text:
# 148 ChemActivity 10: Oxidation and Reduction Okandy hot of Foto Heo a Exercises for Part A...

###### Question:

148 ChemActivity 10: Oxidation and Reduction Okandy hot of Foto Heo a Exercises for Part A A acolo acetic acid to Cartooth, 1. Is the second step of Synth. Transf. 10.1 an oxidation of the organic product or a reduction Explain your reasoning. Synth. Transf. 10.1 is an updated version of Synth Transf. 8.3. What new information about this transformation was added in this activity? Find your note cards for 8.3, and update them with this new information 3. Draw a product that does not form in Synth. Transf. 10.1 that has the correct stereochemistry, but the wrong regiochemistry. 4. Draw a product that does not form in Synth. Transf. 10.1 that has the correct regiochemistry, but the wrong stereochemistry. 5. Identify each product below as Markov, or Anti-Markov, and syn or anti, as appropriate. Then place one of the following over each reaction now. a. 1) Hg(OAc), THF, H, 0 2) NaBH, NOOH b. 1) BHS, THF 2) H.O,/NaOH/H.0 c. not possible to get this product with reagents a orb H, OH 6. Draw the major product of each reaction. 1) THE 1) ROTH 3) NB NOH 2) HONORIO

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# Revision history [back]

### tshark programmatic capture

I'm trying to use tshark programmatically in Qt. I need to start tshark with -w capture.pcapng to capture to a file and a use a capture filter for a particular host. I then make some HTTP Qt network requests to the host and let tshark capture the communication.

I first tried launching tshark using QProcess() in Qt and waiting on the started() signal before making a network request in Qt. When doing this I was getting no packets captured at all. I figured it was because the started() signal likely only happens when the tshark process has successfully launched, but not necessarily when it's actually ready to capture packets. I noticed in the tshark docs that it will write a "Capturing On Interface..." message to stderr at the beginning of a capture, so instead I listened for a readyReadStandardError signal from the QProcess before making my network requests. This ALMOST works. It's just missing the initial SYN packet and instead only is capturing forward from the SYN-ACK coming back from the host.

I can of course delay the network requests by an arbitrary second or so to make sure tshark really is ready and that of course works and finally captures everything, but I'm wondering if anyone has any other ideas on how to make sure tshark is ready to capture requests?

Another question I have is regarding stopping the capture. I only need to capture until I get the full HTTP response from the host. My first try was that after my network requests in Qt finished, was to immediately tell tshark to terminate, but when doing this I lose a majority of the capture since I'm guessing tshark was still processing the packets when a terminate() was issued on it. In fact I had to use kill() in Windows to make it actually stop and get a non zero exit code this way.

So instead I again just went the timer route and added -a duration:5 to tshark and hope the webserver request completes in that allotted time.

I'd appreciate any other ideas/tips from others who might also use tshark programmatically.

Thanks

### tshark programmatic capture

I'm trying to use tshark programmatically in Qt. I need to start tshark with -w capture.pcapng to capture to a file and a use a capture filter for a particular host. I then make some HTTP Qt network requests to the host and let tshark capture the communication.

I first tried launching tshark using QProcess() in Qt and waiting on the started() signal before making a network request in Qt. When doing this I was getting no packets captured at all. I figured it was because the started() signal likely only happens when the tshark process has successfully launched, but not necessarily when it's actually ready to capture packets. I noticed in the tshark docs that it will write a "Capturing On Interface..." message to stderr at the beginning of a capture, so instead I listened for a readyReadStandardError signal from the QProcess before making my network requests. This ALMOST works. It's just missing the initial SYN packet and instead only is capturing forward from the SYN-ACK coming back from the host.

I can of course delay the network requests by an arbitrary second or so to make sure tshark really is ready and that of course works and finally captures everything, but I'm wondering if anyone has any other ideas on how to make sure tshark is ready to capture requests?

Another question I have is regarding stopping the capture. I only need to capture until I get the full HTTP response from the host. My first try was that after my network requests in Qt finished, was to immediately tell tshark to terminate, but when doing this I lose a majority of the capture since I'm guessing tshark was still processing the packets when a terminate() was issued on it. In fact I had to use kill() in Windows to make it actually stop and get a non zero exit code this way.

So instead I again just went the timer route and added -a duration:5 to tshark and hope the webserver request completes in that allotted time.

EDIT: Using dumpcap instead I'm able to capture the initial SYN packet, but still looking for a clean way to kill dumpcap. I'm tried killing the process with various SIG* signals and ones specific to Windows like CTRL_C_EVENT, etc, but some of them do nothing, and other end up killing the process non gracefully with a non zero exit code and when opened up in Wireshark usually give a message of "packet capture was interrupted" message, and with a lot of data missing.

I'd appreciate any other ideas/tips from others who might also use tshark programmatically.

Thanks

### tshark programmatic capture

I'm trying to use tshark programmatically in Qt. I need to start tshark with -w capture.pcapng to capture to a file and a use a capture filter for a particular host. I then make some HTTP Qt network requests to the host and let tshark capture the communication.

I first tried launching tshark using QProcess() in Qt and waiting on the started() signal before making a network request in Qt. When doing this I was getting no packets captured at all. I figured it was because the started() signal likely only happens when the tshark process has successfully launched, but not necessarily when it's actually ready to capture packets. I noticed in the tshark docs that it will write a "Capturing On Interface..." message to stderr at the beginning of a capture, so instead I listened for a readyReadStandardError signal from the QProcess before making my network requests. This ALMOST works. It's just missing the initial SYN packet and instead only is capturing forward from the SYN-ACK coming back from the host.

I can of course delay the network requests by an arbitrary second or so to make sure tshark really is ready and that of course works and finally captures everything, but I'm wondering if anyone has any other ideas on how to make sure tshark is ready to capture requests?

Another question I have is regarding stopping the capture. I only need to capture until I get the full HTTP response from the host. My first try was that after my network requests in Qt finished, was to immediately tell tshark to terminate, but when doing this I lose a majority of the capture since I'm guessing tshark was still processing the packets when a terminate() was issued on it. In fact I had to use kill() in Windows to make it actually stop and get a non zero exit code this way.

So instead I again just went the timer route and added -a duration:5 to tshark and hope the webserver request completes in that allotted time.

EDIT: Using dumpcap instead I'm able to capture the initial SYN packet, but still looking for a clean way to kill dumpcap. I'm tried killing the process with various SIG* signals and ones specific to Windows like CTRL_C_EVENT, etc, but some of them do nothing, and other end up killing the process non gracefully with a non zero exit code and when opened up in Wireshark usually give a message of "packet capture was interrupted" message, and with a lot of data missing.

EDIT2: CTRL_C_EVENT signal is now working correctly. Not sure why it wasn't working initially.

I'd appreciate any other ideas/tips from others who might also use tshark programmatically.

Thanks

### tshark programmatic capture

I'm trying to use tshark programmatically in Qt. I need to start tshark with -w capture.pcapng to capture to a file and a use a capture filter for a particular host. I then make some HTTP Qt network requests to the host and let tshark capture the communication.

I first tried launching tshark using QProcess() in Qt and waiting on the started() signal before making a network request in Qt. When doing this I was getting no packets captured at all. I figured it was because the started() signal likely only happens when the tshark process has successfully launched, but not necessarily when it's actually ready to capture packets. I noticed in the tshark docs that it will write a "Capturing On Interface..." message to stderr at the beginning of a capture, so instead I listened for a readyReadStandardError signal from the QProcess before making my network requests. This ALMOST works. It's just missing the initial SYN packet and instead only is capturing forward from the SYN-ACK coming back from the host.

I can of course delay the network requests by an arbitrary second or so to make sure tshark really is ready and that of course works and finally captures everything, but I'm wondering if anyone has any other ideas on how to make sure tshark is ready to capture requests?

Another question I have is regarding stopping the capture. I only need to capture until I get the full HTTP response from the host. My first try was that after my network requests in Qt finished, was to immediately tell tshark to terminate, but when doing this I lose a majority of the capture since I'm guessing tshark was still processing the packets when a terminate() was issued on it. In fact I had to use kill() in Windows to make it actually stop and get a non zero exit code this way.

So instead I again just went the timer route and added -a duration:5 to tshark and hope the webserver request completes in that allotted time.

EDIT: Using dumpcap instead I'm able to capture the initial SYN packet, but still looking for a clean way to kill dumpcap. I'm tried killing the process with various SIG* signals and ones specific to Windows like CTRL_C_EVENT, etc, but some of them do nothing, and other end up killing the process non gracefully with a non zero exit code and when opened up in Wireshark usually give a message of "packet capture was interrupted" message, and with a lot of data missing.

EDIT2: CTRL_C_EVENT signal is now working correctly. Not sure why it wasn't working initially.initially. Also starting to get the missing initial SYN packet sometimes. Looking at the source I see the stderr capturing on message happens just before the actual capture_loop function is called so I can't rely on that message. So I'm back to now just waiting an arbitrary second.

I'd appreciate any other ideas/tips from others who might also use tshark programmatically.

Thanks

format: plain


text:
# IVP- Uniqueness Theorem and Newton's Law of Cooling

Printable View

• Aug 11th 2014, 03:42 AM
GirandoniAirRifle
IVP- Uniqueness Theorem and Newton's Law of Cooling
Hi guys,
I would like some help on these problems please.
Attachment 31416Attachment 31417

I have a quiz soon, which is on this problem set. I can't get help tomorrow as I have classes all day. I have been really busy during the last few weeks on the family farm. I have hired a personal tutor, but we don't start until after the quiz. Thus help would be much much appreciated.I had a read through the uniqueness theorem, but don't really know what to do.

I'm not exactly sure how to model this.

Thanks in advance,

GirandoniAirRifle
• Aug 11th 2014, 04:07 AM
HallsofIvy
Re: IVP- Uniqueness Theorem and Newton's Law of Cooling
Well, what, exactly, does the "uniqueness theorem" say? In particular, what hypotheses does it have and are they true for this problem?

For the second, do you know what "Newton's law of cooling" is? It says that the rate at which heat travels from a warmer body to a cooler is proportional to the difference in temperatures. That is, $\frac{dT}{dt}= k(T- T_a(t))$.


text:
# Difference between revisions of "Temperature"

The temperature keyword is a required keyword in the input file. It takes 1 argument, the system temperature in K.
Summary for temperature keyword
temperature = 298.


text:
Contents\@mkbothCONTENTSCONTENTS

EMPG–12–01

HWM–12–01

DMUS–MP–12/01

A Twistor Description of Six-Dimensional

Super Yang–Mills Theory

Christian Sämann, Robert Wimmer, and Martin Wolf ***E-mail addresses: c.saemann@hw.ac.uk

Maxwell Institute for Mathematical Sciences

Department of Mathematics, Heriot–Watt University

Edinburgh EH14 4AS, United Kingdom

[.5cm]

Université de Lyon, Laboratoire de Physique, UMR 5672, CNRS

École Normale Supérieure de Lyon

46, allée d’Italie, F-69364 Lyon cedex 07, France

[.5cm]

Department of Mathematics, University of Surrey

Guildford GU2 7XH, United Kingdom

Abstract

We present a twistor space that describes super null-lines on six-dimensional superspace. We then show that there is a one-to-one correspondence between holomorphic vector bundles over this twistor space and solutions to the field equations of super Yang–Mills theory. Our constructions naturally reduce to those of the twistorial description of maximally supersymmetric Yang–Mills theory in four dimensions. 15th May 2012

\@starttoc

toc

## 1 Introduction and results

The twistor description of solutions to chiral field equations in six dimensions goes back to the work of Hughston [Hughston:1987he]. For recent works in this direction, see also [Saemann:2011nb, Mason:2011nw] and references therein. A corresponding twistor description of solutions to non-chiral field equations in six dimensions, such as the equations of motion of Yang–Mills theory with maximal supersymmetry, has only been developed partially [Devchand:1985au, Harnad:1987xq, Harnad:1995zy].

The purpose of this letter is to give a complete twistor description of the maximally or supersymmetric Yang–Mills (MSYM) equations in six dimensions with an emphasis on the underlying geometries.111Notice that twistor methods have recently been applied in the description of scattering amplitudes in this theory, see e.g. [Cheung:2009dc, Dennen:2009vk, Brandhuber:2010mm, Dennen:2010dh]. Our way of describing ambitwistor space might prove useful in this context. It is known that these equations can be encoded in constraint equations for a connection on superspace [Harnad:1985bc], which in turn correspond to the integrability condition of this connection along super null-lines [Devchand:1985au, Samtleben:2009ts]. We start by describing a twistor correspondence for null-lines in six-dimensional space-time in some detail. We then present the corresponding supersymmetric extension for maximal supersymmetry. The resulting twistor space, denoted by , turns out to be a rank- holomorphic supervector bundle over the four-dimensional Graßmannian . Next, we derive a Penrose–Ward transform to establish a one-to-one correspondence between equivalence classes of certain holomorphic vector bundles over and gauge equivalence classes of solutions to the equations of motion of six-dimensional MSYM theory. We end by demonstrating how our constructions reduce to those appearing in the twistorial description of MSYM theory in four dimensions [Witten:1978xx, Isenberg:1978kk, Isenberg:1978qd].

Throughout this letter, we shall be working in the complex setting. Concretely, our six-dimensional space-time is a copy of . If desired, however, reality conditions can be imposed at any point of our constructions, cf. [Saemann:2011nb, Mason:2011nw].

## 2 Ambitwistor space L9|8 of N=(1,1) superspace

In this section, we shall construct an ambidextrous twistor space (or ambitwistor space for short) of six-dimensional superspace. This twistor space is very similar in spirit to the ambitwistor space of four-dimensional superspace [Witten:1978xx, Isenberg:1978kk, Isenberg:1978qd]: while the latter parametrises super null-lines in four dimensions, parametrises certain super null-lines in six dimensions. We shall first describe the body of the supermanifold in detail, before we come to the supersymmetric extension. Our notation and conventions are close to those of [Saemann:2011nb].

### 2.1 Construction of the body L9 of L9|8

#### Outline of the construction.

As usual in twistor geometry, we would like to establish a double fibration in which a correspondence space is simultaneously fibred over both twistor space and complexified flat space-time . The correspondence space in such a twistor fibration is a direct product of two manifolds.222If we are considering a compactified space-time, this direct product has to be compactified appropriately. The first factor in this product is space-time itself. The second factor is the moduli space of linear subspaces of space-time that we wish to describe with the twistor correspondence, restricted to those through the origin. Note that this makes the correspondence space the space of such linear subspaces with a given base point. Different base points may describe the same subspace, and modding out the dependence on equivalent base points, we obtain twistor space. In this letter, we are interested in light-rays or null-lines in six dimensions. We shall see below that the space of null-lines through the origin is given by the four-dimensional Graßmannian , which is the space of two-planes in . The correspondence space, which we shall denote by , is therefore ten-dimensional and we have . Modding out the dependence on equivalent base points amounts to quotenting the correspondence space by an (integrable) rank-one distribution known as a twistor distribution. This yields a nine-dimensional complex manifold which we denote by . Altogether, we have the following double fibration:

 (2.1)

Here, the projection is the quotient map by the distribution and is the trivial projection. In the following, we shall discuss this double fibration, and in particular the structure of the space , in more detail.

#### Null-lines in six dimensions.

For simplicity, we shall work in spinor notation on , that is, we identify the tangent bundle with the antisymmetric tensor product of the rank-four bundle of anti-chiral spinors over . Correspondingly, we shall use local coordinates with and take the (flat) metric , where is the Levi-Civita symbol in four dimensions.

A null-vector in satisfies the equation

 12εABCDλABλCD = 0\leavevmode\nobreak . (2.2)

Null-lines are then obtained from null-vectors via the identification with . The resulting equivalences classes describe points on the Graßmannian , the homogeneous coordinates are called Plücker coordinates, and (2.2) is called the Plücker relation. The space features prominently in four-dimensional twistor correspondences, and a detailed account can be found, e.g., in [Ward:1990vs]. In the following, we merely recall a few facts necessary for our discussion.

Plücker coordinates provide an embedding of into via the quadric (2.2) with the being the six homogeneous coordinates on . Furthermore, as a coset space, the Graßmannian is given by

 G2,4 ≅ SL(4,\mathbbmC)SL(2,\mathbbmC)×˜SL(2,\mathbbmC)\leavevmode\nobreak , (2.3)

where is the little group of a null vector in . The relation (2.2) implies that the Plücker coordinates factorise according to

 λAB = 12εABCDλCaλDbεab\leavevmode\nobreak , (2.4)

where and is the invariant tensor for with . We therefore have homogeneous coordinates and a coset description

 G2,4 ≅ Mat4×2(\mathbbmC)SL(2,\mathbbmC)×\mathbbmC∗\leavevmode\nobreak . (2.5)

Every plane in has a natural dual , which is spanned by a pair of chiral spinors for with . The represent homogeneous coordinates on a dual Graßmannian and they define a set of dual Plücker coordinates according to

 μAB = 12εABCDμC˙cμD˙dε˙c˙d\leavevmode\nobreak . (2.6)

The indices are to be understood as indices of the subgroup of the little group and is the invariant tensor of with . Furthermore, the two Graßmannians and can be identified via

 λAB = 12εABCDμCD⟺12εABCDλCaλDbεab = μA˙aμB˙bε˙a˙b\leavevmode\nobreak . (2.7)

The above equality, like all equalities in the following involving homogeneous coordinates, is to be understood as an equality of equivalence classes.

#### Double fibration.

So far, we have seen that the correspondence space is topologically and it is trivially fibred over space-time. We may coordinatise by either or . To mod out the dependence of the null-lines on equivalent base points, we quotient the correspondence space by the rank-one twistor distribution that is generated by the vector field

 V := λAB∂∂xAB = 12εABCDλCaλDbεab∂∂xAB\leavevmode\nobreak . (2.8)

The resulting space is a nine-dimensional complex manifold . Note that by construction, the twistor space is a rank-five holomorphic vector bundle over . Let us now give a few more details about the geometry of .

To this end, consider the dual tautological bundle333For more details on Graßmannians and bundles over them, see e.g. [Manin:1988ds]. over . This rank-two holomorphic vector bundle is generated by its global sections. We can parametrise the latter by moduli according to , where is the anti-chiral spin bundle over space-time. This allows us to write down the following short exact sequence:

 0 ⟶ E ⟶ \mathbbmC4⊗T∨ κ:vAa↦vA(aλAb)−−−−−−−−−→ ⊙2T∨ ⟶ 0\leavevmode\nobreak . (2.9)

Notice that has rank three such that is a rank-five holomorphic vector bundle, and its sections obey

 vA(aλAb) = 0\leavevmode\nobreak . (2.10)

The short exact sequence (2.9) induces a long exact sequence of cohomology groups and since all higher cohomology groups of and vanish, we conclude that and for . Global holomorphic sections of are of the form with , and is generated by these sections.

In fact, we can identify with the twistor space provided we identify the moduli with the space-time coordinates : the projection is given by with

 vAa = xABλBa\leavevmode\nobreak , (2.11)

and the vector fields (2.8) generating the twistor distribution indeed annihilate the . We shall refer to the relation (2.11) as the incidence relation. This relation implies a geometric twistor correspondence: points in correspond to null-lines in space-time given by , where is a particular solution to the incidence relation (2.11) and . Vice versa, points in space-time correspond to submanifolds bi-holomorphic to the Graßmannian . Note that the above null-lines are the null-lines obtained by intersecting two three-planes which are totally null (so-called -planes) [Hughston:1986hb]. Note also that we have just derived twistor space from space-time. Inversely, one can derive space-time from twistor space using Kodaira’s theorem of relative deformation theory: because of , there are no obstructions to relative deformations of inside . Thus, we have a family of such deformations whose moduli space can indeed be identified with space-time .

#### Other descriptions.

In addition to using , we may also use the dual Graßmannian with homogeneous coordinates (or Plücker coordinates ). Equipping (dual) space-time with coordinates , we can associate a correspondence space with coordinates and introduce a twistor distribution generated by

 ~V := μAB∂∂yAB = 12εABCDμC˙cμD˙dε˙c˙d∂∂yAB\leavevmode\nobreak . (2.12)

Altogether, we have a dual double fibration

 (2.13)

where is the trivial projection and is given by with and . This incidence relation yields an analogous geometric twistor correspondence to the one above: points in correspond to null-lines in and points in correspond to embedding of in . Geometrically, the null-lines just described arise from intersecting two dual three-planes which are totally null (so-called -planes) [Hughston:1986hb]. Note that also can be described by a short exact sequence of the form (2.9).

The manifolds and yield a description of null-lines in terms of chiral spinors and anti-chiral spinors , respectively. To obtain an ambidextrous description, that is, a description involving both and simultaneously, we identify and via (2.7) and write and introduce as the zero-locus

 (vAavBbεab−12εABCDw˙aCw˙bDε˙a˙b) mod λAB = 0\leavevmode\nobreak . (2.14)

Because of (2.14), global holomorphic sections are of the form

 vAa = xABλBa\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak w˙aA = 12εABCDxCDμB˙a\leavevmode\nobreak , (2.15)

and therefore is a rank-five holomorphic vector bundle over . Altogether, we have a double fibration

 (2.16)

with the same space-time manifold as in (2.1) and

 π1:(xAB,{λAa,μA˙a}) ↦ ({vAa,w˙aA},{λAa,μA˙a}) (2.17)

is given by (2.15).

#### Double fibrations in Plücker coordinates.

In order to extend the above discussion to the supersymmetric setting with manifest maximal R-symmetry, we shall find it more convenient to work directly with Plücker coordinates. The advantage of these coordinates is that one can easily switch between chiral and anti-chiral descriptions by virtue of (2.7). The results of [Harnad:1987xq, Harnad:1995zy] seem to suggest that a description using the homogeneous coordinates and and having the full R-symmetry for supersymmetry manifest at the same time is not possible. Note, however, that in principle we can always substitute the Plücker coordinates by the homogeneous coordinates.

Recall that the Plücker coordinates define an embedding as the quadric (2.2). The bundle can be identified with [Manin:1988ds], and global sections of this bundle are given by , , and . This bundle appears in the short exact sequence analogue to (2.9) involving Plücker coordinates,

 0 ⟶ E ⟶ \mathbbmC16⊗detT∨ κ:vAB↦(vABλBC,vA(BλC)A)−−−−−−−−−−−−−−−−→ E′ ⟶0\leavevmode\nobreak , (2.18)

where has rank444To analyse the rank of such maps, it is helpful to consider them over a convenient point on the base manifold , e.g. . 11 and therefore has rank five. In more detail, the bundle is generated by its global sections and we can write . The eleven linear equations and in the fibre coordinates reduces the rank-16 bundle to a rank-five bundle . In Plücker coordinates, global holomorphic sections of are of the form with .

As before, identifying with , we can write the projection in the double fibration (2.1) as

 π1:(xAB,λAB) ↦ (vBA,λAB)\leavevmode\nobreak (2.19a) with an incidence relation of the form vAB=xACλCB\leavevmode\nobreak . (2.19b)

In a similar manner, we may repeat this analysis for and .

### 2.2 Supersymmetric extension

Let us now come to the supersymmetric extension of the twistor space . We shall first construct twistor spaces for chiral and anti-chiral super null-lines, before extending these spaces to the twistor space of super null-lines. We shall use Plücker coordinates on all the Graßmannians.

#### Twistor space for chiral super null-lines.

Let be the Graßmann parity changing operator. We start from superspace , which we describe by Graßmann even (bosonic) coordinates and Graßmann odd (fermionic) coordinates . Here, the index is an index of , the chiral subgroup of the R-symmetry group555not to be confused with the little group . On , we introduce the vector fields

 PAB := ∂∂xAB\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak DmA := ∂∂θmA+εmnθnC∂∂xAC\leavevmode\nobreak , (2.20)

which satisfy the relation

 {DmA,DnB} = 2εmnPAB\leavevmode\nobreak . (2.21)

Chiral super null-lines are linear -dimensional subspaces . The moduli space of such linear superspaces through the origin is still so that the correspondence space is . To obtain a twistor space, we have to mod out the dependence on equivalent base points. Here, this amounts to quotienting by a twistor distribution generated by the vector fields

 V := λABPAB\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak VAm := λABDmB\leavevmode\nobreak . (2.22)

For each , we have four independent equations and hence the twistor distribution is of rank-. Moreover, it is integrable since while . Therefore, we have a foliation of by -dimensional complex supermanifolds . By construction, is a rank- holomorphic supervector bundle over , which we describe as a subbundle of :

 0 ⟶ L9|4 ⟶ \mathbbmC16|8⊗detT∨ \lx@stackrelκ⟶ E′ ⟶ 0\leavevmode\nobreak . (2.23)

Using coordinates in the fibres of and the usual Plücker coordinates on the base, the map is implicitly given by the relations

 vACλBC = 0\leavevmode\nobreak ,vC(AλB)C−12ϑmAϑnBεmn =0\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak ϑmBλBA = 0\leavevmode\nobreak . (2.24)

We can define a projection with

 vAB = (xAC−12θmAθnCεmn)λCB\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak ϑmA = θmBλBA\leavevmode\nobreak , (2.25)

and the vector fields (2.22) generating the twistor distribution indeed annihilate and . Equations (2.25) represent a chiral super extension of the incidence relation (2.19b).

Because of the projection given in (2.25) and the trivial projection , we have the double fibration

 (2.26)

The geometric twistor correspondence here is between points on and chiral super null-lines in as well as between points on and holomorphic embeddings of into . Explicitly, for any fixed point , the incidence relation (2.25) yields a (-dimensional) chiral super null-line

 xAB = xAB0+τλAB+τmCλC[AθnB]0εmn\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak θmA = θmA0+τmBλBA\leavevmode\nobreak , (2.27)

where represent a particular solution to the incidence relation while and constitute one free bosonic parameter and four fermionic parameters (note that the matrix is of rank two, so only four out of the initial eight fermionic parameters enter).

#### Twistor space for anti-chiral super null-lines.

A twistor space for anti-chiral super null-lines is constructed analogously. Here, we start from , with bosonic coordinates and fermionic coordinates , . The vector fields generating supertranslations read as

 ~PAB := ∂∂yAB\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak DA˙m := ∂∂θ˙mA+ε˙m˙nθ˙nC∂∂yAC\leavevmode\nobreak , (2.28)

and they satisfy the relation

 {DA˙m,DB˙n} = 2ε˙m˙n~PAB\leavevmode\nobreak . (2.29)

The correspondence space is given by , and to mod out the dependence on equivalent base points, we have to quotient by the vector fields

 V := μAB~PAB\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak V˙mA := μABDB˙m\leavevmode\nobreak . (2.30)

The resulting vector bundle can be regarded as a subbundle of over , which we coordinatise by in the fibres and the Plücker coordinates on the base. The relations satisfied by the fibre coordinates are

 wCAμBC = 0\leavevmode\nobreak ,w(ACμB)C−12ϑ˙mAϑ˙nBε˙m˙n = 0\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak ϑ˙mBμBA = 0\leavevmode\nobreak , (2.31)

and we have a double fibration

 (2.32)

Here, the projection reads as

 wAB = (yBC−12θ˙mBθ˙nCε˙m˙n)μCA\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak ϑ˙mA = θ˙mBμBA\leavevmode\nobreak . (2.33)

This incidence relation yields again a geometric twistor correspondence between points in and (-dimensional) anti-chiral super null-lines in as well as points in and submanifolds in bi-holomorphic to .

#### Ambitwistor space.

Let us now come to the discussion of full supersymmetry. In particular, consider superspace equipped with coordinates . On this space, we have both the vector fields (2.20) and (2.28) with the identification . They generate the supersymmetry algebra in six dimensions,

 {DmA,DnB} = 2εmnPAB\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak {DA˙m,DB˙n} = 2ε˙m˙nPAB\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak {DmA,DB˙n} = 0\leavevmode\nobreak , (2.34)

where . Note that here, the metric appears explicitly.

The correspondence space is then topologically and coordinatised by . On , we introduce a rank- distribution generated by the vector fields

 V := λABPAB = λABPAB\leavevmode\nobreak ,VAm := λABDmB\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak V˙mA := λABDB˙m\leavevmode\nobreak . (2.35)

This distribution is integrable, with the non-vanishing Lie brackets given by and . The quotient of by this distribution is the ambitwistor space . It is a rank- supervector bundle over and its body is . We describe as a subbundle of with fibre coordinates and the map implicitly given by its kernel:

 vACλBC+12ϑ˙mAϑ˙nBε˙m˙n = 0\leavevmode\nobreak , vC(AλB)C−12ϑmAϑnBεmn = 0\leavevmode\nobreak , (2.36) λABϑmB = 0\leavevmode\nobreak , λABϑ˙mB = 0\leavevmode\nobreak .

Note that has indeed rank , and we we have constructed a double fibration

 (2.37)

As before, the projection is the trivial projection, while reads as

 vAB = (xAC−12θmAθnCεmn)λCB+12θ˙mBθ˙nCε˙m˙nλCA\leavevmode\nobreak , (2.38) ϑmA = θBmλBA\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak ϑA˙m = θ˙mBλBA\leavevmode\nobreak ,

which describe global holomorphic sections of the bundle .

The geometric twistor correspondence induced by the incidence relation (2.38) is between points on and super null-lines in as well as between points on and holomorphic embeddings of into . Explicitly, for a fixed point , the above incidence relations determine a (-dimensional) super null-line by

 xAB = xAB0+τλAB+τmCλC[AθnB]0εmn+12εABCDτ˙mEλE[Cθ0˙nD]ε˙m˙n\leavevmode\nobreak , (2.39) θmA = θmA0+τmBλBA\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak θ˙mA = θ0˙mA+τ˙mBλBA\leavevmode\nobreak ,

where represent a particular solution to the incidence relation while and constitute one free bosonic parameter and eight fermionic parameters.

## 3 Twistor construction of the MSYM equations in six dimensions

#### Constraint equations.

We now come to the description of classical solutions to the equations of motion of MSYM theory on by means of holomorphic data on the ambitwistor space . The key fact here is that these equations are equivalent to certain constraint equations for a connection on the superspace [Harnad:1985bc] and furthermore, that these constraint equations can in turn be interpreted as integrability conditions along certain null-lines [Devchand:1985au, Samtleben:2009ts]. Concretely, the equations of motion of MSYM theory in six dimensions are equivalent to the following set of constraint equations [Harnad:1985bc]:

 {∇mA,∇nB} = 2εmn∇AB\leavevmode\nobreak , (3.1) {∇mA,∇B˙n}−14δAB{∇mC,∇C˙n} = 0\leavevmode\nobreak , {∇A˙m,∇B˙n} = ε˙m˙nεABCD∇CD = 2ε˙m˙n∇AB\leavevmode\nobreak .

Here, the covariant derivatives are given in terms of a gauge potential with components , and as

 ∇AB := ∂AB+AAB\leavevmode\nobreak ,∇mA := DmA+AmA\leavevmode\nobreak ,\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak and\leavevmode\nobreak \leavevmode\nobreak \leavevmode\nobreak ∇A˙m := DA˙m+AA˙m\leavevmode\nobreak , (3.2)

where the derivatives


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The Basics of Event Probability

In this lesson, some basics concepts of probability will be introduced and connected to real-life situations.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Possible Outcomes of Target Shooting

Magdalena is in a target shooting competition. For her next test, the target has three zones and she can shoot three arrows.
Assuming that Magdalena always hit the target, list all the possible results that Magdalena can get on the test.

Probability of Winning a Prize

On a late-night game show, there are three closed doors. Behind one of the doors is a car, and there are sheep behind the other two. The host invites Tearrik to win the car by choosing only one door.
a What are the chances that Tearrik will choose the winning door? Round the answer to two decimal places.
b What are the chances of choosing a door with a sheep? Round the answer to two decimal places.

Defining Experiments in Probability

Below, some basic definitions of probability are examined.

Experiment - Probability

In probability, an experiment is an action that can be repeated infinitely many times and has various results called outcomes. In other words, an experiment is a process used to determine the probability of an event occurring in the future. For example, rolling a die can be considered an experiment.
Experiments are repeated several times to collect some data, and each repetition is called a trial. For example, if a die is thrown times, each throw can be considered a trial. An experiment is usually performed to estimate the theoretical probability of a particular outcome.

Outcome

An outcome is a possible result of a probability experiment. For example, when rolling a six-sided die, getting a is one possible outcome.

Note that when performing an experiment, each possible outcome is unique — that is, on each trial, only one outcome will occur.

Event

An event is a combination of one or more specific outcomes. For example, when playing cards, an event might be drawing a spade or a heart. For this event, one possible outcome is drawing the or drawing the

However, these are not the only outcomes of this event. All the possible outcomes that satisfy the event are listed below.

Sample Space

The sample space of an experiment is the set of all possible outcomes. For example, when flipping a coin, there are two possible outcomes: heads, H, or tails, T. Therefore, the sample space is

Here, the sample space is shown in a tree diagram. Each row represents the possible outcomes of a toss. When the coin is flipped another time, the tree diagram gets another row with the possible outcomes.

In this case, the sample space has possible outcomes.

Sample Space of Rolling Dice

For each of the following experiments, list the possible outcomes in the sample space and count the total number of outcomes.

a Roll one die.
b Roll two dice at the same time.

Answer

a Number of Outcomes:

Outcomes:

b Number of Outcomes:

Outcomes:

Hint

a What are the possible outcomes when a die is rolled?
b For each outcome of the first die, there are six different outcomes for the second die.

Solution

a When a die is rolled, there are different possible outcomes.

Consequently, the sample space is the set

b As was found in Part A, each possible outcome of rolling one die is just one number. However, when rolling two dice, one possible outcome is getting a on the first die and a on the second die.

Therefore, each outcome in the sample space will consist of two numbers — one for each die.

To list all the possible outcomes, create all possible combinations by determining one outcome for the first die and then varying the outcome of the second die. Then, change the outcome of the first die and repeat the process.

Since each die has possible outcomes, the total number of possible outcomes for this experiment is

Sample Space: Drawing Marbles From a Bag

Paulina bought two white and three black marbles, all of different sizes, and put them in a bag. When she got home, her little brother Diego and sister Emily asked her to give them two marbles. Paulina agreed but told them to draw one marble each without looking inside the bag. Diego drew the first marble, then Emily.

a Assuming all the marbles are distinguishable, write down the sample space for the event of Diego and Emily randomly drawing two marbles from the bag. What is the total number of outcomes?
b How many of the outcomes in the sample space represent the case where Diego and Emily both drew white marbles?
c How many of the outcomes in the sample space represent the case where Diego and Emily both drew black marbles?
d How many of the outcomes in the sample space represent the case where Diego and Emily drew marbles of different colors?
e In a few words, write a different event for the situation presented.

Answer

a Number of Outcomes:

Outcomes: Here, and represent the white marbles, and and represent the black marbles.

b Number of Outcomes:

Outcomes: and

c Number of Outcomes:

Outcomes:

d Number of Outcomes:

Outcomes:

e Example Event: Diego and Emily ask Paulina to give them two marbles each.

Hint

a Give each marble a label. How many outcomes does the first drawn marble have? How many outcomes does the second marble have?
b Look for the outcomes containing only the labels assigned to the white marbles.
c Look for the outcomes containing only the labels assigned to the black marbles.
d Look for the outcomes containing one label assigned to the white marbles and one label assigned to the black marbles.
e If Diego and Emily ask for a different number of marbles, this is a different event.

Solution

a Despite having only two colors, all five marbles are said to be distinguishable. That is why starting by labeling each of the marbles is a good option.

Since both marbles are randomly drawn, each outcome of the event will consist of two labels. For the first marble, there are possible outcomes. For the second marble, there are possible outcomes because one marble has already been removed from the bag.

Since the Diego draws a marble first and Emily draws a marble after him, the order in which the marbles are drawn matters. Therefore, the outcomes and are different. The following table lists all the possible outcomes for the event of drawing two marbles from the bag. Consequently, there are a total of possible outcomes in the sample space.

b The case in which two white marbles are drawn is represented by the outcomes containing only and From Part A, it can be seen that there are only two such outcomes — and
c The case in which two black marbles are drawn is represented by the outcomes containing neither nor Analyzing the sample space from Part A, six such outcomes can be found.

d The case in which one white and one black marble are drawn is represented by the outcomes containing one label and one label. From Part A, there are twelve such outcomes.

e There are an infinite number events that can be considered. For example, Diego and Emily could ask Paulina to give them two marbles each. In that case, the event would be the drawing of four marbles.

Union, Intersection, and Complement of Events

Sometimes more than one event can be involved in an experiment. In such cases, it is important to know how to determine the union or intersection of the events.

Union - Probability

The union of two events and is the set of all the outcomes that are in or are in or in both and The union of and is usually written as or
The probability of the union of and is the probability that event or event will occur. It can be found using the Addition Rule of Probability.

Intersection - Probability

The intersection of two events and is the set of all the outcomes that satisfy both events and simultaneously. The intersection of and is usually written as or
The probability of the intersection of and is the probability that and will occur and can be found using the Multiplication Rule of Probability.

Also, there may be situations where it is easier to determine which outcomes do not satisfy an event rather than determining which outcomes do. For such cases, the following concept will be useful.

Complement - Probability

The complement of an event is the set of all the possible outcomes in the sample space that are not in event The complement of event is usually written as or Notice that the union of an event and its complement is the entire sample space.
For example, let be the event of getting an odd number when a six-sided die is rolled. The complement of is the event of getting an even number when the die is rolled.

Operating With Events

Let be the set of all integers from to
Consider the experiment of randomly picking a number from
a Write the complement of each event in words.
b List the outcomes of each event and its complement.
c Write the event in words and list its outcomes.
d Write the event in words and list its outcomes.
e Make a Venn diagram illustrating the set and the events and

Answer

a Complements:
• is the event of picking a non-prime number.
• is the event of picking an even number.
• is the event of picking a number that is not a multiple of
b Outcomes of the Events:

Outcomes of the Complements:

c Intersection of Events: is the event of picking a prime, odd number.

Outcomes:

d Union of Events: is the event of picking either an odd number or a number that is not a multiple of

Outcomes:

e Venn Diagram:

Hint

a To write the complement of an event, consider the opposite event.
b The complement of an event is the set of all the possible outcomes that are not in that event.
c Remember how the intersection of events is defined.
d Use the definition of a union of events.
e First, look for the numbers that are common for the three events. Then, look for the numbers that are in two events but not in the third one. Finally, look for the numbers from the sample space that do not occur in any of the events.

Solution

a By definition, the complement of an event is the set of all possible outcomes that are not in that event. Therefore, the complement of picking a prime number is picking a non-prime number. The complement of events and can be similarly written.
Event Complement
picking a prime number picking a non-prime number
picking an odd number picking an even number
picking a multiple of picking a number that is not a multiple of
b The outcomes of the event are all numbers in that are prime.

In a similar way, the outcomes of the remaining events can be written. Finally, the outcomes of are the outcomes in that are not in Similarly for the outcomes of and

c By definition, the intersection of two events is the set of all outcomes that satisfy both events simultaneously. To write start by recalling what events and represent.

- event of picking a prime number
- event of picking an odd number

Therefore, is the event of picking a prime and odd number.

Number Is it prime? Is it odd?
No Yes
Yes No
Yes Yes
No No
Yes Yes
No No
Yes Yes
No No
No Yes

Consequently,

d By definition, the union of two events is the set of all outcomes that satisfy either of the events. Before writing what is, begin by remembering what each of these events represents.

- event of picking an odd number
- event of picking a number that is not a multiple of

Therefore, is the event of picking either an odd number or a number that is not multiple of

Number Is it odd? Is not a multiple of
Yes Yes
No Yes
Yes No
No Yes
Yes Yes
No No
Yes Yes
No Yes
Yes No

Consequently, Notice that is the only element of that is not in since it satisfies neither nor

e In order to make a Venn diagram, start by considering the sample space which is a set of integers from to

Next, draw three sets representing the events and and write down the outcomes of each event inside the corresponding set.

Comparing the outcomes of the three events, some conclusions can be drawn.

• The number is common for the three events. Thus, the three sets intersect each other.
• The numbers and are common for events and but they are not in
• The number is common for events and but it is not in
• The events and have nothing in common other than the number
• The numbers and do not belong to any of the events.

With this information, the Venn diagram can be drawn.

In the following diagram, the events and their complements can be appreciated separately.

Theoretical and Experimental Probability

Since an event is a combination of possible outcomes of an experiment, in some cases the event happens rarely, while in others it happens frequently. This frequency depends on the experiment and the event itself.

Probability

Probability measures the likelihood that something will occur. It can be any value from or to or inclusive. When it is certain that the situation will not occur, the probability is Further, when it is certain that the situation will occur, the probability is

The probability of an event occurring can be determined both theoretically and experimentally. Theoretical probability shows the expected probability when all outcomes in a sample space are equally likely, whereas experimental probability is based on data collected from repeated trials of an experiment.

Theoretical Probability

When all outcomes in a sample space are equally likely, the theoretical probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.

For example, suppose a standard six-sided die is rolled. There are six equally likely outcomes in the sample space. For the event of rolling an even number, there are favorable outcomes.
Therefore, the probability of event is found as follows. After simplification, it is obtained that is equal to Consequently, there is a chance an even number is rolled.

When an experiment is performed, the results may be a little different from what was expected. In other words, slightly different results may be obtained from what the theoretical probability predicted.

Experimental Probability

Experimental probability is the probability of an event occurring based on data collected from repeated trials of a probability experiment. For each trial, the outcome is noted. When all trials are performed, the experimental probability of an event is calculated by dividing the number of times the event occurs by the number of trials.

By repeating an experiment many times, the result will tend to the theoretical probability of the event. For example, consider the event of getting tails when a coin is tossed many times. The following applet simulates the outcomes and calculates the experimental probability.
It can be seen that the experimental probability of flipping tail is close to the theoretical probability, which is

Theoretical Probability vs. Experimental Probability

Ramsha and Mark conducted an experiment consisting of rolling two dice and adding their results. The following diagram shows the numbers obtained in each roll.

Consider the event of getting a result greater than or equal to

a What is the theoretical probability of this event? Round the answer to two decimal places.
b Find the experimental probability of the given event according to Ramsha's data. Round the answer to two decimal places.
c Find the experimental probability of the given event according to Mark's data. Round the answer to two decimal places.

Hint

a Determine all the possible outputs of rolling two dice. Then, calculate the sum of the outputs and count how many of them are greater than or equal to
b Start by finding the number of successes, the number of times the event was fulfilled.
c Divide the number of successes by the number of trials.

Solution

a To find the theoretical probability, the number of favorable outcomes should be divided by the total number of possible outcomes.

Since the experiment consists of rolling two dice and each die has possible outcomes, there is a total of possible combinations. Calculate the sum of the outputs for each combination.

The in the second row and third column represents the event of rolling a on the first die and a on the second die. The other outcomes can be calculated in the same fashion. Next, highlight the outcomes satisfying the given event, that the sum of the dice is greater than or equal to

There is a total of favorable outcomes. With this information, the theoretical probability can be calculated.
Substitute values and simplify
Consequently, the probability of getting a sum greater than or equal to when two dice are rolled is about or Notice that the sample space of the experiment contains only outcomes. However, the outcomes are not equally likely. From the above table, there is greater chance of rolling a than rolling a
b The experimental probability of an event equals the number of successes obtained divided by the number of trials.

Therefore, to find the experimental probability obtained by Ramsha, divide the number of successes she got by the number of trials she conducted. These two numbers can be deduced from the given diagram.

As can be seen, Ramsha got successes in trials. With this information, the experimental probability can be found.
Substitute values and simplify
Consequently, for the given event, Ramsha got an experimental probability of or
c Similar to Part B, to find the experimental probability obtained by Mark, divide the number of successes he got by the number of trials he conducted. As before, this information can be found in the given diagram.
Mark got successes in trials. With this information, the experimental probability can be found.
Substitute values and simplify
Therefore, for the given event, Mark got an experimental probability of or

Probability of the Complement of an Event

The probability of drawing a club from a standard deck of cards is Knowing this, what is the probability of drawing a spade, heart, or diamond if a card is drawn randomly?

To figure it out, instead of counting the favorable outcomes, the complement rule can be used.

Complement Rule - Probability

The sum of the probability of an event and the probability of its complement is equal to

This formula is useful when calculating the probability of the complement of an event is easier than calculating the probability of the event itself. Then, the probability of the event is calculated as follows.

Proof

Complement Rule

Let be an event, be its complement, and be the sample space. By the definition of the complement, the union of an event and its complement is equal to the entire sample space. Because and represent the same event, their probabilities are equal. Since the complement of consists of the outcomes that are not in events and are disjoint. By the Addition Rule of Probability, the probability of the union is the sum of the individual probabilities of each of the events. Now the Transitive Property of Equality can be applied to the equalities. Additionally, the probability of the entire sample space is equal to By applying the Transitive Property of Equality once more, the rule is proven.

Using the Subtraction Property of Equality, the formula for the probability of is obtained.

Applying this formula, the probability of drawing a spade, heart, or diamond can be computed.

Counting Not Multiples of 5 vs. Counting Multiples of 5

Dylan cut out squares of paper and wrote a number from to on each square. He then put the papers in a bag and asked his dad to choose a paper at random.

What is the probability that Dylan's father picks a number that is not a multiple of

Hint

What is easier to count, the numbers that are not multiples of or the numbers that are multiples of Note that these sets of numbers are the complements of each other.

Solution

Start by remembering what the probability of an event is. Let be the given event, picking a number that is not a multiple of However, counting the numbers from to that are not multiples of can be tedious. In this case, it is worth considering the complement of

Event Complement
Picking a number that is not a multiple of Picking a number that is a multiple of
Counting how many numbers are multiples of is an easier task. There is a total of favorable outcomes for Knowing this, the probability of can be calculated. The probability of picking a multiple of is Applying the Complement Rule, the probability of can be computed.
Consequently, the probability that Dylan's father picks a number that is not a multiple of is

Chances of Picking the Right Door

At the beginning of this lesson, Tearrik was invited by the host of a late-night game show to pick one door out of thee to win a car. However, behind two of the doors are sheep.
a What are the chances that Tearrik will choose the winning door? Round the answer to two decimal places.
b What are the chances of choosing a door with a sheep? Round the answer to two decimal places.
c If there were doors and prizes, what is the chance that Tearrik picks a door with a sheep behind it? Round the answer to two decimal places.

Hint

a Divide the number of favorable outcomes by the number of possible outcomes.
b Use the Complement Rule.
c How many of the doors have sheep behind them?

Solution

a The probability of an event is the quotient of the number of favorable outcomes to the number of possible outcomes.

Since there are three doors, there are total possible outcomes, of which only is favorable. Knowing this, the chance that Tearrik will choose the wining door can be calculated.

b Note that the event of picking a door with a sheep is the complement of picking the winning door. Therefore, the Complement Rule can be used.

In Part A, the probability of picking the winning door was found to be about Substituting this value into the previous equation, the probability of picking a door with a sheep will be obtained.

c In this case, there are doors and prizes. This time finding a sheep is the favorable outcome.
Since there are prizes, there are doors with sheep. With this information, the required probability can be found.
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text:
# Right Cylinder

Written by Jerry Ratzlaff on . Posted in Solid Geometry

### Height of a Right Cylinder formula

$$\large{ h = \frac{V}{\pi \; r^2} }$$

Where:

$$\large{ h }$$ = height

$$\large{ r }$$ = radius

$$\large{ V }$$ = volume

$$\large{ \pi }$$ = Pi

### Lateral Surface Area of a Right Cylinder formula

$$\large{ A_l = 2\; \pi\; r\; h }$$

Where:

$$\large{ A_l }$$ = lateral surface area (side)

$$\large{ r }$$ = radius

$$\large{ h }$$ = height

### Radius of a Right Cylinder formula

$$\large{ r = \sqrt{ \frac{V}{\pi \; h} } }$$

Where:

$$\large{ r }$$ = radius

$$\large{ h }$$ = height

$$\large{ V }$$ = volume

$$\large{ \pi }$$ = Pi

### Surface Area of a Right cylinder formula

$$\large{ A_s = 2\; \pi\; r\;h+2\; \pi\; r^2 }$$

Where:

$$\large{ A_s }$$ = surface area (bottom, top, side)

$$\large{ r }$$ = radius

$$\large{ h }$$ = height

### Volume of a Right cylinder formula

$$\large{ V = \pi\; r^2\;h }$$

Where:

$$\large{ V }$$ = volume

$$\large{ r }$$ = radius

$$\large{ h }$$ = height

Tags: Equations for Volume


text:
# Effect of TiO2 Nanoparticles on Polymer-Based Bulk Heterojunction Solar Cells

## Abstract

This work addresses the enhanced power conversion efficiency $\eta$ of a hybrid organic solar cell in which the active layer comprises poly(3-hexylthiophene) (P3HT), [6,6]-phenyl-C61-butyric acid methyl ester (PCBM), and titanium dioxide (TiO2) nanoparticles. Under the illumination, the open circuit voltage increases, the fill factor is improved and the series resistance reduces as the concentration of TiO2 nanoparticles increases. Highly refractive TiO2 nanoparticles are assumed to promote the internal trapping of photons by scattering (redirecting) the light reflected from the metallic electrode in the active layer. The transport of charge carriers through the active layer is also improved due to the formation of TiO2 nanoparticles/Al interfaces. $\eta$ is doubled and the short circuit current is maximized by optimizing the blending concentration of TiO2 nanoparticles.

## Journal

• Jpn J Appl Phys

Jpn J Appl Phys 45(49), L1314-L1316, 2006-12-25

INSTITUTE OF PURE AND APPLIED PHYSICS

## References:  14

You must have a user ID to see the references.If you already have a user ID, please click "Login" to access the info.New users can click "Sign Up" to register for an user ID.

## Codes

• NII Article ID (NAID)
10018461227
• NII NACSIS-CAT ID (NCID)
AA10650595
• Text Lang
EN
• Article Type
SHO
• ISSN
0021-4922
• NDL Article ID
8590252
• NDL Source Classification
ZM35(科学技術--物理学)
• NDL Call No.
Z54-J337
• Data Source
CJP  NDL  JSAP

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text:
# Reciprocal strategies in a language without reciprocal markers

There was a very enlightening small workshop on reciprocals in Utrecht just now. I was invited to talk about reciprocals in Daakaka, which was an interesting assignment since Daakaka does not have reciprocal pronouns or verbal reciprocal morphology. Speakers do not have to distinguish between reciprocal, reflexive and regular transitive structures. There are however things they can do to facilitate, or force, reciprocal interpretations. Look at my slides to find out more.


text:
#### Function Estimation Using Data Adaptive Kernel Estimation - How Much Smoothing?

##### Kurt S. Riedel, A. Sidorenko

We determine the expected error by smoothing the data locally. Then we optimize the shape of the kernel smoother to minimize the error. Because the optimal estimator depends on the unknown function, our scheme automatically adjusts to the unknown function. By self-consistently adjusting the kernel smoother, the total estimator adapts to the data. Goodness of fit estimators select a kernel halfwidth by minimizing a function of the halfwidth which is based on the average square residual fit error: $ASR(h)$. A penalty term is included to adjust for using the same data to estimate the function and to evaluate the mean square error. Goodness of fit estimators are relatively simple to implement, but the minimum (of the goodness of fit functional) tends to be sensitive to small perturbations. To remedy this sensitivity problem, we fit the mean square error %goodness of fit functional to a two parameter model prior to determining the optimal halfwidth. Plug-in derivative estimators estimate the second derivative of the unknown function in an initial step, and then substitute this estimate into the asymptotic formula.

arrow_drop_up


text:
### Logic Colloquium: Sabrina Ouazzani (Paris)

Friday, November 9, 2018
2:00pm – 3:30pm

Storrs Campus
Oak 112

Sabrina Ouazzani (École Polytechnique, Paris)

"How to compute with an infinite time Turing machine?"

In this talk, we present infinite time Turing machines (ITTM), from the original definition of the model to some new infinite time algorithms. We will present algorithmic techniques that allow to highlight some properties of the ITTM-computable ordinals. In particular, we will study gaps in ordinal computation times, that is to say, ordinal times at which no infinite time program halts.

https://logic.uconn.edu/

Contact:


text:
**First snake equation**.

We can show that the diagrams for the snake equation are equal in terms of feasibility relations by expanding the right hand side (RHS) and showing it equals the identity on \$$X\$$.

$\left(1_X \times \cup_X \right)\left(\cap_X \times 1_X\right) = 1_X$

Each of these definitions a function from the product of their inputs and outputs to \$$\mathbf{Bool}\$$.

The first part is cap and the identity: \$$(\cap_X \times 1_X ) : 1 \times X \nrightarrow X \times X^\mathrm{op} \times X\$$
$(\cap_X \times 1_X )((\bullet,x),(x',x^\ast,x'')) = (x^\ast \le x') \ \mathrm{and} \ (x \le x'')$
where \$$\bullet \in 1\$$.

Dually, we get \$$(1_X \times \cup_X ) : X \times X^\mathrm{op} \times X \nrightarrow X \times 1\$$
$(1_X \times \cup_X )((x',x^\ast,x''),(x,\bullet)) = (x'' \le x^\ast) \ \mathrm{and} \ (x' \le x)$

Composing, we get
$(1_X \times \cup_X ) (\cap_X \times 1_X )((\bullet,x),(x,\bullet)) = x \le x'' \le x^\ast \le x' \le x$
where the composite is true only if we can find one \$$x'',x^\ast,x'\$$ in \$$X\$$ such that \$$x\le x\$$, but this is always true since we can set \$$x = x' = x'' = x^\ast\$$.

Therefore \$$(1_X \times \cup_X ) (\cap_X \times 1_X ) \simeq 1_X\$$. Equality holds if \$$X \otimes 1 = X = 1 \otimes X\$$ because \$$(1_X \times \cup_X ) (\cap_X \times 1_X ) : 1\otimes X \nrightarrow X \otimes 1 \$$.

**Second snake equation**

Same as above, but with opposite preorders (i.e., turn the arrows around).

$(1_{X^\mathrm{op}} \times \cap_{X^\mathrm{op}})((x,\bullet),(x',x^\ast,x'')) = x'' \le x^\ast \ \mathrm{and} \ x' \le x$

and

$(\cup_{X^\mathrm{op}} \times 1_{X^\mathrm{op}})((x',x^\ast,x''),(\bullet,x)) = x^\ast \le x' \ \mathrm{and} \ x \le x''$

Composing gives
$(\cup_{X^\mathrm{op}} \times 1_{X^\mathrm{op}})(1_{X^\mathrm{op}} \times \cap_{X^\mathrm{op}})((x,\bullet),(\bullet,x)) = x \le x'' \le x^\ast \le x' \le x$
for some choice of \$$x'',x^\ast,x'\$$ in \$$X\$$, but again this is always possible since we can choose \$$x = x' = x'' = x^\ast\$$.

This composite behaves the same as the identity on \$$X^\mathrm{op}\$$, the dual object of \$$X\$$.

**Edits:** Changed the name of the element in 1 from 1 to \$$\bullet\$$. Also see Yoav's comment since I only prove that the two profunctors are equal along their diagonal.


text:
Proof of "Entropy of preparation" in Von Neumann entropy

How should I prove this?

From John Preskill's quantum computation & quantum information lecture notes(chapter 5)

If a pure state is drawn randomly from the ensemble{|φx〉,px}, so that the density matrix is ρ = ∑pxx〉<φx|
Then, H(X)≥S(ρ)
where H stands for Shannon entropy of probability {px} and S stands for Von Neumann entropy.

In short, you express $|\phi_{x}\rangle$ in terms of the eigenstates of the density matrix $\rho$. Since the entropy of a mixture of states is larger than the corresponding mixture of entropies, you can eventually find:
$S(\rho)\leq H(X)+\sum_{x}p_{x}S(\rho_{x})$


text:
2019-01-21

# Building the PIXELCUBE (Part 1)

I’ve spent the last few weeks on building a cube made out of a bunch of PVC vertices and acrylic edges but with no face, where the edges are filled with over 60 individually addressable LEDs per meter. It’s not done yet, but i felt that sharing an update on why I’m doing this and why it turned out to be harder than i thought would be nice.

## Idea & Background

To be honest here, I don’t know how this really started. There is a possibility i was watching something, together with an influx of other ideas that led me to the conclusion that i just had to build a large light fixture. I talked with my friends from Sweden about building something interesting for ANDERSTORPSFESTIVALEN and i guess this is the idea i subconsciously came up with. This is in a way very representative for how my mind works, once it is clear what I want the idea is the result is almost finished inside my head, the end result is clear to me early on but mapping this onto a process of developing is the hard part.

This cube would be a huge physical light fixture that behaved like an LED wall but without the somewhat boring LED wall form factor. If you’ve ever worked with LED walls you know that after a while the 2-dimensional plate tends to become very flat, even if you build intricate shapes with mapping, as light is only projected at a 180 degree angle. There is no volume to it, meaning that perspective is hard to achieve. Since a LED wall maps to a screen, content tend to be 2-dimensional clips, pre-rendered and then applied in a manner to sync to the music. While there is some available software to do generative works, most of this caps out at 60 FPS + has a visible latency due to the processing needed to generate -> feed over HDMI -> go into processor -> get fed to wall.

I wanted to build something that’s different to this, a volumetric cube with the same LEDs that’s in a good LED wall. Having high dynamic range, fast dimming and aggressive driving speed allows for driving speeds over 240 FPS, meaning you’re able to do aggressive PoV (persistence of vision) effects and stroboscopic patterns. All of this while not projecting 2D content onto a 3D fixture, rather building something that’s inherently volumetric

Here’s an early concept around how the cube could look:

## Building this monster

Here’s the thing: I had no idea of how to construct this cube when starting. Absolutely zero knowledge about building something physical except for electronics. So in order to prototype how this would even look, I glued together an old Amazon cardboard box and hanged in my kitchen in order to make it easier to visualize what’s needed in order to build it.

I went through a couple of iterations on building materials and eventually decided to buy the acrylic rods needed and just commit to experimenting instead of trying to solve this on paper. After a long internal debate, i ended up deciding the cube would be 50 inches per side (+ some for the fittings), since 50 inches is exactly 127 cm (very pleasing). I ordered the acrylic rods and once they arrived i immediately realized i had a couple of issues.

• Turns out that these acrylic tubes are measured in outside diameter, whereas PVC pipe fittings are measured in inside diameter with a standard thickness, meaning my 1/2 rods had no great counterpart. I ended up trying to solve this problem for a fairly long time, trying different approaches with heat shrink tubing, electrical tape and some other experiments in order to adapt them with no good solution.

• No idea of how to make the LED strip float” inside the acrylic tube. I thought i could mount it on a L angle aluminium bar, but how would i fixate the bar with such tight clearances inside the fittings?

• No good way of actually fixating the rods inside the fittings. Since the cube is to be suspended at 45/45 degrees rotation, it needs to have a strong rigidity in the edges, something that friction alone can’t provide. I experimented with a bunch of different solutions to this problem with no really good outcome.

With all these problem stacking up i started to get a bit negative about this project. It was hard to solve, at least if i wanted it to be close to what i imagined it and not take shortcuts in regards to the build quality. I did not want to have multiple support wires inside the cube and I could not find a good way of actually mounting it. In times like these it turns out that an outside perspective is key. I had two good friends visiting (Love & Pajlada) and they came over to my house in order to hang out and look at the cube. Something Love immediately suggested is that i should 3D print the parts i needed, both to hold the fittings and to hold the L angle. I knew basically nothing about 3D printing except from a small part that I sent away for 3D print on a previous project. This did not stop me however, i purchased a Monoprice MP Select Mini 3D Printer V2 and started printing some test parts I modelled. Turns out this is exactly what i needed.

This just shows how important it is to have an outside perspective when getting stuck in projects. Having Love vet my ideas and add his perspective basically made this project possible. It’s a possibility that i would have come to this conclusion eventually but never as fast as i did here. The fittings are now tight and I’ve modelled in holders for bolts so i can use it as a mounting plate for the screws. On top of this the parts also hold the L angle which has the LED strip mounted on it.

After experimenting together over the weekend, we put together the cube temporarily, just feeding the strips through the rods and taping it with electrical tape in order to hold it’s form. After the first power-up it was obvious that this was i was striving after from the start:

## LED & Processing hardware

The vision puts a couple of different requirements in place, namely a driver that can generate patterns at 240 FPS and do music analysis and a LED strip that’s powerful enough to deliver on this. With these restrictions in place, i had to start looking into what could be used to build this piece. For LEDs it turned out to be rather obvious, the chip / strip that has what i need is the APA102, also referred to as the superled”. The APA102 has a lot of features in contrast to the WS2812 with the primary one being a separate Data / Clock line, meaning the timing of the data is less sensitive. You also get a much higher dimming frequency + a global intensity allowing for LUT lookup to get a higher dynamic range. There’s really no negative part about this LED except for the price.

In order to drive these LEDs i need a platform that’s fast enough to do super granular FFT analysis, has audio input, networking on-board and preferably some groundwork already done. There’s a bunch of directions that could be taken here:

• TouchDesigner

Since i wanted this to be a no hassle setup, the TouchDesigner setup goes away immediately. On top of this, TouchDesigner has a node based layout and for some reason this never plays well with my way of thinking. I’ve tried these node based UIs multiple times and never felt that they’ve been powerful compared to just writing it.

• Arduino

An Arduino is just too slow.

• Raspberry Pi

Raspberry Pi is what i ended up prototyping all this on. It’s a great device but sadly it just doesn’t hold up when you start trying to push multiple volumetric patterns at 200 FPS.

• PixelBlaze (ESP8266 & ESP32)

PixelBlaze is a great alternative. Ben Hecke has built a very neat small controller on the ESP8266 platform (with a ESP32 upgrade in the works) that’s the easiest to setup and get started with. It basically runs itself however i felt a tad limited by the CPU speed and general pattern blending. It was hard to re-use components of patterns or mix multiple patterns.

• SnickerDoodle (ARM/FPGA)

The SnickerDoodle is a good alternative but coding FPGA is a nightmare. On top of this the actual ARM core isn’t that fast, it runs a 866 MHz Dual-Core ARM Cortex-A9 which is slower than the Raspberry Pi so a lot of functionality would have to be shifted to the FPGA, even more headache. It’s also not really rapid prototyping working with FPGAs. I have one of these sitting around but this is a in case of emergency” solution.

With all this considered, this is the actual solution:

• Intel NUC with a FT232H chip attached

This is the route that I will end up taking with the finished cube. It gives me enough CPU speed in a small package, less of a headache in terms of setup and can be mounted inside enclosures. The FT232H chip speaks SPI at up to 30MHz, which is perfect since i only need 20. However, this requires me to write most of the software to drive it.

## Wiring and electrical

How does this cube even get powered? After looking at the datasheet of the LEDs it turns out they can pull up to 50mA per LED. With 75 LEDs per rod and 12 rods, that’s about 900 LEDs to drive at 5V, not exactly something you can do with a regular USB port.

$(75: pixels: x: 12: rods): x: 50: mA: ==:$45 000: mA

The PSU to drive this needs to be at minimum 45A at 5V in order to drive the LEDs at peak load. Mean Well has exactly what i need. These PSUs are commonly sighted within LED walls so I felt safe with this option. Still, pushing 45A at 5V requires a thicker cable, since the drop over distance will be substantial. Now this is a question of how the cube gets wired, since the APA102 LEDs also needs to be connected in sequence. In order to figure this out, i drew it out in Sketchup in order to visualize the connection order. Once i drew it i realized this is common graph theory issue, how can you walk all the edges in the shortest amount of trips. (entry -> a - ab - bc - cd - da - ae - ef - fg - gh - he - ef (through skip) - fb - bc (through skip) - cg - gh (through skip) - hd )

When building the rods I add an extra signal wire and 5V carrier on the back side of the L angle, that way i can create a common voltage rail for the 5V stuff and have backfeed for signal in order to wire it up as seen in the Sketchup.

In order to connect all this, i’ll use regular Molex connectors in the ends with a custom split one for 5V and serial ones for data in order to wire it all.

## Software & Design philosophy

So as mentioned in the beginning of this post, in order to break free from the 2D video mapped onto 3D textures, the cube has to support volumetric patterns. Think of volumetric patterns as resolving pixels inside an XYZ space, you could for example use a 3D model have the intersecting LEDs light up on start. The software should be able to seamlessly mix between patterns and needs input from multiple sources, such as Websocket, OSC, Audio and Time. After discussing the structure with Love & Pajlada, the processing pipeline ended up looking like this:

This is the layout of the intended processing pipeline. The pipeline is almost like a node based editor, in the sense that it uses multiple generators producing floats for every single pixel between 0.0 and 1.0, the operators you chain together with various operations is what creates the effect. You could for example use a stroboscope effect, chain it to a colorlookup that changes the RGB output to red and in that way create a red strobe, a hue shift or colorizer becomes powerful especially when using pre-defined palettes. The generator & color lookup can also make use of the FFT data and the beat in order to generate dynamic, music-driven effects with high resolution.

It helps to start with the layout of the flow, since it makes it easy to figure out what software has to be written and what tools to use. In this case i decided on Go for a couple of different reasons.

• I know Go. Use what you know, the intent here is not to build distributed software, but to make this cube look good.

• Go has great SPI support. Go supports the SPI chip on the Raspberry Pi, the FT232H chip and the APA102 led through periph.io. This makes it easy to prototype and move between different hardware stacks. I can do the prototyping on the Raspberry Pi, developing on my Mac and effortlessly move this to a Intel based Linux box without having to deal with compiling for different archs.

• Go is fast enough. Compared to Node.JS and the other languages i evaluated, Go hits the sweet spot between rapid iteration and execution speed, which is needed for achieving a solid 200 FPS.

• There are bindings for aubio and PortAudio. Getting audio in and performing analysis on it turns out to be very easy with good bindings available for these two great libraries. Aubio itself provides you with tools to analyze and make sense out of a PCM audio stream, detecting interesting data like tempo, onset, beat and loudness. On top of that you also get a good FFT implementation. PortAudio makes it easy to acquire audio across all platforms, lending into the goal of portability between platforms.

## Timeline

I’ve been working on this since late October and intend to finish this before June 30th 2019 in order to bring it to ANDERSTORPSFESTIVALEN. As you can see, at the time of writing this post the cube is not yet done. I will post updates here on the blog once more progress has been made. I will end this post with a video that I shot when we tested the cube:

Mt. Pepega
##### Next post
Building a weird lamp (Pixelcube Part 2) I needed a way to test my setup and have something small visible when i work on the Pixelcube in order to see what sort of mounting will work for


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 Mat. Zametki: Year: Volume: Issue: Page: Find

 Mat. Zametki, 1996, Volume 60, Issue 4, Pages 511–518 (Mi mz1859)

Optimal control of a nonlinear singular system with state constraints

S. Ya. Serovaĭskiĭ

Al-Farabi Kazakh National University

Abstract: A control system described by a nonlinear equation of parabolic type is considered in the situation where there may be no global solution. A particular optimal control problem subject to state constraints is studied. A proof of the existence of an optimal control is presented. The penalty method is used to obtain necessary conditions for optimal control. A proof of the convergence of this method is given. The successive approximation method is used to obtain an approximate solution for the conditions derived.

DOI: https://doi.org/10.4213/mzm1859

Full text: PDF file (160 kB)
References: PDF file   HTML file

English version:
Mathematical Notes, 1996, 60:4, 383–388

Bibliographic databases:

UDC: 517.977

Citation: S. Ya. Serovaǐskiǐ, “Optimal control of a nonlinear singular system with state constraints”, Mat. Zametki, 60:4 (1996), 511–518; Math. Notes, 60:4 (1996), 383–388

Citation in format AMSBIB
\Bibitem{Ser96} \by S.~Ya.~Serova{\v\i}ski{\v\i} \paper Optimal control of a~nonlinear singular system with state constraints \jour Mat. Zametki \yr 1996 \vol 60 \issue 4 \pages 511--518 \mathnet{http://mi.mathnet.ru/mz1859} \crossref{https://doi.org/10.4213/mzm1859} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=1619431} \zmath{https://zbmath.org/?q=an:0910.49004} \transl \jour Math. Notes \yr 1996 \vol 60 \issue 4 \pages 383--388 \crossref{https://doi.org/10.1007/BF02305421} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=A1996WN90400027}


text:
TCAP Success Grade 7 MATH Chapter 3

### TCAP Success Grade 7 MATH Chapter 3 Sample

1pt
2.

The day starts at 72◦ F. By noon, it is 23◦ warmer. At 3:00 p.m. it is 14.5◦ hotter than at noon. At 6:00 p.m. it is 7.5◦ cooler than at 3:00 p.m. At dusk it is 11.5◦ cooler than it was at 6:00 p.m. What is the temperature at dusk?

1pt
4.

Solve $4\frac{9}{10}-1\frac{1}{5}$ using the number line below.

1 pt
5.

Simplify: $\frac{52}{3}$

1pt
7.

977.62 − 128.41 =

1pt
11.

Jackson bought a 12 pound container of nails. He used 3.45 pounds for the addition on his deck and $1\frac{1}{4}$ pounds for his kids’ treehouse. If Jackson had $2\frac{1}{2}$ pounds of nails before he bought the 12 pounds, how many pounds of nails does he have now? Write your answer in the space provided and label the container with the correct amount.

lbs. of nails