TITLE: The lattice of ideals of a distribituve lattice is itself distributive
QUESTION [0 upvotes]: I have found myself stuck on a problem and would appreciate a hint.  The problem is to show that the lattice of ideals $I(L)$ of a distributive lattice $L$ is itself distributive.  This is question 2 in section 3 of the first chapter in the universal algebra book by Burris and Sankappanavar.

REPLY [2 votes]: Hint: to show that
$$(I \vee J) \wedge K = (I \wedge K) \vee (J \wedge K),$$
where $I, J$ and $K$ are ideals of the lattice $L$, notice that meet is given by intersection, and
$$I \vee J = \{ a \in L : \exists i\in I\;\exists j \in J\;(a \leq i \vee j) \}.$$
Additionally, you might find useful the fact that if $k \leq i \vee j$ then $k = k \wedge (i \vee j)$, and use distributivity.
I don't think that I can give you any more hints without spoiling it. Good luck!


TITLE: Can we find a function with describing condition?
QUESTION [0 upvotes]: can we find a function with the following condition?
$g^{\prime}(1)=0$
$g^{\prime}(2)=0$
$g^{\prime}(3)=0$
$g^{\prime}(4)=0$
$g(1)=69$
$g(2)=34$
$g(3)=35$
$g(4)=0$

REPLY [1 votes]: As said in comments, since you have $8$ conditions, a polynomial of degree $7$ could fit all the requirements.
So, let
$$g(x)=\sum_{i=0}^7 a_i \,x^i \qquad \text{and} \qquad g'(x)=\sum_{i=i}^7 i\,a_i\, x^{i-1}$$ and write all equations. You have a system of $8$ linear equations for $8$ unknown coefficients. Write it in matrix form and solve.
If I am not mistaken, you should get
$$a_0= -1928\qquad a_1= \frac{21380}{3}\qquad a_2=
   -\frac{184865}{18}\qquad a_3= \frac{140357}{18}$$ $$a_4= -\frac{30700}{9}\qquad a_5=
   \frac{15587}{18}\qquad a_6= -\frac{2135}{18}\qquad a_7= \frac{61}{9}$$ which makes
$$g(x)=\frac{1}{18} (x-4)^2\left(122 x^5-1159 x^4+4363 x^3-7952 x^2+6933 x-2169\right)$$ Plot it to see how it works.
But other functions could also make the job.


TITLE: Is tensor product of local algebras local?
QUESTION [12 upvotes]: In general, the tensor product of two local rings is not local. For example, $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}\ $ is not a local ring.
Let $\mathbb{F}_{p}$ denote the finite field with $p$ elements. Let $A,B$ be two complete local noetherian $\mathbb{Z}_p$-algebras with residue field $\mathbb{F}_p$. Let $m_A, m_B$ denote the maximal ideals of $A,B$, respectively.
Question:
Is it true that $A \otimes_{\mathbb{Z}_p} B\ $ is a local ring?
Clearly, the ideal $m_A \otimes B + A \otimes m_B$ is a maximal ideal of $A \otimes_{\mathbb{Z}_p} B\ $ with the residue field $F_p$. Is it the only maximal ideal of $A \otimes_{\mathbb{Z}_p} B\ $?

REPLY [7 votes]: Let $A=\mathbb F_p[[t]],B=\mathbb F_p[[u]]$. Then, $1\otimes1-t\otimes u$ is neither in $\mathfrak m_A\otimes B+A\otimes\mathfrak m_B$ nor a unit, so it is contained in some other maximal ideal of $A\otimes B$. (Proof that $1\otimes1-t\otimes u$ is not a unit: An element of $\mathbb F_p[[t]][[u]]$ coming from $A\otimes B$ has the property that its coefficients with respect to $u$ span a finite-dimensional subspace of $\mathbb F_p[[t]]$, but this fails for the coefficients $t^k$ of $(1-tu)^{-1}=\sum t^ku^k$.)


TITLE: Is there an integer z such that $255z\equiv 7\pmod {633}$?
QUESTION [1 upvotes]: I used the extended euclidean algorithm to
"Find integers x and y such that $633x + 255y = 6$, or explain why none exist."
And found that $6x = -58$ and $y = 144$. Now I'm stuck on the follow up question in the title. How do I proceed to answer a question like this?
Thanks

REPLY [4 votes]: Note that $255=3\times85$ and that $633=3\times211$. As such $255z\pmod{633}$ will always be a multiple of $3$. As $7$ is not a multiple of $3$ then there is no solution.


TITLE: Differential cross-section for a 2-particle process in the LAB frame
QUESTION [3 upvotes]: This should really be a straightforward calculation, but somehow, I keep confusing myself and failing over and over again. I did the calculation so many times that I don't even know what I'm looking at anymore. Some help would be greatly appreciated.
Let me show you what I tried to do, so you can check my understanding (or lack thereof).
(1 + 2 $\rightarrow$ 1' + 2') elastic scattering in the LAB frame
This means:
$$ m_1 = m_1'$$
$$m_2 = m_2'$$
$$\mathbf{p_2} = \mathbf{ 0}$$
As you can see, we choose to work in the rest frame of particle 2 before the scattering.
I managed to calculate the differential cross section in the CM (center of mass) frame, which is (in natural units):
$$\left( \frac{d\sigma}{d \Omega }\right)_{CM} = \frac{|\mathcal{M}|^2}{(8 \pi)^2 E_{CM}^2}$$ Here, $\mathcal{M}$ is the invariant amplitude and $E_{CM}=E_1 + E_2 = E_1' + E_2'$.
Following the technique used in the CM frame, I use these known relations:
$$ d\sigma = \frac{V}{T} \frac{dP}{|\mathbf{v_1} - \mathbf{v_2}|}$$
$$dP = \frac{|\left< f | S|i\right>|^2}{\left< f | f\right> \left<i | i\right>} d\Pi$$
$$\implies d\sigma = \frac{|\mathcal{M}|^2}{(2E_1)(2E_2)|\mathbf{v_1} - \mathbf{v_2}|}d\Pi_{LIPS}$$
This leads me to
$$d\sigma = \frac{|\mathcal{M}|^2}{4 E_1 E_2|\mathbf{v_1} - \mathbf{v_2}|} (2\pi)^4 \delta^4(p_1' + p_2' - p_1 - p_2)  \frac{d^3 \mathbf{p}_1'}{(2\pi)^3 2 E_1'}  \frac{d^3 \mathbf{p}_2'}{(2\pi)^3 2 E_2'} $$
Using $p_2 = m_2$ and integrating over the momenta for particle 2,
$$d\sigma = \frac{|\mathcal{M}|^2}{(8 \pi)^2 E_1 m_2|\mathbf{v_1} - \mathbf{v_2}|} \delta(E_1' + E_2' - E_1 - m_2)  \frac{d^3 \mathbf{p}_1'}{E_1' E_2'} $$
Next, I switch to spherical coordinates, $d^3 \mathbf{p}_1' = d\Omega \, p^2 dp$ and I factor out the solid angle differential. Finally, I plug in the integral for the remaining degree of freedom $(p)$, to get this:
$$\left( \frac{d\sigma}{d \Omega }\right)_{CM} = \frac{1}{(8 \pi)^2 E_1 m_2|\mathbf{v_1} - \mathbf{v_2}|} \int_0^\infty \frac{dp \,  p^2 }{E_1' E_2'} |\mathcal{M}|^2 \delta(E_1' + E_2' - E_1 - m_2) $$
This is where I get confused... I tried defining a new variable $x=E_1' + E_2' - E_1 - m_2$ and substituting it in the integral, but everything always gets messy because both $E_1'$ and $E_2'$ depend on $p=|\mathbf{p}_1'|$:
$$E_1' = \sqrt{m_1^2 + p^2}$$
$$E_2' = \sqrt{m_2^2 + (\mathbf{p}_1 - \mathbf{p}_1')^2}$$
Messy, messy, messy... so I tried to do it another way:
I recast the CM formula in a Lorentz-covariant form (for known $\mathbf{p}_1$) using Mandelstam variables, i.e.
$$ s = (p_1 + p_2)^2 = (p_1' + p_2')^2$$
$$ t = (p_1 - p_1')^2 = (p_2 - p_2')^2$$
$$ dt = 2 | \mathbf{p}_1| \, |\mathbf{p}_1'| \frac{d \Omega_{CM}}{2 \pi}$$
$$\implies \frac{d \sigma}{dt} = \frac{|\mathcal{M}|^2}{64 \pi s |\mathbf{p_1}|^2}$$
Next, I write $s$ and $t$ in the LAB frame in hopes of getting the correct formula, but to no avail... I get something extremely complicated or something simply wrong...
THIS is what I should get at the end, but I can't seem to get it:
$$\left( \frac{d\sigma}{d \Omega }\right)_{LAB} = \frac{|\mathcal{M}|^2}{(8 \pi)^2} \frac{|\mathbf{p}_1'|^2}{|\mathbf{p}_1|} \frac{1}{m_2 (E_1 + m_2) \big| |\mathbf{p}_1'| - |\mathbf{p}_1| E_1' \cos{\theta}  \big|}$$
So, what do you think?

REPLY [6 votes]: I consider the scattering process $A+B \to 1 + 2$. The differential cross-section is always given by
\begin{equation}
\begin{split}\label{eq1}
d\sigma &= \frac{1}{(2E_A)(2E_B)|v_A - v_B|} \frac{d^3p_1}{(2\pi)^3}  \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2)
\end{split}
\end{equation}
The take $B$ to be at rest. Thus $p_B^\mu = (m_B,0,0,0)$ and $p_A^\mu = (E_A, \vec{p}_A)$. Plugging this into the equation above, we get
\begin{equation}
\begin{split}
d\sigma &= \frac{1}{(2p_A)(2m_B)} \frac{d^3p_1}{(2\pi)^3}  \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2)
\end{split}
\end{equation}
where we have used $|\vec{v}_A| = \frac{p_A}{E_A}$. We can integrate over $\vec{p}_2$ to get
\begin{equation}
\begin{split}
d\sigma &= \frac{1}{(2p_A)(2m_B)} \frac{d^3p_1}{(2\pi)^2}  \frac{1}{2E_1} \left| {\cal M} \right|^2   \frac{1}{2E_2}  \delta(E_A + m_B - E_1 - E_2) \\
\frac{d\sigma}{d\Omega} &= \frac{1}{(2\pi)^2(2p_A)(2m_B)} dp_1  p_1^2 \frac{1}{2E_1} \left| {\cal M} \right|^2   \frac{1}{2E_2}  \delta(E_A + m_B - E_1 - E_2)
\end{split}
\end{equation}
where $E_2^2 = (\vec{p}_1 - \vec{p}_A)^2 + m_2^2 = p_1^2 + p_A^2  - 2 p_1 p_A \cos\theta  + m_2^2 $ and $E_1^2 = p_1^2 + m_1^2$. To integrate over $p_1$, we see that the $\delta$-function becomes
\begin{equation}
\begin{split}
 \delta & \left(E_A + m_B - \sqrt{p_1^2  + m_1^2} - \sqrt{p_1^2   - 2 p_1 p_A \cos\theta + p_A^2  + m_2^2 }\right) \\
 &= \delta (p_1 - p_f) \left(\frac{p_f}{E_1}  + \frac{p_f - p_A\cos\theta}{E_2} \right)^{-1}
\end{split}
\end{equation}
where $p_1 = p_f$ is the solution to the argument in the delta function. Integrating over $p_1$, we get
\begin{equation}
\begin{split}
\boxed { \frac{d\sigma}{d\Omega} = \frac{1 }{64\pi^2m_B}   \left[ E_2   + E_1\left(1 - \frac{p_i}{p_f} \cos\theta\right)\right]^{-1} \frac{p_f}{p_i}\left| {\cal M} \right|^2 }
\end{split}
\end{equation}
where $E_1 = \sqrt{p_f^2 + m_1^2}$, $E_2 = \sqrt{(\vec{p}_f - \vec{p}_i)^2 + m_2^2}$


TITLE: Proof of a relation in Dirac's 1927 paper
QUESTION [0 upvotes]: I was reading Dirac's paper "The quantum theory of the emission and absorption of radiation".
I am confused regarding the following relation in the paper:
$
e^{{\pm i\theta_r \over h}} f(N_1',N_2', ... N_r', ... ) = e^{\mp \partial \over \partial N_r'}f(N_1', N_2',.. . N_r',... )
=f (N_1', N_2',..., N_r' \mp 1,... )$
I'm only confused regarding
$ e^{\mp \partial \over \partial N_r'}f(N_1', N_2',.. . N_r',... )
=f (N_1', N_2',..., N_r' \mp 1,... )$
pg 10 in his paper
https://royalsocietypublishing.org/doi/10.1098/rspa.1927.0039
Is there a proof for this statement?
Dirac hasn't given a proof, is it proved in any earlier paper? Is it a definition?

REPLY [3 votes]: This is an application of a mathematical theorem (which is not nearly as bad as it looks).
Let's start of with a function of one variable, $f(x)$, for simplicity.
First, we define the derivative operator as
\begin{equation}
D \equiv \frac{d}{dx}
\end{equation}
Then I claim that $D$ obeys the following identity for any (sufficiently well-behaved) function $f$ and constant $\delta$:
\begin{equation}
e^{\delta D} f(x) = f(x+\delta)
\end{equation}
This can be proven by expanding the exponential as a Taylor series, and realizing that the result is the Taylor series expansion of $f(x+\delta)$:
\begin{eqnarray}
e^{\delta D}f(x)  &=& \left[\sum_{n=0}^\infty \frac{\delta^n D^n}{n!} \right] f(x)  \\
&=& \left[\sum_{n=0}^\infty \frac{\delta^n}{n!} \left(\frac{d}{dx}\right)^n \right] f(x) \\
&=& \sum_{n=0}^\infty \frac{\delta^n f^{(n)}(x)}{n!} \\
&=& f(x+\delta)
\end{eqnarray}
As a special case of this identity, note that
\begin{equation}
e^{\pm D}f(x) = f(x\pm 1)
\end{equation}
Now we turn to the multivariate case. The generalization is straightforward. Defining the partial derivative operator
\begin{equation}
\mathcal{D}_r \equiv \frac{\partial }{\partial N_r}
\end{equation}
we have that
\begin{equation}
e^{\delta \mathcal{D}_r} f(N_1, N_2, ..., N_k) = f(N_1,N_2, ..., N_r+\delta,...,N_k)
\end{equation}
Or, as a special case,
\begin{equation}
e^{\pm \mathcal{D}_r} f(N_1, N_2, ..., N_k) = f(N_1,N_2, ..., N_r\pm 1,...,N_k)
\end{equation}
which is the identity you wanted.


TITLE: Volume of the bounded by the region $z=9-x^2-y^2$ and the plane $z=0$
QUESTION [0 upvotes]: I have to find the volume bounded by the equations $z=9-x^2-y^2$ and $z=0$, but I do not know how it is supposed to be done correctly.
I have right now:
$\int_0^{2\pi}\int_0^3 (9-r^2)rdrd\phi = \frac{243}{2}\pi$
But I do not know if this is correct or if it is supposed to be done in this way.
Thanks for the help.

REPLY [0 votes]: In cylindrical coordinates, the computation of that volume becomes$$\int_0^{2\pi}\int_0^9\int_0^{\sqrt{9-z}}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta=\frac{81\pi}2.$$


TITLE: Relation between surface integral and lebesgue integral
QUESTION [1 upvotes]: Let $f :\mathbb{R}^{N}\to\mathbb{R}$ be a smooth function and $B_{R}$ be the $N$-dimensional ball. I consider
\begin{align*}\tag{1.1}\int_{\partial B_{R}}f(x)dS\end{align*}
as the surface integral of $f$ on the surface of $B_{R}$. However, I can also define
\begin{align*}
\chi_{\partial_{B_{R}}}:=
\begin{cases}
1 \quad&\text{ if }x\in\partial B_{R}\\
0 \quad&\text{ if }x\not\in\partial B_{R}
\end{cases}
\end{align*}
Then, I consider
\begin{align*}\tag{1.2}\int_{\mathbb{R}^{N}}f(x)\chi_{\partial B_{R}}dx \end{align*}
So, my question is to consider whether (1.1) and (1.2) are the same integral or not. Specifically, I would like to know also whether
$\int_{\mathbb{R}^{N}}f(x)\chi_{\partial B_{R}}dx$ and $\int_{\mathbb{R}^{N}}f(x)\chi_{\partial B_{R}}dS$ can be considered the same integral or not in this case.
Any help is greatly appreciated! Thank you

REPLY [1 votes]: They are not. For the surface $\partial B_{R}$, it is of measure zero by taking the Lebesgue measure, so
\begin{align*}
\int_{\mathbb{R}^{n}}f(x)\chi_{\partial B_{R}}(x)dx=0.
\end{align*}