TITLE: Working out atomic clock accuracy using measured frequency Hz, error and beat Hz
QUESTION [0 upvotes]: I've got a question where $200,000\,{\rm Hz}$ applied to the Cesium atoms and from the mismatch with the atom's natural frequency, we get a "beat" of $2\,{\rm Hz}$.  The error is $1\%$.  The accuracy of the clock is then $1$ in $10,000,000$.
How did they get the $10,000,000$?
My guess is, since $1\%$ of $2\,{\rm Hz}$ is $0.02\,{\rm Hz}$:
$$\frac{0.02\,{\rm Hz}}{200,000\,{\rm Hz}}  = \frac{2}{20,000,000} = \frac{1}{10,000,000}$$
If the beat is $150\,{\rm Hz}$ and error is $0.002\%$ then what is the accuracy of the clock? I'm confused about the frequency applied to the atom, should I reuse the $200,000\,{\rm Hz}$?  The only other frequency is the natural frequency of the Cesium atom - $9200\,{\rm MHz}$.

REPLY [1 votes]: The example on the first page is a bit confusing because it has nothing to do with caesium. It's giving a hypothetical example of measuring a frequency $F$ against a hypothetical standard 200kHz oscillator. If the beat frequency is $F_b$, the unknown frequency $F$ is given by:
$$ F = 200000 \pm F_b \tag{1} $$
It isn't clear to me how the article resolves the ambiguity or whether to use $+$ or $-$, but let's ignore that for now and assume it's $+$ so $F \approx 200002$ Hz. Anyhow, if the error in measuring $F_b$ is $\Delta F_b$ then equation (1) tells us the error in measuring $F$ will also be $\Delta F_b$ because we assume the 200,000 Hz oscillator has no error. The fractional error is therefore given by:
$$ \frac{\Delta F_b}{F} $$
The article says we can measure $F_b$ to be 2Hz with an error of 1%, i.e. 0.02Hz, so $\Delta F_b = 0.02$ Hz and the fractional error is:
$$ \frac{0.02}{200000} $$
which is 1 in ten million.
Question 3 is doing a similar calculation using caesium as the reference frequency. In that case we have:
$$ F = F_{Cs} + F_b $$
where $F_{Cs}$ is exactly 9,192,631,770 Hz by definition and $F_b$ is 150 Hz. We are told we can measure $F_b$ to 0.002% so the error is $\Delta F_b = 150 \times 0.00002 = 0.003$ Hz.


TITLE: Use the Maximum Likelihood Estimation approach to find an estimator for $\alpha.$ given the Pareto distribution
QUESTION [0 upvotes]: Let $Y_1,...Y_n$ be a random sample from the Pareto distribution with parameters $\alpha$ and $\beta$, where $\beta$ is known. Then, if $\alpha > 0$, $$f(y|\alpha, \beta) = \alpha \beta^\alpha y^{-(\alpha +1)}, y \ge \beta.$$
Goal: Use the Maximum Likelihood Estimation approach to find an estimator for $\alpha.$
MY APPROACH:
$L(\alpha) = \alpha^n \beta^{\alpha n} (\prod_{i=1}^n y_i)^{-(\alpha+1)}$
Taking log for $L(\alpha)$ gives
$ln L(\alpha) = n ln(\alpha) + \alpha n ln(\beta) + \sum_{i=1}^n -(\alpha+1) ln(y_i)$
Taking derivatives of $ln L(\alpha)$ with respect to $\alpha$ gives $n/\alpha + nln(\beta) - \sum_{i=1}^n ln(y_i)$
setting the above equation to zero give,$$ \hat{\alpha} = \frac{n}{\sum_{i=1}^n ln(y_i) - nln(\beta)}$$
Am I right?

REPLY [1 votes]: Yes, you've carried out the steps correctly. I really don't have much more to add since your request was just to confirm.


TITLE: find distance using a light line
QUESTION [0 upvotes]: Let's suppose we have a particular light frequency emitter and relative sensor array, and that there is no external source of this light.
This emitter has a know angle respect the receiver, and emit a point of line. Because the angle is known, we can determine the distance just by looking at which point in the sensor we read the point. This is the triangulation method, used by , click here for an explanation
Now, I want to understand if it possible to use a line instead of a point, and a sensor matrix instead of an array, so we can get 2D reading.
This should work until the "wall" has a obliquity on the axis perpendicular to the light line, but then I can't find a way to "fix" this problem. I may use different pattern instead of a line, like point but then the problem rise if one or more of the point are out of range for the receiver. any thoughts?

REPLY [1 votes]: Your question is a bit muddy, but I believe the answer to everything is "yes, but..."  Here's what's described in the referenced article:  The laser produces a small spot on the target and the lens in the camera creates an image of that spot on one (ideally) pixel.  By using simple lens equations, or by similar triangles, if you know the lens focal length and the pixel dimensions (and which pixel is on the camera's (0,0) optic axis location), you can determine the distance to the target.
Since we never made an explicit definition of which axis in the plane perpendicular to the laser beam we just made this measurement, clearly we could make this measurement in "x" or "y".  However,  the only dimension in which we're determining a value is along the laser beam, i.e. perpendicular to the target.   That's the only thing you can measure.  If your question is whether you could move the laser+camera in "x" and "y" and measure the distance at various locations on the target, most certainly you could.


TITLE: Show that $W\oplus W'=V.$
QUESTION [1 upvotes]: Let $V$ be an inner product space and $W$ be a subspace of $V$ then show that $W\oplus W'=V.$ Where $W'=\{x\in V|\langle x,y\rangle = 0, \forall y\in W\}.$
My Attempt: It is easy to observe that $W\cap W'=\{0\}.$ Next, I want to show that $V=W+W'.$ I was wondering if I can do the following. Say that for a given vector $v\in V$ find the best approximation $u\in W$ or the projection of $v$ onto $W$ such that $\|u-x\|\leq \|w-x\|, \forall w\in W.$ Then we clearly have that for all $v\in V$ $$v=u+v',v'\in W'.$$ And so $V=W+W'$ Thus we can conclude that $V=W\oplus W'.$
Does this work? I am asking this because I read a remark somewhere which mentioned that the best approximation might not always exist if the norm is not induced by the inner product or something like that. So there could be a flaw in my argument. In any case, I would be grateful to get some feedback.

REPLY [1 votes]: As noted in the comments, for the best approximation to exist $W$ has to be a closed subspace of $V$.
Other than that, you actually failed to show that $v' = w-v$ is contained in $W'$.
A way to show this is the following lemma:

Let $b, v \in V$ be two vectors. Then $b \perp v$ if and only if $\|b\| \le \|b + \lambda v\|$ for all scalars $\lambda$.

Proof:
$$\|b + \lambda v\|^2 = \|b\|^2 + 2\operatorname{Re} \overline{\lambda}\langle b, v\rangle + |\lambda|^2\|v\|^2$$
If $b \perp v$ then $\|b + \lambda v\|^2 = \|b\|^2 + |\lambda|^2\|v\|^2 \ge \|b\|$.
Conversely, set $\lambda = - \frac{\langle b, v\rangle}{\|v\|^2}$:
$$\|b\|^2 \le \|b + \lambda v\|^2 = \|b\|^2 + 2\operatorname{Re} \overline{\lambda}\langle b, v\rangle + |\lambda|^2\|v\|^2 = \|b\|^2 - \frac{\left|\langle b, v\rangle\right|}{\|v\|^2} \implies \langle b, v\rangle = 0$$

Now for any $v \in V$ let $w \in W$ be the best approximation of $v$ onto $W$. We wish to show $v-w \in W'$, that is $w - v \perp W$.
For any scalar $\lambda$ and $x \in W$ we have:
$$\|v -w - \lambda x \| = \|v - \underbrace{(w + \lambda x)}_{\in W} \| \ge \|v-w\|$$
Therefore $v-w \perp x$. Since $x \in W$ was arbitrary we obtain $v - w \in W'$.


TITLE: Confusion with abstract tensor notation
QUESTION [2 upvotes]: I am currently going through the abstract tensor notation in Wald's "General Relativity". I understand the purpose of it, but I need help understanding some of the conventions and definitions. So, first Wald writes "Since a metric [tensor] g is a tensor of type $(0,2)$, it is denoted $g_{ab}$. If we apply the metric to a vector, $v^a$, we get the dual vector $g_{ab}v^b$. It is convenient to denote this vector as simply $v_a$, thus making notationally explicit the isomorphic between $V_p$ [the tangent space at $p$] and $V_p^*$ defined by $g_{ab}$".
Let me try to unpack what he just said here: The metric tensor $g_{ab}$ is a covariant $2$-tensor on $V_p$, so it does not make sense to apply the metric to a single vector $v^a$ like Wald says, but I suppose he means that $g_{ab}(v^a, \cdot)$ is a dual vector if $v^a$ is held fixed (by the way, does $v^a$ have to be in the first slot because the metric tensor is denoted as $g_{ab}$ and not $g_{ba}$?). Is this correct? Wald goes on to say that this dual vector is $g_{ab}v^b$. Now, in a previous paragraph he wrote that writing two tensors side by side like that denotes the tensor product. So $g_{ab}v^b$ is short for $g_{ab} \otimes v^b$, according to Wald. But surely he does not mean that the dual vector $g_{ab}(v^a, \cdot)$ is equal to $g_{ab}v^b$? Because that would be a $(1,2)$ type tensor. Could someone please clear up this confusion for me?

REPLY [3 votes]: Ok, lets formalise all that have been said in the comments:
Let $(v_1,\dots,v_n)$ and $(v^1,\dots,v^n)$ be basis of $V_p$ and $V^*_p$, respectively. Take a tensor $\tau \in \mathcal{T}^r_s(V_p)$, and pick indexes $k \leq r, l \leq s$. Then we define the contraction $C^k_l\tau \in \mathcal{T}^{r-1}_{s-1}(V_p)$ as
$$C^k_l\tau(f^1,\dots,f^{r-1},w_1,\dots,w_{s-1}):=\sum_{a=1}^{n}\tau(f^1,\dots,\underbrace {\hbox{$v^a$}}_{\hbox{$k$-th position}},f^{r-1},w_1,\dots,\underbrace {\hbox{$v_a$}}_{\hbox{$l$-th position}},\dots,w_{s-1}).$$
Recall that a $(r,s)$-tensor is a sum over all $(r,s)$-combinations of vectors of the respective basis. So, take indexes $1 \leq i_1 < i_2 < \dots < i_r \leq n$ and $1 \leq j_1 < j_2 < \dots < j_r \leq n$, and denote them by $I = i_1\dots i_r$ and $J = j_1 \dots j_s$, respectively. Then an $(r,s)$-tensor $\tau$ is the sum over all such $I$'s and $J$'s
$$\tau := \sum_J \sum_I \tau^I_J \, v_I \otimes v^J := \tau^{i_1\dots i_r}_{j_1\dots j_s} \, v_{i_1} \otimes \dots \otimes v_{i_r} \otimes v^{\, j_1} \otimes \dots \otimes v^{\, j_s},$$
$\tau^{i_1\dots i_r}_{j_1\dots j_s} = \tau (v^{\, i_1}, \dots, v^{\, i_r}, v_{j_1}, \dots, v_{j_s}).$ The last equality is just the $Einstein \, summation \, convention.$
Returning to contraction, since it is a $(r-1,s-1)$-tensor the components of it are
$$\left( C^k_l \tau \right) ^{i_1 \dots {\hat{i_k}}\dots i_r}_{j_1 \dots {\hat{j_l}}\dots j_s} := \tau^{i_1 \dots a \dots i_r}_{j_1 \dots a \dots j_r} \, (\text{sum over $a$}),$$
where $\hat{i_k}, \, \hat{j_l}$ means omission.
Tackling your example, we have $g \in \mathcal{T}_2(V_p), \, j_1= a, \, j_2 = b,$ that written in component form is $g = g_{ab}$. The act of "applying" a vector to the metric then, is equivalent of building a contraction $C^c_b g \nu \in \mathcal{T}_1(V_p)=V_p^*$, to the tensor product $g_{ab}\nu^c \in \mathcal{T}^1_2(V_p)$. Using the component notation I introduced above we get that
$$v_a := \left(C^c_bg\nu \right)_a= g_{ab} \nu^b \, \text{(sum over $b$)}.$$
you can also get $v_b$ by contracting $a$, so
$$v_b := \left( C^c_a g\nu \right) _b = g_{ab}\nu^a. $$
edit: There is a natural map $V \to V^{**}$ that sends $v\in V \mapsto \hat{v} \in V^{**}$, where $\hat{v}$ is such that for any $f \in V^*, \hat{v}(f) := f(v).$ In the case of finite dimensional vector spaces, this map is an isomorphism. Now, since $v \in V$, we can express it as a linear combination of elements of a basis of $V$, lets say $v = \sum_{k=1}^n \alpha_k e_k$. It turns out that if we evaluate the elements of the corresponding dual basis $\{e^1, \dots, e^n \} $ of $V^*$ to $\hat{v}$, we get the coefficients of $v$
$$\hat{v}(e^k):= e^k(v)= \alpha_k.$$
Now, $\hat{v} \in \mathcal{T}^1(V)$, so we can make the tensor product $g \otimes \hat{v} \in \mathcal{T}^1_2(V)$, given by:
$$g \otimes \hat{v} (f^1, w_1, w_2) := g\hat{v} (f^1, w_1, w_2) = g(w_1,w_2) \cdot \hat{v}(f^1).$$
Now, if we denote $g=g_{ab}, \, \hat{v}= \hat{v}^c$, and make the contraction $C^c_b g\hat{v}$, we obtain:
$$C^c_b g \hat{v}(-):= \sum_{b=1}^n g\hat{v}(e^b,-,e_b) = \sum_{b=1}^n g(-, e_b)\cdot \hat{v}(e^b) = \sum_{b=1}^n g(-, e_b) \cdot \alpha_b$$
$$= \sum_{b=1}^n g(-,\alpha_b e_b) = g(-,v),  $$
 which is the evaluation of a vector in the second entry of $g$.


TITLE: Intuition for comma category and slice category?
QUESTION [1 upvotes]: As usual, I am not happy with the definitions until I dont gain some intuition. Could you help me and maybe provide some examples of the concepts? Thank you! My questions are:
1 Comma category = "Morphism category". Category where objects are morphisms. And morphisms are.. Morphisms between those morphisms, but in a different category (say C).
1.1 Why we need comma categories?
1.2 How to explain comma category in layman terms?
1.3 Is comma category a special type of 2-category? Or an object in a 2-category?
2 Slice category = Special case of comma category.
2.1 Why we need comma categories?
2.2 How to explain comma category in layman terms?
2.3 Simplest examples?
3 Limits and colimits - what is the intuition?

REPLY [5 votes]: Addressing all of these topics adequately is about 2 chapters in a good textbook. So, I'll concentrate on the idea for these constructions and some example that I hope will be illuminating.
The "morphism category", also known as the arrow category. In a category, you can think of the objects as meaningless entities whose sole purpose in life is to serve as the domain and codomain of morphisms. Whatever interest, or structure, we perceive the objects to have is a by-product of the morphisms they carry, and the composition of such morphisms. In a sense, the morphisms give 'life' to the objects. Once the morphisms are there, and we have a category, we can (and should) turn our attention to the morphisms themselves. The arrow category does just that. The original category embeds nicely inside the arrow category by viewing each object as the identity morphism on it. Now, the more general comma category allows us to mix this idea between different categories. It says that given some functors between categories, we can similarly 'promote' the objects into morphisms and 'enlarge' the category. It's a way to understand the functors involved.
Why we need comma categories? Because it's a simple way to build new categories out of old ones (and some functors between them).
How to explain them in layman terms? Depends on how layman you wish to be, but the above may help.
Is it a special type of 2-category, or something of that sort. Yes, and probably in more ways than one. There is a notion of comma object in a 2-category, and it captures the usual notion as a special case.
Now for slice categories. Yes, they are a special case of comma categories, but I'm going to suggest that forgetting that for a second is a good idea. An important example of comma category is as important as it is elementary: the slice category $Set/S$ of sets over a given set $S$. An object $X\to S$, just a function, can be thought of as a set $X$ of objects, each of which is assigned a colour from $S$. So, we fix the set $S$ of colours and look at all sets of $S$-coloured elements. If $S$ is a terminal object, then $Set/S\cong Set$. Rather boring. But, if $S$ is a doubleton, say $S=\{T,F\}$, the boolean truth values, then $Set/S$ is very interesting. An object is a set $X\to S$, which we may think of as a characteristic function. Thus, the objects are essentially pairs $(Y,X)$ of a set $X$ and a subset $Y$. Now, clearly there is a forgetful functor sending $X\to S$ to $X$. Now, consider an arbitrary function $f\colon X\to X'$. It is common to consider the direct image function and the inverse image function associated with $f$. The inverse image function thus takes an object $(X',Y')$ to some object $(X,Y)$. But, then this can be phrased also in the language of the comma category $Set/S$. It's thus a process that carries $X'\to S$ to some $X\to S$. The way this is done is precisely by computing the pullback. So, inverse image = pullback. Similarly, direct image = composition. This shows that very elementary operations are captured well within a suitable comma category. Much more is true; this is just the beginning of categorical logic.
So, why we need comma categories? They encode much of classical logic and extend it much beyond the classical realm. This is only one reason. How to explain them in layman terms? Again, perhaps the above helps, as well as some further examples below. Simplest examples? How about the slice category $*/Set$ of a singleton set over $Set$? It is essentially the category of pointed sets, so again we see that it is a general principle for constructing new categories from old.
Limits and colimits: the intuition is that a diagram in a category 'sees' the rest of the category in some way. The way the diagram as a whole looks at the rest of the world is by constructing a cone from it, namely each member of the diagram looks at a specific object via a morphism, and all these individual points-of-view must collate nicely in the sense of the commutativity defining the cone. Therefore, a diagram sees an object, typically, in a much more intricate way than any one of its members does (since they somehow need to agree on what they see). It's a bit like collectively a group of people who believe in $B$, see the world differently than the way they do individually. The collective (encoded perhaps the morphisms in the diagram) dictates the behaviour. Back to maths, it may be that the way the diagram perceives the world is identical to the way some object somewhere perceives the world. That object represents the diagram then. It is it's colimit. Limits are similar, but it's more the way the rest of the words sees the diagram (arrows are reversed).


TITLE: Do topological semantics of SOL satisfy the compactness theorem?
QUESTION [1 upvotes]: For a language $\tau$, let a topological $\tau$-model be a connected Hausdorff topological space $X$ equipped with a $\tau$-structure $\mathcal{M}$ on the set of points of $X$ such that:

For any $n$-ary relation symbol $R$, $R^\mathcal{M}$ is a closed subset of $X^n$
For any $n$-ary function symbol $f$, $f^\mathcal{M}$ : $X^n\rightarrow X$ is continuous

Note that, by the Hausdorff condition, equality (the diagonal) is a closed subset of $X^2$ and so we may add it to our topological theories as we like. Also note that this is a generalization of topological algebras.

This question is about the semantics for second-order logic arising when predicate symbols are interpreted as open subsets. For example, consider the following axiom:
"For any nonequal points $x$ and $y$ there exist open sets $U$, $V$, $U'$, and $V'$, such that $x$ is in $U$, $y$ is in $V$, no point in $U$ is in $U'$, no point in $V$ is in $V'$, and no point is in both $U$ and $V$."
A topological model satisfies the above axiom if and only if it is Urysohn.
To be more precise:

The sentences of the logic are precisely the second-order sentences.
If $\varphi$ is $P(x)$ for a predicate $P$, then for a topological model $X$, you have $X\models\varphi[U, x]$ iff $U$ is an open set and $x\in U$
$\land$, $\neg$, and first order quantifiers behave exactly as they do in first-order logic, via the underlying structure $\mathcal{M}$
If $\varphi$ is $\exists P(\psi(P, x_1...x_n))$ then $X\models\varphi(x_1...x_n)$ iff there is some open set $U$ such that $X\models\psi[U, x_1...x_n]$.

Does this logic satisfy the compactness theorem?

The above logic is a lot tamer than full second-order logic; in fact, you can think of any theory in the logic as being a theory in Henkin semantics with 2 more axioms in first-order logic (stating that the empty set is an open set and that the intersection of open sets is again open), as well as a third-order axiom stating that the union of any collection of open sets is again open. Due to this third-order axiom though, it's not immediately clear whether or not this logic is compact.
The main problem with this interpretation is that I'm not sure there's any way to state that a relation is closed in $X^n$, or that a function from $X^n$ to $X$ is continuous, when $n>1$. These reasons only serve to obfuscate this logic even more.
A friend proposed to me that a modified form of the ultraproduct argument could be applied here, where "$\{i : x_i = y_i\}$ is in the filter" was replaced with "For every basic open $U$ in $X$ containing $[x_i]$, $\{i : b_i\in U_i\}$ is in the filter" but we found an issue in this argument that made it unusable.
We both are under the impression that this logic may be compact, but are unable to prove it. How should we go about doing this?

Bonus questions!

Does this logic satisfy any analogues to the Lowenheim-Skolem theorem?
What model-theoretic properties does this logic have?
Can you find a complete, effective deductive system for this logic?

REPLY [3 votes]: Unless I'm misunderstanding this, this is in fact as complicated as standard second-order logic.

First, as a warm-up let's look at the not-necessarily-connected version.
Consider the sentence $$(*):\quad \forall x\exists U\forall y(y\in U\leftrightarrow y=x).$$ The topological models of $(*)$ are precisely the topological structures whose topology is discrete. The map sending a standard structure $\mathcal{M}$ to the topological structure gotten by slapping the discrete topology onto $\mathcal{M}$ then appropriately embeds standard second-order logic into not-necessarily-connected topological second-order logic.
This embedding in turn means that all the usual pathologies of second-order logic - the failures of compactness, Lowenheim-Skolem, computable enumerability for validity, and set-theoretic absoluteness - carry over to this topological second-order logic.

OK, now what about the connectedness requirement?
Well, the "slap on the discrete topology" idea no longer works, but the broader intuition is still valuable. We'll find a way to associate to each standard structure $\mathcal{M}$ a connected topological structure $\hat{\mathcal{M}}$ in some "information-preserving" way, and then argue that this gives rise to the desired embedding. One step of this will be to show that the set of topological structures associated to classical structures in the manner above is actually definable; in the not-necessarily-connected version, this was accomplished by $(*)$.
Here's a sketch of one approach which works. Suppose for simplicity our language consists of a single binary function symbol. Given a structure $\mathcal{M}=(M, f)$, consider the "$\mathcal{M}$-octopus:"

First, we consider the following function on $M\times[0,1]$: $$\hat{f}((m,i),(n,j))=(f(m,n), \min\{i,j\}).$$

Next, consider the equivalence relation $\sim$ on $M\times[0,1]$ given by $(a,i)\sim (b,j)$ iff $(a,i)=(b,j)$ or $i=j=0$. Note that $\hat{f}$ respects $\sim$, so we get an induced $\underline{f}:(M\times[0,1]/\sim)^2\rightarrow(M\times[0,1]/\sim)$.

Finally, equip $M$ with the discrete topology and $[0,1]$ with the usual topology, and $M\times[0,1]/\sim$ with the induced topology. Let $\hat{\mathcal{M}}$ be the topological structure in the same language as $\mathcal{M}$ with underlying space $M\times[0,1]/\sim$  and the binary function symbol interpreted as $\underline{f}$.

The set of topological structures isomorphic in the appropriate sense to some $\hat{\mathcal{M}}$ is indeed definable via a topological second-order sentence, although this is a bit tedious. Meanwhile, we can locate $\mathcal{M}$ inside $\hat{\mathcal{M}}$ in a precise way: consider the non-cut points of $\hat{\mathcal{M}}$. So the resulting translation is to send $\varphi$ to "the structure is an octopus and $\varphi$ holds with all element-quantifiers relativized to the non-cut-points."
So again, we wind up with the full terribleness of standard second-order logic.

Actually, I think the really valuable idea along these lines is to look at a specific topological space. E.g. what happens when we demand that our structure be built on $\mathbb{R}$ with the usual topology?
It turns out that even for equational first-order logic things get quite complicated; see e.g. this paper by Taylor. To be honest, I think that no reasonably rich space will yield a second-order logic which is compact or has a good deductive system, and Lowenheim-Skolem is of course dead on arrival. A certain amount of set-theoretic absoluteness, however, will hold for some spaces under appropriate set-theoretic assumptions like large cardinals - specifically, for spaces like $\mathbb{R}$, large cardinals will ensure that the corresponding logic is forcing-invariant.