TITLE: Boolean Expression simplification help
QUESTION [1 upvotes]: Hi I am new to the board.  Taking a computer architecture course and I am having trouble understanding further simplification on a question I got on a previous quiz.  When I type in the expression after setting up my K map and simplifying I come to this:
(A AND B) OR (NOT A AND NOT B).
Through the rules of simplification A+ NOT A=1.  Does this then mean that AB +A'B'=1? and if so how is that represented as a logic gate?
Thanks.

REPLY [0 votes]: A'B' would be.
A---o|---|
     | & |----A'B'
B---o|---|

The truth tables for (AB)' is:
(AB)'
FF=T,
FT=T,
TF=T,
TT=F.

Where as the truth table for A'B' is
A'B'
FF=T,
FT=F,
TF=F,
TT=F.


TITLE: Tensor product of Spinorial Left/Right Lorentz Representation
QUESTION [1 upvotes]: I searched in various textbook (Zee Group and QFT, Weinberg 1-3, Cornwell, Maggiore) and all the similar question on this site but I didn’t find a final answer.
My question is simple:

What is the decomposition of the tensor product of two Spinorial Representation $(a,b) \otimes (c,d)$ of the Lorentz Group?

I know that $(a,b) \otimes (c,d)= (a\otimes c, b \otimes d)$, and I also know how to compose angular momentum, so I am trying to find how to express $(a\otimes c, b \otimes d)$ as a direct sum of different terms.
Since we have 2 tensor product, I would naively say that:
$$ (a\otimes c, b \otimes d) = (a+c,b+d) \oplus (a+c-1,b+d) \oplus \dots \oplus (|a-c|,b+d) \oplus (a+c,b+d-1) \oplus \dots \oplus (|a-c|,b+d-1) \oplus (a+c,|b-d|) \oplus \dots \oplus (|a-c|, |b-d|) $$
But since
$$(\frac{1}{2},\frac{1}{2})\otimes \left((0,\frac{1}{2}) \oplus (\frac{1}{2},0)\right)=(1, \frac{1}{2} )\oplus (\frac{1}{2},0)\oplus( \frac{1}{2},1) \oplus(0, \frac{1}{2})$$
My formula seems wrong.
So I am asking which is the right answer?

REPLY [0 votes]: Apparently the formula is right, and the result also coincide with it, I just did an error thinking the result was different from the one from the formula.


TITLE: Are (functions of) sets naturally continuous in some way which real numbers aren't?
QUESTION [0 upvotes]: I am reading some introductory notes on probability theory, and I came across the following claim here (page 7/8), :
If some sequence of sets $A_n$ has the limit $A$, then
$\lim_{n \rightarrow \infty} \mu(A_n) = \mu(A)$ for any measure $\mu$.
This really surprised me because it looks just like the definition for continuous functions (of real numbers), and it's not immediately apparent why all measures should be "continuous". The fact that it's possible to have a discontinuous cumulative distribution function makes this even more unintuitive.
It seems that one property of sets which makes this work is that the limit of a sequence of real numbers can lie outside the "range" of that sequence (for example, $a_n = 1/n$ has limit $L=0$), whereas if i take $A_n$ to be the interval $(a_n, \infty)$ or  $[a_n, \infty)$, I get $(L, \infty)$ as the limit, which doesn't contain the limit ($0$) of $a_n$. Otherwise, one could set $\mu(\{0\}) = \infty$ and cause a discontinuity.
My question is 1). is there some obvious intuition I'm missing with respect to how unintuitive this seems? 2). Are sets/real numbers a special case, or do other objects also naturally "cause" functions to be continuous?

REPLY [2 votes]: The measure of a set captures the abstract notion of "how big a set is". Measure theory deals with the construction of measures and the proofs of their properties. So the precise answer to your question can only be given once that background is properly established.
In the absence of that, I'll try to convey some of the intuition behind these concepts. Measures are closely related to integrals; If you have a definition of the integral, the measure of the set is just the integral of the indicator function of the set:
$$ \mu( (-1,3)) = \int_{-1}^3 1 \,dx = 4.$$
Conversely, you can define the integral if you start from a measure.
From that point of view, it shouldn't be surprising that $\mu( (0,1) ) = \mu([0,1))$. Indeed, when you write the familiar (Riemann) integral, you don't differentiate between them! This is because a single point is too "thin" with respect to the line and the area under a point - a line segment - is too "thin" with respect to area under a curve.
Setting $\mu(\{0\}) =\infty$ can be done, but it needs a lot of extra care. Doing this leads to the definition of the Dirac delta function, which isn't really a function, but a measure! Besides this, giving any other "finite value" to $\mu(\{0\})$ doesn't change the integral.
You can think of a function $f(x):[0,1]\to \mathbb{R}$ and a function $g$ with $f(x)=g(x)$ for $0\leq x <1$. Then
$$\int_0^1 f(x)dx= \int_0^1 g(x) dx,$$
as we learn in calculus, even though $f(1)\neq g(1)$. For that reason, when dealing with the "usual" measure in $\mathbb{R}$ (called the Lebesgue measure) we have $\mu(\{p\}) =0$ for every single point $p\in\mathbb{R}$. You can think of that as
$$\int_p^p f(x) dx = 0,$$
for every function $f$. That way $\mu ( [0,1]) = \mu ([0,1)) + \mu(\{1\}) = \mu ([0,1))$.
Finally, to answer your question, the property that you found unintuitive is called the continuity of a measure, and it roughly says that if you have (any) true measure and you are measuring a set, if you measure things that approximate the set, then you get the measure of the set.
Of course "approximating the set" needs to be done in a "nice" way, that's why the sequence of sets that you take need to be "increasing" or "decreasing". See the wiki article for details.


TITLE: 5UDH Tandem electrostatic accelerator
QUESTION [0 upvotes]: What does UDH stands for in the name of an electrostatic accelerator 5UDH-2 Tandem accelerator? Does it is related to its working or just a trade mark?

REPLY [0 votes]: That is a naming scheme of National Electrostatic Corp. (NEC), and is likely trademarked. The number has to do indirectly with the maximum terminal potential. Their website (pelletron.com) has more info. Not employed by NEC, long time user of their machines (and many others).


TITLE: For which values of the parameter $ a \in \mathbb{R}$ is the transformation $T(x,y,z)=(x+ay,x-ay,x+z)$ an isometry?
QUESTION [0 upvotes]: Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal.

$$T(x,y,z)=\begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ \end{pmatrix}$$
\begin{equation*}
\begin{split}
det(T(x,y,z)-I_n \lambda ) & =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =(1-\lambda)\begin{vmatrix} -1-\lambda & 0 \\ 0 & 1-\lambda \\ \end{vmatrix}-\begin{vmatrix} 1 & 0 \\ 1 & 1-\lambda \\ \end{vmatrix} \\
& =(1-\lambda)^2(-1-\lambda)-(1-\lambda) \\
& =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =-(1-\lambda)[(1-\lambda)(1+\lambda)+1] \\
& =-(1-\lambda)[1-\lambda^2+1] \\
& =(\lambda-1)[2-\lambda^2]=0 \\
\end{split}
\end{equation*}
Hence $\lambda=1, \pm \sqrt{2}$.
$\lambda=1$: $\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\  0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1,0)^T, (0,0,1)^T \}$.
$\lambda=\sqrt{2}$: $\begin{pmatrix} 1-\sqrt{2} & 1 & 0 \\ 1 & -1-\sqrt{2} & 0 \\ 1 & 0 & 1-\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1-\sqrt{2} \\  0 & 1-\sqrt{2} & 1-\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,\sqrt{2}-1,0)^T, (1-\sqrt{2},1+\sqrt{2},0)^T \}$.
$\lambda=-\sqrt{2}$: $\begin{pmatrix} 1+\sqrt{2} & 1 & 0 \\ 1 & -1+\sqrt{2} & 0 \\ 1 & 0 & 1+\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1+\sqrt{2} \\  0 & -1+\sqrt{2} & 1+\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1-\sqrt{2},0)^T, (1+\sqrt{2},1-\sqrt{2},0)^T \}$.

For which values of the parameter $ a \in \mathbb{R}$ is the transformation $T(x,y,z)=(x+ay,x-ay,x+z)$ an isometry?

$$T(x,y,z)=\begin{pmatrix} 1 & a & 0 \\ 1 & -a & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ \end{pmatrix}$$
How do I find the values? I know the definition of isometry is $\|Tx\|=\|x\|$.

REPLY [1 votes]: Just evaluate $Tx$ and expand out the norms in the equation $\| Tx \| = \|x \|$ (and square to make it nicer); you need precisely that for all $x,y,z$,
$$(x+ay)^2 + (x-ay)^2 + (x+z)^2 = x^2 + y^2 + z^2.$$
Also note that part (a) implies that $a=1$ can't be a correct answer, as the eigenvalues of an isometry have modulus one.


TITLE: Rings with the same additive law
QUESTION [1 upvotes]: Are there rings, on the same set, which have also the same additive law, but not the same multiplicative law ?
If the answer is yes, are there rings of cardinal $p$ (a prime) other than $\mathbb{Z}/p\mathbb{Z}$ ?
[Update] Here is an interesting link : Rings of Prime Cardinality

REPLY [3 votes]: The case of prime order $p$ having been already discussed in the answer by Sarango and the comments by OP and joriki, note that $\operatorname{GF}(p^{2})$ and $\mathbf{Z}_{p} \times \mathbf{Z}_{p}$ are unital rings with isomorphic additive structure, but one is a field, while the other is not an integral domain.


TITLE: Definition of graph, using half-edges.
QUESTION [2 upvotes]: There are several valid definitions of a (multi)graph. Let me quote the one I use (ref.1 §1.1):
Definition: A graph is a triple $(V,E,\psi)$ where $V$ is the vertex set, $E$ the edge set, and $\psi\colon E\to V^2$ a relation that associates with each edge an unordered pair consisting of two vertices (not necessarily distinct).
An alternative definition is that a graph is a set of half-edges (e.g., ref.2, §5.1):
Definition: A graph is a set of half-edges along with a set $V$ of disjoint subsets of half edges known as vertices which partition the set of half edges, and a set $E$ of disjoint pairs of half edges known as internal edges.
How is this latter definition understood in the context of the former? What is the precise definition of a half-edge, when we regard a graph as a triple $(V,E,\psi)$?
References.

Bondy, Murty - Graph Theory with Applications.
Yeats - A Combinatorial Perspective on Quantum Field Theory.

REPLY [4 votes]: Starting from the triple $(V,E,\psi)$, or from basically any other definition of graphs, we have a half-edge $h_{v,e}$ for every pair $(v,e) \in V \times E$ such that $v$ is incident to $e$. (In the formal definition with $\psi$, a pair $(v,e)$ such that $v \in \psi(e)$.)
If $e$ is a loop, we should probably be more careful and have two half-edges $h_{v,e}^1$ and $h_{v,e}^2$, because we still want the edge to correspond to a pair of half-edges.
Then a vertex $v$ in the original definition corresponds to the set of half-edges with vertex $v$, and an edge $e$ in the original definition corresponds to the set of (exactly two) half-edges with edge $e$.


TITLE: In a thermal equilibrium, why is the energy of individual photons proportional to the temperature?
QUESTION [0 upvotes]: In the book of The First Three Minutes, by Weinberg, he argues that

(p.65-66)when radiation and matter were in thermal equilibrium, the universe
  must have been filled with black-body radiation with a temperature
  equal to that of the material contents of the universe.

$$$$

(p.68)The photon picture allows us easily to understand the chief
  qualitative properties of black-body radiation. First, the principles
  of statistical mechanics tell us that the typical photon energy is
  proportional to the temperature, while Einstein's rule tells us that
  any photon's wavelength is inversely proportional to the photon
  energy. Hence, putting these two rules together, the typical
  wavelength of photons in black-body radiation is inversely
  proportional to the temperature.

However, I do not understand how the conclusion is arrived at, that typical photons (i.e photons emitted with maximum intensity at given temperature) have an energy that is proportional to the temperature ? Is there any explanation that shows that $E \propto T$ for a photon in thermal equilibrium ?
Note that, I have seen this question, but that does not contain any answer to my question.
Please also do note that, the book is making arguments using the historical development of the subject, it is mostly based on the experimental facts, so please do not throw any quantum theoretical formulas at us.

REPLY [2 votes]: If the radiation and matter are in complete thermodynamic equilibrium, then the radiation has a blackbody spectrum.
The frequency at which the flux density (in frequency space) is maximised is given by Wien's law:
$$ f_{\rm max} = \frac{2.82}{h} k_{B}T$$
Thus a photon at the frequency at which the radiation spectrum peaks has an energy $h f_{\rm max} = 2.82 k_B T$.
You can do some similar analysis to come up with the average energy of a photon, and it comes to $2.70 k_B T$.
If you are asking for an explanation that does not involve Wien's law, then you have to go back a step and find out why the blackbody spectrum has the form it does and why Wien's law results from that. This is well-trodden ground covered in any textbook on the subject.


TITLE: Limit of a non monotone sequence of subsets of a set
QUESTION [0 upvotes]: In book " Real analysis Theory of measure and integration " . There's a definition of limit of a monotone sequence of subsets of a set X . Then it gives the definition of limit of a sequence of subsets of a set X using limit inferior and limit superior . Where $\lim_{n\to\infty} inf A_n= \bigcup_{n\in \mathbb{N} } \bigcap_{k\geqslant n} A_k $
I'm trying to evaluate the limit of this sequence .
$A_{2n-1}=[0,\frac{1}{2n-1} ]$ ,,
$A_{2n}=[0,2n]$.
I expanded some terms , i found that $\lim_{n\to\infty} inf A_n=\{0\}$ and $\lim_{n\to\infty} sup A_n =[0,\infty]$ .
The problem is how can i mathematically write this ?

REPLY [0 votes]: Your notation isn’t quite right: the limit inferior is the set $\{0\}$, not the number $0$, and the limit superior doesn’t include $\infty$: it’s just $[0,\to)$ (or, in the notation that you’re using, $[0,\infty)$).
You have
$$A_n=\begin{cases}
[0,n],&\text{if }n\text{ is even}\\
\left[0,\frac1n\right],&\text{if }n\text{ is odd}\;.
\end{cases}$$
Notice that
$$\ldots\subseteq A_5\subseteq A_3\subseteq A_1\subseteq A_2\subseteq A_4\subseteq A_6\subseteq\ldots\;,$$
This helps to give a picture of what’s going on.
Let $n$ be any positive integer. Then certainly $2k\ge n$ when $k\ge n$, so
$$[0,2m]=A_{2k}\subseteq\bigcup_{k\ge n}A_k$$
for each $m\ge n$, and therefore
$$[0,\to)=\bigcup_{k\ge n}[0,2k]\subseteq\bigcup_{k\ge n}A_k\subseteq[0,\to)\;,$$
where the last inclusion follows from the fact that $A_k\subseteq[0,\to)$ for each $k\ge 1$. Thus,
$$\bigcup_{k\ge n}A_k=[0,\to)$$
for each $n\ge 1$, and therefore
$$\limsup_{n\to\infty}A_n=\bigcap_{n\ge 1}\bigcup_{k\ge n}A_k=\bigcap_{n\ge 1}[0,\to)=[0,\to)\;,$$
the set of non-negative real numbers.
Once again we’ll let $n$ be any positive integer, but this time we’ll use the fact that $2k+1\ge n$ when $k\ge n$. This means that
$$\bigcap_{k\ge n}A_k\subseteq A_{2m+1}=\left[0,\frac1{2m+1}\right]$$
for each $m\ge n$ and hence that
$$\bigcap_{k\ge n}A_k\subseteq\bigcap_{k\ge n}\left[0,\frac1{2k+1}\right]=\{0\}\;.$$
Since $0\in A_k$ for all $k\ge 1$, this implies that
$$\bigcap_{k\ge n}A_k=\{0\}$$
for each $n\ge 1$ and hence that
$$\liminf_{n\to\infty}A_n=\bigcup_{n\ge 1}\bigcap_{k\ge n}A_k=\bigcup_{n\ge 1}\{0\}=\{0\}\;.$$