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# Implicit differentiation and piston engine speed

Question:

The crankshaft rotates with the constant angular velocity $$\omega$$ rad/s. Calculate the piston speed when

$$\omega t = \frac{\pi}{2}$$

Attempted solution:

Here is an image of the problem:

I have absolutely no knowledge whatsoever about engines or pistons. So I am far outside my comfort zone. But I have worked on a few implicit differentiation problems, so I get the overarching strategy: find relevant function, differentiate both sides, plug in the values for derivatives and variables/constants and solve for whatever you need to get.

From the image, I am guessing that finding $$x(t)$$ seems like a productive start since it is a straight distance from the center. Then taking the derivative and plugging in the expression given in the question.

The inner part seems like a triangle with sides $$x$$, $$b$$ and $$R$$, then the angle velocity $$\omega t$$. Perhaps the law of cosines will give us something:

$$x(t) = \sqrt{R^2 + b^2 -2Rb \cos \omega t}$$

Taking the derivative:

$$x'(t) = \frac{1}{2\sqrt{R^2 + b^2 -2Rb \cos \omega t}} 2Rb\omega \sin(\omega t)$$

Putting in the value for \omega t:

$$x'(t) = \frac{1}{2\sqrt{R^2 + b^2 -2Rb \cos \frac{\pi}{2}}} 2Rb\omega \sin \frac{\pi}{2}$$

Simplfying gives:

$$x'(t) = \frac{Rb\omega}{\sqrt{R^2 + b^2}}$$

However, I am not able to get it to go anywhere from here. The answer is $$R \omega$$ downwards. I am doubtful that I have even gotten the basic idea of the question right.

Edit:

Using Aretino's equation:

$$b^2 = x^2 + R^2 - 2Rx \cos{\omega t}$$

Implicit differentiation gives:

$$0 = 2x + 2Rx\omega \sin{\omega t}$$

Putting in $$\omega t = \frac{\pi}{2}$$:

$$0 = 2x + 2Rx\omega =2x(1+R\omega)$$

...but I think this is wrong and I cannot get any further.

I am looking to finish this question off using a calculus-based answer.

• I personally vote that this should be in Physics StackExchange. Aug 1, 2016 at 13:56
• The question is from a general first-year single variable calculus textbook that does not involve any emphasis on physics or engineering. Aug 1, 2016 at 14:01
• Aretino’s answer points out where you went wrong in setting up the equation. However, for this particular problem, you can skip the calculus. When $\omega t=\frac\pi2$, the short arm of length $R$ is perpendicular to the direction of the piston’s motion (there’s no “wasted” sideways motion) so the piston’s velocity must be equal to the linear velocity of the end of the rotating arm.
– amd
Aug 1, 2016 at 18:39
• The bounty should be granted to @amd.
– user65203
Jul 11, 2019 at 11:54
• That answer is in a comment, was posted two years before the bounty and does not fulfill the stated bounty requirements (answer needed to use implicit differentiation, even if simpler answers were possible). I have awarded the bounty to the first answer that fulfilled the bounty, which was yours. Jul 13, 2019 at 10:59

From

$$b^2=x^2+R^2-2xR\cos(\omega t),$$ by differentiation on time $$0=2x\dot x-2\dot xR\cos(\omega t)+2xR\omega\sin(\omega t).$$

Then at $$\omega t=\dfrac \pi2$$,

$$0=\dot x+R\omega.$$

More generally,

$$\dot x=\frac{xR\omega\sin(\omega t)}{R\cos(\omega t)-x}=\frac{\left(R\cos(\omega t)-\sqrt{b^2-R^2\sin^2(\omega t)}\right)R\omega\sin(\omega t)}{\sqrt{b^2-R^2\sin^2(\omega t)}}.$$

You must be careful with the angle: $\omega t$ is the angle between $R$ and $x$. You should then use: $b^2=x^2+R^2-2Rx\cos\omega t$ and differentiate this equation.

I would solve the crankshaft kinematics through the sines law $${{\sin \alpha } \over R} = {{\sin \omega \,t} \over b} = {{\sin \left( {\pi - \omega \,t - \alpha } \right)} \over x} = {{\sin \left( {\omega \,t + \alpha } \right)} \over x} = {{\sin \omega \,t\cos \alpha + \cos \omega \,t\sin \alpha } \over x}$$ which is valid also for $$\pi < \omega \, t < 2 \pi$$, that is for negative values of the sines.

From there it is easy to deduce $$x = b{{\sin \omega \,t\cos \alpha + \cos \omega \,t\sin \alpha } \over {\sin \omega \,t}} = b\sqrt {1 - \left( {R/b} \right)^{\,2} \sin ^{\,2} \omega \,t} + R\cos \omega \,t$$ which for $$R< correctly reduces to $$b+R \cos \omega \, t$$.

The derivate is then $$x' = - \,\omega \,R\sin \omega \,t\left( {1 + {{R\cos \omega \,t} \over {b\sqrt {1 - \left( {R/b} \right)^{\,2} \sin ^{\,2} \omega \,t} }}} \right)$$

An example of the dispacement - speed pattern is sketched here

Engineering-wise then, without any calculation, it is clear that at $$\omega\,t = \pi / 2$$, the speed of the piston will be same as the radial speed $$\omega R$$ (with the minus sign as giving a diminuishing $$x$$), because the two are parallel which means that the crank has only translatory speed with no rotation.


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0 energy points

# Proportion word problems: dimensional analysis

Sal solves two related proportion word problems about a squirrel crossing the road and a car that approaches it. He does that using dimensional analysis. Created by Sal Khan.
Video transcript
A squirrel is running across the road at 12 feet per second. It needs to run 9 feet to get across the road. How long will it take the squirrel to run 9 feet? Round to the nearest hundredth of a second. Fair enough. A car is 50 feet away from the squirrel-- OK, this is a high-stakes word problem-- driving toward it at a speed of 100 feet per second. How long will it take the car to drive 50 feet? Round to the nearest hundredth of a second. Will the squirrel make it 9 feet across the road before the car gets there? So this definitely is high stakes, at least for the squirrel. So let's answer the first question. Let's figure out how long will it take the squirrel to run 9 feet. So let's think about it. So the squirrel's got to go 9 feet, and we want to figure out how many seconds it's going to take. So would we divide or multiply this by 12? Well, to think about that, you could think about the units where we want to get an answer in terms of seconds. We want to figure out time, so it'd be great if we could multiply this times seconds per foot. Then the feet will cancel out, and I'll be left with seconds. Now, right over here, we're told that the squirrel can run at 12 feet per second, but we want seconds per foot. So the squirrel, every second, so they go 12 feet per second, then we could also say 1 second per every 12 feet. So let's write it that way. So it's essentially the reciprocal of this because the units are the reciprocal of this. So, it's 1 second for every 12 feet. Notice, all I did is I took this information right over here, 12 feet per second, and I wrote it as second per foot-- 12 feet for every 1 second, 1 second for every 12 feet. What's useful about this is this will now give me the time it takes for the squirrel in seconds. So the feet cancel out with the feet, and I am left with 9 times 1/12, which is 9/12 seconds. And 9/12 seconds is the same thing as 3/4 seconds, which is the same thing as 0.75 seconds for the squirrel to cross the street. Now let's think about the car. So now let's think about the car. And it's the exact same logic. They tell us that the car is 50 feet away. So the squirrel is trying to cross the road like that, and the car is 50 feet away coming in like that, and we want to figure out if the squirrel will survive. So the car is 50 feet away. So it's 50 feet away. We want to figure out the time it'll take to travel that 50 feet. Once again, we would want it in seconds. So we would want seconds per feet. So we would want to multiply by seconds per foot. They give us the speed in feet per second, 100 feet per second. And so we just have to realize that this is 100 feet for every 1 second, or 1/100 seconds per feet. This is once again just this information, but we took the reciprocal of it, because we don't want feet per second, we want seconds per feet. And if we do that, that cancels with that, and we're left with 50/100 seconds. So this is 50/100 is 0.50 seconds. And so now let's answer the question, this life and death situation for the squirrel. Will the squirrel make it 9 feet across the road before the car gets there? Well, it's going to take the squirrel 0.75 seconds to cross, and it's going to take the car only half a second. So the car is going to get to where the squirrel is crossing before the squirrel has a chance to get all the way across the road. So unfortunately for the squirrel, the answer is no.

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# Quadrupolar correlations and deformation effect on two neutrino $\varepsilon \beta ^{+}$ and $\varepsilon \varepsilon$ modes of $^{156}$Dy isotope

Authors
Type
Published Article
Publication Date
Submission Date
Identifiers
DOI: 10.1088/0954-3899/37/5/055108
Source
arXiv
The two-neutrino positron double-$\beta$ decay modes of $^{156}$Dy isotope are studied in the Projected Hartree-Fock-Bogoliubov framework for the $0^{+}\to 0^{+}$ transition. Theoretically calculated half-lives of electron-positron conversion and double-electron capture modes are presented. The effect of the quadrupolar deformation on nuclear transition matrix element $M_{2\nu}$ is also investigated.


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# how to use a table

In this entry, how to use a table to multiply numbers will be explained and illustrated with examples. The general rule is as follows:

Suppose you want to multiply two numbers. Find the first number in the very first column. Follow the row in which this number appears until you land in the column which is labelled with the second number on top. The number on which you land is your answer.

Let us illustrate with an example. To keep things simple, we will use a 5 by 5 multiplication table

 $\times$ 1 2 3 4 1 2 3 4 5 1 2 3 4 5 2 4 6 8 10 3 6 9 12 15 4 8 12 16 20 5 10 15 20 25

Suppose we want to figure out $3\times 4$. We first find the row which starts with “3”. That row is highlighted below:

 $\times$ 1 2 3 4 1 2 3 4 5 1 2 3 4 5 2 4 6 8 10 3 6 9 12 15 4 8 12 16 20 5 10 15 20 25

Next we follow this row until we find the column which has “4” on top. This column is highlighted below:

 $\times$ 1 2 3 4 1 2 3 4 5 1 2 3 4 5 2 4 6 8 10 3 6 9 12 15 4 8 12 16 20 5 10 15 20 25

Note that “12” appears where the highlighted row and the highlighted column cross. (For convenience, it has been underlined.) Therefore, we found that $3\times 4=12$.

Title how to use a table HowToUseATable 2013-03-22 16:36:25 2013-03-22 16:36:25 rspuzio (6075) rspuzio (6075) 5 rspuzio (6075) Example msc 11B25


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# Cons to UV printing

I've been curious about the various UV/Laser printers in (or coming into) market that use liquid resin. I've seen the samples of the Pegasus Touch, Form1, and the Carbon3D as examples. I like the specifications of the quality that machines can put out. However, in my experience with FDM printing, there almost always seems to be something not quite right about the print.

So, what are some major maintenance considerations for these types of 3D printing? Also, specifically, are supports and overhangs as much an issue in these types of printers as with FDM/FFF?

Here are some things I consider major maintenance considerations in FDM:

• Extruder Clogging
• Build platform conditions (i.e. levelness, clean, type of tape, bubbles in tape)
• Variances in material quality (i.e. diameter, purity, physical conditions)
• Mechanics of the machine (i.e. belts, rods, gear teeth, etc.)
• Build environment (i.e. ensuring steady temperature in the build environment, minimize draft)

I'm not necessarily looking for printer recommendations, more so technical insight on the technology.

Maintenance for a resin printer means keeping the vat or tray clean, using appropriate methods to remove the unused resin (or leaving it in the vat per manufacturer's directions). Cleaning the tray should be done also per manufacturer's spec, although each printer's user forum may provide better or more effective options.

The Pegasus Touch has a caution regarding dripping resin on the mirrors, so there's operational care considerations for these types of printers.

There is a build platform for these printers. The flatness and level are as critical or more so for resin printers, as the resolution can be astonishingly high. If any portion of a print does not bond to the platform, that entire print will have a failed section, creating an entirely failed print. Gravity is not particularly helpful in that respect, at least with the Pegasus Touch.

The release medium varies from device to device. The Pegasus Touch originally used PDMS (silicon release compound) and now uses what's called a SuperVat. The plastic material in the SuperVat is purported to provide better release and fewer failures, along with increased lifespan. PDMS becomes cloudy from repeated printing in the same location and can be torn away from the vat if the print does not properly release.

I've become aware of a product from Australia that has had good reports from use in a B9 Creator resin printer. The report indicates that it releases the model quite easily and barely turns cloudy. I have an order pending for this material, as I am hopeful it will perform as described.

The mechanics are also varied. One expects a system to raise and lower the build platform and to direct the laser or illumination system (DLP), but generally, this type of printer is somewhat simpler mechanically.

Because I live in a hot humid climate, my Pegasus Touch remains in the box, and my brain is about to explode with what I've learned of using it. Environmental conditions are likely to vary with different machines. I've seen references that 70 degrees F is too cold, others that say 70-75 degrees F is just fine, anything higher is too hot. Another user says that 65 degrees is good. The type of resin also becomes an important factor for environmental conditions.

The laser will create heat in the resin, so I'm inclined to believe that cooler is better. Different colors require different durations of laser light, somewhat akin to various plastics having different temperatures.

supports and overhangs are important considerations in an SLA or DLP printer, just as they are in FDM.

Expect also that many of the resin printers require that the user purchase only the product provided by the manufacturer. This isn't necessarily a negative as most of the resin sources are priced similarly.

If I've missed any part of your question, let me know.

Despite how many vendors make it appear, resin-curing SLA/DLP printers are industrial or commercial tools that are really not suitable for home desktop use. Here are the major downsides:

• Significantly more expensive to operate than FDM printers, in most cases.
• The resin is seriously toxic until fully cured. Fumes can be an issue for users handling raw resin, and you should NEVER put a photopolymer print into a chemically-sensitive environment like an aquarium or children's toy.
• Prints require messy post-processing to rinse off excess resin (usually with rubbing alcohol) and additional UV light exposure to finish hardening the photopolymer. The used alcohol/resin rinse mix is basically hazmat waste.
• In bottom-up printers, the window in the print vat is typically a consumable. Some printers require replacing the vat () after every liter or two of cured resin. (Technology is advancing rapidly here though.)
• The peel mechanism in bottom-up printers is often a major source of print flaws, due to the need to rock/tilt/slide the print to free it from the vat window.
• In top-down printers, you have to pay a large up-front consumables cost to initially fill the resin tank. (There are workarounds here like floating a layer of resin on brine, but these have their own technical issues.)
• If you leave the resin in the printer for an extended period, you'll probably find a hardened layer on the surface from stray light exposure and have to clean out or replace the vat.
• Resin vats/tanks need to be kept clean and free of cured resin debris from failed prints or stray light.
• Every combination of resin chemistry, printer light source, and printer optics requires specific tuning to dial in the photopolymer curing behavior. This means it's somewhat difficult to change resin brands, and you may effectively be locked into the printer manufacturer's resin. Many light sources will change in intensity or develop dim regions over time as they age, which will either harm print quality, require period recalibration, or require frequent light source replacement.
• There is a limited number of options for print materials. Technology here is advancing rapidly, but for the most part, SLA/DLP prints are non-load-bearing models with a limited range of color options.

These are some pretty significant "user experience" downsides compared to a consumer desktop FDM printer. It's more hazard, more work, and more cost than FDM. SLA/DLP is primarily advantageous where high resolution or high print speeds are required.


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# diagonal times rank 1 matrix still rank 1?

First let me make two statements to give my question the proper context.

1. Consider $D$ a diagonal matrix and $u v^T$ a rank 1 matrix. From my current knowledge there exist computationally cheap algorithms to diagonalize $D+uv^T$ as discussed here: maximum eigenvalue of a diagonal plus rank-one matrix

2. Diagonalizing a matrix of the form $Duv^T$ is straightforward, see here: Eigenvector of a diagonal matrix times a rank-1 matrix

Now to my question(s):

1. I want to diagonalize $D^2+Duv^T$. Is $Duv^T$ still a rank 1 matrix? I considered some examples but I am not really sure if this is a general statement.

2. This question depends on the answer of the first. If the statement is true the algorithms in statement 1 should be applicable to diagonalization of $D^2+Duv^T$, no? If it is not true are there others ways to tackle this?

Let $u=v=(1,0)^T$, and

$$D=\left(\begin{matrix}0&0\\0&1\end{matrix}\right)$$

Then

$$uv^T=\left(\begin{matrix}1&0\\0&0\end{matrix}\right)$$

And $Duv^T=0$.

Of course, if $D$ has only nonzero diagonal elements, then it has full rank and the product has rank $1$, see matrix rank properties on Wikipedia. And if $Duv^T$ has not rank $1$, it's necessarily the null matrix, so it's not a very interesting case.

• Ok thanks, so its actually very easy to write a proof from these properties... let D be of full rank ($rank(D)=n$) and use sylvester rank inequality: $1 = rank(uv^T) = rank(D)+rank(uv^T)-n \le rank(Duv^T) \le min(rank(D),rank(uv^T)) =1$ – J. Goe Apr 6 '16 at 14:02


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ISSN : 1225-0562(Print)
ISSN : 2287-7258(Online)
Korean Journal of Materials Research Vol.30 No.2 pp.57-60
DOI : https://doi.org/10.3740/MRSK.2020.30.2.57

# High-Temperature Deformation Behavior of Ti3Al Prepared by Mechanical Alloying and Hot Pressing

Chang-Suk Han1, Sung-Yooun Jin1, Hyuk-Ku Kwon2
1Dept. of ICT Automotive Engineering, Hoseo University, 201, Sandan7-ro, Seongmun-myeon, Dangjin City, Chungnam 31702, Republic of Korea
2Dept. of Environmental Engineering, Hoseo University, 20, Hoseo-ro 79beon-gil, Baebang-eup, Asan City, Chungnam 31499, Republic of Korea
Corresponding author E-Mail : hancs@hoseo.edu (C. -S. Han, Hoseo Univ.)
November 15, 2019 November 5, 2019 January 22, 2020

## Abstract

Titanium aluminides have attracted special interest as light-weight/high-temperature materials for structural applications. The major problem limiting practical use of these compounds is their poor ductility and formability. The powder metallurgy processing route has been an attractive alternative for such materials. A mixture of Ti and Al elemental powders was fabricated to a mechanical alloying process. The processed powder was hot pressed in a vacuum, and a fully densified compact with ultra-fine grain structure consisting of Ti3Al intermetallic compound was obtained. During the compressive deformation of the compact at 1173 K, typical dynamic recrystallization (DR), which introduces a certain extent of grain refinement, was observed. The compact had high density and consisted of an ultra-fine equiaxial grain structure. Average grain diameter was 1.5 μm. Typical TEM micrographs depicting the internal structure of the specimen deformed to 0.09 true strain are provided, in which it can be seen that many small recrystallized grains having no apparent dislocation structure are generated at grain boundaries where well-developed dislocations with high density are observed in the neighboring grains. The compact showed a large m-value such as 0.44 at 1173 K. Moreover, the grain structure remained equiaxed during deformation at this temperature. Therefore, the compressive deformation of the compact was presumed to progress by superplastic flow, primarily controlled by DR.

## 초록

© Materials Research Society of Korea. All rights reserved.

This is an Open-Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

## 1. Introduction

Titanium aluminides (TiAl and Ti3Al) have attracted special interest as a light-weigh/high-temperature material for structural applications. The major problem limiting the practical use of these compounds is their poor ductility and formability. The powder metallurgy processing route has been attractive alternative for such materials. Mechanical alloying (MA) was proposed about scores of years ago as a technique for combining alloys.1) Many researchers have applied this technique for producing a compact of TiAl (γ) intermetallic compound, and have reported its characteristics in the structure and the mechanical properties.2-5) The purpose of the present study is to make a compact of Ti3Al (α2) by similar procedure, and examine the structure and the high-temperature deformation behavior of the compact.

## 2. Experimental Procedure

Elemental titanium and aluminum powders having an average particle size of 31 μm and 26 μm in diameter respectively, were mixed to give the starting composition of Ti-30 at.%Al which approximately corresponds to the center of the composition range of α2 phase in the Al- Ti equilibrium diagram.6) The MA by ball-milling was performed in a stainless steel chamber on argon atmosphere. The ratio of the ball (stainless steel) to powder weight was 60 to 1. The process of MA was followed by scanning electron microscopy (SEM) and X-ray diffractometer (XRD).

The compact (approximately 12 mm φ × 6 mm) was produced by vacuum hot pressing (VHP) of the MAprocessed powder at 1173 K for 1 hr under 100 MPa. Transmission electron microscopy (TEM) of the compact was performed with a JEM-2000FX equipment combined with an energy-dispersive X-ray (EDX) analyzer. Thin foils used for the TEM observations were prepared by ion milling.

Compression tests of the specimens (3 mm × 3 mm × 6 mm) machined from the compact was carried out at 1173 K. The specimens quenched during and after compressive deformation were offered to TEM observation. Compressive strain-rate-change test at l173 K was also performed for the specimen, in order to examine the high temperature deformation behavior of the compact.

## 3. Results and Discussion

### 3.1 MA process

The changes in the XRD patterns of the mixture of Ti- 30 at.%Al powder with MA processing time are shown in Fig. 1. In the as-mixed state, all the expected lines of Ti and Al elemental powders are observed. After 720 ks of MA processing, these sharp peaks have almost disappeared and a broad maximum near 2θ = 40° has appeared.

Superimposed on this broad maximum are relatively sharper Bragg peaks characteristic of the microcrystallized elemental phases. No notable changes in the XRD patterns with further increase in MA processing time up to 1800 ks were observed. The morphology and the microstructure of the powder MA-processed for 1800 ks are shown in Fig. 2. In this stage, the shape of the particles of the MAprocessed powder become nearly spherical (average particle size: 25 μm in diameter) as seen in Fig. 2(a). The microstructure of the powder particles typically shown in Fig. 2(b) presumably consists of a mixture of amorphous and microcrystallized elemental phases coincident with the results of the XRD.

### 3.2 Structure of the compact produced by VHP

A compact having relative density of 99.8 % was obtained. The results of the XRD measurement shown in Fig. 3 indicate that the compact consists of α2 single phase. This fact was also supported by these results of the TEM observation combined with the EDX analysis, as seen in Fig. 4. At the same time, the TEM micrograph shown in Fig. 4(a) indicates that the compact has an ultra-fine α2 grain structure. The average grain diameter is approximately 1.5 μm.

### 3.3 High-temperature deformation behavior of the compact

True stress-true strain curve obtained from the results of the compression test of the compact at 1173 K and initial strain rate of 2.8 × 10−3 s−1 was shown in Fig. 5. During the initial stages of the compressive deformation, a sharp stress peak which indicates the occurrence of a typical dynamic recrystallization (DR) is observed.7) In order to confirm such a fact, we carried out TEM observation of the specimens immediately quenched during and after the compression test.

Typical TEM micrographs depicting the internal structure of the specimen deformed to 0.09 true strain are shown in Fig. 6, where (b) is the high magnification photograph of the white square line area of (a). Many small recrystallized grains having no apparent dislocation structure are generated at grain boundaries where the well developed dislocations with high density are observed in the neighboring grains.

Fig, 7 shows typical TEM micrographs observed in the specimen deformed to 0.28 true strain. In this figure, (b) is the high magnification photograph of the white square line area of (a). A little amount of recrystallized small grains which presumably generated in the later stages of the deformation is similarly observed.

These results of the TEM observation evidently support the occurrence of DR during the deformation. Such a DR introduces a certain extent of gain refinement, which can be recognized by comparison between Fig. 4(a) and Fig. 7(a) (average grain diameter, 1.0 μm). In addition to these results, the grain structure of the specimen remains approximately equiaxed during the deformation [Fig. 7(a)]. This fact coincides with the general characteristic of the superplastically deformed alloys. The specimen of the compact was offered to the compressive strain-ratechange test at 1173 K. The relationship between true stress-true strain rate obtained from the results of the test was shown in Fig. 8. The results give the value of strain- rate sensitivity exponent, m, for the compact to be 0.44 according to the general equation:

$σ = C ε ′ m$
(1)

where σ, ε and C are true stress, true strain-rate and a constant, respectively. The relationship obtained from the similar experimental results of the specimen of TiAl compact, which showed no evidence of DR, and was confirmed to be deformed by superplastic flow controlled by grain boundary sliding,3) is also shown in Fig. 8.

The value of m for the compact (α2) of the present study is considerably larger than that of TiAl, and sufficiently satisfies a general requirement for the superplastic material, namely m > 0.3.8) Kim et al. have reported that an ultrafine grained (grain size = 1.6 μm) boron-doped Ni3Al showed m > 0.3 and exhibits superplasticity accompanied by DR during tensile deformation at 973 K.9) From these results, they have proposed that DR plays an important role for relieving a stress concentration arising from grain boundary sliding, and suppressing cavitation during the superplastic deformation.9) The similar interpretation seems to be adaptable to the deformation behavior of Ti3Al compact in the present study. Namely its compressive deformation process presumably progressed by superplastic flow primary controlled by DR.

## 4. Conclusions

A Ti3Al compact was produced by hot pressing of the mechanically alloyed powder, and compressive hightemperature deformation behavior of the compact was investigated. The results obtained can be summarized as follows;

• (1) The compact had high density and was consisted of ultra-fine equiaxial grain (average grain diameter =1.5 μm) structure.

• (2) During compressive deformation of the compact at 1173 K, a typical dynamic recrystallization (DR) which introduces a certain extent of grain refinement was observed.

• (3) The compact showed a large m-value such as 0.44 at 1173 K. Moreover, the grain structure remained equiaxed during deformation at this temperature. Therefore, the compressive deformation of the compact presumed to be progressed by superplastic flow primary controlled by DR.

## Figure

XRD patterns of Ti-30 at%Al powder ball-milled for various times.

SEM micrographs of Ti-30 at.%Al powder ball-milled for 1800 ks. (a): Shape of the powder particles. (b): Microstructure of a sectioned particle.

XRD pattern of the compact.

TEM micrograph of the compact (a) and the results of EDX analysis of the typical grains from 1 to 10 (b): Ti and Al composition of corresponding circled areas.

True stress-true strain curves at l173 K and initial strain rate of 2.8 × 10−3 s−1 by compression test.

(a) TEM micrographs of the specimen water quenched after compressive deformation to the true strain, έ = 0.09. (b) is the high magnification micrograph of the white square line area.

(a) TEM micrographs of the specimen water quenched after compressive deformation to the true strain, έ = 0.28. (b) is the high magnification micrograph of the white square line area.

True stress-true strain rate curve obtained from the results of compressive strain rate change tests of the compact at 1173 K.

## Reference

1. J. S. Benjamin, Sci. Am., 234, 40 (1976).
2. M. R. Farhang, A. R. Kamali and S. M. Nazarian, Mater. Sci. Eng., B, 168, 136 (2010).
3. S. X. Mao, N. A. McMinn and N. Q. Wu, Mater. Sci. Eng., A, 363, 275 (2003).
4. C. S. Han, J. Korean Soc. Heat Treatment, 18, 281 (2005).
5. C. S. Han and K. W. Koo, Korean J. Mater. Res., 18, 51 (2008).
6. H. Zhang and S. Wang, J. Mater. Sci. & Tech., 26, 1071 (2010).
7. B. Paul, A. Sarkar and J. K. Chakravartty, Metall. Mater. Trans. A, 41, 1474 (2010).
8. B. Mintz and W. B. Morrison, Mater. Sci. Technol., 4, 719 (1988).
9. M. S. Kim, S. Hanada, S. Watannbe and O. Izumi, Mater. Trans. JIM, 30, 77 (1989).


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# Multiple integral involving trigonometric functions over a hypercube.

Let $d\ge 1$ be an integer and $\vec{A} := \left(A_j\right)_{j=1}^d \in {\mathbb R}^d$ subject to $\sum\limits_{j=1}^d A_j^2 \le 1$. Define a following integral: $${\mathfrak I}^{(d)}(\vec{A}) := \frac{1}{(2 \pi)^d} \cdot \int_{[0,2\pi]^d} \frac{\prod\limits_{j=1}^d d\phi_j}{\left(1+\sum\limits_{j=1}^d A_j \cos(\phi_j)\right)} = \int\limits_0^\infty \exp(-s) \left(\prod\limits_{j=1}^d I_0(s A_j)\right) ds$$ Now using the Cauchy' residue theorem and then the definition of elliptic integrals we found the following: \begin{eqnarray} {\mathfrak J}^{(1)}(\vec{A})&=& \frac{1}{\sqrt{1-A_1^2}}\\ {\mathfrak J}^{(2)}(\vec{A})&=& \frac{2}{\pi} \frac{1}{\sqrt{1-(A_1-A_2)^2}} \cdot K\left( \frac{4 A_1 A_2}{1-(A_1-A_2)^2}\right)\\ {\mathfrak J}^{(3)}(\vec{A})&=& \frac{2}{\pi^2} \cdot \\ &&\int\limits_{-\infty}^\infty \frac{K(\frac{4 A_2 A_3 (1+t^2)^2}{(1+A_1+A_2-A_3+(1-A_1+A_2-A_3) t^2)((1+A_1-A_2+A_3+(1-A_1-A_2+A_3) t^2)})} {\sqrt{(1+A_1+A_2-A_3+(1-A_1+A_2-A_3) t^2)((1+A_1-A_2+A_3+(1-A_1-A_2+A_3) t^2)}} dt \end{eqnarray} Here $K()$ is the complete elliptic integral of the first kind. Now my question is is it possible to find a closed form expression for the quantities in question when $d\ge 3$?

• Are you sure the listed condition $\sum_{j=1}^{d}A_{j}^{2}\le1$ shouldn't actually be $\sum_{j=1}^{d}|A_{j}|<1$? I'm fairly sure this alternative condition is a necessary and sufficient one for your integral to converge. – David H Jul 28 '18 at 9:38
• Alright, thank you for this comment.I'd be grateful if you could tell me where your condition comes from. This would give me more insight into this integral. – Przemo Jul 30 '18 at 11:38


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# Graphing Data

## Key Questions

Se explanantion

#### Explanation:

Consider the equation $y = m x + c$

You can give any value you chose to $x$ and the value of $y$ depends on what value you give to $x$

So $y$ is the dependant variable and $x$ is the independent variable.

What you looking for is this: if the independent variable only maps to one value in the dependant variable then the equation/graph is that of a function.

• Well, this is a difficult one! I am not sure this is going to help, but I try anyway.

When I have to analyze a phenomenon (I am a physicist...!) I collect data that characterize the phenomenon (for example, I measure the height of a kid each week) and then I plot them.

The result is a graph with points on it that hopefully shows a "tendency".

This can be a linear tendency, for example, so that I can use a line to represent them (the line that "best fit" all the data points).

Next I evaluate the equation of the line in the general form: $y = a x + b$

Now, even if I do not know what happens for a certain value of the variable $x$ (because I didn't measure it directly) I can put it in the equation of the line and get the corresponding value for $y$.

In the example of the height of a kid, I can measure it along, say, a period of 5 weeks, get my graph and evaluate the linear equation as:
$h e i g h t = 2 \cdot t i m e + 40$ in cm for the eight and week for the time (2 will have units of cm/week and 40 of cm).

If I want the height of the kid in the 7th week I put time=7 and find the projected height.

Remember:

You can have different curves that fit your data (parabolas. hyperbolae, sinusoidals...);

The fit depends upon the "goodness" of you experiment in the first place. If you try to correlate the height of a kid to the number of red cars you observe ...it won't probably work very well!

(hope it helps!)

• To visualize it (and eventually, for example, to find out patterns in the data that otherwise would be hard to find).


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# Pumping Tests in Bounded Aquifers

by Glenn M. Duffield, President, HydroSOLVE, Inc.

## What Is ABounded Aquifer?

Mathematical solutions for pumping test analysis such as the well-known Theis solution typically assume unbounded aquifers of infinite lateral extent.

In reality, however, the areal extent of an aquifer may be limited by one or more boundaries. Aquifer boundaries may not impact the results of short-term pumping tests, but the effects of distant boundaries become more evident as the duration of pumping increases.

Boundaries frequently encountered in pumping test analysis include no-flow boundaries, constant-head boundaries and channel aquifers.

## No-Flow (Barrier) Boundary

When an aquifer composed of permeable materials abuts impermeable or confining materials, it is possible to simulate the contact by a no-flow boundary. Lithologic changes (e.g., glacial outwash deposited in a low-permeability bedrock valley) or impermeable vertical faults can result in no-flow boundaries that limit the lateral extent of aquifers. No-flow boundaries are sometimes called barrier or negative boundaries.

A discharging well located near a no-flow boundary exhibits more drawdown than the same well pumping from an infinitely extensive aquifer. One may use the method of images to represent the bounded aquifer in mathematical terms by placing a discharging image well on the opposite side of the boundary from the real well (Figure 1). The image well has the same discharge rate and distance to the boundary as the real well. Thus, the drawdown at the no-flow boundary is exactly twice the drawdown observed an infinite aquifer.

Example

Consider a pumping well discharging at a constant rate, $Q$, from a nonleaky confined aquifer located at a distance $R$ from an impermeable boundary formed by an impermeable fault. To simulate the effect of the barrier on groundwater flow, position a discharging image well at a distance $2R$ from the pumping well on the opposite side of the barrier.

If an observation well is located at a distance ${r}_{1}$ from the real well, the drawdown at the observation well is given by the Theis formula as follows:

$s1 = Q4πTw u1$ $u1=r1 2S 4 Tt$

where $T$ is transmissivity [L²/T], $S$ is storativity [-], $t$ is time since pumping began [T] and $\text{w(}u\text{)}$ is the Theis well function [-].

Similarly, drawdown due to the image well located at a distance ${r}_{2}$ from the observation well is given by:

$s2 = Q4πTw u2$ $u2=r2 2S 4 Tt$

Using the principle of superposition, total drawdown at the observation well, ${s}_{\text{t}}$, is simply the sum of the two preceding drawdown components:

$st=s1+s2$

The example shown in Figure 2 illustrates two diagnostics for a single no-flow boundary. After the barrier becomes evident at 10 to 20 minutes, the semilog slope of the drawdown ultimately doubles and the magnitude of the drawdown derivative also increases by a factor of two. Prior to the influence of a boundary, look for sufficient data to estimate aquifer properties using standard methods of analysis for extensive aquifers. In Figure 2, the first derivative plateau between 0.4 and 4 minutes indicates the infinite-acting period where one could use the Cooper and Jacob (1946) method to estimate T and S.

When a surface water body is hydraulically connected to an aquifer, one may represent the feature by a constant-head boundary. Lakes, streams and rivers are examples of constant-head boundaries encountered in pumping test analyses. Constant-head boundaries are alternately known as recharge or positive boundaries.

A discharging well located near a constant-head boundary experiences less drawdown than the same well pumping from an extensive aquifer. One may use the method of images to represent the bounded aquifer in mathematical terms by placing a recharging image well on the opposite side of the boundary from the real well (Figure 3). The image well has the same pumping rate and distance to the boundary as the real discharging well except that the image well recharges the aquifer. Thus, drawdown in the aquifer is exactly zero at the constant-head boundary.

Example

Consider a pumping well discharging at a constant rate, $Q$, from a nonleaky confined aquifer located at a distance $R$ from a perennial stream that fully penetrates the aquifer. To simulate the effect of the stream on groundwater flow, position a recharging image well with rate $-Q$ at a distance $2R$ from the pumping well on the opposite side of the stream.

If an observation well is located at a distance ${r}_{1}$ from the real well, the drawdown at the observation well is given by the Theis formula as follows:

$s1 = Q4πTw u1$ $u1=r1 2S 4 Tt$

where $T$ is transmissivity [L²/T], $S$ is storativity [-], $t$ is time since pumping began [T] and $\text{w(}u\text{)}$ is the Theis well function [-].

Similarly, head buildup due to the recharging image well located at a distance ${r}_{2}$ from the observation well is given by:

$s2=-Q4πTw u2$ $u2=r2 2S 4 Tt$

Using the principle of superposition, total drawdown at the observation well, ${s}_{\text{t}}$, is simply the sum of the two preceding drawdown components:

$st=s1+s2$

The influence of a constant-head boundary eventually produces a drawdown plateau on any plot of drawdown vs. time (linear, semilog or log-log) while the derivative during this period approaches zero (Figure 4). If sufficient drawdown data are available prior to the emergence of the plateau, one may estimate aquifer properties using standard methods of analysis for extensive aquifers.

## ChannelAquifers

Channel aquifers (also known as strip aquifers) may occur in long narrow valleys or buried channels. One may conceptualize such a channel aquifer as a permeable body between two parallel no-flow boundaries. Simulation of channel aquifer configurations requires the use of an infinite array of image wells (Figure 5).

Figure 6 shows the drawdown response measured during a pumping test in a nonleaky confined buried channel aquifer near Estevan, Saskatchewan in Canada (Walton 1970). By the end of the test, each well exhibits rapidly increasing drawdown owing to the channel boundaries. Prior to the full impact of channel boundaries when the aquifer is still infinite acting, one may match the Cooper and Jacob (1946) straight-line solution to the data to determine the transmissivity and storativity of the aquifer.


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# How do we calculate energy of food?

Here is the problem, according to me:

In a classic hamburger, according to my internet research, approximately 906 kJ are contained.

Now take a car of 1 000 kilograms, and push it to 30 m/s. It has a certain kinetic energy, which is: 1 000 * 30² = 900 kJ

So does it mean that the work (in a physical sense) needed to stop the car is equivalent to the energy brought by the hamburger, so a human just have to eat an hamburger to be able to stop the car?

It is very hard to imagine, as even with the strength of an human and the possibility to brake the car very slowly, I think a human will need a lot of energy.

So how it is possible to compare the energy brought by food and the mechanical/kinetic energy?

Because on the other hand, it could also mean that with the energy brought by one hamburger (approximately 15% of the food energy is used by the human body, and the value I gave take into account this 85% factor), I am able to put push a car to 30 m/s!

• youtu.be/49EaMd1fu-E – Gert Aug 23 '19 at 19:06
• " It has a certain kinetic energy, which is: 1 000 * 30² = 900 kJ" The kinetic energy is actually given by $K=\frac12 mv^2$. So it's 450 kJ, not 900. – Gert Aug 23 '19 at 19:11
• Note that strength isn't the same as work. Also, work doesn't specify a time period to perform that work. – BioPhysicist Aug 23 '19 at 20:04
• "the work (in a physical sense) needed to stop the car" This part might hide a deep misunderstanding of how energy and work work: Generally, it does not require work to stop a car: just crash the car into something, and the something actually gains energy from the car, often in the form of breaking or starting to move. Humans need energy to push a car to a stop because human muscles are not stiff when excerting a force: there's small movement all the time which heats the muscles. But that energy doesn't need to have anything to do with how much energy the car loses by – JiK Aug 24 '19 at 12:51
• @totalMongot Work has a sign. A force on an object in the direction opposite to the movement of the object does negative work on the object. – JiK Aug 24 '19 at 13:19

A calorimeter is a device that you can use to measure the energy of food. You put a small amount of the food in a calorimeter. Then you submerge the calorimeter in a known quantity of water. Now you burn the food and measure the increase in temperature of the water. By careful measurements, one can determine the energy required to heat the water. That amount of heat is the caloric content of that food.

Once you know the energy of the food you can compare it to other material objects just by calculating the energy of the objects. Kinetic energy is just $$(1/2) mv^2$$

• Is it a fact? Humans cannot oxidize substantial parts of food, cows some other parts, whereas this method oxidizes everything. The result won't be very relevant for us. – kubanczyk Aug 24 '19 at 13:18
• @kubanczyk I may be wrong on this, but I guess the process described in this answer is used to determine the amount of energy provided by the digestible constituents of food. I.e. run one test for fat, another for the protein mix in meat, another for the protein mix in wheat, another for the starch in potatoes. Then analyse the nutrient mixture of the food in question and add up the respective energy contents. Thus, indigestible parts like fibers are naturally excluded from the calculation. – cmaster - reinstate monica Aug 24 '19 at 14:13
• This; you burn it. Which begs the question that's already answered in it: "approximately 15% of the food energy is used by the human body" – Mazura Aug 24 '19 at 18:43

It isn't so far fetched as it might appear. A gallon of gas will accelerate the car to speed and keep it there for perhaps 30 miles. The energy in gas comes from chemical bonds. The bonds in gas are not totally different from the bonds in a hamburger.

Burning a gallon of gas produces 132,000 BTU, or $$1.39 \times 10^8$$ J. That is ~150 times more than the 900,000 J from burning a hamburger. Given the relative sizes, it sounds like gas has more energy per pound. So a hamburger has enough energy to run a car for about 0.2 miles.

Eating a hamburger for lunch produces enough energy to run a human until dinner. Over the course of an afternoon, a human could do enough work to push the car up to speed. Perhaps he could pedal a bicycle connected to a generator, and store the energy in batteries. Then the batteries might well have enough charge to get an electric car up to speed.

• Interesting. But the way the hamburger is "burnt" in the human body is not the same as a gallon of gas is literrally "burnt" in a motor But my understanding is that there are some bounds more or less stable: gas bounds are less stable than hamburger bounds so more energy could be extracted from. So is there any reason that could be added to say that the energy at the end would not be the same? – totalMongot Aug 23 '19 at 19:47
• It actually isn't that different. In both cases, fuel is combined with oxygen to produce CO2 and H2O. It just happens slower in a human. It produces heat, which keeps our body temperature at 98.6. But not the flame you get from running the reactions faster, or the explosion from faster still. – mmesser314 Aug 23 '19 at 19:52
• But is a human able to concentrate all the heat/energy Not strictly all of it because some is needed for vital functions. But energy is energy is energy.... – Gert Aug 23 '19 at 20:11
• But is a human able to concentrate all the heat/energy Well, is any engine able to convert all the energy into a useful work? – kubanczyk Aug 24 '19 at 13:22
• "Perhaps he could pedal a bicycle connected to a generator [...]" Or he could just forget about the car and use the 900kJ to ride about 7km on his bike... Nothing beats the fuel efficiency of a cyclist. – cmaster - reinstate monica Aug 24 '19 at 14:05

There's something you're overlooking that makes the whole thing somewhat counter-intuitive: the maximum force the human can exert on the car.

In order to reach the kinetic energy of $$450\mathrm{kJ}$$ that same amount of work $$W$$ has to be done on the car by the human, according to:

$$W=K$$

where:

$$\mathbf{d}W=F\mathbf{d}x$$

Here $$F$$ is the force exerted by the human (we'll assume it constant, for simplicity's sake) on the car and $$\mathbf{d}x$$ the displacement (linear, for simplicity's sake). Integrating we get:

$$W=F\Delta x$$

and:

$$F\Delta x=\frac12 mv^2$$

Or:

$$\Delta x=\frac{m v^2}{2F}$$

Obviously if $$F$$ is relatively small compared to $$mv^2$$, $$\Delta x$$ will be large. However that doesn't mean the human won't get the car up to speed: it just takes quite a long time.

That time can be calculated from Newton's second Law:

$$F=ma \to a=\frac{F}{m}$$

and:

$$v=a\Delta t \to \Delta t=\frac{v}{a}$$

So:

$$\Delta t=\frac{mv}{F}$$

It's similar if we made the car tow a fully loaded $$20$$ tonnes HGV: the car can do it but it would take an impractical amount of time because the maximum amount of force it can deliver is small compared to that of an HGV tractor.

• I think you should mention friction somewhere. – infinitezero Aug 24 '19 at 8:47
• +1 but it's not only about maximum force, it's also about maximum power. With rope and pulley systems, the problem of maximum force can be avoided, but still we'd have the problem of maximum power. With a good quality and ridiculously big system, a human could actually pull a car to fast speeds, but it still requires time because there's maximum power. But as $P=Fv$, the initial force could be quite big with an ideal rope and pulley system. – JiK Aug 24 '19 at 12:55
• ...and obviously, the inefficiency of human muscles needs mentioning. – JiK Aug 24 '19 at 13:25

The whole scenario seems to be implausible when you consider that in order for the person to accelerate the vehicle to a speed of 30 m/s at the end the person would have to run at a pace of 70 mph!

A more reasonable speed to push the car is about 10 mph or 4.5 m/s which is a kinetic energy of about 10 kJ. That’s only 10% of your hamburger’s calories.

By the way, according to one site an average person's straight pushing force is 60-80 N. At 80 N assuming negligible friction the car's acceleration would be 0.08$$\frac{m}{s^2}$$. So If would take about a minute to reach 4.5 m/s.

Hope this helps.

• You can easily overcome the problem that a human can't run at 30 m/s by changing the problem to a human-powered vehicle. And the "straight pushing force" you quote ignores the fact that the human can use their own weight to apply force to the car. A normally fit person can accelerate a car up to walking pace (say 2 m/s) in a lot less time than 30 seconds. – alephzero Aug 24 '19 at 23:52
• @alephzero Couple of issues here. First, the OP was clearly talking about a person pushing the vehicle, not a human powered vehicle. Second, the force quoted on the site didn't discuss how the person applies it, just that it is horizontal. So it probably includes both the force of leaning into the vehicle, and the force applied due to the friction between the person's feet and the ground. In any case, the amount of force is not the main issue. It's the feasibility of burning the hamburger calories by pushing the car. – Bob D Aug 25 '19 at 16:55

I am able to put push a car to 30 m/s!

You are ignoring the inefficiency of your body burning those 900 kJ here, your actual mechanical output will be much lower.

Ok, so let's assume that you can actually produce 25 % of the burned energy as mechanical work, then, yes, you can accelerate the car to 15 m/s (half the speed, quarter the energy). If the car is suspended frictionlessly in vacuum, that is. It'll take you a while, because you will be feeding this energy into the car at about 200 to 250 watts, so it'll take you about $$225\ \mathrm{kWs}/225\ \mathrm W = 1000\ \mathrm s = 16\ \mathrm{min} + 20\ \mathrm s$$ to get the car up to speed.

Unfortunately, car's loose much more than the 250 watts that you can output to wind resistance when going 54 km/h, so you won't reach that speed without the vacuum. You'll only reach the speed at which the losses to friction and wind resistance match your energy input, which is a rather meager figure. Better get on your bike and ride about 7 km at 25 km/h with the energy your hamburger provided.


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# Multiple moons but only one of them is tidally locked to its planet?

A habitable planet with multiple moons, only one of which is tidally locked to the planet: is this general scenario possible?

I understand that tidal locking takes a very long time to happen. So maybe this setup could be achieved by having one very old moon and some newer moons, where the older moon is tidally locked to the planet (and the newer moons haven't had the chance yet).

Presumably (?), if the new moons were created in an impact event like Earth's moon, this would destabilise the older moon causing it to unlock from the planet. But if the new moons were formed by e.g. asteroids being captured by the planet's gravity, this wouldn't be a problem.

• Hello daisy, and welcome to Worldbuilding. Just to clarify, do you have a number of additional, non-locked moons in mind, and do you have specific sizes and/or distances of any of the moons, or do you just want to know if the general idea is plausible? If the former, please edit your question to include that information. Please take the tour and visit the help center to learn more about the site. Have a nice day! – Gryphon Jun 25 '18 at 23:24

The time it takes a moon to become tidally locked to a planet depends on a number of factors. In particular, $$t_{\text{lock}}\propto\frac{\omega a^6}{m_p^{2}}$$ where $\omega$ is the moon's initial angular speed, $a$ is its semi-major axis, and $m_p$ is the mass of the planet. There's a very strong dependence on $a$ - double it, and we increase $t_{\text{lock}}$ by a factor of 64! So if it takes, say, 10 million years for a moon with semi-major axis $a_1$ to become tidally locked with its planet, then a moon with semi-major axis $a_2=2a_1$ could take 640 million years to become tidally locked - even if they started with the same period of rotation.

Let's crunch some numbers. The initial conditions of the Moon after its formation are unknown, but there's been some modeling of them. Cuk et al. (2016) did some simulations of the rotation of both Earth and the Moon. Using their modeling, I'm going to assume several things:

• The Moon's initial angular speed was $\sim10^{-5}$ radians/second.
• The initial semi-major axis was about 25 Earth radii, or $\sim1.6\times10^8$ meters.
• The time until tidal locking was about 20 million years, give or take.

Therefore, using the mass of the Earth, $M_{\oplus}$, we find that $$t_{\text{lock}}\approx2\times10^7\left(\frac{\omega}{10^{-5}\text{ rad s}^{-1}}\right)\left(\frac{a}{1.5\times10^8\text{ m}}\right)^6\left(\frac{m_p}{M_{\oplus}}\right)^{-2}\text{ years}$$ This does assume that certain physical properties of the satellite, such as its radius, mass, and love number are all the same as the Moon's. However, it's simple to add more terms to the scaling relation to take this into account. Notice that if we take the ratio of the tidal locking timescales for two moons orbiting the same planet, the term involving $m_p$ drops out.

Now, this kind of variation is possible. The moons of Mars have significantly different orbits; Deimos has a semi-major axis two and a half times as large as Phobos'! We see even greater variation in the moons of giant planets, such as the moons of Saturn, but I doubt there would be that sort of orbital difference around terrestrial planets, for the simple reasons that 1) it's hard for lower-mass bodies to form many moons, and 2) the Hill sphere - the region in which an object can have satellites - is smaller for low-mass planets.

It seems likely - possible, at the least - that the moons around the giant planets formed at different times. Some of the ones with smaller orbits could have formed with the planet, out of the disk of dust and rock surrounding it. The outer moons might have been captured later on, from asteroids or bits of protoplanets. Additionally, some may have formed in collisional families, meaning that some groups of moons could have formed at different times. Each group would have formed from the fragmentation of an asteroid captured by the planet - the Ananke and Carme families of moons around Jupiter are prime examples. Of course, less-massive terrestrial planets are less likely to capture asteroids, but then again, the moons of Mars were likely once asteroids.

• nice formula, but is it independent of the mass of the moon? – user Aug 17 '18 at 22:06
• @user Yep, I believe it's independent of the moon's mass. – HDE 226868 Aug 17 '18 at 22:08

The Pluto system is almost exactly what you want, in the terms you are interested, with two caveats:

1. Pluto is mutually tidally locked to Charon, implying a long day (6 Earth days), which could affect habitability

2. Pluto has a small in mass and hence its gravity is small (6% that of Earth), which is bad for human habitability at least

It is not difficult to imagine this configuration with more Earth-like characteristics: closer to the primary, the two main objects with a bigger mass, a more life-friendly composition, etc.

• Charon is about half the mass of Pluto. If it was much smaller I'd assume Pluto would not be tidally locked with Charon. The other three moons of the system are only few kilometers in diameter and much further apart; I'd expect them to disrupt the tidal lock if they were bigger causing enough gravitational pull on Charon. – Otto Abnormalverbraucher Jun 26 '18 at 14:02
• @OttoAbnormalverbraucher I think there is a wide range of not-very-likely-yet-with-a-non-zero-probability solutions that can be explained by phenomena like the late-heavy bombardment (late captures, an impact-driven rotation slow-down here, a body rotating a bit slower than expected there). I mean, even Earth's moon is kind of exceptional, but there you have it – Rafael Jun 26 '18 at 16:11


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# Permutation Problem with a Variable

13P5 = 1287(xPx)

I simplify the above to: 13!/8! = 1287(x!)

The expression on the left simplifies to 154,440.

I divide both sides by 1287 to get: 120=x!

At this point I'm stumped.

-
If you simply calculate $n!$ for the first few positive integers $n$, you should find your $x$ pretty quickly. –  Brian M. Scott Jul 24 '11 at 1:55

I am not sure that there is really a better solution than to just calculate the first few factorials at this point and see which one gives you $120$... You should find your answer pretty fast.
It is not possible to invert the "factorial" function in general since it is not injective, i.e. $0!=1!=1$
There is another way to go about it, though: divide $120$ by $2$, then divide the quotient by $3$, then that quotient by $4$, and so on: either you reach a quotient of $1$ with divisor the desired $x$, or you reach a quotient less than $1$ without ever getting $1$, in which case there’s no solution. –  Brian M. Scott Jul 24 '11 at 1:59


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# 地球不需要咱保护

“环保” 如今是个伟大光明、政治正确的理念，但首先人们应该认清环保的意义是什么—— 是为了保护地球吗？

"It is not sufficient to contrast the imperfect adjustments of unfettered enterprise with the best adjustment that economists in their studies can imagine. For we cannot expect that any state authority will attain, or even wholeheartedly seek that ideal. Such authorities are liable alike to ignorance, to sectional pressure, and to personal corruption by private interests. A loud-voiced part of their constituents, if organized for votes, may easily outweigh the whole."

—— Arthur Pigou

“针对无所拘束的企业而施行的非完美调节，是不足以与经济学家们在学术上想象的最优调节对比的，因为我们不能期望任何政府权力可以实现，或甚至是真心地在追求那个目标。政府当局同样易遭受无知、各方压力、为私牟利的腐败等诸多因素的影响。一部分呼声大的选民如果被组织起来可能会很容易地超越全体、成为主宰。”

—— 亚瑟·庇古


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# Two interpretations of the Lagrange Multiplier

If we have a standard utility maximizing problem $$\max_{x,y} U(x,y)$$ with the constraint $$p_{x}x + p_{y}y = B$$, where $$p_{x}, p_{y}$$ are the prices of $$x,y$$, and $$B$$ is the budget/income. The Lagriangian would work as $$U(x,y) + \lambda(B-p_{x}x - p_{y}y)$$ and we would proceed to take partial derivatives and find critical points. My question is on the interpretation of the Lagrangian function above and specifically the multiplier $$\lambda$$:

1) In what way is $$\lambda$$ the "price we pay for not obeying our constraint"? (as one of the explanations in the linked post has it)

2) Intuitively, not mathematically, how is this equivalent to the interpretation of $$\lambda$$ as a "shadow price"? What exactly is a shadow price?

3) What difference does it make to $$\lambda$$ if our constraint was written as $$p_{x}x + p_{y}y - B$$ instead? (i.e. in reverse order)

If you would NOT obey the constraint and spend 1\\$ more (or more precisely an infinitesimal amount more) your utility would go up by $$\lambda$$. As such $$\lambda$$ is what you give up for obeying the constraint, it is the price you pay indirectly and hence it is called the shadow price.
Writing the constraint in reverse has no effect other than changing the sign of $$\lambda$$, but that depends on the way you set-up your problem anyway (I.e. whether you add the constraint with a plus or a minus).


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# Derivatives Calculator

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• To derive $$f(x)=2x^2+3x-2$$, enter 2*x^2+3*x-2 and the variable x.
• To derive $$f(t)=\frac{1}{3}t^2+\frac{4}{3}t$$, enter 1/3*t^2+4/3*t and the variable t.
• To derive $$f(x)=\frac{1}{2*x}+\frac{1}{3x^2}$$, enter 1/(2*x)+1/(3*x^2) and the variable x.

With this calculator, you can get the derivative of the expression entered. It is possible to find derivatives of polynomials, trigonometric functions, exponential functions, among others. The expression must be entered along with the variable for which to derive.

## How to use the derivatives calculator?

Step 1: Enter the expression to be derived in the first input box. Use * to indicate multiplication between variables and coefficients. For example, instead of entering 2x or 5x^2, enter 2*x or 5*x^2.

Step 2: Enter the variable for which to derive in the second input box. In most cases, the variable is x.

Step 3: Click “Derive” to get the derivative of the entered expression.

Step4: The derivative along with the original expression will be displayed at the bottom.

## How to enter expressions in the calculator?

To enter expressions, we just have to enter an expression without the equals sign. For example, if we want to derive the function f(x)=x+2, we just have to enter x+2.

Also, we must use the * sign to indicate multiplication between variables and coefficients and the ^ sign to indicate an exponent. Therefore, to input the expression $$2x^2+3x$$, we need to enter 2*x^2+3*x.

Lastly, we can enter fractions by using the / sign. To input $$\frac{1}{2}x^2+\frac{1}{3}x$$, we would type 1/2*x^2+1/3*x. The following are some examples of how to use the calculator.

• To derive $$g(x)=3x^2+2x-4$$, enter 3*x^2+2*x-4 and the variable x.
• To derive $$f(x)=\frac{1}{2}x^2+\frac{4}{3}x$$, enter 1/2*x^2+4/3*x and the variable x.
• To derive $$h(t)=\frac{1}{2*t}+\frac{1}{3t^2}$$, enter 1/(2*t)+1/(3*t^2) and the variable t.

## Why find derivatives?

Derivatives have many applications in mathematics, some of the most important are:

• We can find the rate of change of a quantity. This is generally the most important application of the derivative. For example, we can find the rate of change of the volume of a sphere with respect to the change in radius.
• We can determine increasing and decreasing functions. The derivative can tell us whether a function is increasing or decreasing at a certain point.
• The derivative also allows us to find the tangent to a curve.
• Maximum points, minimum points, and other inflection points can be found using the derivative.

## Related calculators

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# Second Derivatives Using Implicit Differentiation

According to my textbook, the second derivative of

\begin{equation*} y^{2}+xy-x^{2}=9 \end{equation*}

is

\begin{equation*} \frac{90}{(2y+x)^{3}}. \end{equation*}

The problem states "Express $\frac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$." I've tried for two days straight now, and I can't get that answer. I am convinced the book has a typo.

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What is the question? or what would you like to know? – anon Sep 18 '10 at 15:13
muad: "express $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ in terms of $x$ and $y$", per the OP. In other words, what is $y^{\prime\prime}(x)$ in terms of $x$ and $y$. – J. M. Sep 18 '10 at 16:06

Based on your work as linked in the comments on J.M.'s answer, you've very nearly got it (except for a typo in differentiating *: there's a dy/dx where there should be a d/dx, but the mathematics that follows is correct as if it were d/dx). The numerator you have is \begin{align} -2 ((2 x - y)^2 &+ (2 x - y) (2 y + x) - (2 y + x)^2) \\ &=-10x^2+10xy+10y^2 \\ &=10(y^2+xy-x^2) \\ &=10\cdot9 \\ &=90. \end{align}

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One little question: How did you get from the first step to the second step? When I factor it out, I get $2(5y^2+4xy-5x^2)$ – G.P. Burdell Sep 18 '10 at 19:50
I threw it into Mathematica and applied Expand[]. Ignoring the -2 outside, the first item is $(2x-y)^2=4x^2-4xy+y^2$, the second is $(2x-y)(2y+x)=3xy-2y^2+2x^2$, and the third is $-(2y+x)^2=-4y^2-4xy-x^2$. All together, $(4+2-1)x^2+(-4+3-4)xy+(1-2-4)y^2=5(x^2-xy-y^2)$. – Isaac Sep 18 '10 at 20:02
I hate basic errors. Thanks Isaac! – G.P. Burdell Sep 18 '10 at 20:09

Hint: treat $y$ as a function $y(x)$, so differentiating $xy(x)$ should give something like $x y^{\prime}(x)+y(x)$. Differentiate expressions twice, and solve for $y^{\prime\prime}(x)$

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I've tried that but I get a different result than the answer. I'll upload my work, just gimme an hour or so in LaTeX – G.P. Burdell Sep 18 '10 at 16:37
Hmm, if you didn't get $\frac{10(y(x+y)-x^2)}{(2y+x)^3}$ ... something has gone horribly wrong. – J. M. Sep 18 '10 at 16:53
Ok, so here's my work: imgur.com/4DDa7.png I'm still not getting what's supposed to be the right answer. – G.P. Burdell Sep 18 '10 at 18:06
M.: the numerator in your comment is $10(y(x+y)-x^2)=10(y^2+xy-x^2)=90$. – Isaac Sep 18 '10 at 18:32


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Can you solve without modifying the integrand? (My twelfth integration problem)

Calculus Level 3

$\large \int_{0}^{1} \sqrt{1+\dfrac{x^{2}}{1-x^{2}}} \, dx$

If this integral can be expressed in the really nice form of $$\dfrac{a \pi}{b}$$, where $$a, b$$ are coprime positive integers, find $$a+b$$.

×


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# SAGE (Soviet–American Gallium Experiment)

SAGE (Soviet–American Gallium Experiment, or sometimes Russian-American Gallium Experiment) is a collaborative experiment devised by several prominent physicists to measure the solar neutrino flux.

## The Experiment

SAGE was devised to measure the radio-chemical solar neutrino flux based on the inverse beta decay reaction, 71Ga${\displaystyle +\nu _{e}\rightarrow e^{-}+}$ 71Ge. The target for the reaction was 50-57 tonnes of liquid gallium metal stored deep (2100 meters) underground at the Baksan Neutrino Observatory in the Caucasus mountains in Russia. The laboratory containing the SAGE-experiment is called gallium-germanium neutrino telescope (GGNT) laboratory, GGNT being the name of the SAGE experimental apparatus. About once a month, the neutrino induced Ge is extracted from the Ga. 71Ge is unstable with respect to electron capture (${\displaystyle t_{1/2}=11.43}$  days) and, therefore, the amount of extracted germanium can be determined from its activity as measured in small proportional counters.

The experiment had begun to measure the solar neutrino capture rate with a target of gallium metal in December 1989 and continued to run in August 2011 with only a few brief interruptions in the timespan. As of 2013 is the experiment was described as "being continued"[1] with the latest published data from August 2011. As of 2014 it was stated that the SAGE experiment continues the once-a-month extractions.[2] The SAGE experiment continued in 2016.[3] As of 2017, the SAGE-experiment continues [4].

The experiment has measured the solar neutrino flux in 168 extractions between January 1990 and December 2007. The result of the experiment based on the whole 1990-2007 set of data is 65.4+3.1
−3.0
(stat.) +2.6
−2.8
(syst.) SNU. This represents only 56%-60% of the capture rate predicted by different Standard Solar Models, which predict 138 SNU. The difference is in agreement with neutrino oscillations.

The collaboration has used a 518 kCi 51Cr neutrino source to test the experimental operation. The energy of these neutrinos is similar to the solar 7Be neutrinos and thus makes an ideal check on the experimental procedure. The extractions for the Cr experiment took place between January and May 1995 and the counting of the samples lasted until fall. The result, expressed in terms of a ratio of the measured production rate to the expected production rate, is 1.0±0.15. This indicates that the discrepancy between the solar model predictions and the SAGE flux measurement cannot be an experimental artifact. Also calibrations with a 37Ar neutrino source had been performed.

In 2014, the SAGE-experiment's GGNT-apparatus (gallium-germanium neutrino telescope) was upgraded to perform a very-short-baseline neutrino oscillation experiment BEST (Baksan Experiment on Sterile Transitions) with an intense artificial neutrino source based on 51Cr.[5] In 2017, the BEST apparatus was completed, but the artificial neutrino source was missing.[6] As of 2018, the BEST experiment was underway.[7] As of 2018, a follow-up experiment BEST-2 where the source would be changed to 65Zn was under consideration.[8]

## Members of SAGE

SAGE is led by the following physicists:

• GALLEX/GNO was the second (of two) large gallium-germanium radiochemical experiment. It was running in 1991-2003.
• Hans Bethe was the architect of the theory of nuclear fusion reactions in stars.
• The University of Washington is playing a major role in the statistical analysis of the SAGE data and in the determination of systematic uncertainties. They are very active in the remaining analysis of the Cr experiment data as well as the solar neutrino data.

## References

1. ^ Gavrin, V. N. (October 2013). "Contribution of gallium experiments to the understanding of solar physics and neutrino physics". Physics of Atomic Nuclei. 76 (10): 1238–1243. Bibcode:2013PAN....76.1238G. doi:10.1134/S106377881309007X.
2. ^ https://www.snolab.ca/sites/default/files/Chen3_EvidenceOsc.pdf
3. ^ www.hephy.at/user/mjeitler/TALKS/Baksan_Hephy5.pdf
4. ^
5. ^ Gavrin, V.; Cleveland, B.; Danshin, S.; Elliott, S.; Gorbachev, V.; Ibragimova, T.; Kalikhov, A.; Knodel, T.; Kozlova, Yu.; Malyshkin, Yu.; Matveev, V.; Mirmov, I.; Nico, J.; Robertson, R. G. H.; Shikhin, A.; Sinclair, D.; Veretenkin, E.; Wilkerson, J. (2015). "Current status of new SAGE project with 51Cr neutrino source". Physics of Particles and Nuclei. 46 (2): 131. Bibcode:2015PPN....46..131G. doi:10.1134/S1063779615020100.
6. ^
7. ^ Babenko, Maxim; Overbye, Dennis (2018-07-16). "The Neutrino Trappers". The New York Times.
8. ^ Gavrin, V. N.; Gorbachev, V. V.; Ibragimova, T. V.; Kornoukhov, V. N.; Dzhanelidze, A. A.; Zlokazov, S. B.; Kotelnikov, N. A.; Izhutov, A. L.; Mainskov, S. V.; Pimenov, V. V.; Borisenko, V. P.; Kiselev, K. B.; Tsevelev, M. P. (2018). "On the gallium experiment BEST-2 with a 65Zn source to search for neutrino oscillations on a short baseline". arXiv:1807.02977 [physics.ins-det].


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# In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre, The graph which

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in Physics
In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre, The graph which would correspond to the above will be:
A.
B.
C.
D.

by (24.2k points)


text:
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Classify the Following Polynomials as Linear, Quadratic, Cubic and Biquadratic Polynomials: X+X^2 +4 - CBSE Class 9 - Mathematics

ConceptIntroduction of Polynomials

x+x^2 +4

Solution

Given polynomial

x+x^2 +4 is quadratic as degree of polynomial is 2.

Is there an error in this question or solution?

APPEARS IN

RD Sharma Solution for Mathematics for Class 9 by R D Sharma (2018-19 Session) (2018 to Current)
Chapter 6: Factorisation of Polynomials
Ex.6.10 | Q: 4.1 | Page no. 3
Solution Classify the Following Polynomials as Linear, Quadratic, Cubic and Biquadratic Polynomials: X+X^2 +4 Concept: Introduction of Polynomials.
S


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Question 156

# ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. From any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and says there. Let $$a_n$$ be the number of distinct paths of exactly n jumps ending in E. Then, what is the value of $$a_{2n-1}$$?

Solution

The number of vertices between A and E is 3. So, a minimum of 4 steps is needed for the frog to jump from A to E. Also, the frog can go to E from A along any path in only an even number of steps. (2n - 1) is odd. So, the frog can never reach E from A in (2n-1) number of steps. So, the answer is 0


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show · ring.euclidean_domain all knowls · up · search:

An Euclidean domain is an integral domain $R$ with a function $\nu: R-\{0\}\to \Z_{\geq 0}$ such that for all $a\in R$ and all $b\in R-\{0\}$, there exists $q,r\in R$ such that $a=bq+r$ and $r=0$ or $\nu(r)<\nu(b)$.

Euclidean domains are principal ideal domains, and hence, unique factorization domains.

Authors:
Knowl status:
• Review status: reviewed
• Last edited by John Jones on 2019-03-20 10:37:05
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# Category Archives: Diary

## CCR in a TV program!!

It is really a pity that I couldn’t go CCR 2016 in Hawaii,
but it’s an honor to have such attention from media.
http://thinktechhawaii.com/bounding-rationality-with-computation/

## (日本語) 乱雑性とは何か

Sorry, this entry is only available in Japanese.

## MathJax test

Let $x$ be a real.
Then
$x^2\ge 0.$

## Understanding randomness

A Japanese journal “Susemi”, which means a seminar of mathematics, features randomness titled “understanding randomness” on the February number in 2011.


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Fluctuations of the total entropy production in stochastic systems

# Fluctuations of the total entropy production in stochastic systems

## Abstract

Fluctuations of the total entropy are experimentally investigated in two stochastic systems in a non-equilibrium steady state : an electric circuit with an imposed mean current and a harmonic oscillator driven out of equilibrium by a periodic torque. In these two linear systems, we study the total entropy production which is the entropy created to maintain the system in the non-equilibrium steady state. Fluctuation theorem holds for the total entropy production in the two experimental systems, both for all observation times and for all fluctuation magnitudes.

05.40.-a
###### pacs:
05.70.-a

Fluctuation phenomena, random processes, noise, and Brownian motion Thermodynamics

Thermodynamics of systems at equilibrium has been developed by defining the internal energy, the injected work, and the dissipated heat. The first law of Thermodynamics describes energy conservation and gives a relation between those three energies. The second law of Thermodynamics imposes that the entropy variation is positive for a closed system. Statistical Physics further gives a microscopic definition of entropy, which allows analytical results on entropy production. The extension of Thermodynamics to systems in non-equilibrium steady states is an active field of research. Within this context, the first law has been extended for stochastic systems described by a Langevin equation [1, 2, 3]. It has been noted that the second law is not verified at all times but only in average, i.e. over macroscopic times : entropy production can have instantaneously negative values. The probabilities of getting positive and negative entropy production are quantitatively related in non-equilibrium systems by the Fluctuation Theorem (FT). This theorem has been first demonstrated in deterministic systems [4, 5, 6] and secondly extended to stochastic dynamics [7, 8, 9, 10]. FTs for work and heat fluctuations have been theoretically and experimentally studied in Brownian systems described by the Langevin equation [11, 12, 13, 14, 15, 16, 17, 3, 2, 18]. For these systems in non-equilibrium steady state, Fluctuation Theorems hold only in the limit of infinite time:

 Φ(a)≡ln(p(Xτ=+a)p(Xτ=−a))→akBTforτ→∞ (1)

where is the Boltzmann constant, the temperature of the heat bath and is the probability density function (PDF) of . is called symmetry function and stands for either injected work or dissipated heat, averaged over a time lag . For the injected work, eq. 1 is valid for all fluctuation magnitudes ; for the dissipated heat, eq. 1 is satisfied for values lower than the mean value [13].

We are interested here in the total entropy production in a non-equilibrium steady state (NESS), introduced in ref. [19, 20, 21] and directly related to previous work on housekeeping heat [22, 23]. Entropy production has been studied both theoretically and experimentally in several systems [24, 25, 26, 27, 28, 29, 30]. In [19], Seifert and Speck have shown that the total entropy production for a stochastic system described by a first order Langevin equation in a NESS satisfies a Detailed Fluctuation Theorem (DFT) :

 Φ(a)=ln(p(Xτ=+a)p(Xτ=−a))=akBT∀τ∀a (2)

The relation (2) is valid for all integration time and all fluctuation magnitudes of the total entropy production. This relation is closely related to the Jarzynski and Crooks non-equilibrium work relations [31, 32] which can be exploited to measure equilibrium free energy in experiments [33, 34, 35, 36, 37] and it has been extended to Markov processes [38].

In this Letter, we measure the total entropy production in two out-of-equilibrium systems and show that this quantity verifies eq. (2). In the first part of the letter, we recall the general definition of dissipated heat, ”trajectory-dependent” entropy and total entropy production. In the second part, we detail our two experimental systems. The first one is an electric circuit maintained in a NESS by an injected mean current. The second system is a torsion pendulum driven in a NESS by forcing it with a periodic torque.

The heat dissipated by the system is the heat given to the thermostat during a time ; we note it . It is related to the work , given to the system during the time , and to the variation of internal energy during this period, thanks to the first law of Thermodynamics:

 Qτ=Wτ−ΔτU. (3)

Expressions of and for our two experimental setups are given below. Following notations of ref [19], we define the entropy variation in the system during a time as :

 Δsm,τ=1TQτ (4)

where is the temperature of the heat bath. For thermostated systems, entropy change in medium behaves like the dissipated heat. The non-equilibrium Gibbs entropy is :

 S(t)=−kB∫d→xp(→x(t),t,λt)lnp(→x(t),t,λt)=⟨s(t)⟩ (5)

where denotes the set of control parameters at time and is the probability density function to find the particle at a position at time , for the state corresponding to . This expression allows the definition of a ”trajectory-dependent” entropy :

 s(t)≡−kBlnp(→x(t),t,λt) (6)

The variation of the total entropy during a time is the sum of the entropy change in the system during and the variation of the ”trajectory-dependent” entropy in a time , :

 Δstot,τ≡stot(t+τ)−stot(t)=Δsm,τ+Δsτ (7)

In this letter, we study fluctuations of computed using (4) and (6). We will show that satisfies a DFT (eq. (2)). In ref. [38], the relevance of boundary terms like has already been pointed out.

Our first experimental system is an electric circuit composed of a resistance  M in parallel with a capacitance  pF [15]. The time constant of the circuit is  ms. The voltage across the dipole fluctuates due to Johnson-Nyquist thermal noise. We drive the system out of equilibrium by injecting a constant controlled current  A. After some , the system is in a NESS. Electric laws give in this setup [15] :

 τ0dVdt+V=RI+ξ, (8)

where is the Gaussian thermal noise, delta-correlated in time of variance .

Multiplying (8) by and integrating it between and , we define the work , injected into the system, and the dissipated heat , together with the internal energy :

 Wτ ≡ ∫t+τtV(t′)Idt′ (9) Qτ ≡ ∫t+τtV(t′)iR(t′)dt′ (10) ΔτU ≡ 12C(V(t+τ)2−V(t)2)=Wτ−Qτ (11)

where is the current flowing in the resistance : . This system has only one degree of freedom, so the trajectory in phase space is defined by the voltage alone, and the only external parameter is the constant current . So the variation of the ”trajectory-dependent” entropy during a time is :

 Δsτ=−kBln(p(V(t+τ))p(V(t))) (12)

Fluctuation relations for the injected work and the dissipated heat (or entropy change in medium) in this system have been reported in [15]. Let us recall the experimental results for the dissipated heat , for several values of the integration time. Its average value is equal to the average of injected work and linear in . The PDFs of are plotted in figure 1a) for four values of . They are not Gaussian for small times and extreme events have an exponential distribution. We now describe new results in this system. The PDF of the ”trajectory-dependent” entropy is plotted in figure 1b) ; we have superposed to it the PDF of at equilibrium ( A). These two PDFs are independent of the integration time. The two curves match perfectly within experimental errors. Therefore the ”trajectory-dependent” entropy is equal in this case to the fluctuations of the entropy exchanged with the thermal bath at equilibrium. The average value of this ”trajectory-dependent” entropy is zero within experimental errors ; so the total entropy production has the same average value than the entropy . The PDFs of the total entropy , computed by adding to , are plotted in figure 1c) for different values of ; they are all Gaussian.

In figure 1d), we have plotted the symmetry functions of the dissipated heat () together with those of the total entropy. is a non linear function of . The linearity is recovered for . The slope tends to for large integration time. Entropy change in medium satisfies relation (1[15]. On the contrary, the symmetry functions for the total entropy are linear with for all integration times and for all values of : . The slope is equal to for all integration times within experimental errors. Measurements can be done for other values of injected current and we find the same results. Thus total entropy satisfies relation (2).

This experimental result can be explained using a first order Langevin equation and noting that fluctuations of the voltage , when a current is applied, are identical to those at equilibrium [15]. Thus the voltage has a Gaussian distribution with mean and variance , whereas its autocorrelation function is the same out of equilibrium and at equilibrium:

 ⟨δV(t+τ)δV(t)⟩=kBTCe(−τ/τ0). (13)

So we can compute the expression of the ”trajectory-dependent” entropy from eq. (12) and we find:

 T.Δsτ = 12CδV(t+τ)2−12CδV(t)2 (14)

Fluctuations of the voltage are identical to those at equilibrium, thus ”trajectory-dependent” entropy is equal to the variation of internal energy divided by at equilibrium and to the opposite of the entropy variation in the system at equilibrium. Using eq. (7),  (10) and (14), the expression of the total entropy is :

 T.Δstot,τ=I∫t+τtV(t′)−Iτ0(δV(t+τ)−δV(t))dt′ (15)

We experimentally see that the PDF of is Gaussian, so fully characterized by its mean value and its variance. Its average value is linear in () and equals the mean injected work divided by the temperature . Its variance is computed using eq. (13) and we obtain . For a Gaussian distribution, the symmetry function of is linear with , that is . The slope is related to the mean and the variance of the total entropy : . Experimentally, we have normalized the total entropy by , and the slope is equal to for any as shown in fig. 1d).

We now consider the second experimental system. It is a torsion pendulum in a viscous fluid which acts as a thermal bath at temperature . The system is driven out of equilibrium by an external deterministic time dependent torque . All the details of the experimental setup can be found in [3]. The torsional stiffness of the oscillator is  N.m.rad, its viscous damping , its total moment of inertia , its resonant frequency  Hz and its relaxation time  ms. The angular motion of the oscillator obeys a second order Langevin equation :

 d2θdt2+2ταdθdt+ω20θ=M+ηC, (16)

where is the thermal noise, delta-correlated in time of variance . The work injected into the system between and is :

 (17)

The dissipated heat is computed according to eq. (3) where, in this case, the internal energy is :

 U(t)=12Cθ(t)2+12Ieff˙θ(t)2 (18)

We investigate a periodic forcing : (  pN.m and  Hz). The integration time is chosen to be a multiple of the period of the forcing : . The average responses of the system and are periodic function of the pulsation and the system is in a steady state. Results for injected work and dissipated heat in this case are reported in [17, 3].

The ”trajectory-dependent” entropy is not as simple as in the case of the resistance. The system has two independent degrees of freedom ( and ) and its expression is :

 Δsτn=−kBln(p(θ(ti+τn),φ).p(˙θ(ti+τn,φ))p(θ(ti+τn),φ).p(˙θ(ti+τn,φ)) (19)

The DFT (eq. (2)) is valid for each fixed starting phase  [21]. For computing correctly the total entropy, we have to calculate the PDFs of the angular position and the angular velocity for each initial phase . Then we compute the ”trajectory-dependent” entropy. As fluctuations of and are independent of  [3]. These distributions correspond to the equilibrium fluctuations of and around the mean trajectory defined by and . As a consequence, we can average over which improves a lot the statistical accuracy. We stress that it is not equivalent to calculate first the PDFs over all values of  — which would correspond here to the convolution of the PDF of the fluctuations with the PDF of a periodic signal — and then compute the trajectory dependent entropy.

In figure 2a), we recall the main results for the dissipated heat . Its average value is linear in and equal to the injected work. The PDFs of are not Gaussian and extreme events have an exponential distribution. Let us now describe new results in this system. The PDF of the ”trajectory-dependent” entropy is plotted in fig. 2b); it is independent of . We superpose to it the PDF of the variation of internal energy divided by at equilibrium : the two curves match perfectly within experimental errors, so the ”trajectory-dependent” entropy can again be considered as the entropy exchanged with the thermostat if the system is at equilibrium. The average value of is zero, so the average value of the total entropy is equal to the average of injected power divided by . In fig. 2c), we plot the PDFs of the normalized total entropy for four typical values of integration time. We find that the PDFs are Gaussian for any time.

The symmetry functions of the dissipated heat and the total entropy are plotted in fig. 2d). For the dissipated heat, the symmetry function is a non linear function of and we observe a linear behavior for with a slope that tends to for large time. For the normalized total entropy, the symmetry functions are linear with for all values of and the slope is equal to for all values of . Note that it is not exactly the case for the first values of because these are the times over which the statistical errors are the largest and the error in the slope is large.

In ref. [3], we have already shown that, when a torque is applied, the fluctuations around have the same statistics and the same dynamics as fluctuations at equilibrium. Using the expression of the distribution of the angular position and the angular velocity, we can compute the ”trajectory-dependent” entropy from eq. (19):

 T.Δsτn = 12C(δθ(ti+τn)2−δθ(ti)2) (20) + 12Ieff(δ˙θ(ti+τn)2−δ˙θ(ti)2)

where and are the fluctuations around the mean trajectory : and . The ”trajectory-dependent” entropy is equivalent to the variation of internal energy for the system at equilibrium divided by the temperature of the bath. The total entropy is :

 T.Δstot,τn = Wτn−12C¯θ(ti)(δθ(ti+τn)−δθ(ti) (21) − 12Ieff˙¯θ(ti)(δ˙θ(ti+τn)−δ˙θ(ti))

Experimentally, we observe that the PDF of the total entropy is Gaussian. We can calculate the mean value and the variance of the total entropy for any . We find that is equal to the injected work. Using the first law of thermodynamics and the equation of motion, we write the dissipated heat as the difference between the viscous dissipation and the work of the thermal noise :

 T.Δsm,τn=∫ti+τntiν˙θ(t′)2dt′−∫ti+τnti˙θ(t′)η(t′)dt′ (22)

This expression holds at equilibrium as well as out of equilibrium. The only difference is that when out of equilibrium and . After some algebra the total entropy is :

 T.Δstot,τn = ν∫ti+τnti⟨˙θ(t′)⟩2dt′ (23) + 2ν∫ti+τnti⟨˙θ(t′)⟩δ˙θ(t′)dt′ − ∫ti+τnti⟨˙θ(t′)⟩η(t′)dt′

The average value of the total entropy is and its variance is :

 ⟨σ2Δstot,τn⟩ = 2kB⟨Δstot,τn⟩+B/T2 B = 4ν∫∫dudv⟨˙θ(u)⟩⟨˙θ(v)⟩ψ(u,v) (24) ψ(u,v) = ν⟨δ˙θ(u)δ˙θ(v)⟩−⟨δ˙θ(u)η(v)⟩ (25)

The first term in the function is the autocorrelation function of the angular velocity. This function is symmetric around . The second term is the correlation function of the angular velocity with the noise. Due to the causality principle, this term vanishes if . Changing variables to and , eq. (24) can be rewritten :

 B=∫ti+τntidr∫τn0⟨˙θ(r+s/2)⟩⟨˙θ(r−s/2)⟩~ψ(r,s) (26) ~ψ(r,s)=2ν⟨δ˙θ(s)δ˙θ(0)⟩−⟨δ˙θ(s)η(0)⟩ (27)

After some algebra we can show that the correlation function is two times the autocorrelation function , therefore . Thus we obtain that the variance of the total entropy is equal to . The Fluctuation Relation for the total entropy is : for all times , for all values of and for any kind of stationary forcing.

For our two experimental systems, we have obtained that the ”trajectory-dependent” entropy can be considered as the entropy variation in the system in a time that one would have if the system was at equilibrium. Therefore the total entropy is the additional entropy due to the presence of the external forcing : this is the part of entropy which is created due to the non-equilibrium stationary process. The derivation we gave for a second order Langevin equation can be extended to a first order Langevin equation : total entropy (or excess entropy) satisfies the Fluctuation Theorem for all times and for all kind of stationary intensity injected in the circuit or all kind of stationary external torque. The ratio between the average value of the total entropy at (relaxation time) and the variance of fluctuations the entropy variation at equilibrium characterizes the distance from equilibrium :

 d2=⟨Δstot,τα⟩σeq=√3nd8σ2Δstot,τασ2eq (28)

where is the number of degrees of freedom. For the second equality, we use the Gaussianity of , i.e. and the variance of of the fluctuation of entropy variation at equilibrium in terms of , i.e. . It turns out that this expression for is the same that was defined in [3] with a completely different approach. Eq. (28) indicates that, when the system is driven far from equilibrium, becomes negligible. As a consequence the fluctuations of the total entropy become equal to the fluctuations of the entropy variation in the system in a time . In other words, the PDFs of the dissipated heat far from equilibrium are Gaussian.

In conclusion, we have studied the total entropy when the system is driven in a non-equilibrium steady state. We have shown that the Fluctuation Theorem for the total entropy is valid not only in the limit of large times, as it is the case for injected work and dissipated heat, but also for all integration times and all fluctuation amplitudes. In the two examples we have discussed, the total entropy corresponds to the difference between the entropy change in medium out of equilibrium and at equilibrium.

We thank U. Seifert and K. Gawedzki for useful discussions. This work has been partially supported by ANR-05-BLAN-0105-01.

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