TITLE: Given the result of a dot product and one of the vectors, is it possible to find the other vector?
QUESTION [0 upvotes]: When processing geometry the dot product or $n\cdot L$ (where $n$ is the normal, $L$ is the light direction) is baked into the $w$ of a vertex, but since the $L$ is constant I wanted to know if there is any way to get $n$ without changing the backend processing?

REPLY [1 votes]: Given a vector $\vec v$ and scalar $a$, there is no unique solution to the equation
$$\vec v\cdot\vec w=a.$$
Consider the case $\vec v=\vec k$, and $a=0$.  Then any vector of the form
$x\vec i+y\vec j$ is a solution to
$$\vec v\cdot \vec w=0.$$

REPLY [1 votes]: No, not even in the two dimensional case. In the 2-D case, however, you can narrow it down to one of two directions which are obtained from a quadratic equation:
$$
\vec{n}\cdot \vec{L} = n_x L_x + n_yL_y = k \implies n_y = \frac{k-n_xL_x}{L_y}
\\n_x^2 + n_y^2 = 1\implies n_x^2 + \left( \frac{k-n_xL_x}{L_y} \right)^2 = 1
$$
which is a quadratic equation giving two possible solutions of $n_x$.
In the 3-D case a whole circle of answers is possible.


TITLE: Translating from exponential to polar representation
QUESTION [0 upvotes]: I am trying to translate the following equation in polar coordinates:
$$ \left|\frac{e^{2jw} - 0.5}{e^{jw}(e^{jw}-0.5)} \right|^2$$
The result should be the following:
$$\frac{1.25 - cos(2\omega)}{1.25+cos(\omega)}$$
Any help on how to do it? I managed to obtain it using $e^{jw} =cos(\omega) -jsen(\omega)$. I am sure there is a faster and easier way.

REPLY [0 votes]: First, for any $\alpha \in \mathbb{R}$:
$$
\begin{align}
\left|2 e^{j \alpha} - 1\right|^2 = \left(2 e^{j \alpha} - 1\right) \overline{\left(2 e^{j \alpha} - 1\right)} = \left(2 e^{j \alpha} - 1\right) \left(2 e^{-j \alpha} - 1\right) = 5 - 4 \operatorname{Re}\left(e^{j\alpha}\right) = 5 - 4 \cos \alpha
\end{align}
$$
Then, assuming $w \in \mathbb{R}$:
$$\left|\frac{e^{2jw} - 0.5}{e^{jw}(e^{jw}-0.5)} \right|^2=\left|\frac{2e^{2jw} - 1}{e^{jw}(2 e^{jw}-1)} \right|^2=\frac{\left|2e^{2jw} - 1\right|^2}{\left|e^{jw}\right|^2\cdot\left|2 e^{jw}-1\right|^2} = \frac{5 - 4 \cos 2w}{1 \cdot \left(5 - 4 \cos w\right)}=\cdots$$


TITLE: Dirac spinors in 2+1 dimensions
QUESTION [1 upvotes]: In 3+1 dimensions, Dirac spinors have four complex components. In 2+1 dimensions, the representation of the Clifford algebra by $\sigma^3$ and $-i\sigma^3\sigma^i$, with $i\in\{1,2\}$ is 2-dimensional, so the dimensional "Dirac spinors" have two complex components. Is there a physical reasons why in 3+1 dimensions there are 4 components, while in one dimension lower there are only two?
I think mathematically one might simply say that the Clifford algebra is always even-dimensional, but I would like to know if there is also some physics behind this.

REPLY [0 votes]: As explained by @Nogueira, the irreducible representations of the Dirac matrices have different dimensions depending on whether d=2n or d=2n-1.
So in d=3+1 dimensions the massive particle Dirac spinor is a 4-component object. It allows for the Dirac equation to be invariant under parity ${\bf r} \to -{\bf r}$.
The massive particle Dirac Spinor in d=2+1 dimensions is a two-component object, but there are two inequivalent irreducible representations!  So if you pile them up (see below) you have a 4 component spinor.
Why might you want to pile them up? Well parity in 2 space dimensions is defined differently than in 3 space because ${\bf r} \to -{\bf r}$ would simply be a rotation in the plane. So define the parity operation in 2 space by flipping only one coordinate: $ x \to -x$ keeping $y \to y$. Then you will find that your 2-component Dirac equation is not invariant under Parity.
Neither of the two different irreducible representations in d=2+1 is invariant under parity but if you combine them (into a  reducible representation) you get a 4 component parity invariant structure.
But it turns out the parity violating 2-component structure is useful and has interesting physics and relations to other structures in the plane. See eg.
https://www.sciencedirect.com/science/article/pii/0003491682901646?via%3Dihub


TITLE: Suppose that $X_i$ are independent random variables, with finite absolute moment. Then $Max(X_1, \ldots, X_n) / n \to 0$ a.s.?
QUESTION [4 upvotes]: Suppose that $X_i$ are i.i.d. random variables, with finite absolute moment: $E|X_1| < \infty$. Then $\max(X_1, \ldots, X_n) / n \to 0$ a.s. ?
Let $M_n = \max(X_1, \ldots, X_n)$.
I know that $M_n$ has a CDF which is $F_n(x) = F_1(x)^n$.
I would like to apply Borel-Cantelli to the sets $A_n = \{ M_n > n \epsilon \}$. So I would like to show that $\Sigma P(A_n) < \infty$.
I suspect that one can obtain a bound that is something like $\Sigma P(A_n) \leq E|X_1|$. However, because one doesn't have $\Sigma 1_{A_n} \leq X_1$, this is unlikely to be literally true. At some point one will need to use the independence.
So I was trying to proceed by analyzing the product $\Pi ( 1 - P(A_n))$ instead. We would like this to converge to a nonzero limit. This means that we want $P( \cap A_n^c) > 0$, at least for small epsilon, because of the assumed independence. But the problem is that $P(A_1) = 0$ is already possible.
I would appreciate a hint to set me on the right track. :)

REPLY [0 votes]: WLOG $X_1,X_2,\dots$ are nonnegative.
Let $M_n =\max(X_1,\dots,X_n)$ and $S_n = X_1+\dots+X_n$ and $S_0=0$.
Now $\frac{X_n}{n} = \frac{S_{n}-S_{n-1}}{n} = \frac{S_n}{n} - \frac{n-1}{n} \frac{S_{n-1}}{n-1}$.
By the Law of Large Numbers, the RHS tends to zero a.s., and so does the LHS, including along all subsequences.
For each $n$, let $k_n\in \{1,\dots,n\}$ be such that $X_{n_k} = \max(X_1,\dots,X_n)$. Then
$$0\le \frac{M_n}{n} = \frac{X_{k_n}}{n} \le \frac{X_{k_n}}{k_n} \frac{k_n}{n}.$$
The RHS tends to zero.