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# Find Cylindrical Co-ordinates of a Point (3𝒊̅ + 4𝒋̅ + 𝒌̅) - Applied Physics 2

Numerical

Find cylindrical co-ordinates of a point (3𝒊̅ + 4𝒋̅ + 𝒌̅)

#### Solution

The point is (3,4,1)
r =sqrt(x^2+y^2)
=sqrt(3^2+4^2)
= 5
θ = tan^-1(y/x)
= tan^-1(4/3)
= 53.13o

= 0.2952𝝅
z = z = 1

Point in cylindrical system is (5,0.2952𝝅,𝟏)

Concept: Cylindrical and Spherical Coordinate System
Is there an error in this question or solution?


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# RD Sharma Solutions Class 12 Differentials Errors And Approximations

## RD Sharma Solutions Class 12 Chapter 14

Let $y=f(x)$ be a function of x, and let \Delta x be a small change in x and $\Delta y$  be the corresponding change in y. Then,

$\Delta y=dy/dx\, \Delta x$ approximately

dy/dx \Delta x is termed as differential of y and is denoted by $dy$

Some of the useful result on differentials

• If $f(x)$ is a constant function, then its differential is zero.
• If $y=cu$, then $dy=c\, du$, $c$ is a constant.

Now explore in depth about Differentials, Errors and Approximations through RD Sharma solutions that is provided in the below section.

 Differentials, Errors and Approximations Class 12th RD Sharma Exercises Differentials, Errors and Approximations Exercise 14.1

#### Practise This Question

Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even is ?

format: plain


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# On the non-wandering set of a map

Let $$f: X \to X$$ be a homeomorphism from a compact metric space $$X$$ to itself.

$$x$$ is said to be a non-wandering point of $$f$$ if for all open neighborhood of $$x$$ say $$U$$ there exists $$n \in \mathbb{Z}$$ such that $$f^n(U) \cap U \neq \emptyset$$. The set of all non-wandering points of $$f$$ is denoted by $$\Omega(f)$$. I'm trying to prove the following exercise :

Show that if $$x \in \Omega(f)$$ and $$U$$ is a neighborhood of $$x$$, there is a sequence of integers $$n_i$$ tending to infinity such that the intersection $$f^{n_i}(U) \cap U$$ is non-empty.

There is a proposition in the book which says if $$U$$ is a neighborhood of $$\Omega(f)$$ then for every $$x \in X$$ there is $$N>0$$ such that for every $$n \geq N$$ , $$f^n(x) \in U$$.

May I use the proposition above to prove the main problem?


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## that1chick 2 years ago how do you write y = 2x2 + 6x + 4 in general form?

1. iheartfood

do u know the general form?

2. that1chick

yes, y=a(x-h)^2+k

3. iheartfood

did u need it in linear form?

4. iheartfood

like the general form of a linear equation?

5. that1chick

just the general form of a quadratic equation

6. iheartfood

okay, so u need to know the quadratic formula i believe..do u know it?

7. that1chick

general form> y=a(x-h)^2+k

8. that1chick

oh, yeah!

9. that1chick

10. that1chick

but how does that help convert it to general form?

11. that1chick

I thought it just helped find the x-intercepts..

12. iheartfood

if i can remember this correctly, i believe u need to first solve to get the values of a,b,c and then you just plug it into the general form... I'm not 100% sure tho... @abb0t, am i remembering this correctly? ;/

13. that1chick

I know how to convert it... the problem is the 2 in 2x^2

14. abb0t

General form of what?

15. that1chick

16. that1chick

2 in 2x^2 is the problem.... you isolate the x variables getting: y-4=2x^2+6x then complete the square and balance the equation: 6/2=3 3^2=9, y-4+9=2x^2+6x+9, y+5=2x^2+6x+9 then convert the trinomial into a binomial... thus lies my dilemma

17. iheartfood

sorry, i'm not very sure, cuz i don't think i remember this correctly ahha and i don't wanna help u wrongly haha @Mertsj probs knows :) good luck!!! :D

18. that1chick

thanks (:

19. Mertsj

$y=2x^2+6x+4$ $y=2(x^2+3x+______ )+4$

20. Mertsj

Now complete the square by adding (3/2)^2 inside the parentheses: $y=2(x^2+3x+(\frac{3}{2})^2)+4-\frac{9}{2}$

21. Mertsj

Now factor: $y=2(x+\frac{3}{2})^2-\frac{1}{2}$

22. that1chick

why 3/2?

23. Mertsj

I want to show you something: $x^2+6x+9=(x+3)^2$

24. Mertsj

$x^2+8x+16=(x+4)^2$

25. that1chick

I don't get what you did with: (3/2^2)+4−92

26. Mertsj

Notice the relationship between the coefficient of x and the constant term in a trinomial square.

27. Mertsj

If you take 1/2 the coefficient and square it, you get the constant term.

28. Mertsj

So I took 1/2 of 3 and got 3/2. Then I squared it and added it to the x^2 + 3x to get a trinomial square.

29. Mertsj

Now the (3/2)^2 was inside a parenthesis which has a 2 in front of it so I was really adding 2 times (3/2)^2 which is 2 times 9/4 which is 9/2. So since I could not change the equation, I then had to subtract 9/2.

30. Mertsj

31. that1chick

I think so

32. that1chick

Okay, yes! I get what you did (: thank you! what were you trying to show me with those other two equations?

33. Mertsj

The relationship between the coefficient of x and the constant term.

34. that1chick

oh, that half of the coefficient of x to the second power = the constant. okay, cool

35. that1chick

Thank you for helping me (:

36. Mertsj

yw


text: Title Measurements of the lineshape of the $Z^{0}$ and determination of electroweak parameters from its hadronic and leptonic decays Author Abreu, P. Adam, W. Adye, T. Verbeure, F. et al. Faculty/Department Faculty of Sciences. Physics Publication type article Publication 1994 Amsterdam , 1994 Subject Physics Source (journal) Nuclear physics: B. - Amsterdam Volume/pages 417(1994) :1-2 , p. 3-57 ISSN 0550-3213 0029-5582 ISI A1994NK18300001 Carrier E Target language English (eng) Full text (Publishers DOI) Affiliation University of Antwerp Abstract During the LEP running periods in 1990 and 1991 DELPHI has accumulated approximately 450000 Z0 decays into hadrons and charged leptons. The increased event statistics coupled with improved analysis techniques and improved knowledge of the LEP beam energies permit significantly better measurements of the mass and width of the Z0 resonance. Model independent fits to the cross sections and leptonic forward-backward asymmetries yield the following Z0 parameters: the mass and total width M(Z) = 91.187 +/- 0.009 GeV, GAMMA(Z) = 2.486 +/- 0.012 GeV, the hadronic and leptonic partial widths GAMMA(had) = 1.725 +/- 0.012GeV, GAMMA(l) = 83.01 +/- 0.52 MeV, the invisible width GAMMA(inv) = 512 +/- 10 MeV, the ratio of hadronic to leptonic partial widths R(l) = 20.78 +/- 0.15, and the Born level hadronic peak cross section sigma0 = 40.90 +/- 0.28 nb. Using these results and the value of alpha(s) determined from DELPHI data, the number of light neutrino species is determined to be 3.08 +/- 0.05. The individual leptonic widths are found to be: GAMMA(e) = 82.93 +/- 0.70 MeV, GAMMA(mu) = 83.20 +/- 1.11 MeV and GAMMA(tau) = 82.89 +/- 1.31 MeV. Using the measured leptonic forward-backward asymmetries and assuming lepton universality, the squared vector and axial-vector couplings of the Z0 to charged leptons are found to be g(V)l2 = (1.47 +/- 0.51) x 10(-3) and g(A)l2 = 0.2483 +/- 0.0016. A full Standard Model fit to the data yields a value of the top mass m(t) = 115(-82)+52 (expt.)-24(+23) (Higgs) GeV, corresponding to a value of the weak mixing angle sin2 theta(eff)lept = 0.2339 +/- 0.0015(expt.)-0.0004(+0.0001) (Higgs). Values are obtained for the variables S and T, or epsilon1 and epsilon3 which parameterize electroweak loop effects. E-info https://repository.uantwerpen.be/docman/iruaauth/eff77a/5d44387.pdf http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:A1994NK18300001&DestLinkType=RelatedRecords&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:A1994NK18300001&DestLinkType=FullRecord&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:A1994NK18300001&DestLinkType=CitingArticles&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 Handle


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• Browse all
Observation of electroweak production of a same-sign $W$ boson pair in association with two jets in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector

The collaboration
Phys.Rev.Lett. 123 (2019) 161801, 2019.

Abstract (data abstract)
CERN-LHC. Observation and fiducial cross-section measurement of electroweak production of a same-sign $W$ boson pair in association with two jets using 36.1 fb$^{-1}$ of data recorded at a center-of-mass energy of $\sqrt{s}=13$ TeV by the ATLAS detector at the Large Hadron Collider. Events containing two same-sign leptons, electron or muon, and at least two jets with a large invariant mass and rapidity difference, are selected.

• #### Table 1

Data from page 7

10.17182/hepdata.84643.v1/t1

Measured fiducial cross section.

• #### Table 2

Data from Figure 2, page 7

10.17182/hepdata.84643.v1/t2

The $m_{jj}$ distribution for events meeting all selection criteria for the signal region. Signal and individual background distributions are shown...

• #### Table 3

Data from Figure 3, page 8

10.17182/hepdata.84643.v1/t3

The $m_{ll}$ distribution for events meeting all selection criteria for the signal region as predicted after the fit. The fitted...

• #### Table 4

Data from Table 2, page 9

10.17182/hepdata.84643.v1/t4

Impact of different components of systematic uncertainty on the measured fiducial cross section, without taking into account correlations. The impact...

• #### Table 5

Data from Table 2 of the auxiliary material

10.17182/hepdata.84643.v1/t5

Efficiency correction factor $C_{WW}$, defined as the ratio of the number of reconstructed $W^{\pm}W^{\pm}jj$ electroweak events in the signal region...

• #### Table 6

Data from Table 2 of the auxiliary material

10.17182/hepdata.84643.v1/t6

Efficiency correction factor $C_{WW}$, defined as the ratio of the number of reconstructed $W^{\pm}W^{\pm}jj$ electroweak events in the signal region...

Your question will be emailed to those involved with the submission. Please mention the relevant table.


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# Pappus's hexagon theorem

#### Dear Wikiwand AI, let's keep it short, summarize this topic like I'm... Ten years old or a College student

In mathematics, Pappus's hexagon theorem (attributed to Pappus of Alexandria) states that

• given one set of collinear points ${\displaystyle A,B,C,}$ and another set of collinear points ${\displaystyle a,b,c,}$ then the intersection points ${\displaystyle X,Y,Z}$ of line pairs ${\displaystyle Ab}$ and ${\displaystyle aB,Ac}$ and ${\displaystyle aC,Bc}$ and ${\displaystyle bC}$ are collinear, lying on the Pappus line. These three points are the points of intersection of the "opposite" sides of the hexagon ${\displaystyle AbCaBc}$.

It holds in a projective plane over any field, but fails for projective planes over any noncommutative division ring.[1] Projective planes in which the "theorem" is valid are called pappian planes.

If one restricts the projective plane such that the Pappus line ${\displaystyle u}$ is the line at infinity, one gets the affine version of Pappus's theorem shown in the second diagram.

If the Pappus line ${\displaystyle u}$ and the lines ${\displaystyle g,h}$ have a point in common, one gets the so-called little version of Pappus's theorem.[2]

The dual of this incidence theorem states that given one set of concurrent lines ${\displaystyle A,B,C}$, and another set of concurrent lines ${\displaystyle a,b,c}$, then the lines ${\displaystyle x,y,z}$ defined by pairs of points resulting from pairs of intersections ${\displaystyle A\cap b}$ and ${\displaystyle a\cap B,\;A\cap c}$ and ${\displaystyle a\cap C,\;B\cap c}$ and ${\displaystyle b\cap C}$ are concurrent. (Concurrent means that the lines pass through one point.)

Pappus's theorem is a special case of Pascal's theorem for a conic—the limiting case when the conic degenerates into 2 straight lines. Pascal's theorem is in turn a special case of the Cayley–Bacharach theorem.

The Pappus configuration is the configuration of 9 lines and 9 points that occurs in Pappus's theorem, with each line meeting 3 of the points and each point meeting 3 lines. In general, the Pappus line does not pass through the point of intersection of ${\displaystyle ABC}$ and ${\displaystyle abc}$.[3] This configuration is self dual. Since, in particular, the lines ${\displaystyle Bc,bC,XY}$ have the properties of the lines ${\displaystyle x,y,z}$ of the dual theorem, and collinearity of ${\displaystyle X,Y,Z}$ is equivalent to concurrence of ${\displaystyle Bc,bC,XY}$, the dual theorem is therefore just the same as the theorem itself. The Levi graph of the Pappus configuration is the Pappus graph, a bipartite distance-regular graph with 18 vertices and 27 edges.


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# Tag Info

## Hot answers tagged ramjet

19

A jet engine compresses air, heats it by mixing it with fuel and burns it, and lets the heated air escape at the end, where it accelerates to more than its initial speed in a convergent-divergent nozzle because the density of the heated gas is lower, thus needing a higher volume at the same pressure. By converting the kinetic energy of the flow into ...

13

Here's a quick estimate for this. The maximum altitude for level flight is when the engine can't produce the thrust required to fly fast enough to generate the lift required to balance the weight. Ramjets work best at around Mach 3 and can operate up to Mach 6 Lift is proportional to speed squared and air density. So doubling the speed allows the plane to ...

13

Tip jet rotors are still being developed, none of them are ramjet powered as far as I know. The Dragonfly DF1 is tip rocket powered, while the Pegasus uses air jets powered by a conventional engine. Ramjet technology has not improved, so the cases for using them on rotor tips has not changed. They are still too heavy, thirsty and noisy.

9

Short answer: No. Or at least, nothing that's been demonstrated that I'm aware of. Whether there have been experimental R&D aircraft or developments, I'm not sure - but I can't find any. The simplest reason is just because there's no need for them - they offer no real advantages over a conventional internal combustion or jet engine, so as with so much ...

9

A rocket-powered missile carries both fuel and oxidiser and oxidiser is the heavier part. For example a gram of jet fuel needs almost 3.4 grams of oxygen to fully burn! Increasing the fuel load requires also increasing the power, which means even more fuel and this limits the practical range of a missile. Ramjets are air-breathing, so they don't need to ...

7

Ramjets could be miniaturized, but at some point, air viscosity and boundary layer thickness will make it impossible for them to work. Simply put, the intake hole will get so small that air will find it easier to flow around the whole thing and ignore all of its complicated inner structure. In more technical terms, as the intake shrinks, intake velocity ...

7

Ramjets can fly at least this high: Scramjet, X-43A flown in 2004: 110,000 feet / 33 km. Ramjet, RJ43-MA-11 on a BOMARC B missile, 1961, 100,000 feet / 30.5 km (photo caption, p. 31). (Ramjet, XRJ47-W-5 tested by NACA in 1955: 73,000 feet / 22 km.) Ramjet, extrapolated from data for flights where the vehicles were still accelerating when they ran out of fuel:...

6

The ramjet is conceptually the simplest jet engine. It is a duct where air is burned creating a hot jet that provides thrust, it is known aero thermodynamic duct as it is no more than a duct where a thermodynamic cycle is performed. They are no more used for airplanes, as it can not provide thrust at zero airspeed -since he does not have any compressor ...

6

Scramjets There's only a couple flying examples. The early Soviet-born Kholod, the HyShot series, and the recent ISRO ATV all flew small scramjet engines, although none were designed for net thrust. By far the most [publicly] successful vehicles have been the X-43A and the X-51. Both were fully controllable, independent flight vehicles (rather than just ...

5

How long does the lithium take to evaporate and then to react with oxygen? Air in a scramjet is moving extremely fast (supersonic by definition!), so there is very little time left for mixing and combustion if you desire to create thrust from the the combustion heat energy. Your lithium engine will most likely trail a bright contrail of burning lithium but ...

5

Well the Pratt & Whitney J58 engines which powered the SR-71 worked in a similar fashion, using a bypass system which allows the engine to operate as a ramjet at high speeds and a turbojet engine at lower speeds. By the very nature of a ramjet, it relies of stagnation pressure at the intake and inlet throat to pressurize the airflow prior to combustion. ...

4

Pressure is the greatest just before the combustor in both jet and ramjet engines - therefore air would have against the pressure differential for this to happen (turbine only has therefore an indirect role)

4

I'm reading a book right now that talks briefly about flight testing the Hiller. As I recall, it was tough to get both engines started at the same time, and if one of them cut out, was a pretty violent affair. It was also an environmental disaster, with horrible fumes and noise. It was an interesting experiment, but it had no real world application at all. ...

4

René Lorin invented the ramjet in 1908, and his ideas were tried in flight for the first time from 1941 onwards. Different sizes were tried; below are two pictures, one of a Do-17 Z with a small Lorin engine (source) and below that one of a Do-217 with a much larger version (source). Speed increases of up to 150 km/h were recorded, but efficiency was poor. ...

4

Are you asking about current production or through out time? The SR-71 was powered by a Pratt & Whitney J-58 engine which operated as a partial ram jet at high speeds. At lower speeds it operated like a turbojet changing its internal shape as it increased its speed. The SR-71 was capable of departure under its own power from the ground. You can find ...

3

Take a look here but the short answer A ramjet, sometimes referred to as a flying stovepipe or an athodyd (an abbreviation of aero thermodynamic duct), is a form of airbreathing jet engine that uses the engine's forward motion to compress incoming air without an axial compressor. In other words it uses the thrust it generates to compress the on coming ...

3

The thrust of a ramjet depends on the dynamic pressure of the airframe since this determines how high the pressure inside the ramjet can become. Thrust grows therefore initially with the square of speed. The efficiency of heat engines depends on the temperature and pressure ratio of their cycle, and again speed helps to improve that. The specific ...

3

Primarily the pressure differential between the stagnation pressure at the inlet and the ambient static pressure. Combustion takes place in an isobaric environment with the enthalpy of the exhaust gases being converted to kinetic energy by exiting the nozzle per Bernoulli’s Principle. Therefore the pressure differential between the combustion section and ...

3

Yes, they have a limit at around 2. And this can only be achieved at Mach 1, as Peter Kampf points out. Increasing the inlet to diffuser area ratio to extreme values will just lead to your engine being filled with slow moving air. The mass and volume of air in your engine will increase, but neither the pressure nor the rate of airflow will increase ...

2

Another way to realise a liquid fuel ramrocket engine, in theory, would be to incorporate the rocket motor into the ramjet air intake centerbody. As the aircraft approached ramjet operating speed, the ratio of fuel to oxidiser being fed into the rocket motor would be increased so that some of the fuel would be expelled from the rocket motor and mixed with ...

2

Jan Hudec’s answer here is pretty good, but the I’d add that it’s more than just oxidizer, it’s the idea with a ramjet that: 1) You can use the air from the atmosphere itself as a working mass. It’s also the same reason airliners and other airplanes don’t make use of rocket engines for propulsion. As opposed to a rocket engine which requires the vehicle ...

2

It's a slightly odd question, as for the most part a ramjet isn't an alternative to a rocket engine for an ASM, but an alternative to a turbojet or turbofan engine. On the whole, rockets are normally used in short-range weapons (typically on semi-ballistic trajectories), whereas cruise missiles (utilising aerodynamic lift for flight) use various types of ...

2

Simplicity of propulsion translates to cost savings, but there are tradeoffs in using a ramjet in a missile. The reason most aircraft-carried missiles do not use ramjet engines is because such a missile has to be accelerated to a sufficient speed for the engine to begin producing thrust. "A ramjet equipped aircraft requires another type of propulsion to ...

2

Why is that the ramjets do not require [variable geometry] to maximize pressure recovery? Turbofans, in aircraft that have variable geometry intakes, such as in an F-111 or F15, operate from standstill, at the start of the runway, to about Mach 2.5. That means, they go from having no shock, to having quite a large (strong) shock. The intake needs to provide ...

2

The Aurora D8 doesn't have traditional jet engines at all. The design incorporates a very large bypass fan which due to its size is not mechanically linked to the engine core. It's more like a turbo-prop engine, but driving a ducted fan instead of a propeller. The fan is specifically designed to cope with the flow disturbance in the boundary layer. Indeed, ...

2

It is simply the rate at which fuel can be burned. For instance, the Saturn V's first stage carried 1.37 million kg of liquid oxygen (along with the kerosene fuel), which it burned in about 165 seconds. Imagine the size of the jet intake you'd need for a comparable amount of air.

2

The Carnot condition notwithstanding, whether a ramjet actually operates at a higher "hot end" temperature than a turbojet is questionable, as well as whether a comparable percentage of thermal energy in the exhaust stream is converted to kinetic energy. Further, just because there's no turbine, doesn't mean there's nothing in the hot part of the ...

1

Well, that would depend on the design parameters, but for example this ideal ramjet would, with stoichiometric combustion process, provide best thrust per unit mass flow at a speed of approx. 2.4 Mach ...the best operating range of a hydrocarbon-fueled ramjet is $2 \leq M_0 \leq 4$ . The parameters used are $\tau_\textrm{max} = 10$ , \$ a_0 \approx 300\...

1

Replacing the turbines with ramjets won't work. If the airplane had ramjet(s), it would still need enough thrust from non-ramjet engines to reach about Mach 0.5, because ramjets can't accelerate from a standstill, and they make hardly any thrust below that speed. See the Nord Griffon, for example.

1

Back in the 1980's I saw a small ram jet from the 1950's to be used in model boats, planes and cars. (1.25-2" diameter and 16-20" length.) The accompanying literature suggested using a baby food jar as a fuel tank. It used gasoline. I expect anything today would utilize a bit more sophisticated (and safer) positive control of the fuel. It was ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# Are pseudo-zeroth order reactions possible?

Are pseudo zeroth-order reactions possible?

I have been taught that pseudo first-order reactions take place and they can be considered pseudo zeroth-order reactions because one reactant is in excess of the other(s).

What are the necessary conditions to label a reaction pseudo-zeroth order?

• I've added this as a comment, below, but it warrants attention here: "Hence, even for elementary reactions, the molecularity of a reaction does not have to be the same as the order of the reaction. When the molecularity is not the same as the order of a reaction because the concentration of one or two species is kept constant either due to the concentration of the species is high or because the concentration of the species is buffered, the reaction order is also referred to as pseudo-order." Jul 21, 2021 at 14:21

I think to some extent, you can think about all zero-order reactions as "pseudo-zero-order" reactions. This is because it's not possible for a reaction to be true zero-order. How can it be that the rate of a reaction does not depend on any one of the reactants? So, anything that is zero-order means that there is some artifact in the system, or that something else is present.

In general, zero-order reactions happen when the reactant is not able to react at the theoretical maximum rate due to something preventing it. For example, for a catalytic process, the reaction happens at the surface of the catalyst, so if you have a large excess of reactant, all of the surface of the catalyst will be covered up with the reactant, and the rate will become constant. When the concentration of the reactant goes down so that all of the catalyst surface is not in use, the reaction will show a first/second or higher order rate equation.

An example of such reaction is the decomposition of nitrous oxide: $$\ce{2N2O\overset{\Delta, Ni}{->}2N2 + O2}$$ When nickel catalyst is used, the rate is limited by the total surface area of the catalyst. Once the catalyst is working at the full capacity, the rate won't increase, even if the concentration of the reactants is increased further.

Without the catalyst, the reaction is slower, but it shows the standard second-order rate equation that we would expect by looking at the equation.

For reactions that are driven only by light (not just initiated by light!), the rate would be limited by the intensity of the light. Once there are enough species that can absorb all of the light, the reaction rate won't go up even if more reactants are added. Then there are biochemical reactions which are diffusion controlled.

So, in summary, all zero-order reactions are zero-order because there is some reactant which is in excess than something else the reaction depends upon (which can be catalyst, light, diffusion etc.). You can say they are probably all pseudo zero-order.

• This is incorrect. There are plenty of reactions where the rate does not depend on one or more of the reactants, which is the defining part of "pseudo zeroth-order." Example: decomposition of $\ce{PH3}$ on hot tungsten at high pressure (taken from this). Jul 21, 2021 at 14:24
• @ToddMinehardt Yes, because the tungsten is the catalyst here, and as long as all the catalyst surface sites are engaged, the rate doesn't depend on the conc. of PH3. Nothing in my answer contradicts anything you say. Did you read my whole answer? Please do so before downvoting. Jul 21, 2021 at 14:28
• @ToddMinehardt I have added another word in the answer, hopefully that makes it clearer. Jul 21, 2021 at 14:29

A standard example in biochemistry are enzyme-catalyzed reactions with a single substrate (e.g. ATPase, where we ignore the second substrate, water, because it is the solvent). At low substrate concentration (low reactant concentration), the reaction is first order (actually, pseudo-first order because of the water). At high susbtrate concentration (high reactant concentration) the reaction is pseudo-zero order. The theoretical expression was derived by Maude Leonora Menten and Leonor Michaelis, and the results are shown as a graph below:

As to whether it is pseudo or not, here is a textbook quote that shares the OPs view:

Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.

Of course, if you disagree with that view, you can always call the source a pseudo-textbook...

In the case of the enzyme-catalyzed reactions, the reaction would proceed at almost constant speed until the substrate concentration is depleted to a concentration in the range of the parameter K$$_\mathrm{M}$$, at which time it would slow down appreciably and continually until the substrate is depleted.

Zero-order kinetics can also appear in certain industrial settings. In some steel strip annealing processes where improved bendability of the product is required, steam is applied to decarburize the steel near the surfaces of the strip according to the reaction

$$\ce{C(s) + H2O(g) -> CO(g) +H2(g)}$$

The decarburization process requires carbon to diffuse out of the steel in order to reach the water molecules at the surface, and this process has a limited rate no matter how much steam is introduced outside the strip. Thus at a high concentration of steam in the atmosphere the decarburization process becomes zero-order in the steam concentration, dependent only on the diffusion of the carbon.

• In this case, wouldn't the diffusion rate of the carbon to the surface be directly proportional to its concentration, making the overall reaction (ideally) first order with respect to it? Jul 21, 2021 at 14:21
• Still zero-order with respect to steam. Jul 21, 2021 at 14:33
• Wouldn't the reaction rate still increase at very high steam concentrations as steam starts to diffuse into the steel? Jul 21, 2021 at 17:04
• Steam would not diffuse into the steel as water molecules, and if individual oxygen atoms were to enter there would be no way to form CO molecules in the solid matrix either. Also I am ignoring a side reaction where the steam oxidizes the iron, which the manufacturers avoid by having hydrogen in the atmosphere. Jul 21, 2021 at 17:54
• @OscarLanzi By that reasoning, every reaction is "zero-order with respect to" one of the reactants that's in excess, which is a fairly uninteresting fact Jul 23, 2021 at 11:47

The best example of a zero-order reaction is the combustion of a candle. If the candle weighs $$m_o$$ grams at the beginning, and $$m$$ at any time afterwards, and if $$a$$ grams of it are burned per minute, the reaction rate $$r$$ is constant from the very beginning to the end of the candle, and it is : $$r = dm/dt = a$$ in grams per minute. The integrated rate law is : $$m = m_o - at$$. This is a zeroth-order kinetics.

• Is this a real candle or a theoretical candle? Seems pretty contrived. Jul 21, 2021 at 13:59
• Contrived or not, a zeroth-order reaction is one where the rate is independent of the quantity of reactants/products. This is a good write-up. I've encountered these (and the defined pseudo-type, which I don't have time to write up now) in the context of geochemical kinetics. Jul 21, 2021 at 14:19
• @ToddMinehardt I am not saying zero-order reactions don't exist. But I have seen real wax candles burn and they don't burn at the same rate from the start to finish. At the beginning it will burn slowly, and after a while as more of the wick comes out, it will burn faster. It's okay if the candle example is given as a thought experiment, but actual candles definitely don't burn at the same rate from start to finish. Jul 21, 2021 at 14:23
• @SRMaiti A bit of calculus might help here. The reaction is 0th order with respect to mass of the candle but non linear with respect to temperature and wax shape. If i give you access to a candle of unknown length who's end you can't and see and its pushed upward such that the flame stays in the same place there is no measurement you can perform even when allowed to turn the candle off and on again that allows you measure the length of the candle without just burning it completely The non linearity with regard to candle warming yields no information about the total length of the candle. => 0th Jul 21, 2021 at 20:04
• @SRMaiti The rate of mass loss (~length loss) of a candle that has been turned on for a while is independent of the remaining fuel. The wick has an influence, as has a lighting a cold vs a hot candle however the wick effect becomes irrelevant really fast. If we compare two identically unlight candles of different lenght and light them we see that the burn rate is independent of the total mass, so the effect of the wick is not order 1 or higher either. The burn rate at the wick therefor does not depend on mass remaining but mass burnt and quickly levels of once the constant thickness is reached Jul 23, 2021 at 8:07

First, be aware that the order of a chemical reaction depends on its mechanism and a great deal of chemical reactions have more than one mechanism, depending on the conditions.

The traditional example for zeroth order reaction is a reaction fully dependent on a catalyst (e.g. enzymes in biochemistry).

When you have enough reactants and a scarce amount of the catalyst, you get a reaction rate that depends only on the concentration of the catalyst. It is zeroth order in regard to reactants.

Another example: topochemical reactions. The site(s) of the reaction may limit the reaction rate.

Related example: reactions limited by the mixing rate.


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# Count distinct possible words without rotations or reflections

Suppose I have an alphabet $${\{A, B, C, D\}}$$ and I want to count all possible words of length $$n$$. Easy: it's $$4^n$$.

What should I do if I want to count all possible words that:

a) are unique given any rotation, and

b) are unique including mirroring

For example:

$$AAAA$$ would definitely be included part of the total (there are no other strings that are rotations or mirrors of it).

Only one of $$BAAA$$ or $$ABAA$$ or $$AABA$$ or $$AAAB$$ would count towards the total (as they're all the same under some rotation).

Only one of $$ABCD$$ or $$CDBA$$ would count towards the total (as they're mirrors of each other).

Only one of $$ABCD$$ or $$ADCB$$ would count towards the total (as you can go between them via a rotation and a mirror).

A small illustration

With the above alphabet and words of length 2, there are 10 possible words:

AA - counted
AB - counted
AC - counted
BA - not counted (a reversal and also rotation of AB)
BB - counted
BC - counted
BD - counted
CA - not counted (a reversal and also rotation of AC)
CB - not counted (a reversal and also rotation of BC)
CC - counted
CD - counted
DA - not counted (a reversal and also rotation of AD)
DB - not counted (a reversal and also rotation of BD)
DC - not counted (a reversal and also rotation of CD)
DD - counted

• Consider the set of ordering of the letter. let $Z/nZ$ act on the set by shifting. We want to count the number of orbit. By Cauchy-Frobinius lemma this is sum of the number of elements fixed by the shifts over n. Then I 'm stuck Apr 19, 2020 at 14:47
• Only 10 words of length 2 as CB == BC. I have not devised a method for counting, but enumeration gives 10, 20, 55 and 136 for lengths 2, 3, 4 and 5. Apr 19, 2020 at 15:32
• Thanks @DanielMathias, fixed my question Apr 19, 2020 at 15:40

To summarise the equation I found at OEIS, for $$k$$ symbols and word length $$n$$, the count is found by:
$$T(n, k) = \frac{k^{\lfloor (n+1)/2 \rfloor} + k^{\lceil (n+1)/2 \rceil}} {4} + \frac{ \sum_{d|n} \phi (d) \cdot k^{n/d} } {2n}$$


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# Eigenvalue accumulation and bounds for non-selfadjoint matrix differential operators related to NLS

• ### Christiane Tretter

Universität Bern, Switzerland

A subscription is required to access this book chapter.

## Abstract

We establish results on the accumulation and location of the non-real spectrum of non-selfadjoint matrix differential operators arising in the study of non-linear Schrödinger equations (NLS) in $\mathbb{R}^d$. In particular, without restrictions on the decay rate of the potentials to $0$ at $\infty$, we show that the non-real spectrum cannot accumulate anywhere on the real axis. Under some weak assumptions satisfied, e.g., by $L_p$-potentials with $p>\frac d2$, $p\ge 2$, we prove that there are only finitely many non-real eigenvalues and that the non-real eigenvalues are located in a bounded lens-shaped region centered at the origin. Our key tool to prove this is a recent result on the existence of $\mathcal{J}$-semi-definite invariant subspaces for $\mathcal{J}$-selfadjoint operators in Krein spaces as well as abstract operator matrix methods.


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Nodal gradient operator (nodes to edges)

This property constructs the 2nd order numerical gradient operator that maps from nodes to edges. The operator is a sparse matrix $$\mathbf{G_n}$$ that can be applied as a matrix-vector product to a discrete scalar quantity $$\boldsymbol{\phi}$$ that lives on the nodes, i.e.:

grad_phi = Gn @ phi

Once constructed, the operator is stored permanently as a property of the mesh.

Returns
(n_edges, n_nodes) scipy.sparse.csr_matrix

The numerical gradient operator from nodes to edges

Notes

In continuous space, the gradient operator is defined as:

$\vec{u} = \nabla \phi = \frac{\partial \phi}{\partial x}\hat{x} + \frac{\partial \phi}{\partial y}\hat{y} + \frac{\partial \phi}{\partial z}\hat{z}$

Where $$\boldsymbol{\phi}$$ is the discrete representation of the continuous variable $$\phi$$ on the nodes and $$\mathbf{u}$$ is the discrete representation of $$\vec{u}$$ on the edges, nodal_gradient constructs a discrete linear operator $$\mathbf{G_n}$$ such that:

$\mathbf{u} = \mathbf{G_n} \, \boldsymbol{\phi}$

The Cartesian components of $$\vec{u}$$ are defined on their corresponding edges (x, y or z) as follows; e.g. the x-component of the gradient is defined on x-edges. For edge $$i$$ which defines a straight path of length $$h_i$$ between adjacent nodes $$n_1$$ and $$n_2$$:

$u_i = \frac{\phi_{n_2} - \phi_{n_1}}{h_i}$

Note that $$u_i \in \mathbf{u}$$ may correspond to a value on an x, y or z edge. See the example below.