text:
Contents

# Contents

## Idea

In quantum field theory with chiral fermions (spinor fields) $\psi$ with chiral version of the Dirac current $J = i \bar \Psi \Gamma \Psi$, a chiral anomaly is a non-conservation of this current

$div J = Anomaly \,.$

(See at Ward identity.)

In the standard model of particle physics this happens and plays a role for pion decay? and for baryogenesis. Here the anomaly term is the Pontryagin 4-form $Anomaly = \langle F_\nabla \wedge F_\nabla\rangle$ of the gauge field $\nabla$, hence the curvature 4-form of the corresponding Chern-Simons line 3-bundle.

If there are instantons, i.e. if the gauge field principal connection $\nabla$ has a nontrivial underlying principal bundle, then also the Chern-Simons line 3-bundle is topologically nontrivial the anomaly term $\langle F_\nabla \wedge F_\nabla\rangle$ is a non-exact integral form, hence the above equation is to be read as the local expression identifying $\star j$ with the local 3-connection on the CS 3-bundle.

## References

The orginal observation is due to

• Stephen Adler. Axial-Vector Vertex in Spinor Electrodynamics Physical Review 177 (5): 2426. (1969)

• John Bell, Roman Jackiw, A PCAC puzzle: $\pi^0 \to \gamma \gamma$ in the σ-model“. Il Nuovo Cimento A 60: 47. (1969)

A detailed mathematical derivation is in

Detailed argument for the theta vacuum (Yang-Mills instanton vacuum) from chiral symmetry breaking is offered in

Review includes

Discussion in the rigorous context of causal perturbation theory/perturbative AQFT is (for $m \gt 0$) in

and (for $m = 0$) in

and reviewed in the context the master Ward identity in

Application to baryogenesis is due to

• Gerard 't Hooft, Symmetry Breaking through Bell-Jackiw Anomalies Phys. Rev. Lett. 37 (1976) (pdf)

• Gerard 't HooftComputation of the quantum effects due to a four-dimensional pseudoparticle, Phys. Rev. D14:3432-3450 (1976).

Last revised on March 24, 2019 at 09:57:03. See the history of this page for a list of all contributions to it.


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By definition, it is impossible to be equally loud on‑axis to the top loudspeaker of a line array, compared to on‑axis to the second‑to‑last loudspeaker. In this short article I will explain why this is, and why overshoot is advised.

For line arrays, the bulk of SPL at a given distance to the array, is primarily caused by those loudspeakers that are coupled, i.e., arrive within 1/3rd cycle or less, and as such, inherently changes with frequency. It is a function of array length, mechanical articulation (splay), and listening distance.

Figure 1Figure 2Figure 1 shows a line array of finite length. On‑axis to the entire array, for the specific wavelength chosen in this example, 7 out of 9 loudspeakers are coupled and cause the bulk of SPL.

In Figure 2, the listener has moved from on‑axis to the entire array, to the edge‑of‑coverage, on‑axis to the upper loudspeaker, while preserving distance.

If the array were not finite but continued to extend, the same amount of loudspeakers would be included in the coupling zone (the green portion of the simplified phase wheel), and SPL would be expected to remain constant.

However, since the array is very much finite, on‑axis to the top loudspeaker, half the required loudspeakers to preserve SPL are missing, and level will drop by 6 dB.

$$p_{edge}\propto\frac{n-1}{2n}$$

Where $$p_{edge}$$ is pressure at the edge‑of‑coverage and $$n$$ is number of loudspeakers in the line array.

When the number of loudspeakers $$n$$ increases,  $$p_{edge}$$ approaches 0,5 or -6 dB in decibels, which is the required amount of angular attenuation that defines coverage angle.

As frequency increases, wavelengths become shorter, and fewer loudspeakers will be included in the primary coupling zone. Regardless, at the edge‑of‑coverage to your finite array, the coupling zone is always short on loudspeakers.

Figure 3Notice that in figure 3, on‑axis to the top (or bottom) loudspeaker, SPL is less than on‑axis to the second‑to‑last loudspeaker. Overshooting by 1 whole loudspeaker would restore the desired sound level.

format: plain


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# Showdown front-end algorithm - Search Search Algorithm - binary search

Index of this series

• Linear table basis
• Tree structure foundation
• Sorting algorithm
• Search Search Algorithm
• To be continued

## Binary search principle

Binary search is an algorithm applied to sequential sequences. In the process of search, we continuously narrow the search range on the premise of ensuring that the target value is in the search range, and finally find the target value

Firstly, the simplest binary search algorithm is introduced. This algorithm is used to locate the given target value. If it cannot be found, it returns - 1

First, set the two pointers at the beginning and end, and then compare them through the intermediate value (head + tail) > > 1. If the intermediate value is equal to the target value, the target value will be returned directly. Otherwise, perform the following operations

{ a r r [ m i d ] < x ， h e a d = m i d + 1 a r r [ m i d ] > x ， t a i l = m i d − 1 \begin{cases} arr[mid]<x，head=mid+1 \\ arr[mid] > x，tail = mid - 1 \end{cases} {arr[mid]<x，head=mid+1arr[mid]>x，tail=mid−1

The operation here is easy to understand. The goal is to ensure that the target is within the search range and narrow the search range

function binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h <= t) {
mid = (t + h) >> 1
if (nums[mid] === target) return mid
if (nums[mid] < target) h = mid + 1
else t = mid - 1
}
return -1
}

## 01 model

00001111

This model is usually used to find the last 0 and the first 1

The last 0 here can be understood as the last value less than or equal to x;
The first 1 can be understood as the first value greater than or equal to x.

Here, take finding the first value greater than or equal to x as an example for reasoning

// 01-1 core part
while (h < t) {
// Middle pointer
mid = (h + t) >> 1
// Narrow selection
if (nums[mid] <= target) h = mid
else t = mid - 1
}

// 01-1 core code
function b_binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = (h + t) >> 1
// mid = h + ((t - h) >> 1)
if (nums[mid] >= target) t = mid
else h = mid + 1
}
return h
}

Similarly, the 01-0 code is obtained

// 01-0 core code
function s_binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = h + ((t - h) >> 1) + 1  // In any case, one to the right
// parseInt(${(h + t + 1) / 2} ) / / find the middle value for odd numbers and the middle right for even numbers if (nums[mid] <= target) h = mid else t = mid - 1 } return h }  Bias of middle pointer In the above code, you can find that the mid of 01-0 and 01-1 are slightly different In the search sequence, if the number of elements is odd, the middle value they locate is the same; However, if it is an even number, 01-0 is positioned to the right of the middle, while 01-1 is positioned to the left of the middle If it is right, there are two cases 1. As mentioned above, odd numbers are positioned in the middle, and even numbers are positioned in the right of the middle mid = parseInt(${(h + t + 1) / 2})

2. The pointer is one bit to the right
mid = ((h + t) >> 1) + 1
mid = h + ((t - h) >> 1) + 1 // In static languages, this way of writing can prevent overflow

Both cases work

## Binary search questions

### Square root of leetCode 69 x

This problem can be transformed into finding the last value less than or equal to x

function mySqrt(x: number): number {
let h = 0, t = x, mid: number
while (h < t) {
mid = ((h + t) >> 1) + 1
if (mid ** 2 <= x) h = mid
else t = mid - 1
}
return h
}

### leetCode 35 search insertion location

Find the first position greater than or equal to x

function searchInsert(nums: number[], target: number): number {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = (h + t) >> 1
if (nums[mid] >= target) t = mid
else h = mid + 1
}
return (h === nums.length - 1 && nums[h] < target) ? h + 1 : h
}

### leetCode 1 sum of two numbers

In this problem, we use simple two-point search to accurately find the target value

First traverse the array, and then find it later t a r g e t − x target - x Value of target − x

Note that the premise of binary search is for ordered sequences, so we have to sort the original array first. Here we use the merge sort we learned before

type dataType = [val: number, ind: number]
function binarySearch (nums: dataType[], l: number, target: number) {
let h = l, t = nums.length - 1, mid: number
while (h <= t) {
mid = (h + t) >> 1
if (nums[mid][0] === target) return nums[mid][1]
if (nums[mid][0] < target) h = mid + 1
else t = mid - 1
}
return -1
}
function twoSum(nums: number[], target: number): number[] {
const data: dataType[] = [], ret: number[] = []
let ind: number
nums.forEach((val, i) => data.push([val, i]))
mergeSort(data, 0, nums.length - 1)
for (let i = 0; i < data.length - 1; i++) {
ind = binarySearch(data, i + 1, target - data[i][0])
if (ind > -1) ret.push(data[i][1], ind)
}
return ret
}
//*Merge sort
function mergeSort (data: dataType[], l: number, r: number) {
if (l >= r) return
const mid = (l + r) >> 1
mergeSort(data, l, mid)
mergeSort(data, mid + 1, r)
const temp: dataType[] = []
let p1 = l, p2 = mid + 1
while (p1 <= mid || p2 <= r) {
if (p2 > r || (p1 <= mid && data[p1][0] < data[p2][0])) temp.push(data[p1++])
else temp.push(data[p2++])
}
temp.forEach((v, i) => data[l + i] = v)
}

Of course, if C + + is used here, the efficiency of using sort through symbolic overloading will be higher

### leetCode 34 finds the first and last positions of elements in a sorted array

This problem is a typical example of the application of model 01

The first position of the element is 01-1, and the last position is 01-0

//* 01-0
function s_binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = ((h + t) >> 1) + 1
if (nums[mid] <= target) h = mid
else t = mid - 1
}
//*When looking for the second value, it is certain that there must be a target value on the field
return h
}
//* 01-1
function b_binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = (h + t) >> 1
if (nums[mid] >= target) t = mid
else h = mid + 1
}
return nums[h] === target ? h : -1
}
function searchRange(nums: number[], target: number): number[] {
let fir: number, sec: number
fir = b_binarySearch(nums, target)
if (fir === -1) return [-1, -1]
sec = s_binarySearch(nums, target)
return [fir, sec]
}

The idea of the main program is also very simple. If the first occurrence position of the target value cannot be found, directly return [- 1, - 1], otherwise continue to find the last occurrence position (the two can coincide)

### leetCode 1658 minimum operand to reduce x to 0

A very intuitive solution to this problem is to first count the prefix sum in the left and right directions. When traversing the left prefix sum, use dichotomy to act on the right prefix and array to find x - sums[l]

But this problem only needs to calculate the prefix sum from left to right once

Here we emphasize that the interval and of our previous study are as follows:
s [ 4 ] − s [ 2 ] = n [ 2 ] + n [ 3 ] Namely s [ i ] − s [ j ] = ∑ n = j n = i − 1 a n s[4] - s[2] = n[2]+n[3] \ \ i.e. s[i]-s[j]=\sum_{n=j}^{n=i-1}a_n s[4] − s[2]=n[2]+n[3], i.e. s[i] − s[j]=n=j Σ n=i − 1 an

And the one used here n [ i ] − s [ j ] n[i]-s[j] n[i] − s[j] is equivalent to statistics ( j , i ) (j, i) Number of elements between (j,i)

So when i and j are found, the result is c n t = n u m s . l e n g t h − ( i − j ) cnt = nums.length - (i - j) cnt=nums.length−(i−j)

function binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h <= t) {
mid = (h + t) >> 1
if (nums[mid] === target) return mid
if (nums[mid] < target) h = mid + 1
else t = mid - 1
}
return -1
}
function minOperations(nums: number[], x: number): number {
//*Building prefixes and arrays
const sums = [0]
nums.forEach((n, i) => sums.push(n + sums[i]))
//*Left and right counting
let j: number, cnt: number, ans = -1
for (let i = nums.length, s = 0; i >= 0; --i, s += nums[Math.max(i, 0)]) {
j = binarySearch(sums, x - s)
if (j === -1) continue
cnt = nums.length - (i - j)
if (cnt > nums.length) continue // This happens when I < = J
if (ans === -1 || cnt < ans) ans = cnt
}
return ans
}

The main function strategy here is to see whether there is a target value (x - s) from the prefix and in each cycle. If not, continue to the next cycle

### leetCode 475 heater

The idea of this problem is very simple. First, sort the address of the heater (the premise of binary search), then traverse the house, find the location of the nearest heater in the house and record the distance

Finally, the maximum distance is the minimum range

//* 01-0
function s_binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = ((h + t) >> 1) + 1
if (nums[mid] <= target) h = mid
else t = mid - 1
}
return nums[h]
}
//* 01-1
function b_binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = (h + t) >> 1
if (nums[mid] >= target) t = mid
else h = mid + 1
}
return nums[h]
}
function findRadius(houses: number[], heaters: number[]): number {
heaters = heaters.sort((i, j) => i - j)
let maxRadius = -1, recentHeater: number
houses.forEach(house => {
recentHeater = Math.min(Math.abs(house - s_binarySearch(heaters, house)), Math.abs(b_binarySearch(heaters, house) - house))
})
}

The practice here is to use two structures of 01 model, but if you master 01 model deeply enough, just use one

function b_binarySearch (nums: number[], target: number) {
let h = 0, t = nums.length - 1, mid: number
while (h < t) {
mid = (h + t) >> 1
if (nums[mid] >= target) t = mid
else h = mid + 1
}
return h
}
function findRadius(houses: number[], heaters: number[]): number {
heaters = heaters.sort((i, j) => i - j)
let maxRadius = -1, a: number, b: number, ind: number
houses.forEach(house => {
ind = b_binarySearch(heaters, house)
a = Math.abs(heaters[ind] - house)
b = Math.abs(ind ? heaters[ind - 1] - house : a + 1)
})
}

The idea here is to first find the heater (01-1) behind the current house. If the heater is not the first, compare it with the previous one. Who is closer to the house

Through this question, we can find that if the 01 model cannot find the result (for example, 01-1 cannot find a value greater than or equal to the target value), the first element subscript will be returned

You can see that the result is also correct

### leetCode 81 search rotation sort array II

This problem is very complicated to read, but the actual situation is shown in the figure below

Each time the middle value is taken, it is only compared with the head / tail. If it is in the non shaded area, just keep the pointers on both sides in the middle

Because of the particularity of this problem, four pointers are required

Two are fixed at the head and tail, and the other two are gradually narrowing intervals

function search(nums: number[], target: number): boolean {
if (!nums.length) return false
if (nums[0] === target || nums[nums.length - 1] === target) return true
//*Set the head and tail of the two sections
let l = 0, r = nums.length - 1
while (l < r && nums[l] === nums[0]) l++
while (l < r && nums[r] === nums[nums.length - 1]) r--
const head = l, tail = r
//*Positioning element
let mid: number
while (l <= r) {
mid = (l + r) >> 1
if (nums[mid] === target) return true
//*If it's in the right section
if (nums[mid] < nums[tail]) {
if (nums[mid] < target && target <= nums[tail]) l = mid + 1
else r = mid - 1
} else {
//*In the left section
if (nums[head] <= target && target < nums[mid]) r = mid - 1
else l = mid + 1
}
}
return false
}

### leetCode 4 finds the median of two positively ordered arrays

To solve this problem, we must first have a consensus

With this consensus, we can start coding

function findK (nums1: number[], nums2: number[], l1: number, l2: number, k: number): number {
//*Boundary condition judgment
if (l1 === nums1.length) return nums2[l2 + k - 1]
if (l2 === nums2.length) return nums1[l1 + k - 1]
if (k === 1) return nums1[l1] < nums2[l2] ? nums1[l1] : nums2[l2]
//*Processing of getting values from two arrays
let cnt1 = Math.min(k >> 1, nums1.length - l1)
let cnt2 = Math.min(k - cnt1, nums2.length - l2)
//*Narrow down
if (nums1[l1 + cnt1 - 1] < nums2[l2 + cnt2 - 1]) {
return findK(nums1, nums2, l1 + cnt1, l2, k - cnt1)
}
return findK(nums1, nums2, l1, l2 + cnt2, k - cnt2)
}
function findMedianSortedArrays(nums1: number[], nums2: number[]): number {
const cnt = nums1.length + nums2.length, mid = (nums1.length + nums2.length + 1) >> 1
const a = findK(nums1, nums2, 0, 0, mid)
if (cnt % 2) return a
const b = findK(nums1, nums2, 0, 0, mid + 1)
return (a + b) / 2
}

First, set a function that can find the k-th element in the two arrays. If the length of the two operation arrays is an odd number, the middle value is the middle number; If it is an even number, you need to find the number to the right and average it

The boundary conditions here are also well understood. When one array is empty, it is equivalent to directly operating on the other array; When k is 1, you only need to compare the header elements of the two arrays to see who is smaller

When taking values from two arrays, it should be noted that the length of the array may not be enough k 2 \frac{k}{2} 2k, so you need to do some compatibility processing

The final narrowing operation is based on the consensus just mentioned

Although the coding of this problem is not like the traditional binary search, it is the same in thought

Binary search determines the target value through the comparison of intermediate values, and here the target value is determined through three boundary conditions;
Binary search reduces the range through left and right pointers. Here, it reduces the range through counting and head pointers.

Xi Wei~

Posted by simenss on Tue, 21 Sep 2021 02:40:42 -0700


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## “Rhythm Training” Micro-Credential Verification

Welcome to the verification page for Ari Hoenig’s “Rhythm Training” Micro-Credential.  Due to the difficulty of this particular course, only very skilled musicians have been certified.  As an added step of verification, we have included the names of the credentialed students below:


text:
# How to prove that $\lim_{(x,y) \to (0,0)} \frac{\sqrt{1+x^2+y^2}-1}{\arcsin{\left(x^2+y^2+x \right) }-\arcsin{x}} =0$?

$$\lim_{(x,y) \to (0,0)} \frac{\sqrt{1+x^2+y^2}-1}{\arcsin{\left(x^2+y^2+x \right) }-\arcsin{x}}=0$$ I have tried using polar coordinates, $x = r\cos\theta$, $x = r\sin\theta$ and I do not get anything.

Can anyone give me a hint on how to approach this?

• I think the Taylor expansion would be fatal to attempt.
– user211599
Jun 24, 2018 at 17:47
• Why do you think the limit is $0?$
– zhw.
Jun 24, 2018 at 17:58

By the binomial theorem, the numerator approximates $(x^2+y^2)/2$. By differentiating the arcsine function, the denominator approximates $(x^2+y^2)/\sqrt{1-x^2}$. The desired limit is $\lim_{x\to 0}\frac{\sqrt{1-x^2}}{2}=\frac{1}{2}$.

Multiply by the conjugate of the numerator over itself and you find the limit is equal to $$\frac12\lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\arcsin{\left(x^2+y^2+x \right) }-\arcsin{x}}$$

[Note that here we go ahead and evaluate $\lim_{(x,y) \to (0,0)} \frac{1}{\sqrt{1+x^2+y^2}+1}=\frac{1}{2}$ rather than keeping such a cumbersome expression along for the ride unnecessarily. This is the main hint I would give but I can't find a way to express that in the form of a hint without just doing it.]

Convert to polar and you have: $$\frac12\lim_{r\to0^+} \frac{r^2}{\arcsin{\left(r^2+r\cos(\theta) \right) }-\arcsin(r\cos(\theta))}$$

There is an identity for summing two arcsin expressions when both arguments are small, as they would be here when taking a limit as $r\to0^+$.

$$\frac12\lim_{r\to0^+} \frac{r^2}{\arcsin\left(\left(r^2+r\cos\theta\right)\sqrt{1-r^2\cos^2\theta} -r\cos(\theta)\sqrt{1-\left(r^2+r\cos\theta\right)^2}\right)}$$

Multiply by $\frac{\arcsin(\text{expression})}{\text{expression}}$ which has limit $1$:

$$\frac12\lim_{r\to0^+} \frac{r^2}{\left(r^2+r\cos\theta\right)\sqrt{1-r^2\cos^2\theta} -r\cos(\theta)\sqrt{1-\left(r^2+r\cos\theta\right)^2}}$$

Rearrange the $r$'s:

$$\frac12\lim_{r\to0^+} \frac{r}{\left(r+\cos\theta\right)\sqrt{1-r^2\cos^2\theta} -\cos(\theta)\sqrt{1-\left(r^2+r\cos\theta\right)^2}}$$

Use a conjugate again:

$$\cos\theta\lim_{r\to0^+} \frac{r}{\left(r+\cos\theta\right)^2\left(1-r^2\cos^2\theta\right) -\cos^2(\theta)\left(1-\left(r^2+r\cos\theta\right)^2\right)}$$

Expand and condense:

$$\cos\theta\lim_{r\to0^+} \frac{r}{r^2+2r\cos\theta}$$

And this works out to $\frac12$, not $0$.

I don't think the limit is $0.$ Let $h=x^2+y^2$ and divide top and bottom by $h.$ The expression becomes

$$\frac { \dfrac{\sqrt{1+h}-1}{h}}{\dfrac{\arcsin(h+x)-\arcsin x}{h}}.$$

As $(x,y)\to (0,0),$ $h\to 0^+.$ So the numerator goes to the derivative of $\sqrt u$ at $u=1,$ which equals $1/2.$ Downstairs we can apply apply the MVT. Doing so shows the denominator equals $\arcsin'(c(x,h)),$ where $c(x,h) \in (x,h+x).$ As $(x,y)\to (0,0),$ both $x,x+h$ are dragged to $0,$ so the limit of the denominator is $\arcsin'(0)=1.$ It follows that the limit in question is $1/2.$

• I considered something like this too, but the $x$ and $h$ in $\arcsin(h+x)$ are not independent, which made me pause with moving on. You can't let $h$ go to $0$ without $x$ going to $0$ as well, so you can't directly get $\arcsin'(x)$ in the denominator. Or maybe you can give an argument that it is OK? Jun 24, 2018 at 18:51


text:
# Don't definite Integrals find area?

1. Dec 16, 2007

### lLovePhysics

I'm confused here.. My definite integral doesn't match by Riemman Sum... and it should right? I think that I have not integrated correctly. Can someone help me spot the problem? Thanks.

Find the Area of the region bounded by:
$$f(x)=5-x^2$$ , [-2, 1]

Using the Riemma Sum idea (limit of the sum of rectangles as the number of rectangles approaches infinity), I got 12 units^2 as my area, which is correct.

However, using definite integrals and the Fundamental Theorem of Calculus, I get:

$$\int_{-2}^{1} (5-x^2)dx}$$

$$=-\frac{x^3}{3}\biggl] ^{1}_{-2}$$

Which equals -3 ???

2. Dec 16, 2007

### rocomath

5x - 1\3 * x^3

3. Dec 16, 2007

### lLovePhysics

Noooo.. I can't believe I made another stupid mistake!! Thanks Roco


text:
Chapter 8 - 8.3 - Vectors in the Plane - 8.3 Exercises - Page 586: 71

$\lt 2, 2 \sqrt 3+4 \gt$

Work Step by Step

The component form of the vector $u$ is: $u=||u|| \lt \cos \theta_{u} , \sin \theta_{u} \gt =4 \lt \cos 60^{\circ},\sin 60^{\circ} \gt \\= 4 \lt \dfrac{1}{2},\dfrac{\sqrt {3}}{2} \gt \\= \lt 2 , 2 \sqrt 3 \gt$ Now, the component form of the vector $v$ is: $v=||v|| \lt \cos \theta_{v} , \sin \theta_{v} \gt =4 \lt \cos 90^{\circ},\sin 90^{\circ} \gt \\= 4 \lt 0,1 \gt \\= \lt 0,4 \gt$ Thus, $u+v=\lt 2 , 2 \sqrt 3 \gt+\lt 0,4 \gt = \lt 2, 2 \sqrt 3+4 \gt$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
[rochester] / trunk / rochester_problib / set0 / prob2.pg Repository: Repository Listing bbplugincoursesdistsnplrochestersystemwww

# View of /trunk/rochester_problib/set0/prob2.pg

Mon Jun 25 19:20:51 2001 UTC (11 years, 11 months ago) by sam
File size: 2015 byte(s)
added problem library as webwork/ww_prob_lib with revision tag prob-lib-2

    1 ##DESCRIPTION
2 ##  sample true-false question
3 ##ENDDESCRIPTION
4
5 ##KEYWORDS('true-false')
6
7 DOCUMENT();        # This should be the first executable line in the problem.
8
10 "PG.pl",
11 "PGbasicmacros.pl",
12 "PGchoicemacros.pl",
14 "PGauxiliaryFunctions.pl"
15 );
16
17 TEXT(&beginproblem);
18 $showPartialCorrectAnswers = 1; 19 20 TEXT(EV2(<<EOT)); 21 This problem demonstrates a WeBWorK True/False question.$PAR
22 Enter a T or an F in each answer space below
23 to indicate whether the corresponding statement is true or
24 false.$BR 25 You must get all of the answers correct to receive credit. 26$BR
27 EOT
28
29 ## First we set up our variables.
30 $a = random(1,5,1); 31$b = random(6,10,1);
32 $c = random(-10,-1,1); 33$d = random(-10,-1,1);
34 $e = random(1,10,1); 35 36$questStr1 = EV2(" $$-a \lt -b$$");
37 $ansStr1 = "F"; 38$questStr2 = EV2(" $$c \leq c$$");
39 $ansStr2 = "T"; 40$questStr3 = EV2(" $$d \lt d$$");
41 $ansStr3 = "F"; 42$questStr4 = EV2(" $$\pi \geq 3.1416$$");
43 $ansStr4 = "F"; 44$questStr5 = EV2(" $$e-1 \leq e$$");
45 $ansStr5 = "T"; 46 47 @questions =($questStr1,$questStr2,$questStr3,$questStr4,$questStr5);
48 @answers =( $ansStr1,$ansStr2,$ansStr3,$ansStr4,$ansStr5); 49 50 ## Now choose randomly 4 questions out of the 5 question strings above. 51 52 @slice = NchooseK(scalar(@questions),4); 53 54 ## Next we output the 4 chosen questions. #match_ 55 TEXT( 56 &match_questions_list(@questions[@slice]) 57 ); 58 59 &ANS(str_cmp([ @answers[@slice] ] )); 60 TEXT(<<EOT); 61$PAR Notice that if one of your answers is wrong then, in this problem, WeBWorK will tell you which
62 parts are wrong and which parts are right.  This is the behavior for most problems, but for true/false
63 or multiple choice questions
64 WeBWorK will usually only tell you whether or not all the answers are correct.
65 It won't tell you which ones are wrong.  The idea is to encourage you think rather than to just try guessing.
66 \$PAR
67 In every case all of the
68 answers must be correct before you get credit for the problem.
69
70
71 EOT
72 ENDDOCUMENT();        # This should be the last executable line in the problem.


text:
# How to generate a xy diagram with multiple grouped data sets?

Based on the answer of Stefan Pinnow in 'How to group several charts into one chart?', I tried to generate a diagram with multiple grouped data. Therefore, I modified the diagram but I have some problems. The resulting diagram now looks:

And here is my MWE (I hope it is as minimal as possible MWE):

% restructured data file:
% - 'n' values are sorted ascending
% - added column 'z' to know which data belong to which measurement
% - added a dummy line at the end of each data set to produce an empty entry
%   for the separation of the blocks
% - put all data in one file
\begin{filecontents}{data.txt}
z   n     val1  val2  val3
10  0.1   6    10      400.0
10  1.0   9    70      300.0
10  10.0 10   120      100.0
10  {}    {}   {}     {}
20  0.1   10   80      500.0
20  1.0   15   95    600.0
20  10.0 20   140      690.0
20  {}    {}   {}     {}
30  0.1  15   100      450.0
30  1.0  18   200      700.0
30  10.0 28   260     850.0
30  {}    {}   {}    {}
40  0.1  13   300      1500.0
40  1.0  19   260      1300.0
40  10.0 21   220    1000.0
40  {}    {}   {}     {}
50  0.1   25   250     3000.0
50  1.0   37   180   2400.0
50  10.0  34   100    2000.0
\end{filecontents}

\documentclass[border=0pt]{standalone}
\usepackage{pgfplots}
\usetikzlibrary{patterns}
\pgfplotsset{
scale only axis,
ymajorgrids=true,
tick align = outside,
yticklabel style={
/pgf/number format/fixed zerofill,
/pgf/number format/precision=1,
},
legend style={
text=black,
legend pos=north west,
legend cell align={left},
},
}

\begin{document}
\begin{tikzpicture}

\begin{axis}[
color=black,
% set the width' of the plot to the maximum length ...
width=\textwidth,
% ... and use half this length for the height'
height=0.5\textwidth,
scale only axis,
/pgf/number format/.cd,
use comma,
1000 sep={\,},
ymin=-0.5,
ymax=30.5,
ytick={0.0, 5.0, ..., 30.0},
% use data' for the positioning of the xticks' ...
xtick=data,
% ... and use table data for labeling the xticks'
xticklabels from table={data.txt}{n},
% add extra ticks "at the empty entries to add the vertical lines
extra x ticks={3,7,11,15},
% this ticks shouldn't be labeled ...
extra x tick labels={},
% ... but grid lines should be drawn without the tick lines
extra x tick style={
grid=major,
major tick length=0pt,
},
xlabel={X Axis 1 Label},
ylabel={Y Axis 1 Label},
ylabel style={blue},
y axis line style={blue},
ytick style={blue},
yticklabel style=blue,
xtick pos=bottom,
axis y line*=left,
% because of the category labels, shift the xlabel' a bit down
xlabel style={
yshift=-4ex,
},
% ... and with that you also have to adjust the x limits
enlarge x limits={abs=1},
% set clip mode' to individual' so the category labels aren't clipped away
clip mode=individual,
]

% plot the "blue" points
color=blue,
only marks,
mark size=2pt,
mark=square*,
mark options={solid,fill=blue},
] table [
% use just the coordindex' as x coordinate,
% the correct labeling is done with xticklabels from table'
x expr=\coordindex,
y=val1,
] {data.txt};
\label{label-plot1}

\begin{scope}[
% because the reference point will be the lower axis line the
% labels have to be moved a bit more down to don't overlap with
% the xticklabels'
every label/.append style={
label distance=3ex,
},
]
\node [label=below:{System A}]
at (axis cs:1,\pgfkeysvalueof{/pgfplots/ymin}) {};
\node [label=below:{System B}]
at (axis cs:5,\pgfkeysvalueof{/pgfplots/ymin}) {};
\node [label=below:{System C}]
at (axis cs:9,\pgfkeysvalueof{/pgfplots/ymin}) {};
\node [label=below:{System D}]
at (axis cs:13,\pgfkeysvalueof{/pgfplots/ymin}) {};
\node [label=below:{System E}]
at (axis cs:17,\pgfkeysvalueof{/pgfplots/ymin}) {};
\end{scope}

\end{axis}

% Frame 2
\begin{axis}[
color = red,
% set the width' of the plot to the maximum length ...
width=\textwidth,
% ... and use half this length for the height'
height=0.5\textwidth,
scale only axis,
/pgf/number format/.cd,
use comma,
1000 sep={\,},
ymin=-5.0,
ymax=305.0,
ytick={0.0, 50.0, ..., 300.0},
% use data' for the positioning of the xticks' ...
xtick=data,
% ... and use table data for labeling the xticks'
xticklabels from table={data.txt}{n},
% add extra ticks "at the empty entries to add the vertical lines
extra x ticks={3,7,11,15},
% this ticks shouldn't be labeled ...
extra x tick labels={},
% ... but grid lines should be drawn without the tick lines
extra x tick style={
grid=major,
major tick length=0pt,
},
ylabel={Y Axis 2 Label},
% because of the category labels, shift the xlabel' a bit down
xlabel style={
yshift=-4ex,
},
ytick style={red},
axis x line*=none,
axis y line*=right,
ylabel style = red,
% ... and with that you also have to adjust the x limits
enlarge x limits={abs=1},
% set clip mode' to individual' so the category labels aren't clipped away
clip mode=individual,
hide x axis
]

% plot the "red" points
color=red,
only marks,
mark size=2pt,
mark=triangle*,
mark options={solid,fill=red}
] table [
x expr=\coordindex,
y=val2,
] {data.txt};
\label{label-plot2}

\end{axis}

% Frame 3
\begin{axis}[
color = green,
% set the width' of the plot to the maximum length ...
width=\textwidth,
% ... and use half this length for the height'
height=0.5\textwidth,
scale only axis,
/pgf/number format/.cd,
use comma,
1000 sep={\,},
ymin=-50,
ymax=3050.,
ytick={0, 500, ..., 3000},
% use data' for the positioning of the xticks' ...
xtick=data,
% ... and use table data for labeling the xticks'
xticklabels from table={data.txt}{n},
% add extra ticks "at the empty entries to add the vertical lines
extra x ticks={3,7,11,15},
% this ticks shouldn't be labeled ...
extra x tick labels={},
% ... but grid lines should be drawn without the tick lines
extra x tick style={
grid=major,
major tick length=0pt,
},
ylabel={Y Axis 3 Label},
% because of the category labels, shift the xlabel' a bit down
xlabel style={
yshift=-4ex,
},
ylabel style = green,
ytick style={green},
axis x line*=none,
axis y line*=right,
% ... and with that you also have to adjust the x limits
enlarge x limits={abs=1},
% set clip mode' to individual' so the category labels aren't clipped away
clip mode=individual,
hide x axis,
every outer y axis line/.style={
green,
xshift=2.0cm
},
every tick/.style={
green,
xshift=2.0cm
},
every y tick label/.append style={
green,
xshift=2.0cm
},
]

% plot the legend

% plot the "green" points
color=green,
only marks,
mark size=2pt,
mark=diamond*,
mark options={solid,fill=green},
] table [
x expr=\coordindex,
y=val3,
] {data.txt};
\label{label-plot3}

\end{axis}

\end{tikzpicture}
\end{document}

My problems/ questions are:

1. The last data set of System E is not vizualized fully. How can I solve this?
2. There is some miss arrangement of the data point on the x-axis. What I am doing wrong?
3. There is some miss alignment of the data points of 'Algorithm B' and 'Algorithm C' regarding the correlation to 'Algorithm A' and the x-axis. Do you have a hint to solve this?
4. I would also like to write a multi-line text as column caption with defined line breaks. In contrast to 'System A' there should be 'System A \n \n Parameter: \n \alpha=5 \n beta=3 \n \lambda=8'. Using of newline (\n) is not working. Can you give me a hint for that?
5. Is it possible to draw interconnection lines (for instance: a dotted line) between the points (here: 0.1 , 1.0, 10.0) of one group (here grouped by color: 'Algorithm A') within another grouping (here: grouped by system, for instance 'System A'), but not to draw interconnection lines between groups (here: between the systems)?
• Didn't see your edit before I first posted my answer, so I just updated it. – Torbjørn T. Jan 19 '19 at 20:13

I think the main problem is that you have ymax=30.5 in the first axis, while the data has a max of 37. Because of that, not all the points are plotted. I also needed to change xtick=data to xtick={0,...,18} in the first axis, not sure why.

I also changed the ylimits and yticks of all three axes.

For the System N nodes, I think it would be easier to not use label, but the node contents directly, and add align=center to the node options. (Which I here did with the every node style, similar to the every label style that was there originally.) With align=center you can use \\ to add line breaks in the nodes.

For the connection lines, replace the {} in your datafile with inf, add unbounded coords=jump to the \pgfplotsset, and replace only marks in the \addplots with dotted.

% restructured data file:
% - 'n' values are sorted ascending
% - added column 'z' to know which data belong to which measurement
% - added a dummy line at the end of each data set to produce an empty entry
%   for the separation of the blocks
% - put all data in one file
\begin{filecontents}{data.txt}
z   n     val1  val2  val3
10  0.1   6    10      400.0
10  1.0   9    70      300.0
10  10.0 10   120      100.0
10  inf    inf   inf     inf
20  0.1   10   80      500.0
20  1.0   15   95    600.0
20  10.0 20   140      690.0
20  inf    inf   inf     inf
30  0.1  15   100      450.0
30  1.0  18   200      700.0
30  10.0 28   260     850.0
30  inf    inf   inf    inf
40  0.1  13   300      1500.0
40  1.0  19   260      1300.0
40  10.0 21   220    1000.0
40  inf    inf   inf     inf
50  0.1   25   250     3000.0
50  1.0   37   180   2400.0
50  10.0  34   100    2000.0
\end{filecontents}

\documentclass[border=0pt]{standalone}
\usepackage{pgfplots}
\usetikzlibrary{patterns}
\pgfplotsset{
scale only axis,
ymajorgrids=true,
tick align = outside,
yticklabel style={
/pgf/number format/fixed zerofill,
/pgf/number format/precision=1,
},
legend style={
text=black,
legend pos=north west,
legend cell align={left},
},
unbounded coords=jump
}

\begin{document}
\begin{tikzpicture}

\begin{axis}[
color=black,
% set the width' of the plot to the maximum length ...
width=\textwidth,
% ... and use half this length for the height'
height=0.5\textwidth,
scale only axis,
/pgf/number format/.cd,
use comma,
1000 sep={\,},
ymin=-0.5,
ymax=37.5,
ytick={0.0, 5.0, ..., 35.0},
% use 0,...,18
xtick={0,...,18},
% ... and use table data for labeling the xticks'
xticklabels from table={data.txt}{n},
% add extra ticks "at the empty entries to add the vertical lines
extra x ticks={3,7,11,15},
% this ticks shouldn't be labeled ...
extra x tick labels={},
% ... but grid lines should be drawn without the tick lines
extra x tick style={
grid=major,
major tick length=0pt,
},
xlabel={X Axis 1 Label},
ylabel={Y Axis 1 Label},
ylabel style={blue},
y axis line style={blue},
ytick style={blue},
yticklabel style=blue,
xtick pos=bottom,
axis y line*=left,
% because of the category labels, shift the xlabel' a bit down
xlabel style={
yshift=-4ex,
},
% ... and with that you also have to adjust the x limits
enlarge x limits={abs=1},
% set clip mode' to individual' so the category labels aren't clipped away
clip mode=individual,
]

% plot the "blue" points
color=blue,
dotted,
mark size=2pt,
mark=square*,
mark options={solid,fill=blue},
] table [
% use just the coordindex' as x coordinate,
% the correct labeling is done with xticklabels from table'
x expr=\coordindex,
y=val1,
] {data.txt};
\label{label-plot1}

\begin{scope}[
% because the reference point will be the lower axis line the
% nodes have to be moved a bit more down to don't overlap with
% the xticklabels'
every node/.append style={
below=3.2ex,align=center,
},
]
\node at (axis cs:1,\pgfkeysvalueof{/pgfplots/ymin}) {System A\\foo};
\node at (axis cs:5,\pgfkeysvalueof{/pgfplots/ymin}) {System B\\foo};
\node at (axis cs:9,\pgfkeysvalueof{/pgfplots/ymin}) {System C\\foo};
\node at (axis cs:13,\pgfkeysvalueof{/pgfplots/ymin}) {System D\\foo};
\node at (axis cs:17,\pgfkeysvalueof{/pgfplots/ymin}) {System E\\foo};
\end{scope}

\end{axis}

% Frame 2
\begin{axis}[
color = red,
% set the width' of the plot to the maximum length ...
width=\textwidth,
% ... and use half this length for the height'
height=0.5\textwidth,
scale only axis,
/pgf/number format/.cd,
use comma,
1000 sep={\,},
ymin=-5.0,
ymax=375.0,
ytick={0.0, 50.0, ..., 350.0},
% use data' for the positioning of the xticks' ...
xtick=data,
% ... and use table data for labeling the xticks'
xticklabels from table={data.txt}{n},
% add extra ticks "at the empty entries to add the vertical lines
extra x ticks={3,7,11,15},
% this ticks shouldn't be labeled ...
extra x tick labels={},
% ... but grid lines should be drawn without the tick lines
extra x tick style={
grid=major,
major tick length=0pt,
},
ylabel={Y Axis 2 Label},
% because of the category labels, shift the xlabel' a bit down
xlabel style={
yshift=-4ex,
},
ytick style={red},
axis x line*=none,
axis y line*=right,
ylabel style = red,
% ... and with that you also have to adjust the x limits
enlarge x limits={abs=1},
% set clip mode' to individual' so the category labels aren't clipped away
clip mode=individual,
hide x axis
]

% plot the "red" points
color=red,
dotted,
mark size=2pt,
mark=triangle*,
mark options={solid,fill=red}
] table [
x expr=\coordindex,
y=val2,
] {data.txt};
\label{label-plot2}

\end{axis}

% Frame 3
\begin{axis}[
color = green,
% set the width' of the plot to the maximum length ...
width=\textwidth,
% ... and use half this length for the height'
height=0.5\textwidth,
scale only axis,
/pgf/number format/.cd,
use comma,
1000 sep={\,},
ymin=-50,
ymax=3750.,
ytick={0, 500, ..., 3500},
% use data' for the positioning of the xticks' ...
xtick=data,
% ... and use table data for labeling the xticks'
xticklabels from table={data.txt}{n},
% add extra ticks "at the empty entries to add the vertical lines
extra x ticks={3,7,11,15},
% this ticks shouldn't be labeled ...
extra x tick labels={},
% ... but grid lines should be drawn without the tick lines
extra x tick style={
grid=major,
major tick length=0pt,
},
ylabel={Y Axis 3 Label},
% because of the category labels, shift the xlabel' a bit down
xlabel style={
yshift=-4ex,
},
ylabel style = green,
ytick style={green},
axis x line*=none,
axis y line*=right,
% ... and with that you also have to adjust the x limits
enlarge x limits={abs=1},
% set clip mode' to individual' so the category labels aren't clipped away
clip mode=individual,
hide x axis,
every outer y axis line/.style={
green,
xshift=2.0cm
},
every tick/.style={
green,
xshift=2.0cm
},
every y tick label/.append style={
green,
xshift=2.0cm
},
]

% plot the legend

% plot the "green" points
color=green,
dotted,
mark size=2pt,
mark=diamond*,
mark options={solid,fill=green},
] table [
x expr=\coordindex,
y=val3,
] {data.txt};
\label{label-plot3}

\end{axis}

\end{tikzpicture}
\end{document}

• Thank you very much! That's what I was looking for! – TomTom Jan 20 '19 at 14:53


text:
# Effect of differencing on prediction intervals in ARIMA models

I am sitting with a couple of time-series that I am analysing using ARIMA models. I have a question regarding prediction intervals. When predicting using a model that takes a first difference (a SARIMA(1,1,0)x(1,0,0) model), I get an increasing size of the prediction interval. Without I get a very constant and narrow band (see below):

The corresponding results are as follows:

Can anyone explain why the band is so constant? First I thought it was because of a large significant MA coefficient. This, however, I removed and the "problem" persisted. Then I though it was because the ARIMA without differencing automatically included an intercept. However, again, when I specified include.mean = FALSE, nothing changed.

Any help would be appreciated.

• When forecasting, you deal with prediction intervals rather than confidence intervals. The two are not the same. – Richard Hardy Aug 17 '16 at 13:46
• Hey Richard. Thanks for pointing out my imprecision. I have now edited the post accordingly – pkpkPPkafa Aug 18 '16 at 7:02


text:
# Lagrangian density

1. May 2, 2014

### Geometry_dude

Now this is a bit of a mix of a math and a physics question, but I think it is best asked here.

Assume we are given a Lorentzian manifold $(Q, g)$ together with a metric connection $\nabla$. Naturally we define geodesics $\gamma$ via
$$\nabla_{\dot \gamma} \dot \gamma = 0 \quad ,$$
$$0=\ddot \gamma^k + \Gamma^k{}_{(ij)} \, \dot \gamma^i \dot \gamma^j \quad ,$$
where $\dot \gamma = \frac{d \gamma}{d \tau}$.
Alternatively, we could have defined the Lagrangian $L \in C^\infty(TQ,\mathbb{R})$, i.e. a function on the tangent bundle, by
$$L(q,\dot q) = \frac{1}{2} g_{q}(\dot q, \dot q)$$
for $(q,\dot q) \in TQ$ and then varied the action
$$S(\gamma) = \int_{\gamma} L (\gamma, \dot \gamma) \, d t$$
leading to the same local geodesic equation.

Now, in physics, one makes the step to "field theory" by considering a function $\phi \in C^\infty(Q, \mathbb R)$ and then taking a new functional
$$\tilde S (\phi) = \int_Q \mathcal L (\phi, d \phi) \, \mu$$
where $\mu \in \Omega^n(Q)$ is the canonical volume form. $\mathcal L$ is a function from some space of smooth functions cartesian product with their exterior derivative to $\mathbb R$ and is called the Lagrange density.
My question is:
How do the two formalisms connect? What justifies us in calling $\tilde S$ the action as well? Is there a mathematical discipline dealing with such systems $(Q,g, \mathcal L)$?

Last edited: May 2, 2014
2. May 2, 2014

### Ben Niehoff

Careful! These two definitions are not always equivalent. The first condition

$$\nabla_{\dot \gamma} \dot \gamma = 0 \quad ,$$
defines "autoparallels", or curves whose tangent vector is parallel-transported along itself.

The second condition minimizing the action

$$S(\gamma) = \int_{\gamma} L (\gamma, \dot \gamma) \, d t$$
defines a "locally length-extremizing curve". If the connection is Levi-Civita (i.e., both metric and torsion-free --- these are independent conditions!), then these two definitions for "geodesics" coincide. But for other connections, they need not define the same sets of curves.

This question has basically nothing to do with the first half of your post, so it's not clear to me why you've put these things together.

An action is any scalar quantity that is a functional from some function space $\mathcal{F}$ into $\mathbb{R}$. Extremizing the action with respect to its arguments is a standard variational problem.

In the length-minimizing curve problem, the function space $\mathcal F$ is the space of functions $\gamma : \mathbb{R} \to M$ where $M$ is the manifold.

In the field theory problem, the function space $\mathcal F$ is the space of functions $\phi : M \to \mathbb{R}$. One can also write actions involving functions from $M$ into $TM$ or $T^*M$, or tensor products of those (i.e., particles with integer spin), or functions into the spin bundle, if it exists* (particles with half-integer spin), or functions into Lie algebras (gauge theories), or even functions into manifolds (sigma models).

* Not every manifold admits a spin structure.

3. May 2, 2014

### Matterwave

I'm not sure what you mean with your question...$\tilde{S}$ is called the action if it is the correct action leading to the correct equations of motion. It is the action for the field, given you have the correct Lagrange density. This is why we call any S "THE action". The $S$ you wrote earlier is, up to a constant prefactor (-mc), the action for a free particle (but I think your Lagrangian is meant to have a square-root, instead of a 1/2?). If your action $\tilde{S}$ is the action for a free field, then one needs one more action term describing the interaction of the field and the particles. For the electro-magnetic field, for example, this action term is:

$S_{int}=-\frac{e}{c}\int A_\mu dx^\mu$

Varying the total action with respect to the fields only gives you the field equations, and varying the total action with respect to the matter particles only gives you the equations of motion of the particles.

I'm not sure if this was your question though.

4. May 3, 2014

### Geometry_dude

The torsion does not have an effect on the shape of geodesics.
Proof: Let $(.)$ and $[.]$ denote the anti-symmetrization w.r.t. enclosed indicies.
$$\Gamma^k{}_{ij} \, \dot \gamma^i \, \dot \gamma^j = \Gamma^k{}_{ij} \, \dot \gamma^{(i} \, \dot \gamma^{j)} = \Gamma^k{}_{(ij)} \, \dot \gamma^i \, \dot \gamma^j$$
Since
$$\Gamma^k{}_{ij}= \Gamma^k{}_{(ij)} + \Gamma^k{}_{[ij]}$$
and
$$T^k{}_{ij}= 2 \Gamma^k{}_{[ij]}$$
by definition of torsion, it is only relevant for parallel transport of vectors, not for geodesics.

Hence the definition of the connection being metric is enough to uniquely define geodesics. The metric assumption gives the conceptual link to straight lines.

The reason is that they are conceptually related and somehow (the how is my question) the second follows by considering "infinite degrees of freedom". For example, a free Klein-Gordon field has the density
$$\mathcal L (q, \phi, d \phi) = \frac{1}{2} g^{-1}_q (d \phi, d \phi) - (m^2+ \xi \mathcal S) \phi^2 (q)$$
where $m$ is the mass (a constant), $\mathcal S$ the Ricci scalar and $\xi$ some parameter. The first summand seems very similar to the free particle Lagrangian I mentioned above indicating a deeper relation.

In the Wikipedia article on Lagrangians they do everything in a Newtonian formulation (take $Q= \mathbb R ^3$ and $g= \delta$, the standard inner product with respect to the frame induced by the standard chart) and s.t. $t= \tau$ (the parameter of $\gamma$ is ordinary Newtonian time $t$) and then they simply set
$$L(t) = \int_{\mathbb R^3} \mathcal L (x,y,z,t) \, dx dy dz$$
but this seems really strange and false from a conceptual point of view to me.
So my question is, what is the conceptual relation between $L$ and $\mathcal L$?

Yes, this makes sense. Then a general "classical (relativistic) field" in physics is just a section of a smooth fiber bundle, where the base space is a "spacetime" (that is a manifold $Q$ equipped with a Lorentzian metric $g$ and a metric connection as discussed above) and a field action is just a real functional on those sections. So what you are suggesting is that the field action and the "curve action" just carry this name, because they are both functionals on spaces of (smooth) functions, is that correct?

Sorry, I forgot to mention that I hid the mass in the metric. If you don't like that, you get
$$L(q,\dot q)= \frac{m}{2} g_q (\dot q, \dot q)$$
for the motion of a mass point on straight lines. When g is unitless, then this has unit energy, as it should have.
What you probably mean is the proper time $\Delta \tau$ for a curve $\gamma$ parametrized by $s$ from $s_0$ to $s_1$
$$\Delta \tau (\gamma) = \frac{1}{c^2}\int_{s_0}^{s_1} \sqrt{g_{\gamma(s)}(\dot \gamma (s), \dot \gamma (s))} \, d s$$
where I wrote $\dot \gamma = \frac{d \gamma}{d s}$ and $s$ has physical dimension of time.
Note that I use the $(+,-,-,-)$ convention.
If you vary this integral and let the variation vanish, you should also get the geodesic equation, but I haven't checked this.
$$d F=0$$
am I correct?

5. May 3, 2014

### Ben Niehoff

The relation between fields and particles is the "eikonal approximation". A particle is a localized excitation of a field that travels along the characteristics (I think) of the field equation. This is the relationship between wave and ray optics.

If you assume a localized ("delta function") excitation of the field and pullback to its worldline, then the field action becomes the particle action. Try it with the free particle.

Yes.

You do. Classically it doesn't matter if you use the length functional, the squared length functional, or even the length-raised-to-any-power functional. However, quantum mechanically, the specific form of the action tells us how the system must be quantized, and these are not necessarily equivalent (even the constant out front makes a difference).

No, he gave the interaction part of the action which should give the Lorentz force. But his integral needs to be pulled back to the worldline of some particle. Try it, it's instructive.

$F$ is typically defined as $dA$, so you won't get that out of an action. The action for Maxwell's equations is

$$S[A] = \int_\mathcal{M} \bigg( -\frac12 \, F \wedge \star F - \frac12 \, A \wedge \star J \bigg),$$
where $F \equiv d A$. If you vary this action with respect to $A$, you should get the other half of Maxwell's equations, that couple $F$ to the external source $J$.

$\star$ is the Hodge dual, and note that I use $(-,+,+,+)$ signature (so all kinetic terms come in with a coefficient of $-1/2$). I have also set the speed of light and the elementary charge to 1.

6. May 3, 2014

### Matterwave

The usual free-particle action is simply:

$$S=-mc\int ds$$

This is good for a free particle in special relativity, general relativity, or classical mechanics, just take the appropriate form of ds. In terms of the metric, for a massive particle, then this action is:

$$S=-mc\int \sqrt{g(T,T)}dt$$

Where T is the tangent vector to the particle's world line, with t being the parametrization of that worldline. Varying this action and setting it equal to 0 will yield the geodesic equations because free particles travel along geodesics.

Now if there is a background E&M field as well as the gravitational field, we have to add the interaction term I showed you earlier to the action, and vary both this term and the interaction term together with respect to the particle's coordinates. This will give you the motion of this particle, which will no longer be geodesic motion due to the presence of the EM field. We have to vary the action:

$$S=-mc\int ds-\frac{e}{c}\int A_\mu dx^\mu$$

It will result in the equations for the 4-velocity of this particle:

$$\nabla_u u=\frac{e}{m}F(\quad,u)$$

At this point, we have only cared about the motion of the (test) particle due to a gravitational field, described by the metric, and a E&M field, described by the field tensor. We have NOT talked at all about how the fields interact! To do that, we look at the Field Lagrangians. The free EM field Action is:

$$S_{em}=-\frac{1}{16\pi c}\int F_{\mu\nu} F^{\mu\nu} \sqrt{-g}d^4 x$$

This is the same equation as Ben gave, but in different units (Gaussian units) and his action included the interaction term I mentioned before. This free EM field action by itself, when varied with respect to it's degrees of freedom (the 4-potential) will give the vacuum Maxwell equations. Incorporating the interaction term gives the non-vacuum Maxwell equations. The last thing we have to add is the action of the gravitational field itself, given by the Einstein-Hilbert action:

$$S_{g}=\frac{c^4}{16\pi G}\int R\sqrt{-g}d^4 x$$

For the gravitational field, we don't have to include some such "coupling" terms as for the EM field, since the couplings were already done by changing the geometry. So our total action is:

$$S=-\sum_{particles}mc\int ds-\sum_{particles}\frac{e}{c}\int A_\mu dx^\mu--\frac{1}{16\pi c}\int F_{\mu\nu} F^{\mu\nu} \sqrt{-g}d^4 x+\frac{c^4}{16\pi G}\int R\sqrt{-g}d^4 x$$

Variations of this total action with respect to all degrees of freedom should yield, 1) the particles' equation of motion in the EM and gravitational fields, 2) the non-vacuum Maxwell's equations in curved spacetime, and 3) the Einstein field equations with stress-energy from the EM field as well as the particles. Do I actually want to try to do this all in one go? Heck no. Heeeeck no.

The main point is. There are actions for the particles, and for the fields, one will give you E.O.M. of the particles, and the other will give you the field equations. For the former, you have that the action is the integral of a Lagrangian. For the latter, the action is the integral of a Lagrangian, which itself is the integral of a Lagrange density.

7. May 4, 2014

### Geometry_dude

While I don't see why one would need quantum theory for this, I think I understand what you mean. The basic idea is taking the step to distributions.
So I was wrong in saying that $\phi$ is a function, but it should be a distribution acting on test functions $f \in C^\infty_{\text{test}}(Q, \mathbb R )$ on the Lorentzian manifold. So the first step is doing this for $1$ particle:
$$\gamma (\tau) \to \phi :=\delta_{\gamma(\tau)}$$
with
$$\delta_{\gamma(\tau)}(f) = f \circ \gamma \, (\tau)$$.
Then $\mathcal L$ becomes a functional on the space of distributions and I somehow need to choose $\mathcal L$ such that whatever I get out of
$$\partial_i \frac{\partial \mathcal L}{\partial(\partial_i \phi)}= \frac{\partial \mathcal L}{\partial \phi}$$
becomes equivalent to
$$\frac{d}{d \tau} \frac{\partial L}{\partial \dot q^i}= \frac{\partial L}{\partial q^i}$$
for the above $\delta$-distribution, right? This would directly give me the continuum generalization of the Lagrangian to the Lagrange density.
How do I get the exterior derivative on the space of distributions? The fact that all this is happening of a general manifold further complicates the matter.

I think I do get the general idea now, though. Now I only have problems with the implementation. Thank you for your help.

Matterwave, thank you for this exposition of the Lagrangian formulation of field theory in classical mechanics. Do you know whether there have there been any successful attempts to include also the self-fields of the particles?

8. May 4, 2014

### Matterwave

It sort of depends on what you mean by including the self-fields. For example, in E&M, one can calculate the effect of a particle's field on itself by energy and momenta considerations. An accelerating particle emits radiation, which carry away energy and momentum. By integrating over the field's energy and momentum, and by conservation of energy and momentum, one can then look at how that particle will react (to its own field).

However, as the field itself usually diverges at the point of the particle's location, it's quite a hairy situation one finds oneself in if one wants to, for example, use the Lorentz force law on the particle for its own field. It is beyond my knowledge of if there are successful ways of consistently removing these infinities.


text:
# VarHT: Variance estimator of Horvitz - Thompson estimator In Frames2: Estimation in Dual Frame Surveys

## Description

Computes the variance estimator of Horvitz - Thompson estimator of population total

## Usage

 1 VarHT(y, pikl)

## Arguments

 y A numeric vector of size n containing information about variable of interest pikl A square numeric matrix of dimension n containing first and second order inclusion probabilities for units included in y

## Details

Variance estimator of Horvitz - Thompson estimator of population total is given by

\hat{Var}(\hat{Y}_{HT}) = ∑_{k \in s}\frac{y_k^2}{π_k^2}(1 - π_k) + ∑_{k \in s}∑_{l \in s, l \neq k} \frac{y_k y_l}{π_k π_l} \frac{π_{kl} - π_k π_l}{π_{kl}}

## Value

A numeric value representing variance estimator of Horvitz - Thompson estimator for population total for considered values

## References

Horvitz, D. G. and Thompson, D. J. (1952) A generalization of sampling without replacement from a finite universe. Journal of the American Statistical Association, 47, 663 - 685

Sarndal, C. E., Swensson, B. and Wretman, J. (1992) Model Assisted Survey Sampling. Springer-Verlag. New York.

HT CovHT
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ########## Example 1 ########## U <- c(13, 18, 20, 14, 9) #A simple random sample of size 2 without replacement is drawn from population s <- sample(U, 2) #Horvitz - Thompson estimator of population total is calculated. ps <- c(0.4, 0.4) HT(s, ps) #Now, we calculate variance estimator of the Horvitz - Thompson estimator. Ps <- matrix(c(0.4,0.1, 0.1,0.4), 2 ,2) VarHT(s, Ps) ########## Example 2 ########## data(DatA) attach(DatA) data(PiklA) #Let calculate Horvitz - Thompson estimator for total of variable Clothing in Frame A. HT(Clo, ProbA) #And now, let compute the variance of the previous estimator VarHT(Clo, PiklA)


text:
# What is the distance between (–6, 3, 4)  and (–1, 1, 3) ?

$\sqrt{30}$
$\sqrt{{\left(x 1 - x 2\right)}^{2} + {\left(y 1 - y 2\right)}^{2} + {\left(z 1 - z 2\right)}^{2}}$
$\sqrt{{\left(- 6 + 1\right)}^{2} + {\left(3 - 1\right)}^{2} + {\left(4 - 3\right)}^{2}}$
=$\sqrt{30}$


text:
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## G = D7×C2×C12order 336 = 24·3·7

### Direct product of C2×C12 and D7

Series: Derived Chief Lower central Upper central

 Derived series C1 — C7 — D7×C2×C12
 Chief series C1 — C7 — C14 — C42 — C6×D7 — C2×C6×D7 — D7×C2×C12
 Lower central C7 — D7×C2×C12
 Upper central C1 — C2×C12

Generators and relations for D7×C2×C12
G = < a,b,c,d | a2=b12=c7=d2=1, ab=ba, ac=ca, ad=da, bc=cb, bd=db, dcd=c-1 >

Subgroups: 336 in 108 conjugacy classes, 70 normal (22 characteristic)
C1, C2, C2, C2, C3, C4, C4, C22, C22, C6, C6, C6, C7, C2×C4, C2×C4, C23, C12, C12, C2×C6, C2×C6, D7, C14, C14, C22×C4, C21, C2×C12, C2×C12, C22×C6, Dic7, C28, D14, C2×C14, C3×D7, C42, C42, C22×C12, C4×D7, C2×Dic7, C2×C28, C22×D7, C3×Dic7, C84, C6×D7, C2×C42, C2×C4×D7, C12×D7, C6×Dic7, C2×C84, C2×C6×D7, D7×C2×C12
Quotients: C1, C2, C3, C4, C22, C6, C2×C4, C23, C12, C2×C6, D7, C22×C4, C2×C12, C22×C6, D14, C3×D7, C22×C12, C4×D7, C22×D7, C6×D7, C2×C4×D7, C12×D7, C2×C6×D7, D7×C2×C12

Smallest permutation representation of D7×C2×C12
On 168 points
Generators in S168
(1 126)(2 127)(3 128)(4 129)(5 130)(6 131)(7 132)(8 121)(9 122)(10 123)(11 124)(12 125)(13 72)(14 61)(15 62)(16 63)(17 64)(18 65)(19 66)(20 67)(21 68)(22 69)(23 70)(24 71)(25 51)(26 52)(27 53)(28 54)(29 55)(30 56)(31 57)(32 58)(33 59)(34 60)(35 49)(36 50)(37 108)(38 97)(39 98)(40 99)(41 100)(42 101)(43 102)(44 103)(45 104)(46 105)(47 106)(48 107)(73 158)(74 159)(75 160)(76 161)(77 162)(78 163)(79 164)(80 165)(81 166)(82 167)(83 168)(84 157)(85 146)(86 147)(87 148)(88 149)(89 150)(90 151)(91 152)(92 153)(93 154)(94 155)(95 156)(96 145)(109 138)(110 139)(111 140)(112 141)(113 142)(114 143)(115 144)(116 133)(117 134)(118 135)(119 136)(120 137)
(1 2 3 4 5 6 7 8 9 10 11 12)(13 14 15 16 17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45 46 47 48)(49 50 51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70 71 72)(73 74 75 76 77 78 79 80 81 82 83 84)(85 86 87 88 89 90 91 92 93 94 95 96)(97 98 99 100 101 102 103 104 105 106 107 108)(109 110 111 112 113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128 129 130 131 132)(133 134 135 136 137 138 139 140 141 142 143 144)(145 146 147 148 149 150 151 152 153 154 155 156)(157 158 159 160 161 162 163 164 165 166 167 168)
(1 115 15 104 161 152 33)(2 116 16 105 162 153 34)(3 117 17 106 163 154 35)(4 118 18 107 164 155 36)(5 119 19 108 165 156 25)(6 120 20 97 166 145 26)(7 109 21 98 167 146 27)(8 110 22 99 168 147 28)(9 111 23 100 157 148 29)(10 112 24 101 158 149 30)(11 113 13 102 159 150 31)(12 114 14 103 160 151 32)(37 80 95 51 130 136 66)(38 81 96 52 131 137 67)(39 82 85 53 132 138 68)(40 83 86 54 121 139 69)(41 84 87 55 122 140 70)(42 73 88 56 123 141 71)(43 74 89 57 124 142 72)(44 75 90 58 125 143 61)(45 76 91 59 126 144 62)(46 77 92 60 127 133 63)(47 78 93 49 128 134 64)(48 79 94 50 129 135 65)
(1 33)(2 34)(3 35)(4 36)(5 25)(6 26)(7 27)(8 28)(9 29)(10 30)(11 31)(12 32)(13 159)(14 160)(15 161)(16 162)(17 163)(18 164)(19 165)(20 166)(21 167)(22 168)(23 157)(24 158)(49 128)(50 129)(51 130)(52 131)(53 132)(54 121)(55 122)(56 123)(57 124)(58 125)(59 126)(60 127)(61 75)(62 76)(63 77)(64 78)(65 79)(66 80)(67 81)(68 82)(69 83)(70 84)(71 73)(72 74)(85 138)(86 139)(87 140)(88 141)(89 142)(90 143)(91 144)(92 133)(93 134)(94 135)(95 136)(96 137)(109 146)(110 147)(111 148)(112 149)(113 150)(114 151)(115 152)(116 153)(117 154)(118 155)(119 156)(120 145)

G:=sub<Sym(168)| (1,126)(2,127)(3,128)(4,129)(5,130)(6,131)(7,132)(8,121)(9,122)(10,123)(11,124)(12,125)(13,72)(14,61)(15,62)(16,63)(17,64)(18,65)(19,66)(20,67)(21,68)(22,69)(23,70)(24,71)(25,51)(26,52)(27,53)(28,54)(29,55)(30,56)(31,57)(32,58)(33,59)(34,60)(35,49)(36,50)(37,108)(38,97)(39,98)(40,99)(41,100)(42,101)(43,102)(44,103)(45,104)(46,105)(47,106)(48,107)(73,158)(74,159)(75,160)(76,161)(77,162)(78,163)(79,164)(80,165)(81,166)(82,167)(83,168)(84,157)(85,146)(86,147)(87,148)(88,149)(89,150)(90,151)(91,152)(92,153)(93,154)(94,155)(95,156)(96,145)(109,138)(110,139)(111,140)(112,141)(113,142)(114,143)(115,144)(116,133)(117,134)(118,135)(119,136)(120,137), (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81,82,83,84)(85,86,87,88,89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132)(133,134,135,136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153,154,155,156)(157,158,159,160,161,162,163,164,165,166,167,168), (1,115,15,104,161,152,33)(2,116,16,105,162,153,34)(3,117,17,106,163,154,35)(4,118,18,107,164,155,36)(5,119,19,108,165,156,25)(6,120,20,97,166,145,26)(7,109,21,98,167,146,27)(8,110,22,99,168,147,28)(9,111,23,100,157,148,29)(10,112,24,101,158,149,30)(11,113,13,102,159,150,31)(12,114,14,103,160,151,32)(37,80,95,51,130,136,66)(38,81,96,52,131,137,67)(39,82,85,53,132,138,68)(40,83,86,54,121,139,69)(41,84,87,55,122,140,70)(42,73,88,56,123,141,71)(43,74,89,57,124,142,72)(44,75,90,58,125,143,61)(45,76,91,59,126,144,62)(46,77,92,60,127,133,63)(47,78,93,49,128,134,64)(48,79,94,50,129,135,65), (1,33)(2,34)(3,35)(4,36)(5,25)(6,26)(7,27)(8,28)(9,29)(10,30)(11,31)(12,32)(13,159)(14,160)(15,161)(16,162)(17,163)(18,164)(19,165)(20,166)(21,167)(22,168)(23,157)(24,158)(49,128)(50,129)(51,130)(52,131)(53,132)(54,121)(55,122)(56,123)(57,124)(58,125)(59,126)(60,127)(61,75)(62,76)(63,77)(64,78)(65,79)(66,80)(67,81)(68,82)(69,83)(70,84)(71,73)(72,74)(85,138)(86,139)(87,140)(88,141)(89,142)(90,143)(91,144)(92,133)(93,134)(94,135)(95,136)(96,137)(109,146)(110,147)(111,148)(112,149)(113,150)(114,151)(115,152)(116,153)(117,154)(118,155)(119,156)(120,145)>;

G:=Group( (1,126)(2,127)(3,128)(4,129)(5,130)(6,131)(7,132)(8,121)(9,122)(10,123)(11,124)(12,125)(13,72)(14,61)(15,62)(16,63)(17,64)(18,65)(19,66)(20,67)(21,68)(22,69)(23,70)(24,71)(25,51)(26,52)(27,53)(28,54)(29,55)(30,56)(31,57)(32,58)(33,59)(34,60)(35,49)(36,50)(37,108)(38,97)(39,98)(40,99)(41,100)(42,101)(43,102)(44,103)(45,104)(46,105)(47,106)(48,107)(73,158)(74,159)(75,160)(76,161)(77,162)(78,163)(79,164)(80,165)(81,166)(82,167)(83,168)(84,157)(85,146)(86,147)(87,148)(88,149)(89,150)(90,151)(91,152)(92,153)(93,154)(94,155)(95,156)(96,145)(109,138)(110,139)(111,140)(112,141)(113,142)(114,143)(115,144)(116,133)(117,134)(118,135)(119,136)(120,137), (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81,82,83,84)(85,86,87,88,89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130,131,132)(133,134,135,136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153,154,155,156)(157,158,159,160,161,162,163,164,165,166,167,168), (1,115,15,104,161,152,33)(2,116,16,105,162,153,34)(3,117,17,106,163,154,35)(4,118,18,107,164,155,36)(5,119,19,108,165,156,25)(6,120,20,97,166,145,26)(7,109,21,98,167,146,27)(8,110,22,99,168,147,28)(9,111,23,100,157,148,29)(10,112,24,101,158,149,30)(11,113,13,102,159,150,31)(12,114,14,103,160,151,32)(37,80,95,51,130,136,66)(38,81,96,52,131,137,67)(39,82,85,53,132,138,68)(40,83,86,54,121,139,69)(41,84,87,55,122,140,70)(42,73,88,56,123,141,71)(43,74,89,57,124,142,72)(44,75,90,58,125,143,61)(45,76,91,59,126,144,62)(46,77,92,60,127,133,63)(47,78,93,49,128,134,64)(48,79,94,50,129,135,65), (1,33)(2,34)(3,35)(4,36)(5,25)(6,26)(7,27)(8,28)(9,29)(10,30)(11,31)(12,32)(13,159)(14,160)(15,161)(16,162)(17,163)(18,164)(19,165)(20,166)(21,167)(22,168)(23,157)(24,158)(49,128)(50,129)(51,130)(52,131)(53,132)(54,121)(55,122)(56,123)(57,124)(58,125)(59,126)(60,127)(61,75)(62,76)(63,77)(64,78)(65,79)(66,80)(67,81)(68,82)(69,83)(70,84)(71,73)(72,74)(85,138)(86,139)(87,140)(88,141)(89,142)(90,143)(91,144)(92,133)(93,134)(94,135)(95,136)(96,137)(109,146)(110,147)(111,148)(112,149)(113,150)(114,151)(115,152)(116,153)(117,154)(118,155)(119,156)(120,145) );

G=PermutationGroup([[(1,126),(2,127),(3,128),(4,129),(5,130),(6,131),(7,132),(8,121),(9,122),(10,123),(11,124),(12,125),(13,72),(14,61),(15,62),(16,63),(17,64),(18,65),(19,66),(20,67),(21,68),(22,69),(23,70),(24,71),(25,51),(26,52),(27,53),(28,54),(29,55),(30,56),(31,57),(32,58),(33,59),(34,60),(35,49),(36,50),(37,108),(38,97),(39,98),(40,99),(41,100),(42,101),(43,102),(44,103),(45,104),(46,105),(47,106),(48,107),(73,158),(74,159),(75,160),(76,161),(77,162),(78,163),(79,164),(80,165),(81,166),(82,167),(83,168),(84,157),(85,146),(86,147),(87,148),(88,149),(89,150),(90,151),(91,152),(92,153),(93,154),(94,155),(95,156),(96,145),(109,138),(110,139),(111,140),(112,141),(113,142),(114,143),(115,144),(116,133),(117,134),(118,135),(119,136),(120,137)], [(1,2,3,4,5,6,7,8,9,10,11,12),(13,14,15,16,17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45,46,47,48),(49,50,51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70,71,72),(73,74,75,76,77,78,79,80,81,82,83,84),(85,86,87,88,89,90,91,92,93,94,95,96),(97,98,99,100,101,102,103,104,105,106,107,108),(109,110,111,112,113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128,129,130,131,132),(133,134,135,136,137,138,139,140,141,142,143,144),(145,146,147,148,149,150,151,152,153,154,155,156),(157,158,159,160,161,162,163,164,165,166,167,168)], [(1,115,15,104,161,152,33),(2,116,16,105,162,153,34),(3,117,17,106,163,154,35),(4,118,18,107,164,155,36),(5,119,19,108,165,156,25),(6,120,20,97,166,145,26),(7,109,21,98,167,146,27),(8,110,22,99,168,147,28),(9,111,23,100,157,148,29),(10,112,24,101,158,149,30),(11,113,13,102,159,150,31),(12,114,14,103,160,151,32),(37,80,95,51,130,136,66),(38,81,96,52,131,137,67),(39,82,85,53,132,138,68),(40,83,86,54,121,139,69),(41,84,87,55,122,140,70),(42,73,88,56,123,141,71),(43,74,89,57,124,142,72),(44,75,90,58,125,143,61),(45,76,91,59,126,144,62),(46,77,92,60,127,133,63),(47,78,93,49,128,134,64),(48,79,94,50,129,135,65)], [(1,33),(2,34),(3,35),(4,36),(5,25),(6,26),(7,27),(8,28),(9,29),(10,30),(11,31),(12,32),(13,159),(14,160),(15,161),(16,162),(17,163),(18,164),(19,165),(20,166),(21,167),(22,168),(23,157),(24,158),(49,128),(50,129),(51,130),(52,131),(53,132),(54,121),(55,122),(56,123),(57,124),(58,125),(59,126),(60,127),(61,75),(62,76),(63,77),(64,78),(65,79),(66,80),(67,81),(68,82),(69,83),(70,84),(71,73),(72,74),(85,138),(86,139),(87,140),(88,141),(89,142),(90,143),(91,144),(92,133),(93,134),(94,135),(95,136),(96,137),(109,146),(110,147),(111,148),(112,149),(113,150),(114,151),(115,152),(116,153),(117,154),(118,155),(119,156),(120,145)]])

120 conjugacy classes

 class 1 2A 2B 2C 2D 2E 2F 2G 3A 3B 4A 4B 4C 4D 4E 4F 4G 4H 6A ··· 6F 6G ··· 6N 7A 7B 7C 12A ··· 12H 12I ··· 12P 14A ··· 14I 21A ··· 21F 28A ··· 28L 42A ··· 42R 84A ··· 84X order 1 2 2 2 2 2 2 2 3 3 4 4 4 4 4 4 4 4 6 ··· 6 6 ··· 6 7 7 7 12 ··· 12 12 ··· 12 14 ··· 14 21 ··· 21 28 ··· 28 42 ··· 42 84 ··· 84 size 1 1 1 1 7 7 7 7 1 1 1 1 1 1 7 7 7 7 1 ··· 1 7 ··· 7 2 2 2 1 ··· 1 7 ··· 7 2 ··· 2 2 ··· 2 2 ··· 2 2 ··· 2 2 ··· 2

120 irreducible representations

 dim 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 type + + + + + + + + image C1 C2 C2 C2 C2 C3 C4 C6 C6 C6 C6 C12 D7 D14 D14 C3×D7 C4×D7 C6×D7 C6×D7 C12×D7 kernel D7×C2×C12 C12×D7 C6×Dic7 C2×C84 C2×C6×D7 C2×C4×D7 C6×D7 C4×D7 C2×Dic7 C2×C28 C22×D7 D14 C2×C12 C12 C2×C6 C2×C4 C6 C4 C22 C2 # reps 1 4 1 1 1 2 8 8 2 2 2 16 3 6 3 6 12 12 6 24

Matrix representation of D7×C2×C12 in GL3(𝔽337) generated by

 336 0 0 0 1 0 0 0 1
,
 208 0 0 0 189 0 0 0 189
,
 1 0 0 0 0 1 0 336 227
,
 336 0 0 0 304 110 0 33 33
G:=sub<GL(3,GF(337))| [336,0,0,0,1,0,0,0,1],[208,0,0,0,189,0,0,0,189],[1,0,0,0,0,336,0,1,227],[336,0,0,0,304,33,0,110,33] >;

D7×C2×C12 in GAP, Magma, Sage, TeX

D_7\times C_2\times C_{12}
% in TeX

G:=Group("D7xC2xC12");
// GroupNames label

G:=SmallGroup(336,175);
// by ID

G=gap.SmallGroup(336,175);
# by ID

G:=PCGroup([6,-2,-2,-2,-3,-2,-7,122,10373]);
// Polycyclic

G:=Group<a,b,c,d|a^2=b^12=c^7=d^2=1,a*b=b*a,a*c=c*a,a*d=d*a,b*c=c*b,b*d=d*b,d*c*d=c^-1>;
// generators/relations

׿
×
𝔽


text:
aliquot

< a quantity that can be divided into another a whole number of time />

Today was my first day at my new lab. Everything went fine, despite a very bad night. At least I have been able to go back home without too much dizziness or paresthesia in the legs (I don’t know where this one comes from). Guess what: For the first time in 10 years, I am able to connect my Macbook on the network! #self


text:
# Contents

## Definition

A binary relation from a set $X$ to a set $Y$ is called functional if every element of $X$ is related to at most one element of $Y$. Notice that this is the same thing as a partial function, seen from a different point of view. A (total) function is precisely a relation that is both functional and entire.

## Properties

Like any relation, a functional relation $r$ can be viewed as a span

$\array { & & \Delta_r \\ & \swarrow_{\iota_r} & & \searrow^{\phi_r} \\ X & & & & Y }$

Such a span is a relation iff the pairing map from the domain $\Delta_r$ to $X \times Y$ is an injection, and such a relation is functional iff the inclusion map $\iota_r$ is also an injection. Such a relation is entire iff the inclusion map $\iota_r$ is a surjection.

(Total) functions are precisely left adjoints in the bicategory of relations, in other words relations $r: X \to Y$ in $Rel$ for which there exists $s: Y \to X$ satisfying

$r \circ s \leq 1_Y; \qquad 1_X \leq s \circ r.$

For if a relation $r: X \to Y$ is a function, then

1. $r r^{op} \leq 1_Y$: this just says that if $r^{op}(y, x)$ and $r(x, y')$, then $y = y'$. Equivalently, if $r(x, y)$ and $r(x, y')$ then $y = y'$: this holds precisely because $r$ is functional or well-defined.

2. $1_X \leq r^{op} r$: this says that if $x = x'$ in $X$, then there exists $y$ in $Y$ such that $r(x, y)$ and $r^{op}(y, x')$, i.e., there exists $y$ such that $r(x, y)$ and $r(x', y)$. This holds precisely because a function is an entire relation.

On the other hand, suppose $r: X \to Y$ is a left adjoint.

1. From $1_X \leq s r$ we deduce that if $x = x'$ in $X$, then there exists $y$ in $Y$ such that $r(x, y)$ and $s(y, x')$; in particular $r$ is entire.

2. Suppose $r(x, y)$. From $x = x$ there exists $y'$ such that $r(x, y')$ and $s(y', x)$. Thus $s(y', x)$ and $r(x, y)$; from $s r \leq 1_Y$ we infer $y' = y$. We conclude at most one $y$ satisfies $r(x, y)$, making $r$ functional.

Since $r$ is a function, it has a right adjoint $r$, and by uniqueness of right adjoints, we may conclude $s = r^{op}$. The monad $t = r^{op} r$ has a unit $1_X \leq t$ ($t$ is reflexive) and a multiplication $t t \leq t$ ($t$ is transitive), and also $t^{op} = (r^{op} r)^{op} = r^{op} r^{op\; op} = r^{op} r = t$ ($t$ is symmetric). So the monad $t$ is an equivalence relation. The above reasoning may be internalized to apply to the bicategory of relations internal to a regular category, in which case this equivalence relation is exactly the kernel pair of the map $r$. The comonad $c = r r^{op}$ has a counit $c \leq 1_Y$ ($c$ is coreflexive), and such coreflective relations in $Set$ correspond to subsets of $Y$. More generally, in a regular category, the subobject of $Y$ named by $c$ is the image of $r$ (the coequalizer of the kernel pair).

Further remarks: surjectivity of a function $r: X \to Y$ can be expressed as the condition $1_Y \leq r \circ r^{op}$, and injectivity as $r^{op} \circ r \leq 1_X$.

## References

Last revised on November 21, 2017 at 02:34:19. See the history of this page for a list of all contributions to it.


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## Section48.3Magnifying Glass

Recall that when an object is placed within a focal length of a convex lens its image is virtual, erect and larger than the object. Therefore, when you look at an object through a convex lens when the object is within a focal length distance behind the lens, you find it magnified. When a convex lens is used for this purpose, it is also called a magnifying glass or a simple microscope.

Our perception of magnification is based on the larger angle that the virtual image from the magnifying glass subtends at the eye. Figure 48.3.1 shows a virtual image IQ of the physical object OP. To define the angular magnification that we would perceive, we virtually move the object OP at the same distance as the virtual image IQ. This is shown as $\text{IP}_\text{Q}\text{.}$ Then, we have angle $\theta_\text{P} = \angle{\text{P}_\text{Q}\text{EI}}$ subtended at the eye by moved object and $\theta_\text{Q}=\angle\text{QEI}$ at the eye by the virtual image.

$$\theta_\text{P} = \angle{\text{P}_\text{Q}\text{EI}},\ \ \theta_\text{Q}=\angle\text{QEI}.\tag{48.3.1}$$

The ratio of these two angles is called the angular magnification of the magnifying glass. We denote it by capital letter $M\text{.}$

$$M = \frac{\theta_\text{Q}}{\theta_\text{P}}.\tag{48.3.2}$$

Below, we will discuss critical role that the eye plays in this magnification.

### Subsection48.3.1Role of Near Point of Eye

#### Subsubsection48.3.1.1Placing the Viewing Object

Stretch your arm and look at the tip of one of your fingers, say the index finger. They try bringing your hand closer. You will notice that the tip of the finger is clearest at about $25\text{ cm}$ from your eye. This is the nearest point of your eye. Your eye muscles are contracted in this state.

When you look through a magnifying glass, for the clearest view, you would want the image by the magnifying glass to be at your near point. So, where should you plce the object?

Refer to Figure 48.3.1 as we figure out a formula for where the object be placed for clear viewing. Let us denote the distance from your eye to the near point by $n_p\text{.}$ For a normal eye, $n_p\approx 25\text{cm}\text{.}$ Let $f$ be the focal length of the magnifying glass and $d$ the distance themagnifying glass is from your eye. This gives the image distance from the lens to be $- |n_p-d|\text{.}$

\begin{equation*} q = - |n_p-d| = -(n_p-d) \text{ since }n_p \gt d. \end{equation*}

With these, we can calculate where we should place the object for clear viewing. Let $p$ be the distance from magnifying glass to the physical object.

\begin{equation*} \frac{1}{p} = \frac{1}{f} - \left( - \frac{1}{n_p - d} \right). \end{equation*}

This gives

\begin{equation*} p = \frac{f(n_p-d)}{n_p-d+f}. \end{equation*}

Here $d \lt\lt n_p\text{,}$ and hence $d$ can be ignored.

$$p \approx \frac{f\;n_p}{n_p+f}.\label{eq-placing-the-viewing-object-magnifying-glass}\tag{48.3.3}$$

#### Subsubsection48.3.1.2Role of Eye in the Angular Magnification

We can work out the angular magnification by exploiting the geometry in Figure 48.3.1. Since both the virtual image and the virtually moved object are at the near point of the eye, we get the following for the angles subtended at the eye.

\begin{align*} \amp \tan\theta_\text{P} = \frac{h_\text{o}}{n_p} \\ \amp \tan\theta_\text{Q} = \frac{h_\text{i}}{n_p} = \frac{h_\text{i}}{n_p} \end{align*}

If we make small angle approximation, then, we will have

\begin{align*} \amp \theta_\text{P} \approx \frac{h_\text{o}}{n_p} \\ \amp \theta_\text{Q} \approx \frac{h_\text{i}}{n_p} \end{align*}

Therefore, the angular magnification will be

\begin{equation*} M = \frac{\theta_\text{Q}}{\theta_\text{P}} = \frac{h_\text{i}}{h_\text{o}}. \end{equation*}

That is, the angular magnification in the small angle approximation is nuercally equal to the the lateral magnification of the magnifying glass itself. We can replace $h_\text{i}/h_\text{o}$ by $-q/p$ by the magnifying glass itself.

\begin{equation*} \frac{h_\text{i}}{h_\text{o}} = -\frac{q}{p} = \frac{n_p+f}{f}. \end{equation*}

Therefore, the angular magnification can also be written as

$$M = \frac{n_p+f}{f}.\label{eq-angular-magnification-by-a-magnifying-glass}\tag{48.3.4}$$

This says that furthest is your near point more angular magnification you will perceive for the same magnifying glass. So, if your friend says he sees the magnification only 5x with a magnifier that has focal length $4.5\text{ cm}$ and you say you see magnification 7x. Both of you may be right. Your $n_p \approx 31.5\text{ cm}$ and your friend's $n_p \approx 22.5\text{ cm}\text{.}$

(a) Find the angular magnification of an image by a mgnifying glass of focal length $f=5\text{ cm}\text{.}$ Assume the magnifying glass is next to the eye and the near point of the eye is $25\text{ cm}$ from the eye.

(b) How far behind the magnifying glass should the physical specimen be for clearest view?

Hint

(a) Use Eq. (48.3.4). (b) Use Eq. (48.3.3).

(a) $6\text{,}$ $4.17\text{ cm}\text{.}$

Solution 1 (a)

(a) From Eq. (48.3.4), we see that

\begin{equation*} M = \frac{n_p+f}{f} = \frac{25+5}{5} = 6. \end{equation*}
Solution 2 (b)

(b) From Eq. (48.3.3), we see that

\begin{equation*} p \approx \frac{f\;n_p}{n_p+f} = \frac{5\times 25}{5+25} = 4.17\text{ cm}. \end{equation*}

You perceive an angular magnification of a magnifying glass to be $10\times\text{.}$ Suppose your near point is $25\text{ cm}\text{.}$ What is the power in unit $\text{diopter}$ of the magnifying glass?

Hint

Use Eq. (48.3.4) and convert $1/f$ to power.

$45\text{ diopters}\text{.}$

Solution

Let's write Eq. (48.3.4) in terms of $1/f$ with $f$ in meter. With near point in meter, we get

\begin{equation*} M = 1 + \frac{0.25\text{ m}}{f} = 1 + .25\, P. \end{equation*}

Therefore,

\begin{equation*} P = \frac{M-1}{0.25} = \frac{10-1}{0.25} = 45\text{ diopters}. \end{equation*}

You can see for higher magnification, $M \approx n_p\, P\text{,}$ where $n_p$ is the near point in $m\text{.}$


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# Understanding JavaScript "this" Keyword

I know there are a number of blogs and articles on this particular topic (no pun intended) and it only takes a quick Google search to find the information on how JavaScript’s this keyword actually works.

However, since this blog was primarily started so I could better understand my craft, I’m going to just add another article to the mix because if I can explain a topic, then I understand it. (Yeah…sorry…it’s all about me.)

The reason I even decided to write this article in the first place is I had never really run across problems with the this keyword. In fact, I will admit that I had mistakenly always thought of this to operate how it would as if I were working in Java. Unfortunately (or fortunately, once you understand it), this was not the case, and I ran into an issue that was non-obvious to me until I started doing some research. So, I now present to you the findings of my research and my better understanding of JavaScript’s this keyword.

Because this is a blog about code, let’s start with an example where I created a simple Contact JavaScript object.

Fairly simple, right?

Note: In my particular situation, I was loading the objects using the jQuery.getJSON() function and mapping to it with the jQuery.map() function. But, it still does not change the fact that I can access the various properties of the object through my display function. I am simplifying this example quite a bit because I did not want to distract from the main point of this blog post.

At this point, I decided that I wanted to create another type of contact which was a little more specific:

Pretty straightforward again, right?

Well, this is where things start to get a little tricky with this – particularly if you come from a Java background.

In my previous code example, you will get:

John Doe email is: johndoe@example.

In memory, BusinessContact.display() and Contact.display() reference the same object. But, the this keyword will be whatever object is before the period in the call. So, in the case of the BuisnessContact, it is the business contact information, and in the case of the Contact, it is the contact information.

This is why I had a particular problem with my code. I did something like this:

So, there is nothing in front of the period here. What happens?

Well, technically, when there is no object explicitly used, there is the implicit, global object (also known as window – I believe named for the browser window in which the JavaScript would run). As a result, I get the following output:

undefined undefined is: undefined

And that is, of course, because there are no firstName, lastName, or email addresses defined for window. And, in actuality, I had saved a global variable called email which meant I got:

undefined undefined is: undefined

This is what started my entire confusion in the first place.

So, what can you do to get around this problem (pun very much intended)? Well, the first thing to do is realize that it’s not a problem. The ability for this to change allows for some very powerful uses in JavaScript.

This is where the call and apply become important. In short, you can call your function with either:

Contact.display.call(BusinessContact);

Person.display.apply(BusinessContact);

Both are very similar, but slightly different in how you pass in your arguments (if you have a generic function instead of the specific ones I am showing in this example). The different would be:

This table shows the value of this depending on the syntax of the function call. (Thank you to JavaScript Web Blog for the table.)

Type Syntax of function call Value of this
Method call myObject.foo(); myObject
Baseless function call foo(); global object (e.g. window) (undefined in strict mode)
Using call foo.call(context, myArg); context
Using apply foo.apply(context, [myArgs]); context
Constructor with new var newFoo = new Foo(); the new instance (e.g. newFoo)

So, as you can see, it’s very understandable as to why this gets so confusing – especially if you are used to it in another language. Hopefully this post helped you under this just a little bit better.


text:
If you find any mistakes, please make a comment! Thank you.

## Exhibit symmetric group as a subgroup of a general linear group

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.9

Solution: We know that if $F$ is a finite field then $\mathsf{Aut}(F^n) \cong GL_n(F)$. This isomorphism $\zeta$ can be defined as follows: given $\theta \in \mathsf{Aut}(F^n)$, $\zeta(\theta)$ is the matrix in $GL_n(F)$ whose $i$-th row is precisely $\theta(e_i)$. (In particular, $\zeta$ is not canonical since it depends on the choice and order of a basis; here we choose the standard basis.) In Exercises 5.1.7 and 5.1.8, we found an injective group homomorphism $\psi : S_n \rightarrow \mathsf{Aut}(F^n)$. Combining these results we have an injective group homomorphism $\Psi : S_n \rightarrow GL_n(F)$, computed as follows: $$\Psi(\pi) = [e_{\pi(1)}\ \cdots\ e_{\pi(n)}]^T.$$ We can see that each $\Psi(\pi)$ is obtained from the identity matrix by permuting the rows, so that each has a single 1 in each row and column and 0 in all other entries. Thus $S_n$ is identified with a subgroup of $GL_n(F)$ consisting of permutation matrices; moreover, by counting we see that all permutation matrices are represented this way.

Thus we have proven the result for a finite field $F$; we began with this case because it was shown previously that $\mathsf{Aut}(F^n) \cong GL_n(F)$. If this is true for arbitrary fields then the same proof carries over to arbitrary fields $F$; however this will not be proven until later in the text. In the meantime we can convince ourselves that the result holds over arbitrary fields $F$ by noting that, in computing the product of two permutation matrices, we never deal with numbers other than 1 or 0. In particular, the question of whether or not $k = 0$ in $F$ for some integer $k$ never arises, so that all computations hold for arbitrary fields (in which $1 \neq 0$) and in fact the set of permutation matrices over any $F$ is closed under matrix multiplication.


text:
# Introduction to magnetic monopoles

@article{Rajantie2012IntroductionTM,
title={Introduction to magnetic monopoles},
author={Arttu Rajantie},
journal={Contemporary Physics},
year={2012},
volume={53},
pages={195 - 211}
}
• A. Rajantie
• Published 13 April 2012
• Physics
• Contemporary Physics
One of the most basic properties of magnetism is that a magnet always has two poles, north and south, which cannot be separated into isolated poles, i.e. magnetic monopoles. There are strong theoretical arguments why magnetic monopoles should exist, but in spite of extensive searches they have never been found. This article gives an introduction to theoretical and experimental aspects of magnetic monopoles and discusses their role in our understanding of fundamental physics, both historically…
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• A. Rajantie
• Physics
Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences
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The role of magnetic monopoles in quantum field theory is reviewed and their implications for particle physics and cosmology are discussed.
Status of Searches for Magnetic Monopoles
• Physics
• 2015
Searches for magnetic monopoles (s) are a fascinating interdisciplinary area with implications for fundamental theories, particle physics, astrophysics, and cosmology. The quantum theory of s, and
Magnetic monopole creation
Elementary magnetic monopoles have never been experimentally observed but there are credible theoretical reasons to believe that they may nonetheless exist. This thesis gives partial answers to the
Magnetic monopoles and Maxwell's equations in N dimensions
• Physics
2013 International Conference on Electromagnetics in Advanced Applications (ICEAA)
• 2013
The possible existence of “magnetic charge” has been widely investigated in the last two centuries, since it would allow to symmetrise Maxwell's equations and to explain charge quantization. Though,
Searches for Magnetic Monopoles: A Review
This report briefly reviews selected searches for magnetic monopoles. The theoretical motivation behind their existence is highlighted. The focus is on the results of the searches and the bounds set
Monopole–antimonopole pair production by magnetic fields
• A. Rajantie
• Physics
Philosophical Transactions of the Royal Society A
• 2019
The bounds on monopole masses arising from the strongest magnetic fields in the universe, which are in neutron stars known as magnetars and in heavy ion collision experiments such as lead–lead collisions carried out in November 2018 in the large Hadron collider at CERN are discussed.
Magnetic Monopole Cannot Be Incorporated into the Dual Framework of Electromagnetic Theory
Through reexamining the intention of duality in classical physics, we will develop the electromagnetic field equations (EFEs) into a new dualized form. In the process, we will see 1) the dual
Searches for magnetic monopoles with IceCube
Particles that carry a magnetic monopole charge are proposed by various theories which go beyond the Standard Model of particle physics. The expected mass of magnetic monopoles varies depending on
Radiation arrow and missing magnetic monopoles
In this paper a new solution for the time-asymmetry of radiation is proposed. The novelty is that the time behaviour of the electromagnetic field is unlinked from the time behaviour of the charge in
1 2 Ju l 2 01 9 Monopole-antimonopole pair production by magnetic fields
Quantum electrodynamics predicts that in a strong electric field, electron-positron pairs are produced by the Schwinger process, which can be interpreted as quantum tunnelling through the Coulomb

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A solitary, uncorroborated event at Stanford is the only evidence that magnetic monopoles might exist. Powerful theoretical motivation for monopoles derives from Dirac's assertion that monopoles
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text:
# Extending the analicity region of a function.

Prove or refute (by means of a counterexample) the following claim:

Let $\,f\,$ be an analytic function in $\,D:=\{z\;\;:\;\;|z-a|<r\,\}\,$ and continuous in $\,\overline{D}\,$ , then: there exists $\,\delta>0\,$ s.t. $\,f\,$ can be analitically continued to $\,\{z\;\;:\;\;|z-a|<r+\delta\}\,$

What I tried: At first I though this can be disproved by means of the example $$\frac{1}{1-z}=\sum_{n=0}^\infty z^n\,\,,\,|z|<1$$ but then I realized that it can't be said the function $\,\displaystyle{\frac{1}{1-z}}\,$ is continuous on $\,\overline D\,$ , so I began suspecting the claim is true.

We can assume $\,a=0\,$ to avoid cumbersome stuff.

Now, we have that $$\forall\,z\in D\,\,,\,\,f(z)=\sum_{n=0}a_nz^n$$ so I argued: suppose $\,z_0\in \partial D\,$ is s.t. $\,f\,$ cannot be continued through it. Let us choose a sequence $\,\{z_k\}\subset D\,\,\,s.t.\,\,\,z_k\xrightarrow [k\to\infty]{}z_0\,$ , then by continuity we get $$f(z_0)=\lim_{n\to\infty}f(z_k)=\lim_{k\to\infty}\sum_{n=0}^\infty a_nz_k^n\stackrel{\color{red}{attention!}}=\sum_{n=0}^\infty a_nz_0^n$$

Of course, the above is possible if we assume we can interchange the limit and the summation, which I'm not sure at all...I can't see here any monotone or dominated convergence theorem that'll allow this, so I'm pretty stuck.

Any hint, idea or solution will be duly appreciated.

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Abel's Theorem seems relevant. – Ragib Zaman Jul 1 '12 at 16:00

Since the hints that were posted in the comments seem to have disappeared, I'll give a solution here.

Let $$f(z) = \sum_{n=1}^\infty \frac{z^n}{n^p}$$ where $p>1$. Then the radius of convergence is 1 and the series converges unformly (hence to a continuous function) on the closed unit disc.

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This is weird: AFAIK, the only way an answer together with the comments after it can disappear is if the answer's poster deletes it...is this what happened here? – DonAntonio Jul 2 '12 at 12:40
There were hints giving (more or less) the above solution, but I guess the poster deleted them for some reason. – mrf Jul 2 '12 at 13:05


text:
## CryptoDB

### Paper: Proxy Re-Signatures: New Definitions, Algorithms, and Applications

Authors: Giuseppe Ateniese Susan Hohenberger URL: http://eprint.iacr.org/2005/433 Search ePrint Search Google In 1998, Blaze, Bleumer, and Strauss (BBS) proposed proxy re-signatures, in which a semi-trusted proxy acts as a translator between Alice and Bob. To translate, the proxy converts a signature from Alice into a signature from Bob on the same message. The proxy, however, does not learn any signing key and cannot sign arbitrary messages on behalf of either Alice or Bob. Since the BBS proposal, the proxy re-signature primitive has been largely ignored, but we show that it is a very useful tool for sharing web certificates, forming weak group signatures, and authenticating a network path. We begin our results by formalizing the definition of security for a proxy re-signature. We next substantiate the need for improved schemes by pointing out certain weaknesses of the original BBS proxy re-signature scheme which make it unfit for most practical applications. We then present two secure proxy re-signature schemes based on bilinear maps. Our first scheme relies on the Computational Diffie-Hellman (CDH) assumption; here the proxy can translate from Alice to Bob and vice-versa. Our second scheme relies on the CDH and 2-Discrete Logarithm (2-DL) assumptions and achieves a stronger security guarantee -- the proxy is only able to translate in one direction. Constructing such a scheme has been an open problem since proposed by BBS in 1998. Furthermore in this second scheme, even if the delegator and the proxy collude, they cannot sign on behalf of the delegatee. Both schemes are efficient and secure in the random oracle model.
##### BibTeX
@misc{eprint-2005-12766,
title={Proxy Re-Signatures: New Definitions, Algorithms, and Applications},
booktitle={IACR Eprint archive},
keywords={public-key cryptography /},
url={http://eprint.iacr.org/2005/433},
note={This is the full version of the paper in ACM CCS 2005. srhohen@mit.edu 13115 received 28 Nov 2005},
author={Giuseppe Ateniese and Susan Hohenberger},
year=2005
}