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# Fluid dynamics question, related to the derivation of Bernoulli's Theorem for steady flow

I'm finding it very challenging to understand a step in the derivation of the 'moment equation', found in these notes

http://www.maths.ox.ac.uk/system/files/coursematerial/2011/989/66/fluids1.pdf

I am struggling with the transition of (1.33) to (1.34). I don't see how:

$$\frac{d}{dt}\int_V \rho \mathbf{u} dV=\int_V \rho \frac{D\mathbf{u} }{Dt}dV$$

This is a result of the transport theorem (1.22) and it's corollary (1.32), but I don't really see how the corollary is derived or how it can be applied to vector functions. If anyone could explain as you would a child I would be very grateful..

EDIT: So this really comes down just the derivation of the corollary: For $f=\rho h$

$$\frac{d}{dt}\int_V \rho h dV=\int_V \rho \frac{Dh }{Dt}dV$$

So this depends on the conservation of mass $$\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{u})=0$$ Therefore from the transport theorem $$\frac{d}{dt}\int_V \rho h dV=\int_V \frac{\partial \rho h}{\partial t} +\nabla \cdot(\rho h \mathbf{u}) dV=\int_V h\frac{\partial \rho }{\partial t} +\rho\frac{\partial h }{\partial t} + \rho h(\nabla\cdot\mathbf{u})+ (\nabla \rho h)\cdot \mathbf{u} dV$$

So all we need to do is show that: $$h\frac{\partial \rho }{\partial t} +\rho\frac{\partial h }{\partial t} + \rho h(\nabla\cdot\mathbf{u})+ (\nabla \rho h)\cdot \mathbf{u} =\star\big[\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{u})\big] +\rho \big[ \frac{\partial h}{\partial t} + (\mathbf{u} \cdot\nabla)h\big]$$

Can anyone help with this step? Many thanks.

-

Your problem stems from the fact that also the volume depends on t. This is similar to the one dimensional case but now you have the following equation to hold (eq.(1.22) in the lectures you cite):

$$\frac{\partial}{\partial t}\int_{V(t)}fdxdydz=\int_{V(t)}\left(\frac{\partial f}{\partial t}+\nabla\cdot(f\bf{u})\right)dxdydz.$$

The proof given in your text use the Jacobian and that is all you need.

In your case you have to prove that

$$\frac{d}{dt}\int_V \rho \mathbf{u} dV=\int_V \rho \frac{D\mathbf{u} }{Dt}dV$$

but the volume is dependent on time so you will write

$$\frac{d}{dt}\int_{V(t)} \rho \mathbf{u} dV=\int_{V(0)} \frac{D}{Dt}(J\rho \mathbf{u}) dV'$$

being $J$ the Jacobian. One has the identity

$$\frac{DJ}{Dt}=J\nabla\cdot\mathbf{u}$$

and so

$$\int_{V(0)} \left(\frac{D}{Dt}(\rho \mathbf{u})+\nabla\cdot\mathbf{u}\rho \mathbf{u}\right)J dV'.$$

Now

$$\int_{V(t)} \left(\frac{D}{Dt}(\rho \mathbf{u})+\nabla\cdot\mathbf{u}\rho \mathbf{u}\right)dV=\int_{V(t)} \left(\frac{D\rho}{Dt}\mathbf{u}+\rho\frac{\partial\mathbf{u}}{\partial t}+\rho\mathbf{u}\cdot\nabla\mathbf{u}+\nabla\cdot\mathbf{u}\rho \mathbf{u}\right)dV.\qquad (\dagger)$$

Using continuity equation

$$\frac{\partial\rho}{\partial t}+\nabla\cdot{(\rho\mathbf{u})}=0.$$

you will have

$$\frac{D\rho}{Dt}=\frac{\partial\rho}{\partial t}+\mathbf{u}\cdot\nabla\rho=\mathbf{u}\cdot\nabla\rho-\nabla\cdot(\rho\mathbf{u})=-\rho\nabla\cdot\mathbf{u}.$$

Putting this into eq.($\dagger$) you get your result.

-
Thanks for your response Jon. Indeed, so in my edit I tried to apply this theorem to $f=\rho h$, but I can't see how this implies that $$\frac{d}{dt}\int_V \rho h dV=\int_V \rho \frac{Dh }{Dt}dV$$ – Freeman Mar 19 '12 at 16:11
@LHS: I have extended the answer. Let me know if you need more. – Jon Mar 19 '12 at 22:55
Thanks so much for this! Could you explain this step: $$\frac{d}{dt}\int_V \rho \mathbf{u} dV= \int_V \frac{D}{Dt} (\rho \mathbf{u} )dV$$ I have not seen this identity before, do you know how it can be justified? Also i'm having a little trouble rearranging your last step into the required result.. but i'm pretty exhausted, I will try again in the morning. – Freeman Mar 19 '12 at 23:19
@LHS: Further improved the answer. I hope that now is more readable. – Jon Mar 20 '12 at 9:47
This is fantastic! thanks so much, this has definitely cleared up a few loose ends that I didn't understand when learning this course. I very much appreciate your help. – Freeman Mar 20 '12 at 12:51

format: plain


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# Publication [4.12] of Tomás Oliveira e Silva

Do not bookmark this page, because its URI may change in the future.
Instead, bookmark its parent page (http://sweet.ua.pt/tos/bib.html).

## Reference

Tomás Oliveira e Silva, Siegfried Herzog and Silvio Pardi, "Empirical verification of the even Goldbach conjecture and computation of prime gaps up to 4· 10^18," Mathematics of Computation, vol. 83, no. 288, pp. 2033-2060, Jul. 2014. Published electronically on November 18, 2013.

## Abstract

This paper describes how the even Goldbach conjecture was confirmed to be true for all even numbers not larger than 4· 10^18. Using a result of Ramaré and Saouter, it follows that the odd Goldbach conjecture is true up to 8.37· 10^26. The empirical data collected during this extensive verification effort, namely, counts and first occurrences of so-called minimal Goldbach partitions with a given smallest prime and of gaps between consecutive primes with a given even gap, are used to test several conjectured formulas related to prime numbers. In particular, the counts of minimal Goldbach partitions and of prime gaps are in excellent accord with the predictions made using the prime k-tuple conjecture of Hardy and Littlewood (with an error that appears to be O(\sqrt{t\log\log t}), where t is the true value of the quantity being estimated). Prime gap moments also show excellent agreement with a generalization of a conjecture made in 1982 by Heath-Brown.

## BibTeX entry

@Article
{
Oliveira.e.Silva-2013-3-EVGC,
author = {Oliveira e Silva, Tom{\'a}s} # { and } # {Herzog, Siegfried} #
{ and } # {Pardi, Silvio},
title = {{E}mpirical verification of the even {G}oldbach conjecture and
computation of prime gaps up to $4\cdot 10^{18}$},
journal = {Mathematics of Computation},
year = {2014},
volume = {83},
number = {288},
pages = {2033--2060},
month = Jul,
note = {Published electronically on November 18, 2013.}
}


text:
Conference paper Open Access

Anticipatory Admission Control and Resource Allocation for Media Streaming in Mobile Networks

Bui, Nicola; Malanchini, Ilaria; Widmer, Joerg

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The exponential growth of media streaming traffic will have a strong impact on the bandwidth consumption of the future wireless infrastructure. One key challenge is to deliverservices taking into account the stringent requirements of mobile video streaming, e.g., the users’ expected Quality-of-Service. Admission control and resource allocation can strongly benefit from the use of anticipatory information such as the prediction of future user’s demand and expected channel gain. In this paper, we use this information to formulate an optimal admission control scheme that maximizes the number of accepted users into the system with the constraint that not only the current but also the expected demand of all users must be satisfied. Together with the optimal set of accepted users, the optimal resource scheduling is derived. In order to have a solution that can be computed in a reasonable time, we propose a low complexity heuristic. Numerical results show the performance of the proposed scheme with respect to the state of the art.
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text:
# My first combinatorics post!

How many $$3$$-element subsets of the set $${1,2,3,\ldots,19,20}$$ are there such that the product of the three numbers in the subset is divisible by $$4$$?

×


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# Availabilty of magnesium ions in phosphate buffer

Phosphate buffer is a common buffer in biological applications, it's especially popular for NMR. Magnesium ions are necessary for many biological systems to work, though the amount of magnesium ions you can add in a phosphate buffer is limited by the relatively low solubility of magnesium phosphate.

What I'm curious about is whether the phosphate buffer can affect the availability of magnesium ions even in concentrations where no visible precipitation occurs? Common phosphate buffer concentrations are around 10-100 mM, and magnesium ions are usually 5 mM or below. At those concentrations I see no visible precipitation, but is the magnesium actually fully available or is the phosphate sequestering it and thereby lowering the effective magnesium ion concentration in the solution?

Even just calculating the solubility of magnesium salts in phosphate buffer is not straightforward, as there are multiple species of magnesium phosphate and possibly even other salts like $\ce{MgKPO4}$.

Does phosphate buffer affect the availability of magnesium ions in solution compared to a non-interacting buffer? If that is so, is there a way to calculate the size of that effect?

• Could you provide a link to what phosphate buffer is and what it contains? – F'x Apr 29 '12 at 18:23
• @F'x Just potassium phosphate, not one of the fancier biological buffers. – Mad Scientist Apr 29 '12 at 18:38
• Technically, phosphate can and probably do form associates with magnesium ions. To find how strong, you should search databooks for relevant equilibrium constants, it is not such common info to keep it in mind if not needed. – permeakra Aug 25 '12 at 10:21
• I am using a buffer which has approximately 50 mM Phosphate and 7 mM Mg, There is no visible precipitate. But my experiments are Mg sensitive and at least 5 mM Mg is needed for proper response. My preliminary experiments are indicating that in the presence of phosphate free Mg is limited. – user862 Nov 19 '12 at 5:30

If you're still interested in this, fluorescence may be a good method to test the phosphate dependence on magnesium activity. First record a fluorescence spectrum of a solution of magnesium and fluorescent dye, then add small volumes of concentrated phosphate and note any changes in the fluorescence. Of course, you would have to do a control experiment to see if the phosphate is directly perturbing the fluorescence. This write-up describes some magnesium-chelating dyes that you could use, though if you don't want to spend a lot of money on a dye, you may already have some chemicals in your lab that will work. Phenol looks like promising alternative as it fluoresces, is decently water soluble (83 g/L), and forms a complex with magnesium which breaks planarity, which would certainly change the fluorescence spectrum.

It would only take a day to get accurate quantitative results if you have access to a fluorometer.

I don't have a source (at the moment) but it is accepted that high phosphate buffers do chelate/sequester Mg2+ ions. Pyrophosphate is particularly infamous for this. During RNA transcription experiments, the solution will get cloudy which is due exactly to this phenomena. It is also seen in cell-free protein synthesis which requires a steady concentration of Mg2+, that the increased concentration of the phosphate buffer requires an increased amount of Mg2+ to maintain the same effective free Mg2+ concentration [1].

• Do you have a source for plain phosphate? I have no pyrophosphate in the buffer. – Mad Scientist Apr 30 '12 at 9:00

If you deal with cell culture its irrelevant; because the magnesium channels located at the surface of the cell will be forcing magnesium ions to flow into the cell, and if the balance between free and chelated magnesium ions is disturbed in favor of chelate – it will produce free magnesium ions due to the chelate constant.

• I'm not dealing with any cells here, just simple, purified macromolecules – Mad Scientist Jul 20 '12 at 19:14
• So you should be more specific asking question – what do you mean by "availability of magnesium ions"? availability to what? if simple solution is involved you can calculate and compare the solubility ratio of your possible reaction – java_xof Jul 21 '12 at 19:50


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# Show number of integers $m$ with $1\leq m\leq n$, divisible by $p^h$ is equal to $\left[\frac{n}{p^h}\right]$

Let $n\in\mathbb{N}$, $p$ a prime number.

Show that for each $h\in\mathbb{N}$, the number of integers $m$, with $1\leq m\leq n$, divisible by $p^h$ is equal to $\left[\frac{n}{p^h}\right]$, where $\left[\theta\right]$ denotes integer part of $\theta$.

I can see how this works for simple examples of $n$ and $p$. I have no real idea as to how to begin proving this.

Any help or hints would be greatly appreciated.

• I've edited the answer some days ago, It's now more clear? – Sabino Di Trani Feb 13 '15 at 23:35

Hint: How many integers $k \in \mathbb{N}$ are such that the product $p^hk$ is smaller than $n$?

Let's $X$ be the set of this integer.

Let's $Y$ be the set of integers $m$ such that $m < n$ and $p^h | m$

Now We have to make some observation:

1. If $m$ is in $Y$ , you can write $m$ as the product $kp^h$ in a unique form, where $k$ is an integer contained in $X$.

2. If $k$ is in $X$, the product $m=kp^h$ is an integer smaller than $n$ and such that $p^h|m$, so contained in $Y$.

You can see how to conclude? (The claim is that there exist a bijection between $X$ and $Y$)


text:
# Category:Pendulums

This category contains results about Pendulums.
Definitions specific to this category can be found in Definitions/Pendulums.

A pendulum is an ideal body consisting of two parts:

$(1): \quad$ a rod attached at one end to a pivot
$(2): \quad$ a point mass at the other end of the rod.

## Pages in category "Pendulums"

The following 3 pages are in this category, out of 3 total.


text:
# Capacitor¶

class qucat.C(node_minus, node_plus, *args)[source]

A class representing a capacitor

Parameters
• node_minus (integer) – Index corresponding to one node of capacitor

• node_minus – Index corresponding to the other node of the capacitor

• args (<float> or <str> or <float>,<str>) – Other arguments should be a float corresponding to the capacitance, a string corresponding to the name of that value (ex: “C”), or both. If only a label is provided, a value for should be passed as a keyword argument in subsequent function calls (ex: C = 1e-9) This is the best way to proceed if one wants to sweep the value of this capacitor. Indeed, the most computationally expensive part of the analysis is performed upon initializing the circuit, subsequently changing the value of a component and re-calculating a quantity such as the anharmonicity can be performed much faster.

zpf(mode, quantity, **kwargs)

Returns contribution of a mode to the zero-point fluctuations of a quantity for this component.

The quantity can be current (in units of Ampere), voltage (in Volts), charge (in electron charge), or flux (in units of the reduced flux quantum, $$\hbar/2e$$).

Parameters
• mode (integer) – Determine what mode to consider, where 0 designates the lowest frequency mode, and the others are arranged in order of increasing frequency

• quantity (string) – One of ‘current’, ‘flux’, ‘charge’, ‘voltage’

• kwargs – Values for un-specified circuit components, ex: L=1e-9.

Returns

contribution of the mode to the zero-point fluctuations of the quantity

Return type

float

Notes

This quantity is calculated by multiplying the voltage transfer function $$T_{rc}$$ (between a reference component $$r$$ and the annotated component $$c$$ ), with $$X_{zpf,m,r}$$, the zero-point fluctuations of $$\hat{X}$$ at the reference component.

Note that resistors make the transfer function $$T_{rc}$$, and hence this quantity, complex.

For more detail on the underlying theory, see https://arxiv.org/pdf/1908.10342.pdf.


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# Introduction

A lot of PDE(s) can be writen in a generic form, and depends mainly on the definition of coefficients. The generic form that we use is describe by the next equation :

$\begin{eqnarray*} d \frac{\partial u}{\partial t} + \nabla \cdot \left( -c \nabla u - \alpha u + \gamma \right) + \beta \cdot \nabla u + a u = f \quad \text{ in } \Omega \end{eqnarray*}$

with $u$ the unknown and $\Omega \in \mathbb{R}^n$ the computation domain . We call this generic form by Coefficient Form PDE and the coefficients are

• $d$ : damping or mass coefficient

• $c$ : diffusion coefficient

• $\alpha$ : conservative flux convection coefficient

• $\gamma$ : conservative flux source term

• $\beta$ : convection coefficient

• $a$ : absorption or reaction coefficient

• $f$ : source term

Many problems are multiphysics (i.e. several unknowns) and the generic form can be extended naturally into a system of generic PDE.

 Coefficients of each equation can be defined by an expression that depends of the current unknown or unknowns of other equations.

The unknown of each PDEs can be defined as a scalar function ($u:\mathbb{R}^n \longrightarrow \mathbb{R}$) or a vectorial function ($u:\mathbb{R}^n \longrightarrow \mathbb{R}^n$). We need also to respect some constraint on the coefficient shape as described in the next table.

Table 1. Shape required by the coefficients
Coefficient Scalar Unknown Vectorial Unknown

$d$

scalar

scalar

$c$

scalar or matrix

scalar or matrix

$\alpha$

vectorial

x

$\gamma$

vectorial

x

$\beta$

vectorial

vectorial

$a$

scalar

scalar

$f$

scalar

vectorial

## 2. Boundary Conditions

• Dirichlet :

• Neumann : generic version which can depend on the unknowns

• Robin :

• BDF

• Theta

## 5. Stabilized finite element

Given a `cfpdes` equation named `eq`, SUPG and GaLS can be used as stabilisation methods. To enable them use, in the command-line or `.cfg` file, the option `cfpdes.eq.stabilitsation=1` and define the stabilisation type `cfpdes.eq.stabilitsation.type=gls #supg#unusual-gls #gls`

Table 2. Stabilisation methods
type

`gls`

default option

`supg`

`unusual-gls`

Examples are available here


text:
The Korean Physical Society 06130 22, Teheran-ro 7-gil, Gangnam-gu, Seoul, Republic of Korea 610 Representation : Tae Won NOH TEL: 02-556-4737 FAX: 02-554-1643 E-mail : webmaster@kps.or.kr Copyright(C) KPS, All rights reserved.

# The XXX International Conference on Neutrino Physics and Astrophysics (Neutrino 2022)

30 May 2022 to 4 June 2022
Virtual Seoul
Asia/Seoul timezone

## Data-Driven cross checks for electron (anti)neutrino selection efficiency in NOvA

Not scheduled
5m
Virtual Seoul

#### Virtual Seoul

Poster Neutrino oscillation

### Description

NOvA is a long-baseline neutrino oscillation experiment, designed to make measurements using $\nu_\mu(\bar{\nu_\mu})$ disappearance and $\nu_e(\bar{\nu_e})$ appearance. It consists of two functionally equivalent detectors and utilizes the Fermilab NuMI beam. NOvA uses a convolutional neural network for particle identification of $\nu_e$ events with a validation process that includes several data-driven techniques. Muon-Removed, Electron-Added studies involve selecting $\nu_\mu$ charged current candidates from data and simulation and replacing the muon with a simulated electron of similar energy. For Muon-Removed Bremsstrahlung and Muon-Removed Decay-in-Flight studies we remove muonic hits from cosmic muons that have either experienced Bremsstrahlung radiation or decayed in flight, producing samples of pure electromagnetic showers. Each of these $\nu_e$-like samples is then evaluated by our classifier to obtain selection efficiencies. Our most recent analysis showed good agreement in the $\nu_e$ selection efficiency using these techniques. These results and future improvements are presented here.

Collaboration NOvA

### Primary author

Anna Hall (University of Virginia)


text:
Thread: QS question View Single Post
 2011-06-23, 16:11 #1 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 5,881 Posts QS question Currently in the quadratic sieve we sieve until we have a number of relations slightly above the number of primes in the factorbase. In nfs, however, we try postprocessing at regular intervals hoping that enough of the primes aren't used. Would it make sense to try some method like the one we use in nfs for qs? Do we have any knowledge of how many primes are not used in an average matrix from qs?


text:
# Difference between revisions of "2015 AMC 12B Problems/Problem 25"

## Problem

A bee starts flying from point $P_0$. She flies $1$ inch due east to point $P_1$. For $j \ge 1$, once the bee reaches point $P_j$, she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$, where $a$, $b$, $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$ ?

$\textbf{(A)}\; ? \qquad\textbf{(B)}\; ? \qquad\textbf{(C)}\; ? \qquad\textbf{(D)}\; ? \qquad\textbf{(E)}\; ?$