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# 16.2: Maxwell’s Equations and Electromagnetic Waves

Learning Objectives

By the end of this section, you will be able to:

• Explain Maxwell’s correction of Ampère’s law by including the displacement current
• State and apply Maxwell’s equations in integral form
• Describe how the symmetry between changing electric and changing magnetic fields explains Maxwell’s prediction of electromagnetic waves
• Describe how Hertz confirmed Maxwell’s prediction of electromagnetic waves

James Clerk Maxwell (1831–1879) was one of the major contributors to physics in the nineteenth century (Figure $$\PageIndex{1}$$). Although he died young, he made major contributions to the development of the kinetic theory of gases, to the understanding of color vision, and to the nature of Saturn’s rings. He is probably best known for having combined existing knowledge of the laws of electricity and of magnetism with insights of his own into a complete overarching electromagnetic theory, represented by Maxwell’s equations.

## Maxwell’s Correction to the Laws of Electricity and Magnetism

The four basic laws of electricity and magnetism had been discovered experimentally through the work of physicists such as Oersted, Coulomb, Gauss, and Faraday. Maxwell discovered logical inconsistencies in these earlier results and identified the incompleteness of Ampère’s law as their cause.

Recall that according to Ampère’s law, the integral of the magnetic field around a closed loop C is proportional to the current I passing through any surface whose boundary is loop C itself:

$\oint \vec{B} \cdot d\vec{s} = \mu_0 I. \label{Eq1}$

There are infinitely many surfaces that can be attached to any loop, and Ampère’s law stated in Equation \ref{Eq1} is independent of the choice of surface.

Consider the set-up in Figure $$\PageIndex{2}$$. A source of emf is abruptly connected across a parallel-plate capacitor so that a time-dependent current I develops in the wire. Suppose we apply Ampère’s law to loop C shown at a time before the capacitor is fully charged, so that $$I \neq 0$$. Surface $$S_1$$ gives a nonzero value for the enclosed current I, whereas surface $$S_2$$ gives zero for the enclosed current because no current passes through it:

$\underbrace{\oint_C \vec{B} \cdot d\vec{s} = \mu_0 I}_{\text{if surface } S_1 \text{is used}}$

$\underbrace{ \, =0 }_{\text{if surface } S_2 \text{is used}}$

Clearly, Ampère’s law in its usual form does not work here. This may not be surprising, because Ampère’s law as applied in earlier chapters required a steady current, whereas the current in this experiment is changing with time and is not steady at all.

How can Ampère’s law be modified so that it works in all situations? Maxwell suggested including an additional contribution, called the displacement current $$I_d$$, to the real current I,

$\boxed{\oint_S \vec{B} \cdot d\vec{s} = \mu_0 (I + I_d)} \label{EQ4}$

where the displacement current is defined to be

$\boxed{I_d = \epsilon_0 \dfrac{d\Phi_E}{dt}.} \label{EQ5}$

Here $$\epsilon_0$$ is the permittivity of free space and $$\Phi_E$$ is the electric flux, defined as

$\Phi_E = \iint_{Surface \, S} \vec{E} \cdot d\vec{A}.$

The displacement current is analogous to a real current in Ampère’s law, entering into Ampère’s law in the same way. It is produced, however, by a changing electric field. It accounts for a changing electric field producing a magnetic field, just as a real current does, but the displacement current can produce a magnetic field even where no real current is present. When this extra term is included, the modified Ampère’s law equation becomes

$\oint_C \vec{B} \cdot d\vec{s} = \mu_0 I + \epsilon_0 \mu_0 \dfrac{d\Phi_E}{dt}$

and is independent of the surface S through which the current I is measured.

We can now examine this modified version of Ampère’s law to confirm that it holds independent of whether the surface $$S_1$$ or the surface $$S_2$$ in Figure $$\PageIndex{2}$$ is chosen. The electric field $$\vec{E}$$ corresponding to the flux $$\Phi_E$$ in Equation \ref{EQ5} is between the capacitor plates. Therefore, the $$\vec{E}$$ field and the displacement current through the surface $$S_1$$ are both zero, and Equation \ref{EQ4} takes the form

$\oint_C \vec{B} \cdot d\vec{s} = \mu_0 I. \label{16.5}$

We must now show that for surface $$S_2$$, through which no actual current flows, the displacement current leads to the same value $$\mu_0 I$$ for the right side of the Ampère’s law equation. For surface $$S_2$$ the equation becomes

$\oint_C \vec{B} \cdot d\vec{s} = \mu_0 \dfrac{d}{dt} \left[ \epsilon_0 \iint_{Surface \, S_2} \vec{E} \cdot d\vec{A} \right].$

Gauss’s law for electric charge requires a closed surface and cannot ordinarily be applied to a surface like $$S_1$$ alone or $$S_2$$ alone. But the two surfaces $$S_1$$ and $$S_2$$ form a closed surface in Figure $$\PageIndex{2}$$ and can be used in Gauss’s law. Because the electric field is zero on $$S_1$$, the flux contribution through $$S_1$$ is zero. This gives us

\begin{align} \oint_{Surface \, S_1 + S_2} \vec{E} \cdot d\vec{A} &= \iint_{Surface \, S_1} \vec{E} \cdot d\vec{A} + \iint_{Surface \, S_2} \vec{E} \cdot d\vec{A} \\[4pt] &= 0 + \iint_{Surface \, S_2} \vec{E} \cdot d\vec{A} \\[4pt] &= \iint_{Surface \, S_2} \vec{E} \cdot d\vec{A}. \end{align}

Therefore, we can replace the integral over $$S_2$$ in Equation \ref{16.5} with the closed Gaussian surface $$S_1 + S_2$$ and apply Gauss’s law to obtain

$\oint_{S_1} \vec{B} \cdot d\vec{s} = \mu_0 \dfrac{dQ_{in}}{dt} = \mu_0 I.$

Thus, the modified Ampère’s law equation is the same using surface $$S_2$$, where the right-hand side results from the displacement current, as it is for the surface $$S_1$$, where the contribution comes from the actual flow of electric charge.

Displacement current in a charging capacitor

A parallel-plate capacitor with capacitance C whose plates have area A and separation distance d is connected to a resistor R and a battery of voltage V. The current starts to flow at $$t = 0$$.

1. Find the displacement current between the capacitor plates at time t.
2. From the properties of the capacitor, find the corresponding real current $$I = \dfrac{dQ}{dt}$$, and compare the answer to the expected current in the wires of the corresponding RC circuit.

Strategy

We can use the equations from the analysis of an RC circuit (Alternating-Current Circuits) plus Maxwell’s version of Ampère’s law.

Solution

1. The voltage between the plates at time t is given by $V_C = \dfrac{1}{C} Q(t) = V_0 \left(1 - e^{-t/RC}\right). \nonumber$ Let the z-axis point from the positive plate to the negative plate. Then the z-component of the electric field between the plates as a function of time t is $E_z (t) = \dfrac{V_0}{d}\left(1 - e^{-t/RC}\right). \nonumber$ Therefore, the z-component of the displacement current $$I_d$$ between the plates is $I_d (t) = \epsilon_0 A\dfrac{\partial E_z (t)}{\partial t} = \epsilon_0 A \dfrac{V_0}{d} \times \dfrac{1}{RC} e^{-t/RC} = \dfrac{V_0}{R} e^{-t/RC}, \nonumber$ where we have used $$C = \epsilon_0 \dfrac{A}{d}$$ for the capacitance.
2. From the expression for $$V_C$$ the charge on the capacitor is $Q(t) = CV_C = CV_0 \left(1 - e^{-t/RC}\right). \nonumber$ The current into the capacitor after the circuit is closed, is therefore $I = \dfrac{dQ}{dt} = \dfrac{V_0}{R}e^{-t/RC}. \nonumber$ This current is the same as $$I_d$$ found in (a).

## Maxwell’s Equations

With the correction for the displacement current, Maxwell’s equations take the form

\begin{align} \oint \vec{E} \cdot d\vec{A} &= \dfrac{Q_{in}}{\epsilon_0} \, (\text{Gauss's law}) \label{eq1} \\[4pt] \oint \vec{B} \cdot d\vec{A} &= 0 \, (\text{Gauss's law for magnetism}) \label{eq2} \\[4pt] \oint \vec{E} \cdot d\vec{s} &= - \dfrac{d\Phi_m}{dt} \, (\text{Faraday's law}) \label{eq3} \\[4pt] \oint \vec{B} \cdot d\vec{s} &= \mu_0 I + \epsilon_0I + \epsilon_0 \mu_0 \dfrac{d\Phi_E}{dt} (\text{Ampere-Maxwell law}). \label{eq4} \end{align}

Once the fields have been calculated using these four equations, the Lorentz force equation

$\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}$

gives the force that the fields exert on a particle with charge q moving with velocity $$\vec{v}$$. The Lorentz force equation combines the force of the electric field and of the magnetic field on the moving charge. The magnetic and electric forces have been examined in earlier modules. These four Maxwell’s equations are, respectively:

Maxwell's Equations

1. Gauss’s law

The electric flux through any closed surface is equal to the electric charge $$Q_{in}$$ enclosed by the surface. Gauss’s law (Equation \ref{eq1}) describes the relation between an electric charge and the electric field it produces. This is often pictured in terms of electric field lines originating from positive charges and terminating on negative charges, and indicating the direction of the electric field at each point in space.

2. Gauss’s law for magnetism

The magnetic field flux through any closed surface is zero (Equation \ref{eq2}). This is equivalent to the statement that magnetic field lines are continuous, having no beginning or end. Any magnetic field line entering the region enclosed by the surface must also leave it. No magnetic monopoles, where magnetic field lines would terminate, are known to exist (see section on Magnetic Fields and Lines).

A changing magnetic field induces an electromotive force (emf) and, hence, an electric field. The direction of the emf opposes the change. Equation \ref{eq3} is Faraday’s law of induction and includes Lenz’s law. The electric field from a changing magnetic field has field lines that form closed loops, without any beginning or end.

4. Ampère-Maxwell law

Magnetic fields are generated by moving charges or by changing electric fields. This fourth of Maxwell’s equations, Equation \ref{eq4}, encompasses Ampère’s law and adds another source of magnetic fields, namely changing electric fields.

Maxwell’s equations and the Lorentz force law together encompass all the laws of electricity and magnetism. The symmetry that Maxwell introduced into his mathematical framework may not be immediately apparent. Faraday’s law describes how changing magnetic fields produce electric fields. The displacement current introduced by Maxwell results instead from a changing electric field and accounts for a changing electric field producing a magnetic field. The equations for the effects of both changing electric fields and changing magnetic fields differ in form only where the absence of magnetic monopoles leads to missing terms. This symmetry between the effects of changing magnetic and electric fields is essential in explaining the nature of electromagnetic waves.

Later application of Einstein’s theory of relativity to Maxwell’s complete and symmetric theory showed that electric and magnetic forces are not separate but are different manifestations of the same thing—the electromagnetic force. The electromagnetic force and weak nuclear force are similarly unified as the electroweak force. This unification of forces has been one motivation for attempts to unify all of the four basic forces in nature—the gravitational, electrical, strong, and weak nuclear forces (see Particle Physics and Cosmology).

## The Mechanism of Electromagnetic Wave Propagation

To see how the symmetry introduced by Maxwell accounts for the existence of combined electric and magnetic waves that propagate through space, imagine a time-varying magnetic field $$\vec{B}_0(t)$$ produced by the high-frequency alternating current seen in Figure $$\PageIndex{3}$$. We represent $$\vec{B}_0(t)$$ in the diagram by one of its field lines. From Faraday’s law, the changing magnetic field through a surface induces a time-varying electric field $$\vec{E}_0(t)$$ at the boundary of that surface. The displacement current source for the electric field, like the Faraday’s law source for the magnetic field, produces only closed loops of field lines, because of the mathematical symmetry involved in the equations for the induced electric and induced magnetic fields. A field line representation of $$\vec{E}_0(t)$$ is shown. In turn, the changing electric field $$\vec{E}_0(t)$$ creates a magnetic field $$\vec{B}_1(t)$$ according to the modified Ampère’s law. This changing field induces $$\vec{E}_1(t)$$ which induces $$\vec{B}_2(t)$$ and so on. We then have a self-continuing process that leads to the creation of time-varying electric and magnetic fields in regions farther and farther away from O. This process may be visualized as the propagation of an electromagnetic wave through space.

In the next section, we show in more precise mathematical terms how Maxwell’s equations lead to the prediction of electromagnetic waves that can travel through space without a material medium, implying a speed of electromagnetic waves equal to the speed of light.

Prior to Maxwell’s work, experiments had already indicated that light was a wave phenomenon, although the nature of the waves was yet unknown. In 1801, Thomas Young (1773–1829) showed that when a light beam was separated by two narrow slits and then recombined, a pattern made up of bright and dark fringes was formed on a screen. Young explained this behavior by assuming that light was composed of waves that added constructively at some points and destructively at others (see Interference). Subsequently, Jean Foucault (1819–1868), with measurements of the speed of light in various media, and Augustin Fresnel (1788–1827), with detailed experiments involving interference and diffraction of light, provided further conclusive evidence that light was a wave. So, light was known to be a wave, and Maxwell had predicted the existence of electromagnetic waves that traveled at the speed of light. The conclusion seemed inescapable: Light must be a form of electromagnetic radiation. But Maxwell’s theory showed that other wavelengths and frequencies than those of light were possible for electromagnetic waves. He showed that electromagnetic radiation with the same fundamental properties as visible light should exist at any frequency. It remained for others to test, and confirm, this prediction.

Exercise $$\PageIndex{1}$$

When the emf across a capacitor is turned on and the capacitor is allowed to charge, when does the magnetic field induced by the displacement current have the greatest magnitude?

Solution

It is greatest immediately after the current is switched on. The displacement current and the magnetic field from it are proportional to the rate of change of electric field between the plates, which is greatest when the plates first begin to charge.

## Hertz’s Observations

The German physicist Heinrich Hertz (1857–1894) was the first to generate and detect certain types of electromagnetic waves in the laboratory. Starting in 1887, he performed a series of experiments that not only confirmed the existence of electromagnetic waves but also verified that they travel at the speed of light.

Hertz used an alternating-current RLC (resistor-inductor-capacitor) circuit that resonates at a known frequency $$f_0 = \dfrac{1}{2\pi \sqrt{LC}}$$ and connected it to a loop of wire, as shown in Figure $$\PageIndex{4}$$. High voltages induced across the gap in the loop produced sparks that were visible evidence of the current in the circuit and helped generate electromagnetic waves.

Across the laboratory, Hertz placed another loop attached to another RLC circuit, which could be tuned (as the dial on a radio) to the same resonant frequency as the first and could thus be made to receive electromagnetic waves. This loop also had a gap across which sparks were generated, giving solid evidence that electromagnetic waves had been received.

Hertz also studied the reflection, refraction, and interference patterns of the electromagnetic waves he generated, confirming their wave character. He was able to determine the wavelengths from the interference patterns, and knowing their frequencies, he could calculate the propagation speed using the equation $$v = f\lambda$$, where v is the speed of a wave, f is its frequency, and $$\lambda$$ is its wavelength. Hertz was thus able to prove that electromagnetic waves travel at the speed of light. The SI unit for frequency, the hertz $$(1 \, Hz = 1 \, cycle/second)$$, is named in his honor.

Exercise $$\PageIndex{2}$$

Could a purely electric field propagate as a wave through a vacuum without a magnetic field? Justify your answer.

Solution

No. The changing electric field according to the modified version of Ampère’s law would necessarily induce a changing magnetic field.


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Following the “solving a mystery” approach from my last post, I will be attempting to solve a couple of problems from my Cryptography Assignment.

Here’s the first one:

Let $\small G : \{0,1\}^4 \to \{0,1\}^4$ be a 4-bit block cipher defined by $G(x_1 x_2 x_3 x_4) = x_4 x_2 x_3 x_1$. What is the CBC-MAC of $10100101$?

It looks like $G$ is a function that takes a 4-bit binary number (a “nibble”) as input and gives another nibble as output. From the definition, it seems that if I give $0001$ as input, I will get $G(0001) = 1000$ as output.

• What is a block cipher?
wikipedia: In cryptography, a block cipher is a deterministic algorithm operating on fixed-length groups of bits, called blocks, with an unvarying transformation that is specified by a symmetric key.

• what is a deterministic algorithm?
wikipedia: A deterministic algorithm computes a mathematical function. In other words, given a particular input, the algorithm will always produce the same output, with the underlying machine always passing through the same sequence of states.

So in this case, $G$ is the deterministic algorithm. Now, given $G$, we have to find the CBC-MAC of $10100101$.

• What is a CBC-MAC?
wikipedia: A cipher block chaining message authentication code (CBC-MAC) is a technique for constructing a message authentication code from a block cipher. The message is encrypted with some block cipher algorithm in CBC mode to create a chain of blocks such that each block depends on the proper encryption of the previous block.

• What is a message authentication code?
wikipedia: A message authentication code (MAC), sometimes known as a tag, is a short piece of information used to authenticate a message—in other words, to confirm that the message came from the stated sender (its authenticity) and has not been changed.

Informally, a message authentication code system consists of three algorithms:

• A key generation algorithm selects a key from the key space uniformly at random.
• A signing algorithm efficiently returns a tag given the key and the message.
• A verifying algorithm efficiently verifies the authenticity of the message given the key and the tag. That is, return accepted when the message and tag are not tampered with or forged, and otherwise return rejected.
• What is the CBC mode for a block cipher algorithm?
wikipedia: In CBC mode, each block of plaintext is XORed with the previous ciphertext block before being encrypted. This way, each ciphertext block depends on all plaintext blocks processed up to that point. To make each message unique, an initialisation vector must be used in the first block.

We have to find the CBC-MAC of $10100101$. Upon breaking it into blocks of 4-bits, we get $1010$ and $0101$.

• What should the initialisation vector be when calculating CBC-MAC?
wikipedia: To calculate the CBC-MAC of message m, one encrypts m in CBC mode with zero initialization vector and keeps the last block.

$G(0000 \oplus 1010) = G(1010) = 0011$
$G(0011 \oplus 0101) = G(0110) = 0110$

The last block is the CBC-MAC. Thus, $0110$ is the solution of this problem.

That was fun. Let’s try another one:

Alice and Bob are following the Diffie–Hellman Secret Key Exchange Protocol with $p = 101$ (the prime number $p$), $g=2$ (the generator of $\mathbb{Z}^*_p$), $x=11$ (the private key of Alice), $y=13$ (the private key of Bob). What is the shared secret key between Alice and Bob?

• What is the Diffie–Hellman Secret Key Exchange Protocol ?

The simplest and the original implementation of the protocol uses the multiplicative group of integers modulo p, where p is prime, and g is a primitive root modulo p. These two values are chosen in this way to ensure that the resulting shared secret can take on any value from 1 to p–1. Here is an example of the protocol

1. Alice and Bob publicly agree to use a modulus $p = 23$ and base $g = 5$ (which is a primitive root modulo 23).
2. Alice chooses a secret integer $a = 4$, then sends Bob $A = g^a \bmod p$
• $A = 5^4 \bmod 23 = 4$
3. Bob chooses a secret integer $b = 3$, then sends Alice $B = g^b \bmod p$
• $B = 5^3 \bmod 23 = 10$
4. Alice computes $s = B^a \bmod p$
• $s = 10^4 \bmod 23 = 18$
5. Bob computes $s = A^b \bmod p$
• $s = 4^3 \bmod 23 = 18$
6. Alice and Bob now share a secret (the number 18).

Let’s apply the above procedure to this problem.

1. Alice and Bob publicly agree to use a modulus $p = 101$ and base $g = 2$ (which is a primitive root modulo 101).
2. Alice chooses a secret integer $a = 11$, then sends Bob $A = g^a \bmod p$
• $A = 2^{11}\bmod101 = 28$
3. Bob chooses a secret integer $b = 13$, then sends Alice $B = g^b \bmod p$
• $B = 2^{13}\bmod101 = 11$
4. Alice computes $s = B^a\bmod p$
• $s = 11^{11}\bmod101 = 86$
5. Bob computes $s = A^b\bmod p$
• $s = 28^{13}\bmod101 = 86$
6. Alice and Bob now share a secret (the number 86).

Hence, the solution of this problem is 86. Here’s some additional information:

• What is a primitive root modulo?
wikipedia: $g$ is a primitive root modulo $n$ if for every integer $a$ coprime to $n$, there is an integer $k$ such that $g^k \equiv a \bmod n$. Note that $g$ is a primitive root modulo $n$ if and only if $g$ is a generator of the multiplicative group of integers modulo $n$.

• What is the multiplicative group of integers modulo $n$?
wikipedia: The integers coprime (relatively prime) to $n$ from the set $\{0,1,\dots ,n-1\}$ of n non-negative integers form a group under multiplication modulo $n$, called the multiplicative group of integers modulo $n$.

• What is a group (algebra)?
wikipedia: A group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that four conditions called group axioms are satisfied, namely closure, associativity, identity and invertibility.

So, an example of a multiplicative group of integers modulo n might be $$(\mathbb{Z}_5^*, *)$$ where $\mathbb{Z}_5^* = \{x : gcd(x, 5) = 1\}$ i.e. $\mathbb{Z}_5^* = \{1, 2, 3, 4\}$ and $a * b = ab \bmod 5$. We can verify that the four group axioms are satisfied. I’ll go into more detail if a future problem requires me to do so. For now, I’ll end here.

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# Spring on Ramp#

In the figure below m = 4 kg, $$\theta$$ = 29$$^\circ$$, and k = 150 N/m. In this problem assume that the ramp never moves and that there is friction between the block and the ramp.

## Part 1#

If the coefficient of static friction between the block and the ramp is $$\mu_s$$ = 0.6, what is the maximum amount that the spring can be stretched beyond its equilibrium length before the block begins to slide up the ramp?

Please enter in a numeric value in m.

## Part 2#

Suppose now that the block is sliding up the ramp at a constant velocity. By what length is the spring stretched? Assume that the coefficient of kinetic friction is $$\mu_k$$ = 0.38.


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# Maximum shear stress in a triangular section is

This question was previously asked in
Gujarat Engineering Service 2017 Official Paper (Civil Part 2)
View all GPSC Engineering Services Papers >
1. 1.33 times the average shear stress at y = h/3
2. 1.5 times the average shear stress at y = h/3
3. 1.33 times the average shear stress at y = h/2
4. 1.5 times the average shear stress at y = h/2

Option 4 : 1.5 times the average shear stress at y = h/2
Free
ST 1: Logical reasoning
5232
20 Questions 20 Marks 20 Mins

## Detailed Solution

Explanation

Maximum shear stress ($${\tau _{max}}$$) in the triangular cross-section is given by:

$${\tau _{\max }} = 1.5$$ $${\tau _{avg}}$$ at h/2 distance

Where,

$${\tau _{avg}}$$ = Average shear stress in the cross-section

Shear stress at the neutral axis of the cross-section is given by:

$${\tau _{N.A}} = 1.33$$  $${\tau _{avg}}$$

Important Points

 Section τmax/τavg τNeutral axis / τavg Rectangular/square 3/2 3/2 Solid circular 4/3 4/3 Triangle 3/2 4/3 Diamond 9/8 1


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Connected Graph Searching - Archive ouverte HAL Access content directly
Journal Articles Information and Computation Year : 2012

## Connected Graph Searching

(1) , (2) , (3) , (4) , (5) , (6) , (7, 8)
1
2
3
4
5
6
7
8
Lali Barrière
• Function : Author
Paola Flocchini
• Function : Author
• PersonId : 855751
Fedor V. Fomin
• Function : Author
• PersonId : 857694
Pierre Fraigniaud
• Function : Author
• PersonId : 840045
Nicolas Nisse
Nicola Santoro
• Function : Author
Dimitrios M. Thilikos

#### Abstract

In the graph searching game the opponents are a set of searchers and a fugitive in a graph. The searchers try to capture the fugitive by applying some sequence of moves that include placement, removal, or sliding of a searcher along an edge. The fugitive tries to avoid capture by moving along unguarded paths. The search number of a graph is the minimum number of searchers required to guarantee the capture of the fugitive. In this paper, we initiate the study of this game under the natural restriction of connectivity where we demand that in each step of the search the locations of the graph that are clean (i.e. non-accessible to the fugitive) remain connected. We give evidence that many of the standard mathematical tools used so far in classic graph searching fail under the connectivity requirement. We also settle the question on ''the price of connectivity'', that is, how many searchers more are required for searching a graph when the connectivity demand is imposed. We make estimations of the price of connectivity on general graphs and we provide tight bounds for the case of trees. In particular, for an $n$-vertex graph the ratio between the connected searching number and the non-connected one is $O(\log n)$ while for trees this ratio is always at most 2. We also conjecture that this constant-ratio upper bound for trees holds also for all graphs. Our combinatorial results imply a complete characterization of connected graph searching on trees. It is based on a forbidden-graph characterization of the connected search number. We prove that the connected search game is monotone for trees, i.e. restricting search strategies to only those where the clean territories increase monotonically does not require more searchers. A consequence of our results is that the connected search number can be computed in polynomial time on trees, moreover, we show how to make this algorithm distributed. Finally, we reveal connections of this parameter to other invariants on trees such as the Horton-Stralher number.

### Dates and versions

hal-00741948 , version 1 (15-10-2012)

### Identifiers

• HAL Id : hal-00741948 , version 1

### Cite

Lali Barrière, Paola Flocchini, Fedor V. Fomin, Pierre Fraigniaud, Nicolas Nisse, et al.. Connected Graph Searching. Information and Computation, 2012, 219, pp.1-16. ⟨hal-00741948⟩

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# Do the adjoint functor theorems usefully dualise?

The special and general adjoint functor theorems exist to construct left adjoints to particular functors given certain conditions on them. However, I've not been able to find much mention – at least, not in my lecture notes nor in a quick Google search – of using them or something like them to construct right adjoints.

I suppose that if $F \dashv G$ then $G^\mathrm{op} \dashv F^\mathrm{op}$, so I can just apply the adjoint functor theorems to the opposites of the categories I'm interested in. Perhaps something like:

General Adjoint Functor Cotheorem: suppose $\mathcal C$ is locally small and cocomplete. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint if and only if it preserves small colimits and, for each object $A$ of $\mathcal C$, the comma category $G \downarrow A$ has a weakly terminal set.

Special Adjoint Functor Cotheorem: suppose $\mathcal C, \mathcal D$ both locally small, $\mathcal C$ cocomplete, co-well-powered1, and with a separating set. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint iff $F$ preserves small colimits.

Is that right? Is it useful?

1 Not sure if this is standard terminology, but I hope it's clear what I mean: for each object $A$, there is a category of epimorphisms with domain $A$, and a category would be co-well-powered if all these epimorphism categories were equivalent to a partially ordered set (they are already preorders, since if $fg = h$ with $g$ and $h$ epimorphisms, then $f$ is unique)

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I'm not sure what the footnote is meant to say, but the answer to your question is 'yes, it dualizes (just like anything else in category theory)'. – Ittay Weiss May 21 '13 at 11:08
Right – the real substance of my question is (a) have I got the dualisations right and (b) are they ever actually used? (why haven't I heard about them?) – Ben Millwood May 21 '13 at 11:12
a) the dualisations are correct. b) which uses of the adjoints functor theorems do you know? – Ittay Weiss May 21 '13 at 11:17
The Stone-Cech compactification from the SAFT, the construction of free groups (and other algebraic structures) from the GAFT (although I'm told that you pretty much need to already know how to build free groups to do that). I know there are more but I can't recall them off the top of my head. I have however just found this mathoverflow.net/questions/17409/… which is very similar to my question. – Ben Millwood May 21 '13 at 11:30
These are theorems, not cotheorems! A theorem starts from some premises and reaches some conclusions. A cotheorem starts from some nclusions and reaches some copremises. – Qiaochu Yuan Dec 12 '15 at 19:42

The dualizations are correct and well-known at least for category theorists. Note that the assumptions in Co-Saft are clearly satisfied if $\mathcal{C}$ is presentable. Thus, left adjoint functors $\mathcal{C} \to \mathcal{D}$ coincide with cocontinuous functors $\mathcal{C} \to \mathcal{D}$. This is a basic and important fact of presentable categories. Here is a nice application: Every presentable monoidal category (i.e. a monoidal category whose underlying category is presentable, and the tensor product should be cocontinuous in each variable) is automatically closed monoidal. In other words, as soon we have a tensor product, we get inner homs for free. In practice, often it is vice versa, but not always.

In my work cocomplete monoidal categories appear where I still have no idea how to describe the inner homs. Nevertheless, they are useful, for example in order to apply Kock's results about monoidal monads. More specifically, if we have a monoidal monad on a cocomplete monoidal category, we can quite easily write down a tensor product on the category of modules, but the proof of associativity is not trivial and actually it seems that we have to impose additional assumptions such as closedness (see arXiv:1205.0101 for details). You surely have seen this for tensor products of modules over a ring in the usual sense: The proof for $M \otimes_R (N \otimes_R K) \cong(M \otimes_R N) \otimes_R K$ only using the definition of the tensor product as a classifying object for bilinear maps is a bit tricky, because first we have to fix some element of $M$ etc, and this implicitly uses the closed structure.

Here is another application: Let $f : X \to Y$ be a morphism of schemes. Then $f^* : \mathsf{Mod}(Y) \to \mathsf{Mod}(X)$ has a right adjoint $f_*$. The functor restricted to quasi-coherent modules $f^* : \mathsf{Qcoh}(Y) \to \mathsf{Qcoh}(X)$ also has a right adjoint. When $f_*(\mathsf{Qcoh}(X)) \subseteq \mathsf{Qcoh}(Y)$, the right adjoint is $f_*$. This happens, for example, when $f$ is quasi-compact and quasi-separated. But even when this is not the case, $f^*$ has a right adjoint, since it is cocontinuous and $\mathsf{Qcoh}(Y)$ is presentable by a result of Ofer Gabber. On the other hand, it is not so easy to write down the right adjoint and probably it's not really useful. Related to that: The inclusion $\mathsf{Qcoh}(X) \hookrightarrow \mathsf{Mod}(X)$ is cocontinuous, hence has a right adjoint. This means that every sheaf of modules has a "quasi-coherator". For example, if $X$ is affine, then the quasi-coherator of $M \in \mathsf{Mod}(X)$ is $\widetilde{\Gamma(X,\mathcal{O}_X)}$. See here for more about that.

If $f : X \to Y$ is a morphism of schemes, which is affine, then $f_* : \mathsf{Qcoh}(X) \to \mathsf{Qcoh}(Y)$ is cocontinuous (actually the converse is also true by a classical result of Serre, generalized by Rydh), hence has a right adjoint. But it seems that you can only write it down when $f$ is finite. Then the right adjoint is $f^{!}(M):=\underline{\hom}_{\mathcal{O}_Y}(f_* \mathcal{O}_X,M)$. If $f$ is not finite, we have to take the quasi-coherator of $f^!(M)$.

It seems that in each application of Co-Saft, the right adjoint is either "pathological" or we can write it down directly anyway and showing the adjunction directly.

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Thanks, this is very useful, will accept if nothing else shows up in the next few days. Although I wonder if in your concluding line, "pathological" means "we cannot write it down directly" in which case it's something of a tautology :) – Ben Millwood May 22 '13 at 13:48

as a sort of meta answer: there are more left adjoints than right adjoint (this is of course nonsense!!). More to the point, quite often we are interested in finding left adjoints to functors (in order to create free objects (which include all of the uses you mentioned)). Also quite often, the functors that typically interest us do not have right adjoints at all. This is related to the fact of life that mathematics is not self-dual. For whatever reason, many of the naturally occurring functors we meet tend to have left adjoint but often they lack right adjoints.

So, while category theory is self-dual, practiced mathematics is not. A similar phenomenon is the common concept of a cofibrantly generated model category, while there are very very few examples of naturally occurring categories that support the dual notion of a fibrantly generated model category.

Notice that this lack of duality in practiced mathematics starts very early on. The concept of cartesian product of sets is much more important than the dual notion of disjoint union. And while noticing this, it is the cartesian product of set that is at the heart of the definition of category. So the bias towards one concept over its dual is very deeply rooted.

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text:
# A reference for the equality $|G:H| = \sum_i \dim(V_i)\dim(V_i^H)$

Let $G$ be a finite group and $H$ a subgroup.
Let $V_1, \dots , V_r$ be (equivalence class representatives for) the irreducible complex representations of $G$. Let the stabilizer subspace $V_i^H = \{ v \in V_i \ | \ hv=v \ , \forall h \in H \}$

What is a reference for the following equality? $$|G:H| = \sum_i \dim(V_i)\dim(V_i^H)$$

If you know a proof but not a reference, then you can post it, your answer will be a reference.

• Hmm, this seems like it should follow just by inducing the trivial representation of $H$ to $G$ and describing the dimension of this in two ways, but I will need a pen to write out the details. – Tobias Kildetoft Apr 13 '16 at 9:46

So expanding on my comment, let $1_H^G$ denote the trivial representation of $H$ induced to $G$ which clearly has dimension $|G:H|$.
On the other hand, denoting by $[V,W]_K$ the usual normalized inner product of representations of the group $K$ the dimension is also (by Frobenius reciprocity) $$\sum_{i=1}\dim(V_i)[V_i,1_H^G]_G = \sum_i\dim(V_i)[V_i,1_H]_H = \sum_i\dim(V_i)\dim(V_i^H)$$
• Can you give more details about the Frobenius reciprocity and the equality $[V_i,1_H^G]_G = [V_i,1_H]_H$? – Sebastien Palcoux Apr 16 '16 at 19:26
• The same argument, but without the use of characters: Let $A = K\left[G\right]$. The Artin-Wedderburn theorem yields that $A \cong \bigoplus_{i=1}^r \left(V_i\right)^{\oplus \dim\left(V_i\right)}$ as left $A$-modules. Hence, $A^H \cong \left(\bigoplus_{i=1}^r \left(V_i\right)^{\oplus \dim\left(V_i\right)}\right)^H \cong \bigoplus_{i=1}^r \left(V_i^H\right)^{\oplus \dim\left(V_i\right)}$ as vector spaces (since the operation of taking $H$-fixed points commutes with direct sums). Taking dimensions on both sides of this isomorphism, we obtain ... – darij grinberg Mar 12 '17 at 0:38
• ... the equality $\dim \left(A^H\right) = \sum_{i=1}^r \dim\left(V_i\right) \dim\left(V_i^H\right)$. Now, it remains to prove that $\dim \left(A^H\right) = \left[G : H\right]$ (where $\left[G : H\right]$ is my notation for what you call $\left|G : H\right|$). But this is easy: Just observe that $\left(\sum_{g \in K} g\right)_{K \in H \backslash G}$ is a basis of the vector space $A^H$, where we regard $H \backslash G$ as the set of all right cosets of $H$ in $G$ (that is, subsets of $G$ having the form $Hk$ for some $k \in G$). – darij grinberg Mar 12 '17 at 0:41


text:
# Creating a Matryoshka Shell world: temperature regulation

In a galaxy far far away, during a fit of inspiration for a super massive marvel of technology, the Imperator dedicates an astounding 10% of energy and materials from his civilization's dyson sphere and solar materials processing units to attempt an innovative (and possibly insane) plan to create a matryoshka world. By attempting a massive 3d printing of multiple world shells, the leader hopes to create many planetary shells within each other.

So the first problems appear, heat regulation.

If you don't want to bake the inner shells nor freeze the outer shell, how do you process heat (or a lack thereof) upward or downward between different shells?

Is this really as simple as putting a bunch of massive vents or would it require more than that?

Edit: To be more specific about what is Habitable for this civilization there should be no extended periods of heat above 40 degrees C nor below -5 degrees C.

Edit 2: Septerra Core is a good analog for the world I'm creating. It is a video game with a similar world. But it does not have a ton of information on how the heat is regulated within the world shells. In addition, the shells don't fully cover the other shells allowing light (and heat) to go into lower shells and heat to be radiated upwards. In this constructed world there would be no such gaps. Shell 1 receives all the light from the star and radiates all the heat from every other shell below. Assuming that's how that works (I don't do a lot of science).

Edit 3: This created planet is put in a convenient location near the star enough to get energy put into it and in orbit around the star. But to clarify, this question is primarily concerned with how to regulate waste heat production created from within the shells to keep the lower shells and upper shells in a habitable state.

Edit 4: I have dispensed with a majority of backstory to just get the question at hand answered.

https://en.wikipedia.org/wiki/Kardashev_scale

• Try watching this: youtube.com/… – Thucydides Nov 9 '18 at 23:28
• I watched almost all of this video and I'm entirely confused as to how it relates to my question. The video does go into waste heat but it deals with nothing about how to regulate the temperature so that different layers are habitable. Again, it's great for Matryoshka Brains but my question pertains to another megastructure. Could you go over the areas where this video that would help me with my question? I did go through and watch the series video on shell worlds which does mention worlds I'm talking about but it doesn't go very in depth. – GuidingOlive Nov 10 '18 at 0:06
• As many posters have suggested, the purpose of the Matryoshka Brain is to use the temperature differentials between each layer as a Carnot engine to power the next. In terms of temperature, essentially you pick an orbit and calculate the effective temperature of an object exposed to sunlight there (if the centre is a star like the Sun, then at 1AU, you have our temperature. The heat radiating from the layer below provides the heat energy to power the next level. Layers corresponding to Venus, Earth and Mars would likely be where a carbon based life would prefer to live. – Thucydides Nov 10 '18 at 0:28
• So if I'm getting you correctly use the center layer of this created world like a managed plasma ball then determine the distance to enter habitability for my civilization and the distance at the maximum end of that. And there's no need for additional regulation? What about the energy getting put into it from the star (let's say at a distance of 1 AU)? Would that affect the calculations? Having a central core plasma generator producing energy expanding out and the sun putting energy in? – GuidingOlive Nov 10 '18 at 0:53
• The solar radiation falling on the outer surface will affect how efficiently that layer can be used to radiate into space (the dark side will obviously be more efficient at radiating heat than the illuminated side). I'm not clear why you believe that using an artificial plasma ball will be more efficient than simply englobing the Sun with multiple shells, but if that is your plan, then the "brain" should be built near the orbit of Neptune to limit outside insolation. – Thucydides Nov 10 '18 at 1:15

You do want to freeze the outer shells and bake the inner shells. That is the entire point of a Matryoshka Brain. The entire point is to be able to create a multi-stage Carnot engine with a massive number of stages, and an incredibly high efficiency. This will implicitly call for hot inner shells and cold outer shells.

The one thing you don't do is put massive vents to let the heat out. If you do that, you aren't getting work out of your difference in heat. You pass all the heat through heat engines.

If you want to minimize this effect, keep the layers all in middle-range orbital distances. The sun produces a fixed amount of energy per second. If you have a larger inner radius, that energy is distributed wider, making it easier to manage the heating. Likewise, small outer radii help minimize cooling into space.

However, you'll have to look at how efficient you want to be. The efficiency of a Carnot engine, multi-stage or single-stage, is goverened by the ratio of the temperature of the cold sink to the heat sink. If your inner layer isn't really close to the sun (getting hot), it limits your maximum efficiency.

• This is an excellent answer for a Matryoshka Brain (en.wikipedia.org/wiki/Matrioshka_brain) as far as I can tell. However, my question isn't about Matryoshka Brains but rather Matryoshka Worlds or Habitats. It is about temperature regulation between different shells of a created world where people would live on the surface of each world shell. An analogue to this would be the Video Game Septerra Core where there are seven World Shells with continents passing over continents that are below them leading inward to the core. How would you suggest I clarify my question? – GuidingOlive Nov 8 '18 at 19:02
• Ahh. How much temerature differential can you accept from the first layer to the last? – Cort Ammon Nov 8 '18 at 19:07
• I will add that in, for common habitability of arachnids specifically spider like beings. Don't want to be too humanistic with this group. – GuidingOlive Nov 8 '18 at 19:12
• @GuidingOlive Septerra core already answers your question. The upper shells are a desert, the top being a polar desert, the second being a desert, and the bottom being a warm dark jungle with plants competing for the little light available and the heat coming from the lava on the core itself. – Trevor D Nov 8 '18 at 19:34

This has been said in some of the comments to the answers, but no one has stated it outright:

Why does a civilization capable of producing a Dyson Sphere require this boondoggle?

If it's a vanity project, because the Imperator wants to build one, sure, no problem. But the inner (or outer, depending on star type) surface of a Dyson sphere can be arranged to be at a distance from the star that makes it habitable. If it is so arranged, you have a surface area equivalent to five hundred and fifty million Earths.

Population pressure will not be a problem for at least 25 generations, assuming no population control. (Assuming doubling of population per generation, with one earth's worth of people to start.)

It's an impossibly vast space to fill... and, importantly, vs. the elaborate shell-game that is the proposed planet, they've already built one.

Gravity could potentially be a problem if you're living on the inside, or with certain kinds of star, but this could be overcome with sufficient engineering, provided your civilization has access to some remarkable materials (which they would have to have anyway).

• Why do they need it? As you said, they already built the Dyson sphere. What if they built it 25 or 26 generations ago. They've had all those generations of "We'll never use up all this space" to take away all interest on the part of the population to control their reproduction, so that might be why the government has had no luck on putting the brakes on overpopulation. They've filled their Dyson sphere and are still going. (Breeding like Energizer bunnies.) It reminds me of computer hard drives, when bigger ones came out and we all thought, I'll never need this much space--until we did. – Cadrac Nov 9 '18 at 15:05
• @Cadrac - If they've filled the surface area of 500M Earths, a seven layer planet won't do anything for their population problem. – jdunlop Nov 9 '18 at 15:59
• At this point it's becoming painfully obvious that I'm not going to get an answer on thermal regulation without fully changing the backstory. So how do we make this a better question to get an answer toward? Is this a Class I civilization? Why does it matter why they're creating this contraption? What do I need to reframe to get an answer that works toward the thermal regulation between layers of a matyroshka shell world? – GuidingOlive Nov 9 '18 at 16:57
• @GuidingOlive - Just make it a Culture level question. A civilization capable of harnessing all the energy from a star is not going to need this construct, so it has to be a "want". Eliminate all the context of "needing more living space", and you're fine. This civilization (or its supreme ruler, or their religion) wants to build an impractical shell-world. – jdunlop Nov 9 '18 at 19:27

The upper shells are a desert, the top being a polar desert, the second being a warmer windy desert, and the bottom being a warm dark jungle with plants competing for the little light available and the heat coming from the lava on the core itself. There is not much water on shell 7 but the moisture is high because water falling from the higher shells turns to mist and steam on the lower shells.

The plants have to give off their own light to attract insects to pollinate, making the bottom shell glow to people on shell 5 and 6. Shell 3 becomes a temperate zone, located high enough to get sun, but low enough to get extra water from the above 2 shells.

Shell 5 will likely have the most people, as they are able to mine the core rather easily for resources, collect water from upper shells easily, and still collect enough sunlight for farming.

Shell 3 will have less access to resources, but stronger farming. This means shell 4 will probably become a trade hub between shell 5 and 3.

You want to insulate the outside layer and let heat through to the inside layer.

The outside layer is going to have a temperature dominated by the cold of space, at something like -270C. It's really cold. Heat is going to constantly flow through it. You need to make sure that the energy from 1 star can pass through this layer at a rate which preserves the -5C minumum you mention. This will likely involve controlling the conductivity of the outside shell, and possibly even putting insulation on it.

You'll want to do similar for the inside layer. It is going to face the star all day, so it will get much hotter than our environment does today with its transitions between day and night. You'll probably also need to put insulation here, but we may approach that differently.

The real trick is to get the thermal resistance of the inner shells low. Think of it like 7 layers of copper sandwiched in styrofoam. No matter how you heat or chill the outside layers of the stryofoam, the 7 layers conduct heat quickyl from one to ther other, so they are all relatively similar in temperature.

I would choose to do this by erecting giant fresnel lenses over the inner layer which focus a large fraction of the light through relatively small holes in the the layer. If you covered 50% of the area this way, only 50% of the light would fall on the inner layer (the part that was not covered). The remaining light could go through holes, which would cause the inner layer to look like a giant constelation of stars, with each hole providing some of the sun's light to the next layer. This would let you rapidly move heat from one layer to another without wasting large amounts of surface area. It also means you can have natural light on the second and third layers, which is a nice perk.

• Now we're cooking. However this is still a little too star centric. Should I clarify more that this world isn't covering my dyson sphere? This is a world that is constructed on an orbit around the star that does not collide with any other major celestial body nor alter orbits significantly from any other world. Or would it be better suited being a world covering the dyson sphere? Granted the empire's homeworld would have to be migrated onto that world lest they all freeze in the absence of a star to heat them. – GuidingOlive Nov 8 '18 at 20:24
• Wait... you have a fully functioning Dyson sphere, and a mere planetary-sized world is going to relieve your population pressures? – Cort Ammon Nov 8 '18 at 20:28
• Several planetary-sized worlds, one inside the other but yes. Look, I'm not a scientist I'm a writer. This civilization has built a functioning dyson sphere (maybe it runs at like 1% efficiency) and is far from its next solar neighbor (in a manner like we are). Its population has been utilizing rotating habitats but a larger scale solution is needed. Currently it claims (and has reasonable claim over) its solar system. But its more band-aid then it is relief. – GuidingOlive Nov 8 '18 at 20:36
• I'd say the right answer is "the engineers put the additional habitation wherever it is convenient." It depends on their construction. But I would note that if you lived on 3 Earthlike worlds, and you had a dyson sphere that runs at 1% efficiency and you consume that 1% on the planets, that's $1.276\cdot10^{24} W$ of power to dissipate per planet. Currently the sun provides Earth $1.74\cdot10^{17} W$ of power. Your three planets each need to be sinking almost $10,000,000$ times more heat into space than the Earth does today! That's some incredible engineering capabilities. – Cort Ammon Nov 8 '18 at 20:41
• If you're a Kardashev Class II civlization, you're basically obliged to live distriuted across either a dyson sphere or similarly massive collection of structurse or across the galaxy, or you'll never dissiapate all that waste heat. It's just too much energy for a handful of planets. – Cort Ammon Nov 8 '18 at 20:42


text:
## RMSprop converges with proper hyper-parameter

### Naichen Shi · Dawei Li · Mingyi Hong · Ruoyu Sun

Keywords: [ convergence ] [ hyperparameter ] [ RMSprop ]

Abstract: Despite the existence of divergence examples, RMSprop remains one of the most popular algorithms in machine learning. Towards closing the gap between theory and practice, we prove that RMSprop converges with proper choice of hyper-parameters under certain conditions. More specifically, we prove that when the hyper-parameter $\beta_2$ is close enough to $1$, RMSprop and its random shuffling version converge to a bounded region in general, and to critical points in the interpolation regime. It is worth mentioning that our results do not depend on bounded gradient" assumption, which is often the key assumption utilized by existing theoretical work for Adam-type adaptive gradient method. Removing this assumption allows us to establish a phase transition from divergence to non-divergence for RMSprop. Finally, based on our theory, we conjecture that in practice there is a critical threshold $\sf{\beta_2^*}$, such that RMSprop generates reasonably good results only if $1>\beta_2\ge \sf{\beta_2^*}$. We provide empirical evidence for such a phase transition in our numerical experiments.


text:
# Is there another way of integrating this?

52.$x=\cos^2 t, y=\cos t, 0\leq t\leq 4\pi$. $(\frac{dx}{dt})^2+(\frac{dy}{dt})^2=(-2\cos(t)\sin(t))^2+(-\sin t)^2=\sin^2 t(4\cos^2 t+1)$ \begin{align} Distance &=\int_0^{4\pi}|\sin t|\sqrt{4\cos^2 t+1}dt=4\int_0^\pi \sin t\sqrt{4\cos^2 t+1}dt\\ &=-4\int_1^{-1}\sqrt{4u^2+1}du [u=\cos t, du=-\sin tdt] = 4\int_{-1}^1\sqrt{4u^2+1}du\\ &=8\int_0^1\sqrt{4u^2+1}du=8\int_0^{\tan^{-1} 2}\sec\theta\cdot \frac{1}{2}\sec^2\theta d\theta [2u=\tan\theta, 2du=\sec^2\theta d\theta ]\\ &=4\int_0^{\tan^{-1} 2}\sec^3\theta d\theta = [2\sec \theta\tan\theta+2\ln |\sec\theta+\tan\theta |]_0^{\tan^{-1} 2}=4\sqrt{5}+2\ln(\sqrt{5}+2) \end{align} Thus, $L=\int_0^\pi |\sin t|\sqrt{4\cos^2 t+1}dt=\sqrt{5}+\frac{1}{2}\ln(\sqrt{5}+2)$.

I decided to factor $4$ out of the square root and replaced $u^2$ by $u^2$ and $1/4$, and then used an integration identity. I don't remember how to integrate by substituting variables by tan and sec, so I was wondering if it was absolutely crucial that I remember it for my mid-term.

I got $4(-\sqrt{(5)} -1/4 \ln(-1\sqrt{(5/4)}/1\sqrt{(5/4)})$ in the end. Is there a simpler way? And did I make any mistake?

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Geez, that ad-ridden page was a surprise! You can post images directly into MSE questions via the button above the question box that looks like a pretty landscape painting. –  Kevin Carlson Oct 11 '12 at 20:57


text:
# As shown in the figure below, a stick of length L = 0.460 m and mass...

###### Question:

As shown in the figure below, a stick of length L = 0.460 m and mass m = 0.195 kg is in contact with a rough floor at one end and a frictionless bowling ball (diameter

d = 19.00 cm) at some other point such that the angle between the stick and the floor is

θ = 30°.Determine the following.

(a) magnitude of the force exerted on the stick by the bowling ball

(i know A is 1.07)

(b) horizontal component of the force exerted on the stick by the floor

(c) vertical component of the force exerted on the stick by the floor

30.0

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# A straight, thin wire carries a constant current of 2 amps_ A 4-inch by 1-inch 2A rectangular loop of wire

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A straight, thin wire carries a constant current of 2 amps_ A 4-inch by 1-inch 2A rectangular loop of wire with a resistance of 10 microohms is moved away from the wire at a constant speed of 1 meter per second. Im/s Estimate the magnetic flux through the loop when its center is 10 cm away from the wire: Estimate the magnetic flux through the wire when its center is 20 cm away from the wire. What is the average current induced in the rectangular loop during this move? Which way does current flow? Is it increasing or decreasing? Does the loop experience a force? What is the direction of the force?

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##### A box is sliding up an incline that makes an angle of 32.9Â°with respect to the horizontal. The coefficient of kinetic frictionbetween the box and the surface of the incline is 0.168. Theinitial speed of the box at the bottom of the inclineis 4.25 m/s. How far does the box travel along theincline before coming to rest?
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##### A long wire in the plane of the screen carries a current of 9amps to the right. The charge of 0.274 coulombs in the plane of thescreen is a distance 0.027 m above the wire moving to the left at4,321 m/s.What is the direction of the magnetic field at the charge?What is the direction of the force on the charge?Enter the magnitude of the force on the charge inNewtons.
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text:
# Before proving that $Aut(G)$ is a subgroup of $S_G$ …

Exercise 37 Ch. 9 of the book Abstract Algebra by T Judson :

We will denote the set of all automorphisms of G by $Aut(G)$. Prove that $Aut(G)$ is a subgroup of $S_G$ , the group of permutations of G.

Let $G= \mathbb {Z_6}$. Then there are two isomorphisms between $\mathbb {Z_6}$ and $\mathbb {Z_6}$ with additive operation. One is $\phi (x)=x$ the identity, i.e. ${\{0,1,2,3,4,5}\} \longleftrightarrow {\{0,1,2,3,4,5}\}$, and the other is $\phi (x)=x+5$, i.e. ${\{0,1,2,3,4,5}\} \longleftrightarrow {\{5,0,1,2,3,4}\}$. But the set ${\{(1), (054321)}\}$ is not a group at all let alone to be a subgroup of $S_{|\mathbb{Z_6}|}=S_6$.

Where am I wrong?

• $x \mapsto x+5$ is not an additive homomorphism. You probably mean $x \mapsto 5x$. – lhf May 22 '16 at 1:56

The automorphisms of $(\mathbb {Z_6},+)$ are $x \mapsto ax$, for $a=\pm1$.

They form a cyclic group of order $2$, which is a subgroup of $S_6$.

• Is there any homomorphism with $x \mapsto x+a$ ? – Liebe May 22 '16 at 2:02
• @Liebe, just for $a \equiv 0 \bmod 6$ because it has to send $0$ to $0$. – lhf May 22 '16 at 2:07

Your second map is not an automorphism, because it doesn’t even take the identity ($0$) to itself. Rather, the nontrivial automorphism is $x\mapsto-x$, so leaves $0$ and $3$ fixed, otherwise $1\leftrightarrow5$, $2\leftrightarrow4$.


text:
# Gray theme : Tab border color is blue for a second then turns gray

kapil4ya
172 2

Guys,

I have a problem specific to the silvergray.jar (gray theme). The whole gray them works great, except for the tabs background and border color.

For initial load, the border and background still show blue and then turn gray after a second.

I tried the following change in the color.css.dsp

.z-tabs .z-tabs-cnt {
border-color:${Color04}; -moz-user-select:none; background:url("/jobs/zkau/web/451766/zul/img/tab/tabs-bg.png") repeat-x scroll 0 top #C7E3F3; border-bottom:1px solid${Color04};
display:block;
list-style:none outside none;
margin:0;
}

But it still first shows the tab in blue color for a fraction of a second and then turns to gray.

Please throw some light on this issue.

Best,
Kapil

delete retag edit

## 5 Replies

jimmyshiau
4921 5
http://www.zkoss.org/

Hi Kapil
What ZK version are you using?
due to some structure of ZK component changed,
You have to use the right version of silvergray.jar ( 5.0.0 or 5.0.1 or 5.0.3 or 5.0.4 )

kapil4ya
172 2

I am using 5.0.3 EE version.

And can I know where I can find those different silvergray.jar (5.0.3).

jeanher
1719 1 6

We will contact you for providing older versions of themes.

kapil4ya
172 2

I have downloaded the 5.0.3 silvergray.jar file , but the issue remains the same.

The interesting thing is that only in case of tabs we are getting this issue. Does attaching a screen shot of what firebug shows help ?

jimmyshiau
4921 5
http://www.zkoss.org/

The blue color was cause by

background:url("/jobs/zkau/web/451766/zul/img/tab/tabs-bg.png") repeat-x scroll 0 top #C7E3F3;

background:url("/jobs/zkau/web/451766/zul/img/tab/tabs-bg.png") repeat-x scroll 0 top #E6E5E1

or just use
background-image:url("/jobs/zkau/web/451766/zul/img/tab/tabs-bg.png");

[hide preview]


text:
How to create both list of abbreviations and list of nomenclature using nomencl package?

I have searched for the whole day online, but all the threads indicate only realizing one of the lists by using nomencl package. I want to create both list of abbreviations AND list of nomenclature using nomencl package. How do I differ two \printnomenclature so that LaTeX could separate them clearly?

-
Welcome to TeX.SE! –  Vivi Dec 12 '12 at 6:27
Perhaps you can use the gloss package for the abbreviations? –  guillem Dec 12 '12 at 7:21
Have a look at glossaries, a package which is capable of dealing with multiple - well - glossaries. –  cgnieder Dec 12 '12 at 9:06
+1 for glossaries, Iused that myself, and it works great. I guess it is up to @cgnieder to make that an answer. –  mafp Dec 12 '12 at 16:36

I'm not aware of a way to do this with nomencl. However, there are other packages which can be used. I'll give two examples, one for my package acro and one for glossaries.

1. acro

The acro package allows to assign acronyms to a class and print lists for each class (also for combined classes...). This fact can be used for the task. Entries are defined with the following syntax:

\DeclareAcronym{<ID>}{
short = <short> ,
long  = <long> ,
class = <class>
}

Here is a full example:

\documentclass{article}
\usepackage{acro}

% probably a good idea for the nomenclature entries:
\acsetup{first-style=short}

% class abbrev': abbreviations:
\DeclareAcronym{ny}{
short = NY ,
long  = New York ,
class = abbrev
}
\DeclareAcronym{la}{
short = LA ,
long  = Los Angeles ,
class = abbrev
}
\DeclareAcronym{un}{
short = UN ,
long  = United Nations ,
class = abbrev
}

% class nomencl': nomenclature
\DeclareAcronym{angelsperarea}{
short = \ensuremath{a} ,
long  = The number of angels per unit area ,
sort  = a ,
class = nomencl
}
\DeclareAcronym{numofangels}{
short = \ensuremath{N} ,
long  = The number of angels per needle point ,
sort  = N ,
class = nomencl
}
\DeclareAcronym{areaofneedle}{
short = \ensuremath{A} ,
long  = The area of the needle point ,
sort  = A ,
class = nomencl
}

\begin{document}

\ac{ny}, \ac{la} and \ac{un} are abbreviations whereas
\ac{angelsperarea}, \ac{numofangels} and \ac{areaofneedle} are part of the
nomenclature

\printacronyms[include-classes=abbrev,name=Abbreviations]

\printacronyms[include-classes=nomencl,name=Nomenclature]

\end{document}

2. glossaries

The glossaries package is much more powerfull. You can define as many glossaries as you like. Here we can use the fact that a glossary for acronyms is already defined. Similar to nomencl it requires you to run a script to sort the entries. If your file is called file.tex this is usually done by calling

makeglossaries file

from the command line. Details can be found in the user manual (page 26, section 1.3.3).

The full example:

\documentclass{article}
\usepackage{longtable}
\usepackage[acronym]{glossaries}

% abbreviations:
\newacronym{ny}{NY}{New York}
\newacronym{la}{LA}{Los Angeles}
\newacronym{un}{UN}{United Nations}

% nomenclature:
\newglossaryentry{angelsperarea}{
name = $a$ ,
description = The number of angels per unit area,
}
\newglossaryentry{numofangels}{
name = $N$ ,
description = The number of angels per needle point
}
\newglossaryentry{areaofneedle}{
name = $A$ ,
description = The area of the needle point
}

\makeglossaries
\begin{document}

\gls{ny}, \gls{la} and \gls{un} are abbreviations whereas
\gls{angelsperarea}, \gls{numofangels} and \gls{areaofneedle} are part of the
nomenclature

\printglossary[type=\acronymtype,title=Abbreviations]

\printglossary[title=Nomenclature]

\end{document}

-
Forgive my ignorance, but what are those numbers in the end of each abbreviation / nomenclature entry ? –  joxnas Feb 3 at 9:14
@joxnas the page the respective acronym appeared –  cgnieder Feb 3 at 9:17

This also works (I couldn't get the above script working for a symbol list with units AND glossary..). Important is the order of lines here!

BasicFile.tex:

%Load the package
\usepackage[
nonumberlist, %do not show page numbers
acronym,      %generate acronym listing   -> Not used in this example (see line with %%% )
section]      %use section level for toc entries
{glossaries}

%Generate a list of symbols
\newglossary[slg]{symbols}{syi}{syg}{List of symbols}

%Remove the dot at the end of glossary descriptions
\renewcommand*{\glspostdescription}{}

%Activate glossary commands
\makeglossaries

%These commands sort the lists
%%%makeindex -s filename.ist -t filename.alg -o filename.acr filename.acn
%makeindex -s filename.ist -t filename.glg -o filename.gls filename.glo
%makeindex -s filename.ist -t filename.slg -o filename.syi filename.syg

\begin{document}

%Print the glossary
\printglossary[style=altlist,title=Glossary]

%%%%Print list of acronyms
%%%\printglossary[type=\acronymtype,style=long]

%Print list of symbols
\printglossary[type=symbols,style=long4col]
\clearpage

bla-bla \gls{h} \gls{nslats} \gls{illuminance}

\end{document}

nomenclature.tex:

%Some entries for the list of symbols
\newglossaryentry{nslats}{
name=$nslats$,
description={The number of slats},
sort=S, type=symbols, symbol=[$-$]}

\newglossaryentry{h}{
name=$h$,
description={Height},
sort=h, type=symbols, symbol=[$m$]}

glossary.tex

%Some glossary terms
\newglossaryentry{illuminance}{
name=Illuminance,
description={Light flux striking a surface, measured in Lux (lx).}
}

%%% In case you also want a list acronyms
%%%\newacronym{MS}{MS}{Microsoft}
%%%\newacronym{CD}{CD}{Compact Disc}

I don't use Perl so I type these lines directly in the command prompt after changes in any file (notice: if you didn't change anything concerning the glossary or the symbol list it is be okay to just pdflatex only once!):

pdflatex BasicFile.tex
pdflatex BasicFile.tex
makeindex -s BasicFile.ist -t BasicFile.slg -o BasicFile.syi BasicFile.syg
makeindex -s BasicFile.ist -t BasicFile.glg -o BasicFile.gls Report_Base_v6.glo
pdflatex BasicFile.tex
BasicFile.pdf

-
Welcome to TeX.sx! –  Kurt Feb 15 '13 at 13:28


text:
Hilbert spectral analysis

Hilbert spectral analysis

Hilbert spectral analysis is a signal analysis method applying the Hilbert transform to compute the instantaneous frequency of signals according to
:$omega=frac\left\{d heta\left(t\right)\right\}\left\{dt\right\}.,$After performing the Hilbert transform on each signal, we can express the data in the following form::$X\left(t\right)=sum_\left\{j=1\right\}^\left\{n\right\}a_j\left(t\right)exp\left(iintomega_j\left(t\right)dt\right).,$

This equation gives both the amplitude and the frequency of each component as functions of time. It also enables us to represent the amplitude and the instantaneous frequency as functions of time in a three-dimensional plot, in which the amplitude can be contoured on the frequency-time plane. This frequency-time distribution of the amplitude is designated as the Hilbert amplitude spectrum, or simply Hilbert spectrum.

Hilbert spectral analysis method is an important part of Hilbert-Huang transform.

*Hilbert transform
*Hilbert-Huang transform
*Hilbert spectrum

References

*Alan V. Oppenheim and Ronald W. Schafer, "Discrete-Time Signal Processing"," Prentice-Hall Signal Processing Series, 2 ed., 1999.

*Huang, et al. "The empirical mode decomposition and the Hilbert spectrum for nonlinear and non-stationary time series analysis." "Proc. R. Soc. Lond. A" (1998) 454, 903–995 ( [http://keck.ucsf.edu/~schenk/Huang_etal98.pdf Link] )

Wikimedia Foundation. 2010.

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