by the increase in Δλ of the photon only depends on the scattering angle. The maximum change in wavelength is 0.04692x10-8 cm. But a constant difference in wavelength does not mean there is a constant difference in energy or a limit on the maximum energy difference.

vanhees71
artis
But a constant difference in wavelength does not mean there is a constant difference in energy or a limit on the maximum energy difference.
Do you mean this in the sense that for two equal scattering angles the photon with a higher frequency would give more energy to the electron?

The higher the photon frequency (energy) the more energy that is transferred to the electron like any collision.

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vanhees71
artis
Thanks @gleem for answering so far , maybe @mfb or anyone else can lend their knowledge with respect to the other questions still left in my post #1?

Meanwhile I want to ask one more thing,
The reason Compton scattered gammas get counted is because they interact with an electron after scattering, so if this interaction is the photoelectric effect for example then the whole leftover energy of the gamma gets registered (whatever that energy is) , but I was wondering whether the change in trajectory/direction after the scatter has any impact on the probability of further interaction for the photon? My reasoning is that for a given size, shape and volume detector a straight line trajectory means less material is traversed by the photon and it spends less time in the detector volume while a scattered trajectory means it spends more time within the volume and has a longer path within it, so in theory the probability of interaction increases?
So it would seem to me that if a photon managed to get scattered then it's probability of interaction increases compared to one that wasn't scattered given both had the same incoming trajectory and energy.

Can a single photon get scattered twice in a row? I would think it should be able but the probability of that happening within a small volume is little?

Mentor
1) as discussed the energy loss is unlimited.
2) The photoeffect happens at an energy of a few eV where the binding energy of electrons is relevant. The Compton effect is in the MeV range where the binding energy is negligible (pretty much by definition, otherwise it's not called Compton effect).

I don't understand questions 3 and 4 and I think they are based on your misconception about (1).

vanhees71
artis
1) as discussed the energy loss is unlimited.
By which I suppose you mean what has already been said that the higher the incoming gamma energy the higher the KE given to the electron even though the total photon energy loss is always limited to the "Compton wavelength" or 0.238 MeV?
I must say this is rather counter intuitive although makes sense given at what energy those 0.238MeV are lost, if they are lost from 10MeV then that gives more energy than if they are lost from a gamma of 5 MeV.

2) The photoeffect happens at an energy of a few eV where the binding energy of electrons is relevant. The Compton effect is in the MeV range where the binding energy is negligible (pretty much by definition, otherwise it's not called Compton effect).
OK I understand, in both cases electron is ejected and it's binding energy is overcome but if that happens at high gamma energy the binding energy being so negligible is simply ignored.

The 4th question was meant like this, if I can clarify, the gamma comes in the detector, Compton scatters and loses some energy, this lost energy is given to an electron , the gamma then either leaves and is not counted further or interacts further within the detector. So if it interacts further , say by photoeffect, then I assume the Compton + resulting photoeffect gets counted together as if it were Photo effect to begin with and added to the photo peak due to the short time scale in which both of these events happen in series (Compton + photo) ?
But in the case where the photon only Compton scattered and then exits the detector without further interaction , why would those Compton electrons get counted at an energy level close to the photopeak?

I think I have an idea , let me see whether it's correct,
The 0.238MeV aka the Compton wavelength is the maximum energy a photon can lose in a single interaction but it is not the maximum energy the electron gains , the electron gains the total incoming gamma energy minus whatever energy the photon lost in the scatter?
So a 1.5MeV gamma that lost say 0.230 MeV would scatter an electron with an energy of 1.270MeV?

. . . . even though the total photon energy loss is always limited to the "Compton wavelength" or 0.238 MeV
Are you saying that the energy loss of the scattered photon is limited to 0.238 MeV? Also, the Compton wavelength is measured in cm, not MeV. The Compton wavelength is the wavelength of a photon whose energy is the same as the rest mass energy of an electron, 0.511 MeV.

vanhees71 and artis
Staff Emeritus
Can a single photon get scattered twice in a row? I would think it should be able but the probability of that happening within a small volume is little?
A scattered photon can scatter again.

As initial photon energy increases, the photon is more likely to be forward scattered.

Consider that for low energy photons, the photoelectron effect is more likely, i.e., a photon is more like to be absorbed, which yields the photopeak.
https://www.nrc.gov/docs/ML1122/ML11229A703.pdf

As photon energy is increased Compton scattering competes with the photoelectron (or photoelectric) effect. Beyond a photon energy of 1.022 MeV, one begins to observe pair production. The energy at which the probability of Compton scattering (any angle) equals that of pair production is dependent on the Z of the nucleus of the atom interacting with the photon.

http://rcwww.kek.jp/research/shield/photon_r.pdf
https://courses.physics.illinois.ed...nteraction of x-ray and gamma ray photons.pdf

If one compares the cross-sections for photon interaction with Fe vs Zr vs W, one can see the shift.

Compton scattering from a free electron is reasonable for a valence/conduction electron (binding energy ~ eV) or a light element, where the ionziation energies are low. For heavier (periods 6 or 7) elements, e.g., error would be introduced if the scattering involves a K- or L- electron depending on the photon energy, and one ignores the binding energy of the electron.

I did some calculations using the Hyperphysics calculator:

For Cs-137, with incident energy Ei = 0.662 MeV, produces backscattered photon energy Es, and electron energy as follows:
Es 0.18435 MeV
Ke 0.47765 MeV

Similarly for 1 MeV photon, 10 MeV photon

Ei 1.0 MeV
Es 0.203504 MeV
Ke 0.796495 MeV

Ei 10 MeV
Es 0.249134 MeV
Ke 9.750865 MeV

See discussion on relativistic electrons
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/photel.html#c1

artis
artis
Are you saying that the energy loss of the scattered photon is limited to 0.238 MeV? Also, the Compton wavelength is measured in cm, not MeV.
Yes, right wavelengths get measured in length measurement units.

I had made a huge mistake it seems, somehow I listened about the Compton wavelength and got the impression that the maximum amount of energy a photon loses is no more than the energy difference between photopeak and Compton edge and that is why it sounded so weird to me why I cam here and asked.

Now I get it and it sounds fine and reasonable finally and logical that the whole Compton spectrum from 0 up to the Compton edge is made by Compton scattered electrons.
Also the link @Astronuc gave explains this clearly.
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/photel.html#c1

So now I see that a photon in a single Compton scatter event can lose almost all of it's energy to the electron except for a tiny bit (the now infamous 0.238MeV) which is why the Compton spectrum falls off after the Compton edge.
I read that for very high energy gammas the "Compton wavelength" energy difference becomes 0.255MeV or half the electron rest mass, what makes this change happen for increased energy ?

Mentor
By which I suppose you mean what has already been said that the higher the incoming gamma energy the higher the KE given to the electron even though the total photon energy loss is always limited to the "Compton wavelength" or 0.238 MeV?
I must say this is rather counter intuitive although makes sense given at what energy those 0.238MeV are lost, if they are lost from 10MeV then that gives more energy than if they are lost from a gamma of 5 MeV.
That makes no sense and you should read the previous answers, in particular post #2. The energy loss is not limited and neither is the electron energy. The wavelength change is limited but that's a completely different statement.

artis
That makes no sense and you should read the previous answers, in particular post #2. The energy loss is not limited and neither is the electron energy. The wavelength change is limited but that's a completely different statement.
Agreed , I already pointed out in my post #10 that I had made a mistake and now it's clear.

artis
A scattered photon can scatter again.
This make me ask, can a photon lose almost all it's energy in scattering? I would think it cannot as losing all it's energy would mean an absorption, but how low can a photon go and still exist as the "original photon"?
Like for example a 10MeV photon undergoing a backscatter event losing almost all except, say 0.240 MeV of it's initial energy to a electron (I read the probability for high energy photons to backscatter is low) and then that photon becomes a 0.240MeV photon (much lower frequency, since energy=frequency) and say it again backscatters, how much could it possibly lose the second time from those 0.240MeV?
Because the way I understand it, there can be no such thing as a photon with 0 energy/frequency, but can a gamma scatter down from initial MeV's of energy to eV's?

There is one misunderstanding I'm having at the moment.

here in the video from MIT, at 1:35 the lecturer talks about the wavelength shift of the photon.
The claim is that "wavelength shift itself is independent from photon energy, just a matter of scattering angle" The formula seems to show it too.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compton.html
The text says
The wavelength change in such scattering depends only upon the angle of scattering for a given target particle.
Now, since wavelength=frequency= energy of photon, then if a 3MeV photon backscatters , it loses say 2.5MeV while if a 300KeV photon backscatters it loses say 250KeV , clearly 2.5Mev is higher than 250Kev in energy/frequency , so would I be correct in saying that for a fixed angle (whether 90 or 180 or any other) a photon can lose a fixed percentage of it's total wavelength/energy but the amplitude of this wavelength/energy gets proportionally larger as the initial photon energy increases or gets smaller as the initial photon energy decreases?

Like for example 2 photons can lose 50% of their energy at a fixed angle of scatter but 50% of 3 MeV is more than 50% of 300KeV.

And one final somewhat unrelated question , since photon's frequency/energy is not dependent on it;s velocity could it be in theory that, for example, if a gamma ray velocity through a known thickness and density material could be slowed down the probability of it's interaction with said material would increase?

Staff Emeritus
Now, since wavelength=frequency= energy of photon, then if a 3MeV photon backscatters , it loses say 2.5MeV while if a 300KeV photon backscatters it loses say 250KeV , clearly 2.5Mev is higher than 250Kev in energy/frequency , so would I be correct in saying that for a fixed angle (whether 90 or 180 or any other) a photon can lose a fixed percentage of it's total wavelength/energy but the amplitude of this wavelength/energy gets proportionally larger as the initial photon energy increases or gets smaller as the initial photon energy decreases?
Bear in mind that the frequency of a photon, ν, is given by c/λ, where c = speed of light (taken as constant) and λ is wavelength, and the photon energy, E, is given by E = hν = hc/λ. As E and ν increase, λ decreases.

Try some examples with the Hyperphysics calculator, or use a spreadsheet or python program and do a sensitivity study, or differentiate the formula for the change in energy (or wavelength) of a photon with respect to the initial photon energy. Also, one can the change-in-wavelength formula and rewrite it in terms of energy of the photon before and after. Actually, the after photon is different than the initial photon (different wavelengths/frequency/energy).

I did 3 examples at 250 keV, 511 keV (~rest mass of e) and 2.5 MeV with a scattering angle 90° and 180°

Code:
Ini. Ph   Scatt. Angle 90°     Scatt. Angle 180
Ei        Ef, keV  Ke, keV     Ef, keV  Ke, keV
250 keV    167.87    82.12      126.36   123.64
511 keV    255.5    255.5       170.33   340.67
2.5 MeV    424.28  2075.72      231.81  2268.19

One could also try changing the rest mass of the electron and see what effect it has.
there can be no such thing as a photon with 0 energy/frequency, but can a gamma scatter down from initial MeV's of energy to eV's?
At low energies, the photon is more likely to be completely absorbed by an electron, at a greater probability than being scattered (as in the Compton effect). Also, at low energies, a photon may experience Rayleigh scattering (aka, elastic or coherent scattering).

See - https://www.utoledo.edu/med/depts/radther/pdf/RDII - Chapter 7n handout.pdf
https://en.wikipedia.org/wiki/Elastic_scattering#Optical_elastic_scattering

As for 'free' electrons, I like the statement "The predominant inelastic scattering process around 1 MeV is the well known Compton effect in which the gamma ray is scattered by anatomic electron, where the photon energies (Eo) are so large compared with the electron binding energies that the electrons can be considered as free." In other words, Eo >> Ee,b, the latter being the binding energy of the electron in an atom.
Ref: https://scholarworks.wmich.edu/cgi/viewcontent.cgi?article=4117&context=masters_theses

I'm interested in the influence of gammas on structural materials, and mainly the effect on chemical potential of specific elements in an alloy, as well as the atomic displacement, besides the development of a so-called point defect in a crystal lattice. During the university program, in course on radiation dosimetry and protection, the emphasis was on the attenuation effects, rather than the material effects.

artis
Bear in mind that the frequency of a photon, ν, is given by c/λ
Right , so the frequency does change as the speed of the photon deviates from c to below that, like a ray of light gets attenuated while passing through water or glass so it's frequency drops and wavelength increases.
So the sun should seem whiter looking at it directly (if one could) than looking at it through bedroom window.

I did 3 examples at 250 keV, 511 keV (~rest mass of e) and 2.5 MeV with a scattering angle 90° and 180°
Thanks, I can see that I was wrong , even for the same angle different photon energies will result in different percentage of energy divided between the scattered photon and electron.
I'm interested in the influence of gammas on structural materials, and mainly the effect on chemical potential of specific elements in an alloy, as well as the atomic displacement, besides the development of a so-called point defect in a crystal lattice. During the university program, in course on radiation dosimetry and protection, the emphasis was on the attenuation effects, rather than the material effects.
Presumably with regards to nuclear reactors and their vessels containment?

Staff Emeritus
Right , so the frequency does change as the speed of the photon deviates from c to below that, like a ray of light gets attenuated while passing through water or glass so it's frequency drops and wavelength increases.
The speed of light (in a vacuum) is constant. The change in wavelength or frequency is due to loss of momentum (and energy) of a photon to an electron it encounters.
http://hyperphysics.phy-astr.gsu.edu/hbase/ems1.html

Presumably with regards to nuclear reactors and their vessels containment?
Pressure vessel steels are one concern. I am more interested in materials (alloys and ceramics) operating in the core, e.g., stainless steels providing support for in-core instrumentation, e.g., thermocouples and neutron flux measurement, or control elements.

There is a comment in a manual for a code, PENELOPE, that simulates electron/positron and gamma radiation behavior in materials.
Atoms are primarily ionised by photon interactions and by electron or positron impact. There is a fundamental difference between the ionising effects of photons and of charged particles. A photon is only able to directly ionise a single atom. In the case of photoabsorption, when the photon energy is larger than the K-shell binding energy, about 80% of photoabsorptions occur in the K shell, i.e., the resulting ion with a vacancy in the K shell is highly excited. Incoherent scattering is not as highly preferential, but still the probability that an inner subshell is ionised is nearly proportional to the number of electrons in the subshell. Conversely, fast electrons and positrons (and other charged particles) ionise many atoms along their paths; the ionisations occur preferentially in the less tightly bound atomic subshells, or the conduction band in the case of metals, so that most of the produced ions are only weakly excited.
In a nuclear reactor, each fission produces 2 or 3 neutrons, on average about 2.3 +/-, as well as 7 or 8 spontaneous (prompt) gammas, then at least two gammas from the two fission products. To maintain the reaction, one neutron must survive to be absorbed by another fissile atom (235U or 239Pu), and the remaining neutrons are absorbed without causing fission in a process called 'radiative capture'. Both 235U or 239Pu can absorb a neutron without fissioning through gamma emission becoming 236U or 240Pu, respectively. Basically, other nuclides (elements) resident in the core have a chance to absorb neutrons (e.g., 58Ni + n -> 59Ni, 59Ni + n -> 60Ni, 59Ni + n -> 61Ni, . . . What is interesting about these three reactions is the energy of the gamma emitted, ~9 MeV, 11.4 MeV and 7.8 MeV, respectively. 59Ni is not stable, or naturally occurring, so must be created by the first reaction (or by (n,p) reaction with 59Co, which is usually minimized in stainless steels). Radionuclides of 239Pu and 241Pu produce gamma of ~5.5-5.6 MeV, and certain other nuclides produce high energy gammas, E > 3 MeV. It is the effect (through pair production and Compton scattering) of such gammas that is of most interest.

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artis
@Astronuc somewhat related question I had in mind, do all radioactive isotopes and elements in general produce gamma radiation so they can be identified by gamma spectroscopy or are there elements that don't produce gamma at all?

Yes. There are a number of isotope decays that do not produce gammas C14, P33, S35, Ar37, , , , , Hg,194, Pb209 to name a few. However, the resulting atoms will still produce x-rays from disrupted electron orbits resulting from the decay processes which can be used to identify them.

vanhees71 and artis
Staff Emeritus
@Astronuc somewhat related question I had in mind, do all radioactive isotopes and elements in general produce gamma radiation so they can be identified by gamma spectroscopy or are there elements that don't produce gamma at all?
Yes. There are a number of isotope decays that do not produce gammas C14, P33, S35, Ar37, , , , , Hg,194, Pb209 to name a few. However, the resulting atoms will still produce x-rays from disrupted electron orbits resulting from the decay processes which can be used to identify them.
My bold for emphasis. These are examples of beta decay with no gamma emission. T decay is another example.

However, when a nuclide absorbs a neutron, it will emit a gamma ray, i.e., prompt gamma with a unique signature. In these cases, a nuclide can be identified by the prompt gammas, which is the basis of prompt-gamma activation analysis.
https://www.nist.gov/laboratories/tools-instruments/prompt-gamma-ray-activation-analysis-pgaa

PGAA requires a neutron beam incident upon a sample, whereby some atoms absorb a neutron. Decay gamma-spectroscopy requires one to wait for the radionulide to decay. In the absence of gamma emission, one would detect the beta particle (and energy) or X-rays from internal conversion. https://www.sciencedirect.com/topics/chemistry/internal-conversion

Some examples of gamma emission.
https://www.nndc.bnl.gov/capgam/byTarget/z015_31P.html (31P + n => 32P)
https://www.nndc.bnl.gov/capgam/byTarget/z016_34S.html (34S + n => 35S)
https://www.nndc.bnl.gov/capgam/byTarget/z082_208Pb.html (208Pb + n => 209Pb)

Unless one separates individual isotopes, one will receive several isotopes from activation analysis, depending on the element.

vanhees71 and artis
artis
Additionally if I may go back a step with regards to the scattering, I have a question
In the post #14
I did 3 examples at 250 keV, 511 keV (~rest mass of e) and 2.5 MeV with a scattering angle 90° and 180°

Code:
Ini. Ph Scatt. Angle 90° Scatt. Angle 180
Ei Ef, keV Ke, keV Ef, keV Ke, keV
250 keV 167.87 82.12 126.36 123.64
511 keV 255.5 255.5 170.33 340.67
2.5 MeV 424.28 2075.72 231.81 2268.19
If I look at the 2.5MeV photon scattering off an electron with an angle 180° (backscatter) it seems the photon loses most of it's energy to the electron which now has a KE of 2.268MeV even though the photon energy was more than 4 times larger than the electron rest mass of 511KeV ,on the other hand, in elastic scattering between massive particles, IIRC, a incident particle with higher KE than the target particle rest mass can lose no more than the target particle rest mass in a single collision?

This as far as I'm aware is the reason for water as moderator in thermal fission reactors since hydrogen proton has roughly the same rest mass as neutron so a neutron can lose almost all it's energy during a single collision , but what if the neutron KE was so large that it was twice the proton rest mass (neglecting proton KE due to thermal energy) would it then at most lose no more than roughly half it's KE which would sum to be the proton rest mass?

Staff Emeritus


text:
# Which particles are classified as PM2.5? How exactly is this defined?

## Question

When discussing "PM2.5", is there any standardized understanding of which particles are or are not included?

Is it everything that's 2.5 microns and smaller? Or Everything between 2.5 and 0.1 microns?

Likewise does PM10 include all of PM2.5?

And by what method is "size" specified? Is this an aerodynamic size from some kind of inertial impactor configuration (virtual or real), or an optical size from scattering, or something else?

## Background

PM2.5 and other classes of particulates are a "hot topic" now. From Wikipedia's Particulates; size, shape and solubility matter:

[...]The 10-micrometer size does not represent a strict boundary between respirable and non-respirable particles, but has been agreed upon for monitoring of airborne particulate matter by most regulatory agencies. Because of their small size, particles on the order of 10 micrometers or less (coarse particulate matter, PM10) can penetrate the deepest part of the lungs such as the bronchioles or alveoli; when asthmatics are exposed to these conditions it can trigger bronchoconstriction.

Similarly, so called fine particulate matter (PM2.5), tend to penetrate into the gas exchange regions of the lung (alveolus), and very small particles (ultrafine particulate matter, PM0.1) may pass through the lungs to affect other organs. Penetration of particles is not wholly dependent on their size; shape and chemical composition also play a part. To avoid this complication, simple nomenclature is used to indicate the different degrees of relative penetration of a PM particle into the cardiovascular system. Inhalable particles penetrate no further than the bronchi as they are filtered out by the cilia. Thoracic particles can penetrate right into terminal bronchioles whereas PM0.1, which can penetrate to alveoli, the gas exchange area, and hence the circulatory system are termed respirable particles. In analogy, the inhalable dust fraction is the fraction of dust entering nose and mouth which may be deposited anywhere in the respiratory tract. The thoracic fraction is the fraction that enters the thorax and is deposited within the lung's airways. The respirable fraction is what is deposited in the gas exchange regions (alveoli).

The smallest particles, less than 100 nanometers (nanoparticles), may be even more damaging to the cardiovascular system. Nanoparticles can pass through cell membranes and migrate into other organs, including the brain. Particles emitted from modern diesel engines (commonly referred to as Diesel Particulate Matter, or DPM) are typically in the size range of 100 nanometers (0.1 micrometer). [...]

From AirNow.gov's Particle Pollution (PM):

• Coarse dust particles (PM10) are 2.5 to 10 micrometers in diameter. Sources include crushing or grinding operations and dust stirred up by vehicles on roads.

• Fine particles (PM2.5) are 2.5 micrometers in diameter or smaller, and can only be seen with an electron microscope. Fine particles are produced from all types of combustion, including motor vehicles, power plants, residential wood burning, forest fires, agricultural burning, and some industrial processes

However, EPA.gov's Particulate Matter (PM) Basics; What is PM, and how does it get into the air? says:

• PM10: inhalable particles, with diameters that are generally 10 micrometers and smaller; and

• PM2.5: fine inhalable particles, with diameters that are generally 2.5 micrometers and smaller.

Source

• Your quote explains that PM 10 includes everything less than 10 microns Apr 9 '19 at 22:12
• @haresfur If you read the question all the way through, you will see that there are two quotes, and the second quote contradicts the first quote. "However, EPA.gov's Particulate Matter (PM) Basics; What is PM, and how does it get into the air? says..."
– uhoh
Apr 9 '19 at 22:46

I think answering your questions in reverse will make more sense. The "size" of PM is typically the aerodynamic/inertial impaction size, as you guessed. My standard reference for this is this paper. Their figure 8 is a rough schematic of where particles of different size deposit in the lungs. Things with very large or very small inertial impaction size get filtered out by the nose and upper respiratory system. (The "very small" inertial impaction I would consider as the fuzzy boundary between particles and gases, and gases adsorb onto mucous.) It's the particles with an aerodynamic size smaller than 2.5 microns that follow the airflow all the way from the nose and mouth into the lungs. So from a health perspective, aerodynamic size is what matters most.

Similarly, most measurements of particulate matter filter them by size first, and those filters are based on how much they deflect when the airflow has to curve around a bend. So when measurements focus on PM2.5, that's also anything with aerodynamic radius less than 2.5 microns. They might later do optical scattering or other analysis, but AFAIK the aerodynamic filtering always comes first.

I would say that PM2.5 is everything 2.5 microns and smaller, and PM10 includes all of PM2.5. When scientists want to differentiate the two, we usually use the terms for the different "modes": coarse mode for ~1-10 micron, accumulation mode for ~.05-1 micron, and nucleation or Aitken mode for anything smaller. (Fine particulates are nucleation+accumulation.) The reason for PM2.5 including everything is that PM2.5 standards are focused on the mass concentrations, usually in micrograms/cubic meter. The total mass in the nucleation mode is typically negligible compared to the other two. (The number concentrations particles in the nucleation and accumulation modes might be similar, but the masses are way different.)

Edit: I don't think I made this clear, but the reported air quality value for PM10 would include PM2.5, but if something is reported as "coarse particulates" it would be (PM10-PM2.5). You may also see a quantity TSP - total suspended particulates - which may include things even larger than PM10 that are being kicked up by particularly strong winds, or smoke and ash from nearby wildfires.

• This is a really helpful perspective. Yes, Fig 8 is instructive, and I like the way you think about when the larger category should or should not include the smaller one. Thank you for your answer! update: Wow, this is a really compelling paper!
– uhoh
Apr 22 '19 at 11:35
• btw have a look at comments below this answer. I wonder if you are able to assign the pm2.5 tag as a synonym to particulates tag, which would become the main tag and appear whenever someone (like me) starts typing pm2.5? Neither of us has sufficient reputation on these particular tags yet.
– uhoh
Apr 22 '19 at 11:37
• and for those who aren't PubMed-savvy, here's a link to the paper on the Journal's own website. It's Open Access, yay! ehp.niehs.nih.gov/doi/10.1289/ehp.7339
– uhoh
Apr 22 '19 at 11:53
• I think it's also important to point out that particulate matter filter samples are dried before being weighed. Thus, if you have a foggy day, it won't be calculated as a high particulate day.
– f.thorpe
Jul 13 '19 at 1:15


text:
# Intuitive meaning of Newton (units)

Is there any intuitive reasoning behind why the units of a "Newton" is $kg \frac{m}{s^{2}}$ and how it represents force? I always wanted to understand why objects of different mass fall at the same rate on earth and I found an explanation that uses Newtons Second Law of Motion and it made sense, but I'm confused as to why the units of force have this weird formula. Did Newton just use those units to make his second law work out mathematically via cancelling? Or is there some way to interpret $kg \frac{m}{s^{2}}$ as meaning force? Any insight would be appreciated.

-

This is precisely the type of question I've started asking my introductory students. There's indeed an underlying intuition with units. Consider a free particle. It moves through space with constant momentum. If a force acts on that particle, its momentum will change. Remember that momentum is a vector and a change in a vector can involve a change in magnitude, a change in direction, or both (any arbitrary change in a vector can be decomposed into a change in magnitude parallel or antiparallel to the vector and a change in direction perpendicular to the vector). If you look at the units of force $\mathrm{kg}\cdot\mathrm{m}/\mathrm{s}^2$ and think of it as $\frac{\mathrm{kg}\cdot\mathrm{m}/\mathrm{s}}{\mathrm{s}}$ you can see it's the quotient of momentum to time (more correctly, duration because it's a time interval rather than a clock reading). So, think of a force of $X\;\mathrm{N}$ acting on a particle as being the how much momentum the particle gains or loses for every duration of one second that elapses. That's easiest to understand and visualize if the force is collinear with the particle's momentum. It's a bit more difficult to understand when only the direction of the particle's momentum changes, as in uniform circular motion.

So, a force of $12\;\mathrm{N}$ acting on a particle means that the particle's momentum changes by $12\;\mathrm{kg}\cdot\mathrm{m}/\mathrm{s}$ for every $1\;\mathrm{s}$ duration that elapses.

-
This relationship is most easily seen through the relation $F = dp/dt$. –  Muphrid Dec 21 '12 at 2:22
Yes, plus it's valid relativistically since $\vec{F}=m\vec{a}$ is a low speed approximation to $\vec{F}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$. –  user11266 Dec 21 '12 at 2:26

The units of a new physical quantity will be defined when you define the quantity itself. So if you define force by $F = ma$, then it will have units of mass (e.g. $\mathrm{kg}$) times acceleration (velocity per unit time, e.g. meters per second squared). If instead you define force as the time-derivative of momentum, then it is the units of momentum (mass times velocity) divided by the units of time.

If you wanted to start with force as primitive, you could do that, calling the unit $\mathrm{N}$. Then perhaps mass would be the thing defined in $F = ma$, and the unit of mass would be that of force per unit acceleration, e.g. $\mathrm{N}/(\mathrm{m}/\mathrm{s}^2)$, which we could abbreviate as $\mathrm{kg}$, and this would serve as a definition of the kilogram.

It all depends on where you take your starting place to be. As a side note, the development of mechanics was muddled at first because people used the word "force" and similar words for a whole variety of related concepts without appreciating the differences. Today we have separated these cleanly into forces, momenta, accelerations, inertia, kinetic energy, potential energy, etc.

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So if I understand you correctly your saying that there is no intuitive meaning behind the units in this case, rather the units are generated as a result of creating a new physical quantity (in this case force). Is this correct? –  Ockham Dec 20 '12 at 23:52
I guess I'm saying don't look at all four parts (kg, m, two copies of seconds) together. Instead decompose it into a product or quotient of things you already understand. –  Chris White Dec 21 '12 at 0:25


text:
# What is the volume ratio

###### Question:

What is the volume ratio

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### Diego measures the number of cups of water in 15 boxes of various sizes you made this line plot to show how many of each size has he has Diego Dental

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### 5 times the sum of s and 7

5 times the sum of s and 7...

### A force of 10N is applied by a boy while lifting a 20,000g mass. How much does it accelerate by?

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### Does Sarita up early in the morning?

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### ; jfaowifnonobnaofnoij jjfoajfaiojoiaoqjaiw [5;jiothjaokefj[iorejg

; jfaowifnonobnaofnoij jjfoajfaiojoiaoqjaiw [5 ;jiothjaokefj[iorejg...

### Slope = 2 over 5 y intercept = 2

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### Which juice container is larger 2l or 1gal

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### Root 12 of Rood 12, see attached document

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### What length of ladder will warren need to buy if his house is 36 feet tall and the base of the ladder is 15 feet from his house ( he wants

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### 9. Which is not a form of electromagnetic radiation?O radio wavesO gravityO gamma raysO visible

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### Help plis

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### 2. Which expression is equivalent to 1/3 divided by 8

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### Jessica is baking a cake. the recipe says that she has to mix 96 grams of sugar into the flour. jessica knows that 1 cup

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text:
# Voronoi diagram

A very important geometric structure in computational geometry, named after G.F. Voronoi. The earliest significant use of Voronoi diagrams seems to have occurred in the work of C.F. Gauss, P.G.L. Dirichlet and Voronoi on the reducibility of positive-definite quadratic forms (cf. Quadratic form).

Let $S = \{ p _ {1} \dots p _ {n} \}$ be a set of $n$ points in $\mathbf R ^ {d}$. The Voronoi diagram generated by $S$ is the partition of the $\mathbf R ^ {d}$ into $n$ convex cells, the Voronoi cells, $V _ {i}$, where each $V _ {i}$ contains all points of $\mathbf R ^ {d}$ closer to $p _ {i}$ than to any other point:

$$V _ {i} = \left \{ x : {\forall j \neq i, d ( x,p _ {i} ) \leq d ( x,p _ {j} ) } \right \} ,$$

where $d ( x,y )$ is the Euclidean distance between $x$ and $y$.


text:
## Hokkaido Mathematical Journal

### Change of variables for weighted Hardy spaces on a domain

Akihiko MIYACHI

#### Abstract

A generalized version of weighted Hardy spaces on a Euclidean domain is introduced and it is proved that the spaces are transformed to the same kind of spaces by certain smooth change of variables. Some related properties of the spaces, including a modified form of atomic decomposition, and some examples are also given.

#### Article information

Source
Hokkaido Math. J., Volume 38, Number 3 (2009), 519-555.

Dates
First available in Project Euclid: 18 November 2009

https://projecteuclid.org/euclid.hokmj/1258553975

Digital Object Identifier
doi:10.14492/hokmj/1258553975

Mathematical Reviews number (MathSciNet)
MR2548234

Zentralblatt MATH identifier
1186.42011

Subjects
Primary: 42B30: $H^p$-spaces
Secondary: 46E99: None of the above, but in this section

#### Citation

MIYACHI, Akihiko. Change of variables for weighted Hardy spaces on a domain. Hokkaido Math. J. 38 (2009), no. 3, 519--555. doi:10.14492/hokmj/1258553975. https://projecteuclid.org/euclid.hokmj/1258553975