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Conservation of momentum and HUP in PET

1. Sep 8, 2005

El Hombre Invisible

"When a positron emitted in beta+ decay annihilates with an electron, the combined mass of the two particles, m = 2me, is converted into an amount of energy 2mec^2 in the form of two 511 keV gamma rays. Momentum conservation requires that the two gamma ray photons are emitted in opposite directions from the point of annihilation. Positron emission tomography exploits the fact that the detection of two oppositely-directed gamma rays defines a line on which the point of annihilation occurs."

I haven't put this in college homework because this was just a side note in my textbook and is not something I imagine I will get asked about. As such, I imagine there are some details skimmed over. What these are will hopefully get cleared up with answers to the following questions. I'm an undergrad, and we've only studied QM to a relatively low level, so please don't baffle me with stuff there's no chance I'll understand.

1. If the total momentum of the photons MUST be zero, then the total momentum of the electron-positron pair must also be zero. Why do such pairs only annihilate when they have equal and opposite momenta? What would happen if an electron and positron met with unequal momenta, either in magnitude or direction?

2. If the total momentum of the pair must be equal and opposite, and we can determine the position of annihilation from the trajectories of the photons, where is the uncertainty?

3. Presupposing an answer to the last question, I assume that proximity, and so precise position, is not an issue for pair annihilation, and that maybe the pair will annihilate if they are in the same quantum state. If so, what is the meaning of the point of annihilation?

Thanks to all in advance.

2. Sep 8, 2005

vanesch

Staff Emeritus
No, this simply means that you should (for your own easiness) work in the center of gravity of the the electron-positron pair. You can always find such a center of gravity reference frame, and all that is written applies in THAT COG.

Your detector is not necessarily in that COG frame ! The uncertainty resides in the uncertainty in what is the COG frame. You could only work it backwards from your photon detections if you knew perfectly the momentum AND impact positions ; but you can't because of the HUP, so you have an intrinsic uncertainty in this calculating back of the COG frame.

See answer to 2.

cheers,
Patrick.

3. Sep 8, 2005

El Hombre Invisible

Hi Patrick. Thanks for the response.

If I follow you correctly (doubtful ), then what you're saying is that in a reference frame chosen such that the centre of gravity of the electron-positron pair is at rest, the momenta of each will be equal and opposite. That, I dig.

But momentum is conserved within a frame of reference. Surely then if the apparatus is set up such that the COG is not at rest in its frame (and this is by far the most likely situation), then the total momenta of the pair will not be zero. As such, the total momenta of the photon pair should not be zero.

The above looks like an argument but it's really a question. I can follow how you can choose your frame such that the COG is at rest, but the only frame of any import I see is the frame of the apparatus.

4. Sep 8, 2005

vanesch

Staff Emeritus
That's correct. And they will thus not be back-to-back either.

Exactly, and in that frame of the apparatus (the lab frame, say) the momenta are not exactly back-to-back.

5. Sep 8, 2005

El Hombre Invisible

Ahhh, I see. So what you're saying is the momenta of the photon pair are not likely to be exactly equal and opposite, but close enough to pin the point of annihilation down with some uncertainty?

6. Sep 8, 2005

imabug

It looks like you've read a little too deeply into the question and inserted things that aren't there.

Total momentum doesn't need to be 0...it just needs to be conserved. Positrons carry some amount of kinetic energy when emitted. Usually by the time a positron interacts with an electron, much of it's initial energy has been lost through collisions and interactions and it ends up with 0 or very little energy. The annihilation photons get emitted 180° to each other.

When the positron annihilates with an electron before losing a substantial fraction of it's energy, then conservation of momentum still requires the annihilation photons be emitted in opposite directions, but the angle becomes less than 180° Whatever momentum the positron had prior to annihilation is converted to the photons' momentum.

7. Sep 8, 2005

El Hombre Invisible

imabug, ur2l8. Patrick had it covered, thanks, and with less dubious presumptions. My suspicion that the text was oversimplifying for succintness was correct, but the issue was confused by a poster on an earlier thread I've been on stating that the total momentum of an annihilating pair had to be zero. I thought that was wrong, but when I read the quoted text I started doubting my sanity.

Anyway, I'm done here. Thanks again for the help.

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# The Even Case

In this section, we prove Theorem 1.1, that is, we are interested in proving that the graph Kn-I can be decomposed into cycles of length m when m and n are even and . The preliminaries have been completed for the proof except for the basic way in which we wish to view the graph Kn-I. We present this, along with a lemma, and then proceed to the proof of the result.

Definition 3.1   If A is a set of positive integers, then .

Definition 3.2   Let , where n is a positive even integer. Let G be the circulant graph of order n-2with connection set , and u and w be two independent vertices, neither of which are vertices of G. Define , that is, , where the vertices of are labeled uand w.

The preceding definitions imply that the graph Kn - I can be thought of as the join of a circulant graph of order n - 2 with , that is, Kn - I = H(n;A), where . Thus the edges of length (n-2)/2in the circulant together with the edge uw make up the edges of the 1-factor I.

Definition 3.3   Let uiuj be an edge of length j-i in a circulant graph of order n. The edge of length j-i diametrically opposed to uiujis the edge joining ui+n/2 and uj+n/2.

Lemma 3.4    Let m and n be positive even integers satisfying , n < 2m, and . If , where are positive integers satisfying , then .

PROOF.Label the vertices of the subcirculant with . We have if and only if . Let denote the permutation . If L is any subgraph of H(n;A), then is defined as .

Let P be the trail of length given by

Note that P uses precisely one edge of each length in A. Also note that alternate vertices starting with ua1 have strictly increasing subscripts, while alternating vertices starting with u0have strictly decreasing subscripts. Thus, the vertices of P are distinct so that P is a path. The last edge used to form P has length so that all the vertices of P lie in the interval containing u0 from to . This interval has length at most (n-4)/2. Therefore, the path is vertex-disjoint from P, and the edge of length ai in P is diametrically opposed to the edge of length ai in

Thus, we can form a cycle C of length m by taking

From the above remarks, it follows that is a partition of the edge set of H(n;A) into m-cycles.

We now present the proof of the first main theorem.

PROOF OF THEOREM 1.1. Let m and n be positive even integers with . As mentioned in the introduction, if n(n-2)/2 is an even multiple of m, then using [6,11] it is not difficult to show that Kn - I can be decomposed into m-cycles. To do so, we think of Kn-I as the wreath product , that is, replace every vertex of Kn/2 by a copy of and every edge of Kn/2 by a copy of K2,2. A path P of length m/2 in Kn/2 becomes in Kn-I, and this graph can be decomposed into two cycles of length m by the main result of [6]. The proof is completed by using one of the main results of [11] to observe that Kn/2 can be decomposed into paths of length m/2.

Because of the preceding paragraph, we may assume that n(n-2)/2 is an odd multiple of m. Next, by Lemma 2.3, we may also assume that n < 2m, say n=m+r for some positive integer r with . Now if r=0, we are done by a construction of Walecki given in [9]. Thus, we assume that 0 < r < m. We have seen that implies that , and since n(n-2)/2 is an odd multiple of m, Lemma 2.4 implies that r(r-2)/2 is an even multiple of m.

Let and label the vertices of the graph H(n;A) with , where u and w are the two vertices of (see Definition 3.2). Note that the edge joining ui and uj has length j-i or length i-j depending on which lies between 1 and (n-2)/2 modulo n-2. The proof of Theorem 1.1 proceeds as follows. Suppose that B is a subset of A such that , and that we can decompose the circulant into m-cycles. Then since and the graph H(n;A-B) can be decomposed into m-cycles by Lemma 3.4, it follows that we have a decomposition of H(n;A)into m-cycles. Thus, the rest of the proof consists of determining a convenient set B of r/2 lengths such that the circulant can be decomposed into m-cycles.

By Lemma 2.4, we know that r is congruent to 0 or 2 modulo 8, and thus the proof now breaks into these two cases.

CASE 1. SUPPOSE THAT #MATH97#. Let r = 2ea where a is odd and . Thus r(r-2)/2=2ea(2e-1a-1). Since m divides r(r-2)/2 evenly, we have that m=2da'b', where , , and . Then

n-2=m+r-2=2da'b'+2ea-2=b'(2da'+(2ea-2)/b').

Partition the vertices of the circulant H into b' segments, each with vertices. Each segment will contribute 2da' edges to an m-cycle. Define the path P0,0 by

Let , where is the permutation from the proof of Lemma 3.4. Since implies that , the vertices of P0,1 are distinct from the vertices of P0,0 except for , which is the last vertex of P0,0 and the first vertex of P0,1. Similarly, the paths

are vertex-disjoint except that the path P0,i begins at the last vertex of P0,i-1 for and P0,b'-1 ends at u0. Thus is an m-cycle and the lengths of the edges of C0 are .

The definition of P0,0 does not make sense when 2d-2a'+2=1, that is, d=2 and a'=1. In this case let

Let b=r/(2d+1a') = 2e-d-1a/a'. If b > 1, then obtain the path P1,0 by adding to the subscripts of the even vertices of P0,0, that is,

Next, obtain the paths by letting powers of act on P1,0 in the same way they acted on P0,0. Furthermore, the path P1,i begins at the last vertex of P1, i-1 for and the last vertex of P1,b'-1 is u0. Thus, is an m-cycle, where the lengths of the edges of C1 are .

Similarly, for each j with , obtain the path Pj,0 by adding to the subcripts of the even vertices of P0,0. Then the paths are obtained by letting powers of act on Pj,0. Note here that Pj,i begins at the last vertex of Pj,i-1 for and the last vertex of Pj,b'-1 is u0. Hence is an m-cycle and the lengths of the edges of Cj are .

For , let Bj denote the set of lengths that Cjuses, that is, . Let . Now the longest edge in B has length . Since 2ea < 2da'b' implies that , it follows that

Therefore, the lengths of B are distinct and hence |B|=r/2. Thus, the collection

is a partition of the edge set of the circulant into m-cycles.

CASE 2. SUPPOSE THAT #MATH137#. Let r=8k+2 for some nonnegative integer k. In this case we have r(r-2)/2 = 8k(4k+1) = 2ea(4k+1) where 2ea=8k, a is odd, and gcd(a,4k+1)=1. Then m=2da'b', where , , and . Let r/2=bb'. Then

n-2=2da'b'+2ea = 2da'(b'+2e-da/a').

Let and note that . The proof now breaks into two subcases depending on the congruence class of b' modulo 4.

SUBCASE 2.1. SUPPOSE THAT #MATH145#. Observe that because b'=1 implies that which is a contradiction. Let and note that s is odd. Consider the walk P0,0 defined by

Since , we have that the vertices of P0,0 are distinct so that P0,0 is a path. In the case that s=3, we have b'=5 and , and P0,0=u-3,u0,u2,u-2,u3. Observe that the lengths of the edges of P0,0 are . Obtain the path P0,1 as before, that is, . Since , the paths P0,0 and P0,1 are vertex-disjoint. Join the first vertex of P0,1 to the last vertex of P0,0 and note that this edge has length 1.

Again, let the powers of act on P0,0 to obtain the paths P0,0, P0,1, , P0, 2da'-1. Form the m-cycle C0by joining the last vertex of P0,i to the first vertex of P0,i+1for and joining the last vertex of P0,2da'-1to the first vertex of P0,0. Necessarily, these edges all have length 1. Let .

If b > 1, then obtain the path P1,0 from P0,0 by subtracting from the subscript of the first vertex and adding to the subscripts of the remaining odd vertices of P0,0, that is,

Then the lengths of the edges of P1,0 are . As before, let the powers of act on P1,0 to obtain the paths . Form the m-cycle C1 by joining the last vertex of P1,i to the first vertex of P1,i+1 for and joining the last vertex of P1,2da'-1 to the first vertex of P1,0. Note that these edges have length .

Similarly, for , obtain the path Pj,0 from P0,0by subtracting from the subscript of the first vertex and adding to the subscripts of the remaining odd vertices of P0,0. Again, letting the powers of act on Pj,0, we form the m-cycle Cj by joining the last vertex of Pj,i to the first vertex of Pj,i+1 for and joining the last vertex of Pj,2da'-1 to the first vertex of Pj,0. Observe that these edges have length and thus the cycle Cj has edges with lengths .

Let for , and let Then the longest edge of B has length either or . Now r= 2bb' and m = 2da'b', and since r < m, we have that 2b< 2da'. Thus, . Therefore, the lengths are all distinct. Next,

so that B consists of r/2 distinct lengths. Hence, the collection

is a partition of the edge set of the circulant into m-cycles.

SUBCASE 2.2. SUPPOSE THAT #MATH178#. Let P0,0 be the path defined by

Note that P0,0 uses exactly one edge of each length and uses no edge whose length is congruent to modulo . Consider the paths

and note that P0,i begins at vertex and ends at .

There are two special cases that must be considered. When b'=7, let

When b'=3, the condition m>r=2ea+2 forces a'=a and d=e-1 which in turn implies . The initial path P0,0=u0,u3,u-4 works in this case.

If b > 1, then obtain the path P1,0 by adding to the subscripts of the even vertices of P0,0, that is,

Next, obtain the paths by letting powers of act on P1,0 as before. Observe that these paths use edges of lengths .

Similarly, for each j with , obtain the path Pj,0by adding to the subscripts of the even vertices of P0,0. Then the paths are obtained by letting powers of act on Pj,0. For , let denote the collection of paths and Bj denote the set of lengths that uses, that is,

Observe that has b(b'-1) lengths, and thus we must add b lengths to B to use r/2 lengths. The longest edge used thus far has length or . Since r < m implies b < 2d-1a', it follows that . Thus, the lengths are distinct. Furthermore, since (n-2)/2 is an even multiple of , we have that these lengths are all less than (n-2)/2 as well. In Subcase 2.1, we saw that regardless of the congruence class of b' modulo 4. Thus B consists of b(b'-1)distinct lengths.

Recall that is the initial vertex of Pj,i and is the terminal vertex of Pj,i for and . Now we shall link the paths of to form an m-cycle. We shall use Hamilton cycles in an auxiliary circulant graph G to do so. Let the vertices of G be labeled . Suppose we are given a Hamilton cycle C in G. Arbitrarily orient the cycle C to obtain a Hamilton directed cycle . If there is an arc from vi to vk in (of length k-i), then insert an edge, necessarily of length , from the terminal vertex of Pj,i to the initial vertex of Pj,k. If we do this for each arc in , it is easy to see that the paths of are joined together to form an m-cycle. Furthermore, since the edge from the terminal vertex of Pj,i to the initial vertex of Pj,k has different length, namely , than the edge from the terminal vertex of Pj,k to the initial vertex of Pj,i, we can use the reverse orientation of to link the paths in another collection , , into an m-cycle. Observe that all the edges used to link together the paths have lengths congruent to modulo and thus none of them have been used in the construction of the families .

There are several points we need to consider. First, if a Hamilton cycle in G uses exactly one length for its edges, then it is easy to see that it can be used to link together the paths of some to form an m-cycle and, in doing so, we use exactly one length in H. However, if a Hamilton cycle in G uses different lengths for its edges, then using either orientation to link together paths of some , results in some edges of lengths congruent to modulo not being used. We handle this as follows.

J.-C. Bermond, O. Favaron and M. Maheo [3] proved that any connected circulant graph of degree 4 can be decomposed into two Hamilton cycles. So we choose two lengths s1 and s2 in G such that gcd (s1,s2)=1, which guarantees that the circulant graph with connection set is connected. We decompose it into two Hamilton cycles and orient each of them in the two possible ways. We then use these four oriented cycles to form four m-cycles in H. This then uses all the required edges in H of lengths congruent to modulo corresponding to the edges used in the Hamilton cycles. Second, when we are left with 1 or 3 families to link together, we use one or two Hamilton cycles from G with edges of the same length to complete the linking process.

We now must show that there are enough edge-disjoint Hamilton cycles in the auxiliary graph G to make the above scheme work. Recall that the number of vertices in G is 2da' and that 2da'>2b. We now proceed depending on the number of families to link together. If b=1, then let so that Gis a (2da')-cycle. In this case, the length is used in H to link into an m-cycle. If b=3, then G has at least eight vertices so that there are at least two lengths s1 and s2 relatively prime to 2da'. Then each of s1 and s2generates a Hamilton cycle in and thus we can easily link together the three families of paths into m-cycles. In this case, the lengths , and are used in H to link into m-cycles.

Finally, if b >3, then write b = 4q + 1 or b = 4q + 3 for some positive integer q. Suppose first that b = 4q + 1. Then can be used to link into an m- cycle using the length in H. Next take q successive pairs , all of which generate connected circulants, to obtain two Hamilton cycles in , for , to link together four families at a time. Then the lengths are used in H. If b = 4q + 3, then find a second integer relatively prime to 2da' to handle the two extra families, and the lengths and are used in H.

Upon obtaining all the m-cycles as described above, we then let act on each m-cycle C according to resulting in a decomposition of into m-cycles. This completes the proof of Theorem 1.1.

Brian &
2000-01-20

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# It's Totally Totative!

Define a positive integer $$n$$ to be totatively prime if the set of all positive integers less than $$n$$ that are relatively prime to $$n$$ contains no composite numbers. What is the largest totatively prime number?

For example, 9 is not totatively prime because 4 is less than 9 and is relatively prime to 9, but 4 is composite.

×


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# Making Your Open Source Project Newcomer-friendly

One reason I really like open source is that it offers a lot of great opportunities for newish programmers to get some hands-on experience with real world problems. There’s only so much one can learn from small personal projects; but in open source one often gets to tackle interesting problems on large codebases — problems which wouldn’t occur in small/personal ones. There are also valuable skills related to collaboration to be learnt.

Because of this, I care quite a bit about making projects welcoming to new contributions, and try to improve this experience on projects I’m involved in. I’ve picked up a few tricks along the way. Most of these aren’t my ideas, I’ve gleaned them from watching people like Josh Matthews, Margaret Leibovic, Mike Conley, and Joel Maher do their thing. If you’re interested, here is a post by Margaret, and here are some of Josh’s slides from a presentation, both on the same subject.

Before I get started, bear in mind that making a project “newbie-friendly” isn’t something that magically happens. Like most things, it takes effort, but often this effort can come to fruition in the form of motivated contributors helping out on your project and eventually even becoming co-maintainers. It’s really worth it!

## The simple stuff

There’s a lot of really easy stuff you can do to kickstart contributions to your own project. Most of this is obvious:

### CONTRIBUTING.md

Add a CONTRIBUTING.md file. Keep it up to date. Link to it prominently from the README. The README should also have clear and detailed instructions for building the project. These two files are different – README is for those who want to use your project (perhaps by building from sources), CONTRIBUTING is for people who want to contribute.

Mention steps for getting involved: how to find something to work on, how to send a patch/make a pull request, a checklist of things to ensure your patch/PR satisfies before submission (e.g. passing tests, commit message guidelines, etc). Additionally, include some tips and tricks (like a link to the internal documentation) that can help new contributors, links to communication channels (IRC, Slack, Gitter, whatever) and anything else you may find helpful for someone considering contributing to your project. If you use some form of issue labeling, explanations of the labeling scheme can help folks find stuff they want to work on. An overview of the directory structure can be similarly helpful.

For some examples, check out the CONTRIBUTING.md files for servo and rust-clippy.

### Maintain a list of easy bugs

More on this later, but try to use some form of tagging to mark easy bugs. I love this slide from Josh’s talk.

How to politely say f*** off

“Choose something to work on from our issue tracker.”
- every project maintainer

Most bugs on issue trackers are nontrivial, steeped with jargon, devoid of actionable information, or otherwise inaccessible to the average new contributor. It’s relatively low effort to recognize bugs which are “easy” for maintainers, but it’s a lot of work for people unfamiliar to this project to figure this out.

A simple label on GitHub is all you need in most cases. Be sure to link to it from your contributing file!

### Communicate!

Have open channels for communication. IRC is often the favorite here, though IRC is pretty alien for people getting involved in open source for the first time. If you’re using IRC, see if you can link to a web client (like Mibbit) with short instructions on how to join. Stuff like Gitter works too.

Mailing lists also work — everyone knows how to email! However, email can be intimidating to newcomers; many have a “omg I can’t ask my silly questions here!” attitude which stops them from progressing.

Explicitly inviting questions in each issue helps here. Clippy doesn’t have a mailing list or IRC channel (too small a project), but I encourage people working on new bugs to ping me on any communication channel they’d like. I’ve mentored people over email, GH issue threads, IRC, even reddit PMs, and it’s worked out fine in each case.

Often folks will PM you for help. Provide help, but encourage them to ask questions in the main venue. This has the twofold benefit of showing everyone that the main channel is open to questions, and it also helps people get quicker answers since someone else can answer if you’re not around.

### Recognition

Celebrate new contributors. Tweet about them. Mention them in blog posts. Getting a two-line patch accepted in an open source project doesn’t sound like much, but when you’re just getting started, it’s a very awesome feeling. Make it more awesome! Both This Week In Rust and This Week In Servo mention new contributors (sometimes we tweet about it too), and I’ve often got very happy messages from these contributors about the mentions.

### Add a code of conduct

A lot of folks have had bad experiences with people online, often in other open source communities and may be wary about joining others. A code of conduct is a statement that unsavory behavior won’t be tolerated, which helps make the project more welcoming and appealing to these people, simultaneously making it a nicer place which is helpful for everyone. Of course, you should be prepared to enforce the code of conduct if the situation requires it.

I use the Rust code of conduct but the Contributor Covenant is good, too. Various language/framework communities often have their own favorite code of conduct. Pick one.

### Empathize!

We often forget how hard it is to jumpstart in something we’re an active part of. For example, for many of us the process of making a pull request is almost second nature.

However, not everyone is used to these things. I’ve seen contributors who can code well but haven’t used Github in the past having lots of trouble making and updating a pull request. The same applies to other workflow things; like code review, version control1, or build system peculiarities.

Keep this in mind when dealing with new contributors. These are skills which can be picked up relatively quickly, but those without them will have a frustrating experience and end up asking you a lot of questions.

## Improving the newcomer experience

Alright, now you’ve gotten all the basics done. People now have a vague idea of how to contribute to your project. Let’s make it more concrete.

### Mentoring

Don’t just leave an easy bug open. Offer to mentor it! This is a very fun and rewarding experience, and of course contributors are more likely to stick around in a project they percieve to be helpful and welcoming.

It’s often better to go one step further and give tips for fixing the issue before anyone even picks it up (example). Communication in open source has latency — the contributor might be on the other side of the planet, or might otherwise be contributing at a different time of the day than you2. Reducing the number of back-and-forth cycles is really helpful here, and giving some info so that a contributor can get started immediately without needing to wait for a response goes a long way in improving the newcomer experience.

Avoid creating a mentored bug where you yourself aren’t certain on how to fix it. Ideally, you should know the exact steps to take to fix a bug before marking it as mentored. Don’t divulge all the steps to the mentee, but the exercise of solving the bug yourself (without writing the code) ensures that there aren’t any hidden traps.

Mostly mentorship just involves answering questions and laying out a path for the mentee. Be sure to encourage questions in the first place! A lot of people, especially students, are intimidated when joining open source and try to stay as quiet as possible. For a healthy mentorship, you want them to ask questions. A lot. Encourage this.

Remember that in many cases the new contributor may be intimidated by you. For example, I’ve often come across new Firefox contributors (who I introduced to the project) asking me questions instead of their assigned mentor because “the mentor works for Mozilla and is way too awesome for me to bug with questions”. This wasn’t something the mentor told them (Firefox mentors are all very nice and helpful people), it was a conclusion they came to on their own — one which would impede their progress on the bug.

One trick that helps mitigate this is encouraging questions in your main channel. When people PM me with questions on IRC, I answer their questions, but also encourage them to ask in the main IRC channel next time. This is good for everyone — It gives the channel an aura of being “okay to ask questions in” (if other people see that questions are being asked and answered in the channel), and it also lets other maintainers jump in to help the new contributor in case I’m not around.

Once a new contributor has fixed a bug, mentorship isn’t over – it’s just started! See if you can find something more involved for them to work in a related area of the codebase. Get to know the contributor too, a sense of familiarity goes a long way in reducing intimidation and other friction.

### Tailoring process for newcomers

Most open source projects have a set of hoops you have to jump through for a pull request to be accepted. These are necessary for the health of the project and pretty straightforward for existing contributors, but can be intimidating for new ones. They also add extra cycles of communication. I’ve often seen people put up almost-working patches, and disappear after a few cycles — even though the bulk of the work was done and there were just process issues (or code nits) left over for merging; which can be quite disheartening. Reducing extra process helps mitigate this.

For example, Servo uses this great tool called Reviewable for code review. Regular contributors don’t have much friction whilst using this, so we use it wherever possible. However, for small pull requests from new contributors I avoid using Reviewable and instead opt to review directly from the GitHub interface. For these pull requests I don’t need Reviewable’s features, so I don’t lose much, but now the contributor has to go through one less hoop.

Similarly, for rust-clippy, I often make minor fixes and run the readme update script on behalf of the contributor instead of asking them to do it themselves. I usually check out the PR locally, run git merge pr-branch --no-commit --no-ff3, make edits, commit and push. This way the PR still gets marked as merged (commit --amend doesn’t do that), and the history stays bisectable.

OpenSSL uses a mailing list for patches, however they allow contributions via GitHub too. Most seasoned contributors probably stick to the mailing list, but new contributors can use the familiar GitHub interface if they want, reducing friction.

Of course, cutting down on (necessary) process should only be done for the first one or two contributions; try to educate the newcomer about your processes as time passes.

An alternate way to tackle this issue is to go the other way around and teach process first. Give newcomers an extremely easy bug that just involves replacing a string or some other simple one-line fix, and help them push it through the process. This way, the next time they work on something, they’ll be familiar with the process and be able to devote more time to the actual code.

### Creating easy bugs

At some point down this road many projects have a problem where there are people who want to contribute, but not enough suitable easy bugs.

One technique that has helped me create a lot of easy bugs is to just look out for separable and non-critical subfeatures when working on something. There often are things like polish or other small features which you don’t need to include in the main pull request, but you do anyway because it’s a few extra seconds of work. If you think it can be split out as an easy bug, go ahead and file it!

For example, whilst working on some form issues, instead of completely implementing something, I implemented just what I needed, and filed an easy bug. This is another bug with a very similar situation; I’d implemented the framework for form submission, made it work with <input>, and filed an easy bug for wiring it up to <button>.

Sometimes you may not find a subfeature that can be split out, but you may notice something else which could be improved. This is an example of such an issue. I was working on something else, and noticed that this area of the code could be designed better. Whilst I could have fixed it myself with very little effort as part of my other changes, I made it into an easy bug instead.

Simple refactorings can be a source of easy bugs too. These require familiarity with the language, but not much more, so they’re ideal for people new to the project.

It’s also possible to take a hard bug and make it easier, either by partially implementing it, or giving enough hints (code links, explanations, etc) that the hard part is already taken care of.

Avoid making “critical” (i.e, needs to land in a week or two) features into easy bugs. Even simple changes can take a while for new contributors (especially due to the nature of asynchronous communication, lack of time, and/or getting bogged down in the process). Easy bugs should be something which you eventually want, but are okay with it taking longer to solve. It’s very disheartening for a new contributor if they are working on something and a maintainer solves it for them because it was needed to land quickly. (Given enough time this will eventually happen for some bug, in such a case see if you can provide a different bug for them to work on and apologize)

### Discoverability

Make it super easy for newcomers to find a bug they want to work on; not just any easy bug!

Bugs Ahoy and What Can I do for Mozilla are both great examples of this. Servo has servo-starters.

There are also various sites where you can list your easy bugs, some of which are listed in Emily’s post.

### Projects and more involved participation

Having easy bugs and mentoring newcomers is just one step. You probably want to have these newcomers working on harder stuff, projects, and perhaps eventually maintianing/reviewing!

For many people these steps may not necessarily require involvement from you; I’ve seen professional software developers move their way to being a maintainer with very little mentorship just because they’re experienced enough to figure out how the project works on their own.

However, many of your contributors may be students or otherwise inexperienced; indeed they may be contributing to your project to gain this experience and become better developers. Such people can become valuable members of the team with some effort.

This mostly involves nudging people towards harder bugs and/or projects. It’s also very valuable to maintain a list of “student projects” (noncritical but large bodies of work). These can be picked up by contributors or sometimes students wishing to do a project for course credit.

It’s important to try and provide a logical series of issues instead of picking things randomly around the project so that the contributor can focus on one part of the codebase while starting out. If the issues all culminate in a large feature, even better.

Joel Maher and the Mozilla Tools team have started a pretty great program called “Quarter of Contribution” which provides focused mentorship for a particular project. It seems to work out well. Programs like Google Summer of Code and Outreachy also provide ways for new contributors to try out your project at a significant level of involvement.

Creating such projects or harder bugs is a nontrivial problem, and I don’t have a clear idea on how this can be done (aside from using similar techniques as listed in the “creating easy bugs” section above). Ideas welcome!

Projects aren’t always necessary here either. Depending on the contributor, sometimes you can present them with a regular (i.e., not “easy” or otherwise earmarked) issue to work on, provide some hints, and tell them to try and figure stuff out without your help (or with less help). Stay involved, and encourage them to ask questions of others or initiate discussions, but try to stick to observing. It’s really fun to watch someone figure stuff out on their own. I did this with a contributor here, where I only provided the initial hint and the code review; as well as here, where I encouraged the contributor to initiate and direct the relevant bikeshedding on various channels without my involvement. The contributor is now more in touch with the Rust community and codebase as a result; and for me I enjoyed watching him figure stuff out and direct discussions on his own.

## Mentor! Share!

I’m still exploring these techniques myself. I’ve had great results applying some of these to clippy, and we already apply many of them to Servo. I’m slowly working on applying these techniques to Rust.

While some of these projects I’m always open to hearing about more ideas for making it easier for newcomers to contribute, so please let me know if you have any ideas or experiences to share!

Mentoring, while a lot of work, is an insanely rewarding experience, and I hope you try to incorporate it into your open source projects!

Thanks to Josh Matthews, James Graham, Emily Dunham, and Joel Maher for feedback on drafts of this post

discuss: HN

1. I’ve lost track of the number of times I’ve helped someone through git rebase and merge conflicts on Servo

2. This often happens with open source projects with paid staff – the staff is around during the workday, but the contributors are around during the evenings, so there’s less overlap.

3. There are various reasons why you should not do this, mainly because non-merge-related changes in merge commits are hard to track down. Be aware of the downsides and use this trick judiciously.


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# Differential drive robot turns on the spot

I have this differential drive robot with two wheels and a ball caster wheel. With this coefficients for the wheels:

    <mu1>1000000.0</mu1>
<mu2>1000000.0</mu2>
<fdir1>1 0 0</fdir1>
<kp>100000000.0</kp>
<kd>1.0</kd>

and this coefficients for the ball

    <mu1>0.000001</mu1>
<mu2>0.000001</mu2>
<kp>100000000.0</kp>
<kd>1.0</kd>

Now after a certain speed at around 0.7-0.8m/s they start turning on the spot, as can be seen in this video: https://youtu.be/5sMLQWsvHqE I give them random angles between -PI/8 and PI/8. But even though some angles are positive and others are negative the robots always seem to spin in one direction, once they started spinning in that direction. It's not always the same, but it always stays the same once they got into the spinning motion. The underground is grass, with a mu1 and mu2 of 0.5 I tried every possible combination of mu1 and mu2 for all the involved components, wheels, ball and ground. But no tuning of the coefficients prevents the spinning motion at around 0.7-0.8m/s

Further the wheeled robot does not change angles(it only drives a straight line), when provided negative velocities. It does change orientation but only for very small negative velocities at around 0.1 above it only drives in a straight line.

Does anybody know how to prevent this? Let me know if I should attach my robot.

edit retag close merge delete


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Metal oxide + Hydrogen When red hot iron reacts with steam to form iron(II,III) oxide and hydrogen. Which of the following is the correct word equation? This was commonly quoted in textbooks, and my sources were a degree level inorganic book, and a later A level inorganic book. Magnesium hydroxide is formed as a very thin layer on the magnesium and this tends to stop the reaction. Magnesium and steam reaction word equation 1 See answer Aura9915 is waiting for your help. 4 years ago. 3Fe(s) + 4H2O(g) steam -----> Fe3O4 (s) + 4H2(g) b) calcium with water: When magnesium ribbon reacts with steam, a solid called magnesium oxide and hydrogen gas are formed. Relevance. Method. Favorite Answer. Chemistry Chemical Reactions Chemical Reactions and Equations. Magnesium does not react with water to any significant extent. Mg +2 H2O---> Mg(OH)2 + H2. It is much easier to split the O2 molecule into bare O than it is to yank two H atoms from a water molecule, which is why we expect MgO to be more prevalent in steam versus if you just dropped the magnesium in water. go on bbc bitzize. And write any two aplications of Henry"s law Lv 7. magnesium plus steam produces magnesium oxide and hydrogen gas. Magnesium burns in steam to produce white magnesium oxide and hydrogen gas. $Mg_{(s)} + H_2O_{(g)} \rightarrow MgO_{(s)} + H_{2(g)} \label{1}$ Very clean magnesium ribbon has a mild reaction with cold water, given below. This is in contrast with calcium, immediately below magnesium in the periodic table, which does react slowly with cold water. 2 1. An equation for the reaction of: a) iron with steam When a metal reacts with steam then the products formed are metal oxide and hydrogen gas. A compound is a material in which atoms of different elements are bonded to one another. Word equation for Magnesium + Steam? Really, both answers are good ones. 4 Answers. How would you write an unbalanced equation for solid magnesium metal reacts with liquid water to form solid magnesium hydroxide and hydrogen gas? "Magnesium has a very slight reaction with cold water, but burns in steam. A very clean coil of magnesium dropped into cold water eventually gets covered in small bubbles of hydrogen which float it to the surface. Materials: magnesium ribbon, hard glass test tube, water, sand, 2 burners, short glass tube, splint, clamp stand. This is a very tricky question, and you've obviously thought about it quite a bit :). Note: Until November 2016, at this point I had the statement: "Beryllium has no reaction with water or steam even at red heat." After it burns, it forms a white powder of the magnesium oxide. This is done by heating the water until it turns to steam. Metals below magnesium in the reactivity series will not react with water unless additional energy is supplied in the form of heat. Philadelphia Cream Cheese Desserts, 365 Kidney Beans Can, 2-methyl-2-propanol Iodoform Test, Guy Meaning In Tamil, Citrus Vodka Jello Shots, Beautyrest Plush Mattress, Best Places To Live In Los Angeles For Singles, Marxist View On Religion Essay Pdf, G Lydian Scale Notes, What Goes With Hummus For Dinner, " />
Nov 28

## magnesium + steam equation

When the magnesium metal burns it reacts with oxygen found in the air to form Magnesium Oxide. 1 0. goblan. However, it reacts vigorously with steam: magnesium + steam → magnesium oxide + hydrogen . (Equation just like the magnesium one below.) Answer Save. New questions in Chemistry. Add your answer and earn points. Magnesium metal does however react with steam to give magnesium oxide (MgO) (or magnesium hydroxide, Mg(OH) 2, with excess steam) and hydrogen gas (H 2). Anonymous. Lv 6. Explanation: Magnesium + Steam→ Magnesium oxide + Hydrogen gas. Oxygen and magnesium combine in a chemical reaction to form this compound. shilpeem32 shilpeem32 Answer: Mg + 2H₂O → MgO + H₂. O magnesium + hydrogen -> magnesium … State Henry"s law. 1 decade ago. 1 decade ago. science teacher. When metals react with steam, the resultants are a metal oxide and hydrogen. Metal + steam -----> Metal oxide + Hydrogen When red hot iron reacts with steam to form iron(II,III) oxide and hydrogen. Which of the following is the correct word equation? This was commonly quoted in textbooks, and my sources were a degree level inorganic book, and a later A level inorganic book. Magnesium hydroxide is formed as a very thin layer on the magnesium and this tends to stop the reaction. Magnesium and steam reaction word equation 1 See answer Aura9915 is waiting for your help. 4 years ago. 3Fe(s) + 4H2O(g) steam -----> Fe3O4 (s) + 4H2(g) b) calcium with water: When magnesium ribbon reacts with steam, a solid called magnesium oxide and hydrogen gas are formed. Relevance. Method. Favorite Answer. Chemistry Chemical Reactions Chemical Reactions and Equations. Magnesium does not react with water to any significant extent. Mg +2 H2O---> Mg(OH)2 + H2. It is much easier to split the O2 molecule into bare O than it is to yank two H atoms from a water molecule, which is why we expect MgO to be more prevalent in steam versus if you just dropped the magnesium in water. go on bbc bitzize. And write any two aplications of Henry"s law Lv 7. magnesium plus steam produces magnesium oxide and hydrogen gas. Magnesium burns in steam to produce white magnesium oxide and hydrogen gas. $Mg_{(s)} + H_2O_{(g)} \rightarrow MgO_{(s)} + H_{2(g)} \label{1}$ Very clean magnesium ribbon has a mild reaction with cold water, given below. This is in contrast with calcium, immediately below magnesium in the periodic table, which does react slowly with cold water. 2 1. An equation for the reaction of: a) iron with steam When a metal reacts with steam then the products formed are metal oxide and hydrogen gas. A compound is a material in which atoms of different elements are bonded to one another. Word equation for Magnesium + Steam? Really, both answers are good ones. 4 Answers. How would you write an unbalanced equation for solid magnesium metal reacts with liquid water to form solid magnesium hydroxide and hydrogen gas? "Magnesium has a very slight reaction with cold water, but burns in steam. A very clean coil of magnesium dropped into cold water eventually gets covered in small bubbles of hydrogen which float it to the surface. Materials: magnesium ribbon, hard glass test tube, water, sand, 2 burners, short glass tube, splint, clamp stand. This is a very tricky question, and you've obviously thought about it quite a bit :). Note: Until November 2016, at this point I had the statement: "Beryllium has no reaction with water or steam even at red heat." After it burns, it forms a white powder of the magnesium oxide. This is done by heating the water until it turns to steam. Metals below magnesium in the reactivity series will not react with water unless additional energy is supplied in the form of heat.

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# Download The Ceh Prep Guide: The Comprehensive Guide To Certified Ethical Hacking 2007

### Download The Ceh Prep Guide: The Comprehensive Guide To Certified Ethical Hacking 2007

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text:
# Strictly positive variance gamma process?

My goal is to obtain a strictly positive variance-gamma process for the variance process such that, $$Y_{t+1} = Y_t + \mu\Delta + \sqrt{v_t\Delta}\,\,\varepsilon^y_{t+1}\\ \qquad \qquad\quad \,\,\qquad v_{t+1} = v_t + \kappa (\theta-v_t)\Delta + \sigma_v\sqrt{v_t\Delta}\,\,\varepsilon^v_{t+1} + J^v_{t+1} \\ J^v_{t+1} = \gamma G_{t+1} + \sigma \sqrt{G_{t+1}}\,\, \varepsilon^g_{t+1} \\ G_{t+1} \sim \Gamma ( \frac{\Delta}{\nu}, \nu)$$

However, $J^v_{t+1}$ has to be non-negative in order to prevent the volatility from becoming negative. Does anyone have an idea about how to obtain such a process?

## 1 Answer

The simplest way to deal with negative variance-gamma process is to override them as they arise. There are at least two ways to do this

• In the full truncation scheme, a negative value for $v_t$ is floored at zero. Hence $v_t$ is replaced by $v_t^+=\max\{0,v_t\}$ everywhere in the discretization.
• In the reflection scheme, a negative value for $v_t$ is reflected with $-v_t$ . Hence $v_t$ is replaced by $|v_t|$ everywhere in the discretization.

The disadvantage of the full truncation scheme is that it creates zero variances, which is unrealistic because $Y_t$ never exhibit zero variance.

The disadvantage of the reflection scheme is that it reflects a large negative variance to a large positive variance. Hence, it transforms realizations of low volatility into high volatility.

Yet another way is to simulate $\ln v_t$ or $\sqrt{v_t}$ and then exponentiate or square the result.


text:
# New arcane rules of trig!

1. Sep 17, 2005

### bayan

does anyone know what they include?

here is a prefrence.

thanks

2. Sep 17, 2005

### dduardo

Staff Emeritus
3. Sep 17, 2005

### amcavoy

Is this actually going to be taken seriously? I mean, will it be taught instead of "classical" trigonometry?

4. Sep 17, 2005

### Hurkyl

Staff Emeritus
While it's interesting, I find it hard to take it seriously.

First off, classical trigonometry (and length) has the huge advantages of being coordinate-free and additive. You can solve problems directly from the diagrams, whereas it could take a good amount of time to simply set up a problem in a coordinate-based approach.

Secondly, it touts as a "feature" that it treats corresponding angles identically. However, now the student is faced with multiple putative solutions (possibly a great many) from which 'e must distill the desired solution.

Finally, we already have well-understood tools for doing coordinate geometry: the dot and cross products. They do at least the same thing, but have the huge, giant, overwhelming advantage of being linear -- we can use much of our algebraic intuition when working with them.

5. Sep 18, 2005

### rachmaninoff

Meh. All he's done is mearly replaced $\theta$ with $\cos \theta$. He's using (normalized) scalar (inner) products, without referring to them by name. I searched the pdf for "inner" and "scalar" - is the author unaware that he's rediscovered well-known math and is claiming it as his own? Is he that stupid?

Just skimming the pdf, his "quadrance" is in no way different from "length"; and his "spread" is the 'ratio of quadrances' - i.e., the cosine, or the normalized scalar product. And there are many basic algebraic errors throughout! :grumpy:

6. Sep 18, 2005

### arildno

Besides, saying that the "quadrance" concept is more basic or natural than the distance concept because it doesn't involve square roots, is just plain silly.
Has he never heard of an unmarked ruler??

Just because analytic geometry will use square roots in order to express distances in terms of Cartesian coordinates, does not in any way change the fact that "length" or "distance" arguably remains (one of) the most fundamental concepts in geometry.

7. Sep 18, 2005

### amcavoy

"Forty-five degrees becomes a spread of 1/2, while thirty and sixty degrees become respectively spreads of 1/4 and 3/4. What could be simpler than that?"

$\sin^2{\left(45^o\right)}=\frac{1}{4}$

$\sin^2{\left(60^o\right)}=\frac{3}{4}$

I don't see why this is new.

8. Sep 18, 2005

### LeonhardEuler

The "quadrance" is just the length squared and the "spread" is defined as the ratio of the quadrance of the opposite leg of a triangle to the quadrance of its hypatenous. In other words it is $sin^2{\theta}$. I see a number of disadvantages to this system. First, you can not specify a unique point in the plane by giving its quadrance from the origin and spread from an axis. This will give four possible points. Second, it seems kind of unweildly for physical situations. If you have a partilcle moving at constant velocity, its "speed" defined as $\frac{dQ}{dt}$ depends on where you place the origin and changes as the particle moves. If you have an object rotating without the influence of external torques, then $\frac{dS}{dt}$ is constantly changing, while $\frac{d\theta}{dt}$ is constant. Third, the "spread" of two lines can not be greater than 1. In other words, no angles of greater than 90° are allowed. Triangles containing these angles must be split into two triangles to be dealt with. The "simple formula" he gives for the spread of two lines is nothing more than the magnitude of the cross-product of two vectors along these lines divided by thier magnitudes and squared.


text:
# VI Reference (G Dataflow)

Passes a VI reference you opened as a parameter to another VI. Creates a VI reference you can use to control the behavior and properties of the VI.

## Using a VI Reference Terminal

Use VI Reference terminals to pass a VI reference you opened as a parameter to another VI. Open VI Reference requires a strictly typed VI reference terminal or constant as a type specifier input when you want a strictly typed reference to a VI. In this case, the value of the reference terminal is unimportant. The function uses only the type.

To configure a VI Reference terminal navigate to the Item tab, click the button next to the Select VI Signature field, and find the VI you want to reference in the Select VI dialog box.

Where This Node Can Run:

Desktop OS: Windows

FPGA: Not supported

Web Server: Not supported in VIs that run in a web application


text:
## Advances in ab-initio theory of Multiferroics. Materials and mechanisms: modelling and understanding    [PDF]

Silvia Picozzi, Alessandro Stroppa
Within the broad class of multiferroics (compounds showing a coexistence of magnetism and ferroelectricity), we focus on the subclass of "improper electronic ferroelectrics", i.e. correlated materials where electronic degrees of freedom (such as spin, charge or orbital) drive ferroelectricity. In particular, in spin-induced ferroelectrics, there is not only a {\em coexistence} of the two intriguing magnetic and dipolar orders; rather, there is such an intimate link that one drives the other, suggesting a giant magnetoelectric coupling. Via first-principles approaches based on density functional theory, we review the microscopic mechanisms at the basis of multiferroicity in several compounds, ranging from transition metal oxides to organic multiferroics (MFs) to organic-inorganic hybrids (i.e. metal-organic frameworks, MOFs)
View original: http://arxiv.org/abs/1207.5918


text:
# Talk

Talks and seminars, possibly part of a conference or series.

## Freezing transition for the Riemann zeta function on a short interval

Posted in
Speaker:
Louis-Pierre Arguin
Affiliation:
CUNY Stony Brook
Date:
Wed, 2018-06-13 16:15 - 17:15
Location:
MPIM Lecture Hall

In this talk, we will present a proof of the freezing transition for the Riemann zeta function as conjectured by Fyodorov, Hiary & Keating. The connection with log-correlated fields will be emphasized. The problem is related to understanding moments of zeta on a typical short interval. The proof relies on techniques developed to understand the leading order of the maximum of zeta. If time permits, we will discuss the “one-step replica symmetry breaking behaviour” (1-RSB) which can be proved for a simplified model of zeta.

## Mini-course: Riemann zeta function and log-correlated fields, Part 2

Posted in
Speaker:
Jon Keating
Affiliation:
Bristol
Date:
Wed, 2018-06-13 14:00 - 15:45

For info, the Part 1 will take place in Lipschitz Saal, Uni Bonn (Endenicher Allee 60),
Tuesday 12th June, 9h00-10h45.

I will review some aspects of the theory of the Riemann zeta-function, especially those aspects relating to the value distribution of the zeta function on the critical line and the connections with Random Matrix Theory.  I will then review conjectural connections between the extreme value statistics of the zeta function and those of log-correlated Gaussian fields, recent progress towards developing a rigorous understanding of these, and open problems.

## Short talk: The Fyodorov-Bouchaud formula and Liouville conformal field theory

Posted in
Speaker:
Guillaume Rèmy
Affiliation:
ENS Paris
Date:
Wed, 2018-06-13 11:15 - 11:45
Location:
MPIM Lecture Hall

Starting from the restriction of a 2d Gaussian free field (GFF) to the unit circle one can define a Gaussian multiplicative chaos (GMC) measure whose density is formally given by the exponential of the GFF. In 2008 Fyodorov and Bouchaud conjectured an exact formula for the density of the total mass of this GMC. In this talk we will explain how to prove rigorously this formula by using the techniques of conformal field theory. The key observation is that the moments of the total mass of GMC on the circle are equal to one-point correlation functions of Liouville theory in the unit disk. The same techniques also allow to derive a similar result on the unit interval [0,1] (in collaboration with Tunan Zhu). Finally we will briefly discuss applications to random matrix theory and to the asymptotics of the maximum of the GFF.

## Mini-course: Random matrices and logarithmically correlated fields, Part 2

Posted in
Speaker:
Affiliation:
NYU
Date:
Wed, 2018-06-13 09:00 - 10:45
Location:
MPIM Lecture Hall

For info: the Part 1 will take place in Lipschitz Saal at Uni Bonn (Endenicher Allee 60), Monday 11th June, 14h00-15h45.

A connection between branching structures and characteristic polynomials of random matrices emerged in the past few years. We will illustrate this for two models of random matrices, corresponding to dimension 1 and 2 spectra: the Circular Unitary Ensemble and Ginibre random matrices. The discussed topics will include the second moment method, extrema, the Gaussian free field and Gaussian multiplicative chaos.

## Short talk: First order asymptotic of Liouville four-point correlation function and the conformal bootstrap

Posted in
Speaker:
Guillaume Baverez
Affiliation:
Cambridge
Date:
Wed, 2018-06-13 11:45 - 12:15
Location:
MPIM Lecture Hall

Based on the rigorous path integral formulation of Liouville conformal field theory introduced by David-Kupiainen-Rhodes-Vargas, we compute the first order asymptotic of the four-point correlation function on the sphere as two insertions get close together, which is expected to describe the density of vertices around the root of large random planar maps. Moreover, our results are consistent with predictions from the framework of theoretical physics known as the conformal bootstrap, featuring the DOZZ formula and logarithmic corrections in the distance between the insertions.

## Basics of the BV formalism - Part 2

Posted in
Speaker:
Michele Schiavina
Affiliation:
University of California, Berkeley/MPIM
Date:
Tue, 2018-05-29 10:15 - 12:00
Location:
MPIM Seminar Room

If the Batalin-Vilkovisky formalism is a cohomological handle for field theories in the Lagrangian formalism, a similar construction can be set up for the associated Hamiltonian picture and goes under the name of Batalin, Fradkin and Vilkovisky (BFV).
The link between the two has been made explicit recently by Cattaneo, Mnev and Reshetikhin (CMR) as a tool to treat field theories on manifolds with boundary.
In this talk I will review the basics of the BFV and CMR constructions and show how they relate to what we have done so far.

## The Hopf-invariant one problem and applications

Posted in
Speaker:
Danica Kosanovic
Affiliation:
IMPRS MPIM
Date:
Wed, 2018-05-30 16:30 - 18:00
Location:
MPIM Seminar Room
Parent event:
IMPRS seminar on K-theory

## tba

Posted in
Speaker:
Christian Pauly
Affiliation:
Nice
Date:
Thu, 2018-06-07 10:30 - 12:00
Location:
MPIM Lecture Hall

## Introduction to elliptic cohomology

Posted in
Speaker:
Augusto Stoffel
Affiliation:
MPIM
Date:
Mon, 2018-05-28 16:30 - 18:00
Location:
MPIM Lecture Hall
Parent event:
MPIM Topology Seminar

Posted in
Speaker:
Apolonia Gottwald
Affiliation:
MPIM
Date:
Fri, 2018-06-01 14:15 - 16:00
Location:
MPIM Lecture Hall

## tba

Posted in
Organiser(s):
Fedor Manin
Affiliation:
Ohio State University
Date:
Fri, 2018-06-22 14:00 - 15:00
Location:
MPIM Lecture Hall

## Classical torsion and L^2-torsion I

Posted in
Speaker:
Wolfgang Lück
Organiser(s):
Wolfgang Lück, Peter Teichner
Affiliation:
Universität Bonn
Date:
Tue, 2018-05-29 15:15 - 17:00
Location:
MPIM Lecture Hall

## Construction of harmonic maps between Riemannian manifolds

Posted in
Speaker:
Anna Siffert
Affiliation:
MPIM/Heidelberg
Date:
Mon, 2018-06-04 13:00 - 14:00
Location:
MPIM Lecture Hall

By definition, harmonic maps between Riemannian manifolds are critical points of the energy functional.
In local coordinates the condition for a map to be harmonic constitutes a system of non-linear,
elliptic partial differential equations of second order.
Clearly, solutions to such differential equations are typically such maps are hard to construct.
In this talk we present a few construction methods of harmonic maps between Riemannian manifolds.
Furthermore, we will briefly explain applications of harmonic maps to other research areas (in Mathematics).

## On a non-abelian analogue of the Polya-Vinogradov inequality

Posted in
Speaker:
C. S. Rajan
Affiliation:
TIFR/MPIM
Date:
Wed, 2018-05-30 16:30 - 17:30
Location:
MPIM Lecture Hall
Parent event:
Number theory lunch seminar

We consider an analogue of the classical Polya-Vinogradov inequality for character sums in
the context of the group GL(2) over finite fields. This is joint work with Satadal Ganguly.

## On a universal Torelli theorem for elliptic surfaces

Posted in
Date:
Fri, 2018-05-18 12:51 - 13:51
Parent event:
MPI-Oberseminar

## On a universal Torelli theorem for elliptic surfaces

Posted in
Date:
Wed, 2018-05-30 14:30 - 15:30
Parent event:
MPI-Oberseminar

Given two semistable elliptic surfaces over a smooth, projective curve $C$ defined over a field of characteristic zero or finitely generated over its prime field,  we show that any compatible family of effective isometries of the N{\'e}ron-Severi lattices of the base changed elliptic surfaces for all finite separable maps $B\to C$ arises from
an isomorphism of the elliptic surfaces. Without the effectivity hypothesis, we show that the two elliptic surfaces are isomorphic.

We also determine the group of universal automorphisms of a semistable elliptic surface. In particular, this includes showing that the Picard-Lefschetz transformations corresponding to an irreducible component of a singular fibre, can be extended as universal isometries. In the process, we get a family of  homomorphisms of the affine Weyl group associated to $\tilde{A}_{n-1}$ to that of $\tilde{A}_{dn-1}$, indexed by natural numbers $d$, which are closed under composition.

This is joint work with S. Subramanian.

## A universal Torelli-Tate theorem for elliptic surfaces

Posted in
Speaker:
C. S. Rajan
Affiliation:
TIFR/MPIM
Date:
Wed, 2018-05-30 14:30 - 15:30
Location:
MPIM Lecture Hall
Parent event:
Number theory lunch seminar

Given two semistable elliptic surfaces over a smooth, projective curve $C$ defined over a field of characteristic zero or finitely generated over its prime field,  we show that any compatible family of effective isometries of the N{\'e}ron-Severi lattices of the base changed elliptic surfaces for all finite separable maps $B\to C$ arises from
an isomorphism of the elliptic surfaces. Without the effectivity hypothesis, we show that the two elliptic surfaces are isomorphic.

We also determine the group of universal automorphisms of a semistable elliptic surface. In particular, this includes showing that the Picard-Lefschetz transformations corresponding to an irreducible component of a singular fibre, can be extended as universal isometries. In the process, we get a family of  homomorphisms of the affine Weyl group associated to $\tilde{A}_{n-1}$ to that of $\tilde{A}_{dn-1}$, indexed by natural numbers $d$, which are closed under composition.

This is joint work with S. Subramanian.

## Computation without Representation -- change of time --

Posted in
Speaker:
Dror Bar-Natan
Affiliation:
University of Toronto
Date:
Fri, 2018-05-25 15:30 - 17:00
Location:
MPIM Lecture Hall
Parent event:
Web talks in Bonn

A major part of "quantum topology" (you don't have to know what's that) is the definition and computation of various knot invariants by carrying out computations in quantum groups (you don't have to know what are these). Traditionally these computations are carried out "in a representation", but this is very slow: one has to use tensor powers of these representations, and the dimensions of powers grow exponentially fast. I will describe a direct-participation method for carrying out these computations without having to choose a representation and explain why in many ways the results are better and faster. The two key points we use are a technique for composing infinite-order "perturbed Gaussian" differential operators, and the little-known fact that every semi-simple Lie algebra can be approximated by solvable Lie algebras, where computations are easier.

This is joint work with Roland van der Veen and continues work by Rozansky and Overbay. For further details, see http://drorbn.net/b18.

## Braids and the Grothendieck-Teichmüller Group -- change of time --

Posted in
Speaker:
Dror Bar-Natan
Affiliation:
University of Toronto
Date:
Thu, 2018-05-24 16:15 - 18:00
Location:
MPIM Lecture Hall
Parent event:
Web talks in Bonn

I will explain what are associators (and why are they useful and natural) and what is the Grothendieck-Teichmüller group, and why it is completely obvious that the Grothendieck-Teichmüller group acts simply transitively on the set of all associators. Not enough will be said about how this can be used to show that "every bounded-degree associator extends", that "rational associators exist", and that "the pentagon implies the hexagon".

In a nutshell: the filtered tower of braid groups (with bells and whistles attached) is isomorphic to its associated graded, but the isomorphism is neither canonical nor unique - such an isomorphism is precisely the thing called "an associator". But the set of isomorphisms between two isomorphic objects *always* has two groups acting simply transitively on it - the group of automorphisms of the first object acting on the right, and the group of automorphisms of the second object acting on the left. In the case of associators, that first group is what Drinfel'd calls the Grothendieck-Teichmüller group GT, and the second group, isomorphic (though not canonically) to the first, is the "graded version" GRT of GT.

Almost everything I will talk about is in my old paper "On Associators and the Grothendieck-Teichmüller Group I" For further details, see http://drorbn.net/b18.

## An example of BV quantization in action

Posted in
Speaker:
Owen Gwilliam
Affiliation:
MPIM
Date:
Tue, 2018-05-22 10:15 - 12:00
Location:
MPIM Seminar Room

The purpose of this talk is to survey the machinery developed
by Costello, with an emphasis on the procedure by which one starts with
an action functional and ends up with higher algebraic structure at the
end. The running example will take as input the curved beta-gamma
system, a 2d sigma model of maps from a Riemann surface to a complex
manifold. I will sketch how to construct its BV quantization and how to
analyze its factorization algebra, which determines a sheaf of vertex
algebras known as "chiral differential operators." The talk aims to be
conversational and not very technical, as I want to communicate the big
picture.

 © MPI f. Mathematik, Bonn Impressum & Datenschutz


text:
# [texhax] special letters in tex

Sylvain Poirier trustforum at gmail.com
Thu Nov 28 15:26:53 CET 2013

Hello. I wrote a text in .tex that I publish in pdf using pdftex, and
I try to have it translated into other languages. I have problems for
recognition of some letters and alphabets by pdftex.
A big problem I face is that every time I make a search, I only get
Here are my specific questions now:

- In Serbian language there is a barred d : đ
I managed to find that I can catch this letter in tex by the command
\catcode145=13  \def^^91{{\dbar}}
However I need to find a definition of this letter \dbar, that in
other places I see called \dj.

Doing an extensive search it seems that in some latex style file there
is a definition
\def\dj{\@ae at compl{158}}

At another place I found

\ifeurotex
...
\def\dj{^^9e}

I would like to know what I can precisely put in the headers of the
file to make it work by pdftex.

Another question is, is it possible for pdftex to recognize the
cyrillic alphabet.


text:
# Properties

 Label 3.13.aq_eq_auw Base Field $\F_{13}$ Dimension $3$ Ordinary Yes $p$-rank $3$ Principally polarizable Yes Contains a Jacobian No

## Invariants

 Base field: $\F_{13}$ Dimension: $3$ L-polynomial: $( 1 - 7 x + 13 x^{2} )( 1 - 6 x + 13 x^{2} )( 1 - 3 x + 13 x^{2} )$ Frobenius angles: $\pm0.0772104791556$, $\pm0.187167041811$, $\pm0.363422825076$ Angle rank: $3$ (numerical)

This isogeny class is not simple.

## Newton polygon

This isogeny class is ordinary.

 $p$-rank: $3$ Slopes: $[0, 0, 0, 1, 1, 1]$

## Point counts

This isogeny class is principally polarizable, but does not contain a Jacobian.

 $r$ 1 2 3 4 5 6 7 8 9 10 $A(\F_{q^r})$ 616 4398240 10789402624 23377613212800 51150534515943016 112452094830577582080 247116075883767141604168 542851238517134881136140800 1192549190986974855482957436928 2619987533557127401286211408631200

 $r$ 1 2 3 4 5 6 7 8 9 10 $C(\F_{q^r})$ -2 154 2236 28658 371038 4826668 62761606 815806562 10604640748 137858065114

## Decomposition and endomorphism algebra

Endomorphism algebra over $\F_{13}$
 The isogeny class factors as 1.13.ah $\times$ 1.13.ag $\times$ 1.13.ad and its endomorphism algebra is a direct product of the endomorphism algebras for each isotypic factor. The endomorphism algebra for each factor is:
All geometric endomorphisms are defined over $\F_{13}$.

## Base change

This is a primitive isogeny class.

## Twists

Below are some of the twists of this isogeny class.
 Twist Extension Degree Common base change 3.13.ak_bq_afe $2$ (not in LMFDB) 3.13.ae_a_w $2$ (not in LMFDB) 3.13.ac_ag_cw $2$ (not in LMFDB) 3.13.c_ag_acw $2$ (not in LMFDB) 3.13.e_a_aw $2$ (not in LMFDB) 3.13.k_bq_fe $2$ (not in LMFDB) 3.13.q_eq_uw $2$ (not in LMFDB) 3.13.ah_bn_afq $3$ (not in LMFDB) 3.13.ae_m_ao $3$ (not in LMFDB)
Below is a list of all twists of this isogeny class.
 Twist Extension Degree Common base change 3.13.ak_bq_afe $2$ (not in LMFDB) 3.13.ae_a_w $2$ (not in LMFDB) 3.13.ac_ag_cw $2$ (not in LMFDB) 3.13.c_ag_acw $2$ (not in LMFDB) 3.13.e_a_aw $2$ (not in LMFDB) 3.13.k_bq_fe $2$ (not in LMFDB) 3.13.q_eq_uw $2$ (not in LMFDB) 3.13.ah_bn_afq $3$ (not in LMFDB) 3.13.ae_m_ao $3$ (not in LMFDB) 3.13.ao_dw_arg $4$ (not in LMFDB) 3.13.ai_bi_aeu $4$ (not in LMFDB) 3.13.ag_u_acu $4$ (not in LMFDB) 3.13.a_c_adg $4$ (not in LMFDB) 3.13.a_c_dg $4$ (not in LMFDB) 3.13.g_u_cu $4$ (not in LMFDB) 3.13.i_bi_eu $4$ (not in LMFDB) 3.13.o_dw_rg $4$ (not in LMFDB) 3.13.ao_dy_arm $6$ (not in LMFDB) 3.13.al_cx_amk $6$ (not in LMFDB) 3.13.ai_bk_aeo $6$ (not in LMFDB) 3.13.af_bb_adq $6$ (not in LMFDB) 3.13.ac_g_bm $6$ (not in LMFDB) 3.13.ab_p_ack $6$ (not in LMFDB) 3.13.b_p_ck $6$ (not in LMFDB) 3.13.c_g_abm $6$ (not in LMFDB) 3.13.e_m_o $6$ (not in LMFDB) 3.13.f_bb_dq $6$ (not in LMFDB) 3.13.h_bn_fq $6$ (not in LMFDB) 3.13.i_bk_eo $6$ (not in LMFDB) 3.13.l_cx_mk $6$ (not in LMFDB) 3.13.o_dy_rm $6$ (not in LMFDB) 3.13.am_di_aoi $12$ (not in LMFDB) 3.13.aj_cn_ajy $12$ (not in LMFDB) 3.13.ag_bg_ads $12$ (not in LMFDB) 3.13.af_bl_aec $12$ (not in LMFDB) 3.13.ae_w_abs $12$ (not in LMFDB) 3.13.ad_bd_acc $12$ (not in LMFDB) 3.13.ac_q_i $12$ (not in LMFDB) 3.13.ab_z_ac $12$ (not in LMFDB) 3.13.b_z_c $12$ (not in LMFDB) 3.13.c_q_ai $12$ (not in LMFDB) 3.13.d_bd_cc $12$ (not in LMFDB) 3.13.e_w_bs $12$ (not in LMFDB) 3.13.f_bl_ec $12$ (not in LMFDB) 3.13.g_bg_ds $12$ (not in LMFDB) 3.13.j_cn_jy $12$ (not in LMFDB) 3.13.m_di_oi $12$ (not in LMFDB)


text:
# Vol. 94

Front:[PDF file] Back:[PDF file]
Latest Volume
All Volumes
All Issues
2020-11-20

#### The Maxwell Stress Tensor and Electromagnetic Momentum

By Artice Davis and Vladimir Onoochin
Progress In Electromagnetics Research Letters, Vol. 94, 151-156, 2020
doi:10.2528/PIERL20081305

## Abstract

In this paper, we discuss two well known definitions of electromagnetic momentum, ρA and \epsilon0[E x B]. We show that the former is preferable to the latter for several reasons which we will discuss. Primarily, we show in detail|and by example|that the usual manipulations used in deriving the expression \epsilon0[E x B] have a serious mathematical flaw. We follow this by presenting a succinct derivation for the former expression. We feel that the fundamental definition of electromagnetic momentum should rely upon the interaction of a single particle with the electromagnetic field. Thus, it contrasts with the definition of momentum as \epsilon0[E x B] which depends upon a (defective) integral over an entire region, usually all space.

## Citation

Artice Davis and Vladimir Onoochin, "The Maxwell Stress Tensor and Electromagnetic Momentum," Progress In Electromagnetics Research Letters, Vol. 94, 151-156, 2020.
doi:10.2528/PIERL20081305
http://www.jpier.org/PIERL/pier.php?paper=20081305

## References

1. Griffiths, D. J., "Resource letter EM-1: Electromagnetic momentum," Am. J. Phys., Vol. 80, 7, 2012.
doi:10.1119/1.3641979

2. Rousseaux, G., "On the physical meaning of the gauge conditions of Classical Electromagnetism: The hydrodynamics analogue viewpoint," Annales de la Fondation Louis de Broglie, Vol. 28, 2, 2003.

3. Semon, M. D. and J. R. Taylor, "Thoughts on the magnetic vector potential," Am. J. Phys., Vol. 64, 1361, 1996.
doi:10.1119/1.18400

4. Pfeifer, R. N. C., et al., "Momentum of an electromagnetic wave in dielectric media," Rev. Mod. Phys., Vol. 79, No. 4, 1197, 2007.
doi:10.1103/RevModPhys.79.1197

5. Griffiths, D. J., Introduction to Electrodynamics, 3rd Ed., Prentice-Hall Inc., 1999.

6. Panofsky, W. K. H. and M. Phillips, Classical Electricity and Magnetism, 2nd Ed., Ch. 14, Addison-Wesley, Reading, MA, 1962.

7. Tikhonov, A. N. and A. A. Samarski, Equations of Mathematical Physics, Ch. IV.5.5, Dover Publications, Inc., New York, 1990.

8. Darwin, C. G., "The dynamical motion of charged particles," Phil. Mag., Vol. 39, 537, 1920.
doi:10.1080/14786440508636066

9. Page, L., N. I. Adams, and Jr., "Action and reaction between moving charges," Am. J. Phys., Vol. 13, 141, 1945.
doi:10.1119/1.1990689

10. Landau, L. D. and E. M. Lifshitz, The Classical Theory of Fields, Section 65, Pergamon Press, 1971.


text:
• 文章目录 •

### 三峡工程库首区及狮子口重力滑动构造系统的构造应力场研究

1. 中国地质大学
• 收稿日期:1995-06-20 修回日期:1995-06-20 出版日期:1995-12-20 发布日期:1995-12-20

### STRESS FIELD OF THE EASTERN AREA OF THE YANGTZE GORGE RESERVOIR AND THE SHIZIKOU GRAVITATIONAL GLIDING SYSTEM

Song Lihao, Yang Sennan, Wu shuren, Ren Jianye

1. China University of Geosciences, Wuhan 430074
• Received:1995-06-20 Revised:1995-06-20 Online:1995-12-20 Published:1995-12-20

Abstract: Systematic Study on the regional Stress field in the eastern area of the Yangtze Gorge Reservoir and the local stress for the Shizikou gravitational gliding system has been carried out by stereographic and statistical analyses on numerous conjugate joints, slickensides, styolites and micostructures including fractures with their veinlets, pressure solution seams and tectonic lenses. The regional principal compressive stress (σ1 )is determined to be horizontal with two stress groups directed S-N and NEE75°－SWW250°respectively. Based upon quantitative analysis on regional paleostress field by rock rupture criterion combined with e-twin ratio and dislocation density methods, the differential stress during S-N compressive deformation is estimated at 200Mpa, whereas that during NEE75°-SWW250°compression at 80-120MPa. The cross-cutting relationships between conjugate joints, fault striae, microstructures and regional faults and folds indicate that the S-N-and NEE75°-SWW250°oriented compressions were active during the main stages of the Yanshanian and the Himalayan movements respectively. It is shown by finite-element simulation that the Yanshanian differential stress in the southern area has higher strength than that in the northern section In contrast, no significant spacial variations in differential stress have been found for the Himalayan compression. Stereographic and statistical analyses on minor structures inside the Shizikou gravitational gliding system reveal that the stress field of that system is generally characterized by front-edge compression and hack-edge extension with gently dipping σ3 towards NE SW and moderately dipping to nearly vertical σ1 at the back edge, and gently dipping σ1 towards NE-SW and moderately dipping to nearly verticalσ3 at the front edge. The strss field of the Shizikou gravitational gliding system is defined to be the local one superimposed on the regional field yielded by the main stage of the Yanshanian movement．


text: $$W=F*d*cos\theta$$ where F=force ,d=distance and $$\theta$$ is the angle between F and d so the W would be maximum when $$cos\theta$$=1 of $$\theta$$=0


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# September LeetCoding Challenge, Day 12: Combination Sum III

September 13, 2020

This is part of a series of posts about the September LeetCoding Challenge. Check the first post for more information.

The problem for September 12 is Combination Sum III. We are interested in finding all possible distinct combinations of $$k$$ numbers that add up to a number $$n$$, given that only numbers from $$1$$ to $$9$$ can be used and each combination should be a unique set of numbers. Since you can only use numbers from $$1$$ to $$9$$, both $$k$$ and $$n$$ can’t be negative; $$k$$ is at most $$9$$ and $$n$$ is at most $$\sum_{i=1}^{9} i = \frac{9 \times (9 + 1)}{2} = 45$$.

An exhaustive search using a DFS is possible given these limits, so it looks like a good candidate for a solution. The following is an implementation of that: