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## Critical phenomena in the bifurcation line's perspective    [PDF]

A. Kashuba
On the phase diagram of a system undergoing a continuous phase transition of the second order, three lines, hyper-surfaces, convergent into the critical point feature prominently: the ordered and disordered phases in the thermodynamic limit, and a third line, extending into a domain of finite-size systems, defined by the bifurcation of the distribution of the order parameter. Unlike critical phenomena in the thermodynamic limit devoid of known thermodynamic potential and described rather by the conformal symmetry, in finite-size systems near the bifurcation line an explicit Hamiltonian for the zero-mode of the order parameter is found. It bears the impress of the universality class of the critical point in terms of the two critical exponents: $\beta$ and $\nu$.
View original: http://arxiv.org/abs/1307.3826

format: plain


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# Kaggle

I recently signed up for Kaggle to participate in the Microsoft Malware Classification Challenge. On the day after I signed up, Kaggle asked me to verify my identity as a data scientist by asking me:

What is the sum of all numbers less than 4910 that are multiples of 3 or 5?

Luckily, I know my list comprehensions and this was a quick one-liner in python:

  sum([x for x in range(1,4910) if x % 3 == 0 or x % 5 == 0])

This identification strategy was the most legit that I’ve encountered (and made me laugh a bit). Good work, Kaggle.

### Written by Aaron Gonzales

I'm a typically atypical guy.

Updated


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You can evaluate the volume of a solid of revolution. A solid of revolution is created by taking a function, or part of a function, and spinning it around an axis — in most cases, either the x-axis or the y-axis.

A solid of revolution of y = 2 sin x around the x-axis.

For example, the left side of the figure shows the function y = 2 sin x between x = 0 and

Every solid of revolution has circular cross sections perpendicular to the axis of revolution. When the axis of revolution is the x-axis (or any other line that’s parallel with the x-axis), you can use the meat-slicer method directly.

However, when the axis of revolution is the y-axis (or any other line that’s parallel with the y-axis), you need to modify the problem.

To find the volume of this solid of revolution, use the meat-slicer method:

1. Find an expression that represents the area of a random cross section of the solid (in terms of x).

This cross section is a circle with a radius of 2 sin x:

2. Use this expression to build a definite integral (in terms of dx) that represents the volume of the solid.

This time, the limits of integration are from 0 to π/2:

3. Evaluate this integral by using the half-angle formula for sines:

Now evaluate:

So the volume of this solid of revolution is approximately 9.8696 cubic units.


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# Lesson 11

Side-Side-Angle (Sometimes) Congruence

## 11.1: Notice and Wonder: Congruence Fail (5 minutes)

### Warm-up

The purpose of this warm-up is to elicit the idea that not every set of three pairs of congruent corresponding parts will guarantee triangle congruence, which will be useful when students explore Side-Side-Angle Triangle Congruence in a later activity. While students may notice and wonder many things about these statements and images, the fact that the triangles are not congruent despite having so many corresponding parts congruent is the important discussion point.

### Launch

First, display the congruence statements without the image for all to see. Ask students to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice with their partner.

Next, display the image for all to see. Give students 1 minute of quiet think time, followed by a whole-class discussion.

### Student Facing

What do you notice? What do you wonder?

In triangles $$GBD$$ and $$KHI$$:

• Angle $$GBD$$ is congruent to angle $$KHI$$.
• Segment $$BD$$ is congruent to segment $$HI$$.
• Segment $$DG$$ is congruent to segment $$IK$$.

### Activity Synthesis

Ask students to share the things they noticed and wondered. Record and display their responses for all to see. If possible, record the relevant reasoning on or near the image. After all responses have been recorded without commentary or editing, ask students, “Is there anything on this list that you are wondering about now?” Encourage students to respectfully disagree, ask for clarification, or point out contradicting information. If students’ surprise at the images does not come up during the conversation, ask students to discuss this idea.

## 11.2: Dare to Be (Even More) Different (10 minutes)

### Optional activity

In this activity, students are arranged in groups of three, given side lengths and an angle measure, and encouraged to make triangles that do not look like one another’s. Given an angle and two sides that are not both adjacent to the given angle, there are three possible triangles that students can make.

As students negotiate how they will make the triangles, listen for different strategies students have for making different triangles. Monitor for students who:

• changed which side was adjacent to the given angle and which side was opposite
• swung the non-adjacent side around to see if there were multiple triangles that could be made
• found that only one triangle could be made when the longer side was the non-adjacent side

### Launch

Arrange students in groups of 3.

Conversing: MLR2 Collect and Display. As students work on this activity, listen for the language students use to describe how they created different triangles given two side lengths and an angle measure. Capture student language that attempts to describe the position of the sides relative to the given angle. Write the students’ words and phrases on a visual display. As students review the visual display, ask students to revise and improve how ideas are communicated. For example, a statement such as, “I created a different triangle by switching the sides” can be improved with the statement, “I created a different triangle changing which side was adjacent to the given angle and which side was opposite.” This will help students use the mathematical language necessary to explain how to create different triangles given two sides and an angle.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness

### Student Facing

Use technology to make a triangle using the given angle and side lengths so that the given angle is not between the 2 given sides. Try to make your triangle different from the triangles created by the other people in your group.

• Angle: $$40^\circ$$
• Side length: 6 cm
• Side length: 8 cm

### Launch

Arrange students in groups of 3.

Conversing: MLR2 Collect and Display. As students work on this activity, listen for the language students use to describe how they created different triangles given two side lengths and an angle measure. Capture student language that attempts to describe the position of the sides relative to the given angle. Write the students’ words and phrases on a visual display. As students review the visual display, ask students to revise and improve how ideas are communicated. For example, a statement such as, “I created a different triangle by switching the sides” can be improved with the statement, “I created a different triangle changing which side was adjacent to the given angle and which side was opposite.” This will help students use the mathematical language necessary to explain how to create different triangles given two sides and an angle.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness

### Student Facing

Copy these segments and use them to make a triangle using the given angle so that the given angle is not between the 2 given sides. Draw your triangle on tracing paper. Try to make your triangle different from the triangles drawn by the other people in your group.

### Anticipated Misconceptions

If a group of students decides it is only possible to make one or two different triangles, encourage them to list all the possible ways to order the given pieces and check that they have tried all of them. (Angle, short side, long side or angle, long side, short side)

### Activity Synthesis

Invite students to share how many different triangles their group was able to create. Select and sequence students to share their strategies who:

• changed which side was adjacent to the given angle and which side was opposite
• swung the non-adjacent side around to see if there were multiple triangles that could be made
• found that only one triangle could be made when the longer side was the non-adjacent side

In this activity, students are noticing and conjecturing. If no students observed when one triangle could be made and when two triangles could be made, that’s fine.

Ask students whether they think any two sides and one angle measure is enough information to guarantee that any copies of that triangle will be congruent. Ask them how their ideas are informed by the warm-up and this lesson.

• No, it’s not enough information because that’s what we knew about the warm-up triangle, and those triangles weren’t congruent.
• Sometimes it’s enough information, because the Side-Angle-Side Triangle Congruence Theorem works.
• Sometimes it’s enough information; when we had the longer side opposite the given angle, we could only make one triangle.
Engagement: Develop Effort and Persistence. Break the class into small group discussion groups and then invite a representative from each group to report back to the whole class.
Supports accessibility for: Attention; Social-emotional skills

## 11.3: Ambiguously Ambiguous? (20 minutes)

### Optional activity

In the previous activity, students may have noticed that once they decided which side would be adjacent to the given angle, they could sometimes make two triangles and sometimes make one triangle. This may lead to questions about whether knowing an angle, a side, and another side (moving clockwise around the triangle) could be enough information to guarantee you can make a congruent copy of the triangle. This activity formalizes that question and answer by giving students instructions to make different triangles and exploring which instructions always lead to a single unique triangle and which are ambiguous.

Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

Identify a way for students to compare all the examples of a given triangle. For example, invite students to place all the triangles labeled $$ABC$$ in a single visual display.

Arrange students in 8 groups. Provide each group with tools to create a visual display. Assign a different card to each group.

Representation: Internalize Comprehension. Activate strategies students have for making different triangles. Allow students to use dynamic geometry software to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing

### Student Facing

Your teacher will give you some sets of information.

• For each set of information, make a triangle using that information.
• If you think you can make more than one triangle, make more than one triangle.
• If you think you can’t make any triangle, note that.

When you are confident they are accurate, create a visual display.

### Student Facing

#### Are you ready for more?

Triangle $$ABC$$ is shown. Use your straightedge and compass to construct a new point $$D$$ on line $$AC$$ so that the length of segment $$BD$$ is the same as the length of segment $$BC$$.

Now use the straightedge and compass to construct the midpoint of $$CD$$. Label that midpoint $$M$$.

1. Explain why triangle $$ABM$$ is a right triangle.
2. Explain why knowing the angle at $$A$$ and the side lengths of $$AB$$ and $$BC$$ was not enough to define a unique triangle, but knowing the angle at $$A$$ and the side lengths of $$AB$$ and $$BM$$ would be enough to define a unique triangle.

### Activity Synthesis

Display the prompt: “When you are given that two pairs of corresponding sides are congruent, and a pair of corresponding angles that are not between the sides are congruent, that is enough to guarantee triangle congruence if $$\underline{\hspace{.5in}}$$, but not enough information if $$\underline{\hspace{.5in}}$$.”

Invite students to do a gallery walk and determine how to fill in the blanks. (The longer side is opposite the angle...the shorter side is opposite the given angle.)

Note that triangles $$GHI$$ and $$STU$$ have the same side lengths and angle measures, just with a different ordering. Comparing these two cases might help students who are struggling to see the difference.

“Only one triangle can be made, and triangle congruence is guaranteed, when you know that the longer of the two given sides is opposite the given angle.” Add this theorem to the display of triangle congruence theorems.

Speaking, Representing: MLR8 Discussion Supports. Use this routine to support whole-class discussion. As students share how to fill in the blanks of the statement, provide the class with the following sentence frames to help them respond: "I agree because _____” or "I disagree because _____.” If necessary, revoice student ideas to demonstrate mathematical language use by restating a statement as a question in order to clarify, apply appropriate language, and involve more students. For example, a statement such as, “The longer side must be opposite the angle” can be restated as the question, “If the longer side is opposite the angle, is this enough to guarantee triangle congruence?”
Design Principle(s): Support sense-making; Cultivate conversation

## Lesson Synthesis

### Lesson Synthesis

The goal of this discussion is for all students to understand this statement: “When you are given that two pairs of corresponding sides are congruent, and a pair of corresponding angles that are not between the sides are congruent, that is enough to guarantee triangle congruence if the longer side is opposite the given angle, but not enough information if the shorter side is opposite the given angle.”

Encourage students to sketch pictures of the ambiguous case, and the non-ambiguous case, and label them in ways that help them understand. Invite students to share their sketches with the class, and work as a class to annotate and understand the sketches. Make sure to discuss the case of the right triangle and how we know the hypotenuse must be the longest side.

A possible sketch might look like this:


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Limited access

Let $\vec{F}=\langle f(x,y,z), g(x,y,z), h(x,y,z)\rangle$ be a vector field, and all of its components are continuously differentiable. Determine which of the follow statements are correct:

\begin{align*} I: &\text{curl }(\nabla f)=\vec{0} \\\ II: &\langle \frac{\partial f}{\partial x}, \frac{\partial g}{\partial x}, \frac{\partial h}{\partial x}\rangle \text{ is also a vector field} \\\ III: &\frac{\partial f}{\partial y} = \frac{\partial g}{\partial x} \implies \vec{F} \text{ is conservative} \end{align*}

A

Only $I$

B

Only $II$

C

Only $III$

D

Only $I, II$

E

$I, II, III$

Select an assignment template


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# Equations satisfied by the Riemann curvature tensor

It is well known that the Riemann curvature tensor of a metric satisfies

\begin{eqnarray} R_{jikl}=-R_{ijkl}=R_{ijlk},(1)\\ R_{klij}=R_{ijkl},(2)\\ R_{i[jkl]}=0 \mbox{(1st Bianchi identity)}.(3) \end{eqnarray}

The question is whether there are more (may be non-linear and/or differential) equations satisfied by $R_{ijkl}?$ (I do not mention here the 2nd Bianchi identity $R_{ij[kl;m]}=0$ since it involves covariant derivatives and hence is equation not only on the tensor $R$ but also on the original metric $g$).

One can ask this question in the following more precise form.

Question 1. Let $R_{ijkl}$ be a collection of numbers satisfying (1)-(3) with the range of indices from 1 to $n$. Does there exist a metric $g$ in a neighborhood of 0 in $\mathbb{R}^n$ such that its curvature tensor at 0 is equal to the given collection of numbers?

Question 2. Let $R_{ijkl}(x)$ be a collection of smooth (real analytic?) functions in a neighborgood of 0 in $\mathbb{R}^n$ satisfying (1)-(3). Does there exist a metric in a neighborhood of 0 such that its curvature tensor is equal to the given collection of functions $R_{ijkl}(x)$?

UPDATE: As mentioned by Peter Michor below, the answer to Q1 is positive. As mentioned by him and Vladimir Matveev, in dimensions $n>3$, $g\mapsto R(g)$ is an overdetermined system of PDE. Hence the answer to Q2 is 'No' for $n>3$, though it is not clear to me how to write explicitly any constrain on the image of the map. In $n=2$ the answer is 'Yes' according to Vladimir Matveev.

ANOTHER UPDATE: According to Robert Bryant's answer below, in $n=3$ the answer to Q2 is 'No' in general. However under some non-degeneracy assumptions it is 'Yes'.

The answer to Q2 for $n=3$ is actually 'no, without some nondegeneracy hypotheses'. The reason is as follows:

The curvature tensor $\mathcal{R}= R_{ijkl}\,(\mathrm{d}x^i\wedge\mathrm{d}x^j)\circ(\mathrm{d}x^k\wedge\mathrm{d}x^l)$, with all its indices lowered, is a section of the subbundle $K(M)\subset S^2\bigl(\Lambda^2(T^*M)\bigr)$ that is the kernel of the natural linear mapping $$S^2\bigl(\Lambda^2(T^*M)\bigr)\longrightarrow \Lambda^4(T^*M).$$ You are asking, for a given section $\mathcal{R}$, whether there exists a metric $g$ such that $\mathrm{Riem}(g) = \mathcal{R}$. There is no pointwise algebraic condition on the section $\mathcal{R}$ imposed by this equation (that was what Q1 was about), but, as you note, there is the second Bianchi identity $$\mu(\nabla^g\mathcal{R}) = 0,$$ where $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is the natural skewsymmetrization operation. Since $\nabla^g$ depends on one derivative of the metric $g$, the above equation with a given $\mathcal{R}$ can be regarded as a first-order system of equations on $g$. When $n=3$, this is at most $3$ equations, the rank of the bundle $\Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$.

Now, for a point $p\in M$ satisfying $\mathcal{R}(p)=0$, the value $\nabla^g\mathcal{R}(p)$ does not depend on $g$ (just look at the formula for $\nabla^g\mathcal{R}$ in local coordinates). Thus, if $\mathcal{R}(p)=0$, but $\mu(\nabla^{g_0}\mathcal{R})(p) \not= 0$ for some metric $g_0$, then $\mu\bigl(\nabla^{g}\mathcal{R}(p)\bigr) \not= 0$ for all metrics $g$ and hence there is no open neighborhood of $p$ on which the equations $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ have a solution $g$. In particular, the original system $\mathrm{Riem}(g) = \mathcal{R}$ has no solution in a neighborhood of such a point $p$.

To convince yourself that such examples exist when $n=3$, just note that the rank of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$ in this case is $6$, so an arbitrary section that vanishes at $p$ will have $6\times 3 = 18$ independent first derivatives. Thus, the map $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is surjective (and the rank of the target bundle is $3$), so that the generic section of $K(M)$ that vanishes at $p$ will not satisfy the second Bianchi identity at $p$ for any metric $g$.

However, suppose that $n=3$ and that $\mathcal{R}$ is a nondegenerate section of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$. I proved (back in the early 1980s) that, when $\mathcal{R}$ is real-analytic, there always exist local solutions to the equation $\mathrm{Riem}(g) = \mathcal{R}$. Specifically, I showed that, in this case, in addition to the $6$ second-order equations that these equations represent on $g$ and the $3$ first-order equations on $g$ that $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ represents, there is one more first-order equation $Q_\mathcal{R}(g)=0$ on $g$ that is satisfied by any metric $g$ that satisfies $\mathrm{Riem}(g) = \mathcal{R}$. Then I proved that the combined overdetermined system of $6$ second-order equations and $4$ first-order equations for $g$ is involutive, so that an application of the Cartan-Kähler Theorem proves local solvability. Unfortunately, the involutive system is never either hyperbolic or elliptic, though it can be of real principal type.

I never published my proof, but later, Dennis DeTurck and Deane Yang studied the overdetermined system that I wrote down and published a proof of its local solvability in the smooth category. See Deturck and Yang, Local existence of smooth metrics with prescribed curvature, Nonlinear problems in geometry (Mobile, Ala., 1985), 37–43, Contemp. Math., 51, Amer. Math. Soc., Providence, RI, 1986.

Q1: Yes. In Riemannian normal coordinates (using the exponential map as chart), the Taylor expansion of the metric at 0 has: identity as constant part, vanishing linear part (Christoffels vanish at 0), and the quadratic part is the curvature. This is the way, Riemann found his tensor. So just write the second order polynomial as indicated, which gives a Riemannian metric near 0 with the prescribed curvature tensor at 0. It is a metric because of equations (1)-(3).

Q2: No, in general. $g\mapsto R(g)$ is an overdetermined system of PDE's; so there are integrability conditions (like the second Bianchi identity and its iterated covariant derivatives)

• To Q2: in small dimensions (n=2,3) the mapping $g\to R(g)$ is not overdetermined and at least in dimension 2 there always exists a metric with a given $R_{ijkl}$ satisfying (1,2,3) – Vladimir S Matveev Apr 7 '15 at 7:51
• @PeterMichor: I think I understand your answer to Q1. Regarding the 2nd Bianchi identity, it involves both tensors $R$ and $g$. Hence it is not an equation on $R$ only, but also on the unknown metric $g$. Hence, I think, it cannot be considered as a constrain on $R$ only. – orbits Apr 7 '15 at 8:04
• Also I agree that in higher dimensions $g\mapsto R(g)$ is overdetermined. But I do not know how to write explicitly any extra equation on the image without using the metric $g$. – orbits Apr 7 '15 at 8:30

EDIT: Attempted to insert correct constants and fix indices.

Might as well write down the explicit answer to Question 1. Everything below is written with respect to local co-ordinates.

Suppose you are given a curvature tensor $R_{ijkl}$ at one point. This satisfies the following symmetries: $$R_{ijkl} = R_{klij} = -R_{jikl} = -R_{ijlk}$$ and the first Bianchi identity $$R_{ijkl} + R_{iklj} + R_{iljk} = 0.$$ It is straightforward using the definition of the curvature tensor to show that if the Christoffel symbols vanish at the point $p$, then the curvature tensor is given (up to a not necessarily positive constant factor) at $p$ by $$R_{ijkl} = \frac{1}{2}(\partial^2_{ik}g_{jl} + \partial^2_{jl}g_{ik} - \partial^2_{il}g_{jk} - \partial^2_{jk}g_{il}).$$ It is also straightforward to check that this is satisfied (up to a constant factor), if you set $$\partial^2_{ik}g_{jl}(p) = \frac{1}{3}(R_{ijkl} + R_{ilkj}).$$ Note that this is not the unique solution. I leave the determination of the correct constant factors to the reader. Note that you do not even need to assume that $g_{ij}(p) = \delta_{ij}$. Only that $\partial_kg_{ij}(0) = 0$, which is equivalent to having the Christoffel symbols vanish at $0$.

To summarize, given any tensor $R_{ijkl}$ satisfying the symmetries stated above, any smooth metric of the form $$g_{jl}(x) = g_{jl}(0) + \frac{1}{3}(R_{ijkl} + R_{ilkj})x^ix^k + O(|x|^3)$$ has curvature tensor $R_{ijkl}$ at $x = 0$.

The answer to Q1 is YES, as explained by Michor, as metrics that realize a prescribed curvature tensor at a point can be constructed using Taylor series. However, for the sake of completeness, I think it is worth supplying a few more details regarding here (this construction is elementary, but has been quite useful to me before, and I guess others may also find it useful). It is also interesting to note that this manifold is realized as a submanifold of Euclidean space with an explicit upper bound on the codimension.

Proposition. Let $V$ be a finite-dimensional real vector space with $\dim V=n$, and let $R\colon\wedge^2 V\to \wedge^2 V$ be an algebraic curvature operator on $V$. [Or, as the OP puts it, $R$ is a collection of numbers $R_{ijkl}$ satisfying equations (1)-(3).] Then there exists a smooth $n$-dimensional submanifold $M\subset V\times\mathbb R^{k}$, $k\leq \tfrac12 n(n-1)$, whose curvature operator at $(0,0)\in M$ is given by $R$.

Proof. Since $R$ satisfies the first Bianchi identity, there exist symmetric linear operators $H_j\colon V\to V$, $1\leq j\leq k\leq \tfrac12 n(n-1)$, such that $R=-\sum_{j=1}^k (H_j\wedge H_j),$ where $(H_j\wedge H_j)(X\wedge Y):=H_j X\wedge H_j Y$, see e.g. Jacobowitz, p. 102 or Steiner, Teufel, Vilms, p. 422. Thus, $$\langle R(X\wedge Y),Z\wedge W\rangle=%-\sum_{j=1}^k \langle (H_j\wedge H_j)(X\wedge Y),Z\wedge W\rangle= \sum_{j=1}^k \langle H_j X,W\rangle\langle H_j Y,Z\rangle -\langle H_j X,Z\rangle\langle H_j Y,W\rangle.\qquad (*)$$ For each $1\leq j\leq k$, consider the quadratic functions $f_j\colon V\times\mathbb R\to\mathbb R$, given by $f_j(v,y_j):=y_j+\tfrac12\langle H_j v,v\rangle$, and set \begin{equation*} f\colon V\times\mathbb R^k\to\mathbb R^k, \quad f(v,y_1,\dots,y_k):=\big(f_1(v,y_1),\dots,f_k(v,y_k)\big). \end{equation*} Then $f(0,0)=0$ and $0\in\mathbb R^k$ is a regular value of $f$, hence $M:=f^{-1}(0)\subset V\times\mathbb R^k$ is a smooth submanifold of codimension $k$. We have $T_{(0,0)}M=V$, and the second fundamental form $II\colon V\times V\to \mathbb R^k$ of $M$ at this point is \begin{equation*} II(v,w)=-\sum_{j=1}^k \big\langle \nabla_v (\operatorname{grad} f_j),w\big\rangle\operatorname{grad} f_j=-\sum_{j=1}^k \big\langle H_j v,w\big\rangle\operatorname{grad} f_j. \end{equation*} Thus, $(*)$ is the Gauss equation of $M\subset V\times\mathbb R^k$ proving that its curvature operator at $(0,0)$, with the induced metric, is exactly $R$.


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# Anime — hermiene.net

"If a large bomb is allowed to detonate in the Engineering Section, the Marathon would be ^&2~<Colloquialism Search Error #F9C>"

These are the anime series that I've watched, and my thoughts about them. If you don't know, anime is Japanese cartoons, and its idiosyncrasies include big eyes, tons of comic relief, and exaggerated movements (characters can often jump superhumanly high and move superhumanly fast, akin to the wuxia genre in movies, but unlike wuxia, there's usually a logical reason for it). Admittedly I haven't watched much anime yet, but these are what I feel sets anime apart from Western cartoons. (This paragraph will be updated and/or corrected when I have watched more, naturally.)

Another thing that should be mentioned is that Japanese surnames come first and their given names last. So Suzuhara Misaki's name following the Western convention would be Misaki Suzuhara, but I'll use the Japanese one.

## The anime series that I have watched

### Angelic Layer

Summary
Suzuhara Misaki, a seventh grader, moves to Tokyo to live with her aunt. On her way she watches a sport called angelic layer, where robotic dolls are controlled telepathically and fight it out on a special layer (it is only on those layers that the dolls function). She becomes really inspired by watching it, and decides to enter the tournament.
Thoughts
I really sympathize with Misaki, and I really like the concept of telepathically controlled robots. As Misaki says, the reason she loves the sport is that age and size don't matter; what matters is how skillfully you can control the dolls. (Incidentally, Misaki is really clumsy in real life, as evident in the scenes featuring her trying to do gymnastics in school.) Some of the fight scenes are pretty repetitive, but this somehow manages not to be too boring (I think because of the comments she and the people around her make during the fights). The plot is continuous, so you should watch it all in order.

### Blue Submarine No.6

Summary
In the future, water levels on Earth have risen and submerged most cities, and Zorndyke, a deluded scientist, has bred an army of marine animals to wipe out humanity. The only ones who can save it are the the blue class of submarines, spearheaded by No. 6.
Thoughts
This series is only four episodes long, so there isn't that much material to comment on. However, I'm not overly excited about this one. The CG water effects are very well done, but that's about all I can say that is positive. The voice acting is really bad, the story is kind of thin, and Zorndyke is more annoying than evil-looking.

### Elfen Lied

Summary
A so-called diclonius is a mutant who is able to manipulate objects by way of what the series calls vectors, invisible “arms” that can easily slice things open or throw them (telekinesis). Scientists keep these locked up in cells, but one of them manages to escape, brutally murdering all security forces in the facility, and falling into the ocean. When she drifts ashore, her personality is split between the innocent and benevolent Nuuy, and the arrogant and violent Lucy.
Thoughts
This is a really cool series. It's extremely gory and uses nudity liberally, but since when were either of those bad things?

### Full Metal Panic!

Summary
A high school girl named Chidori Kaname is targeted for kidnapping by a terrorist organisation, and Sagara Sousuke, a member of the anti-terrorist organization Mithril, is assigned to protect her. He goes undercover and poses as a student at Jindai High School, where his military attitude coupled with his suspision for everything not-Kaname delivers some hilarious scenes.
Thoughts
Sousuke's uptight attitude and obsession with protecting Kaname and Kaname's utter frustration with this attitude and obsession somehow manages to not get boring, and is paced really well.

### Full Metal Panic? Fumoffu!

Summary
This is a companion series to the original Full Metal Panic. Its episodes are mostly stand-alone, with no serious plot to speak of; it's basically exploring the relationship between Kaname and Sousuke in an extremely humorous way.
Thoughts
I really enjoyed this.

### Full Metal Panic: The Second Raid

Summary
This is a continuation of the story from Full Metal Panic. It's much more serious than FMP and more violent.
Thoughts
I really like how most of the characters have more, uh, character than in FMP, and I really like the new characters Xia Yu Fan and Xia Yu Lan. Some of the more serious scenes, such as the paranoid Kaname ones, became a little dull, but overall I very much enjoyed it.

## Anime I want to watch

• Azumanga Daioh
• Cowboy Bebop
• Gantz
• Hand Maid May
• Hellsing
• inuyasha
• loveless
• Love Hina
• The Melancholy of Haruhi Suzumiya
• Mezzo
• Onegai Teacher
• Planetes


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# Object distance and image distance – Concave lens

The object distance is the distance from the object to the centre of the lens. This is denoted by the symbol do.

The image distance (sometimes confused with the focal length) is the distance from the image to the centre line of the lens. This is denoted by the symbol di.

With the principal focal length added, this would look like:


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and converting calories into joules or kJ? Numerous atomic models including the theory proposed by J.J Thompson and the discovery of nucleus by Ernest Rutherford had emerged. First Law of Thermodynamics introduction. For example, Gibbs free energy is quantified as joules per mole. K - Kelvin Temperature to Krypton . Thanks! We measure it using equipment, calorimeter. Energy Level Formula . See the Derivation of the Joule–Thomson coefficient below for the proof of this relation. The “calorie” you are most familiar with is used to measure the energy content of food. Formula to convert and calculate from Joules to Watts, single phase, two phase and three phase: Watts: Power unit (symbol: W). Q j = (S j/g.K)(6.86g)(193.22K) The units will cancel except for what you're looking for. Chemical formulae provide insight into the elements that constitute the molecules of a compound and also the ratio in which the atoms of these elements combine to form such molecules. What is the quantity of heat energy required to raise the temperature of 100 g of gold by 50.0 K? Calorimetry is the part of chemistry which is about the study of the quantity of heat which is absorbed or released with the surrounding during some chemical reaction. Jordan Foster Posts: 22 Joined: Fri Sep 29, 2017 2:06 pm. This example problem demonstrates how to calculate the amount of energy required to melt a sample of water ice. The SI unit of energy, work, and heat is the joule (J). The electron volt is a unit of energy suited to energies involved in atomic studies. Therefore it would be 1.6022 x 10^-16J/ keV . It has a thermometer and measures the variable called has eaten of combustions. Small & large calories.
Also, modern determinations of μ

Mixtures of carbon dioxide and methanol as the mobile phase. Learn more Accept. The symbol for joule is J. The chemical formula of a compound is a symbolic representation of its chemical composition. For example, it may be quoted in joules / gram degrees C, calories / gram degrees C or joules / mol degrees C. A calorie is an alternate unit of energy (1 calorie = 4.184 joules), grams are 1/1000 of a kilogram, and a mole (shortened to mol) is a unit used in chemistry. Work from expansion. Is there a basic formula that can be used to calculate joules? Aug 14 '18 at 14:28 Joule is the SI unit of work done. It explains how to convert joules to calories in the first problem. joule formula chemistry. We use Joules, kilograms, and meters per second as our defaults, although any appropriate units for mass (grams, ounces, etc.) DeltaT = 26.12 - 105.9 = -79.78C, you may have to convert to K, because most specific heats are in Kelvin. Once you find S, Use this formula to find Q. Q=s*m*deltaT. Thus, 1 A ⋅ V = 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. joule - SI unit of energy equal to the kinetic energy of a 1 kg mass moving at 1 m/s. Back in the early 19th century, a British brewer and physicist named James Joule demonstrated that heat and mechanical work were two forms of the same thing: energy. Joules (J) and electron volts (eV) are two common units of energy. A is an exponential factor that is a constant for a given chemical reaction, relating the frequency of collisions of particles; E a is the activation energy of the reaction (usually given in Joules per mole or J/mol) R is the universal gas constant; T is the absolute temperature (in Kelvins) Kelvin = C+273, so it would be 193.22K. Physics Formulas. One pascal, pascals are also in terms of, are also SI units, and if we convert pascals to kilograms, meters, and seconds, we get that one pascal is one kilogram per meter seconds squared. or velocity (miles per hour, millimeters per second, etc.) In other scientific contexts, the term calorie almost always refers to the small calorie. So, joules is equal to, one joule is equal to one kilogram meters squared over seconds squared. The result in joules is displayed below the ‘Calculate' and ‘Reset' buttons. As long as you use consistent units, the formula above will hold. His discovery earned him a lasting place in science history; today, the unit in which energy and heat are measured is named after him. Current Location > Physics Formulas > Electromagnetism > Joule's Law Joule's Law Don't forget to try our free app - Agile Log , which helps you track your time spent on various projects and tasks, :) Top. Joules to calories conversion Small & large calories. joules = specific heat x mass x change in T. Delta T = T(final) - T(initial) m = 6.86g. The Joule is a unit of energy. Many processes occur with energy changes in thousands of joules, so the kilojoule (kJ) is also common. p Isenthalpic plots show how the temperature of an expanding fluid changes with pressure at constant enthalpy. Formula for calculating the Volts (V) to Joules (J) E (J) = V (V) x Q (C), which means that the energy in Joules is calculated by multiplying the voltage in volts by the electrical charge in coulombs. A calorie is defined as the amount of energy needed to raise the temperature of one gram of water one degree Celsius. 0 Likes. Also I think k stands for kilo in this case. Since, Work done (W) = … Krypton is a noble gas. I read on a previous post that Q=m*c*deltaT would work, is this correct? Heat of fusion is the amount of energy in the form of heat needed to change the state of matter from a solid to a liquid (melting.) This website uses cookies to ensure you get the best experience. This example problem demonstrates how to convert joules to electron volts. In chemistry we measure heat using two different units: calories and joules. Chemical Compound Formulas; Formulas. mol −1 or J/mol) is an SI derived unit of energy per amount of material. Using the mean international ohm and volt (1.000 49 Ω, 1.000 34 V), the "international joule" is about 1.000 19 J, using ... Chemistry. The structure of the atom was the much-debated topic in the mid-1920s. Even though it is not an SI unit, it is still used in chemistry. Chemistry Formulas. Q = heat energy (Joules, J) m = mass of a substance (kg) c = specific heat (units J/kg∙K) ∆ is a symbol meaning "the change in" ∆T = change in temperature (Kelvins, K) Specific Heat Formula Questions: 1) The specific heat of gold is 129 J/kg∙K. When working with energy values typical for the atomic scale, the joule is too large of a unit to be effective. A joule is a unit of energy in the International System of Units and named after the British physicist James Prescott Joule. Power has familiar units of watts. Science Picture Co / Getty Images. For example, it takes about 4 J to warm 1 mL of H 2 O by 1°C. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Energy is measured in joules, and the amount of material is measured in moles. A joule is a tiny amount of energy. Paper Boat Creative / Getty Images. Chilling water problem. Joule: It is a unit of energy and its symbol is J. We just entered the joules unit in my basics chemistry class and I'm having trouble understanding it. Joule’s law is a mathematical description of the rate at which resistance in a circuit converts electric energy into heat energy. Therefore, the dimensional formula of Joule is the same as that of the work done. Calculating internal energy and work example. A lot of chemistry is explained by the sharing and trading of electrons between atoms. Science AP®/College Chemistry Thermodynamics Internal energy. Small calorie (cal) is the energy needed to increase 1 gram of water by 1°C at a pressure of 1 atmosphere.. Large calorie (Cal) is the energy needed to increase 1 kg of water by 1°C at a pressure of 1 atmosphere.. Large calorie is also called food calorie and is used as a unit of food energy.. How to convert from calories to joules I just need a solid jumping off point. Free Chemical Reactions calculator - Calculate chemical reactions step-by-step. Internal energy. A joule is equal to the work done when a current of 1 ampere is passed through a resistance of 1 ohm for 1 second. Dimensional Formula of Joule. In real gases, μ {\displaystyle PV} P is a constant under the assumed conditions and approximations and represents the desired isenthalp. How to convert from Joules to Watts in 1 single step: Step 1: It is a very simple conversion, so all you have to do is divide the Joules between the time which will convert the joules into watts. To answer all these questions, Joule gave a formula that describes this phenomenon precisely and called it Joule’s Law. Posted at 15:35h in Uncategorized by 0 Comments. Post by Jordan Foster » Sat Oct 14, 2017 1:09 am . This physics video tutorial explains how to convert units of heat energy. could certainly be used as well - the calculation is the same regardless. Electrons in an atom are contained in specific energy levels (1, 2, 3, and so on) that are different distances from the nucleus. Please note that formulas can be better expressed with \$\ce{}\$ for chemical formulas/equations, \$\mathrm{}\$ for math term/equations, and \$\pu\$ for units. It is also a SI derived unit of molar thermodynamic energy defined as the energy equal to one joule in one mole of substance. Please take a minute to look over the help center to better understand our guidelines and question policies $\endgroup$ – A.K. Have questions or comments? Because of their interrelationships, energy, work, and heat have the same units. Pressure-volume work . By using this website, you agree to our Cookie Policy. Joule’s Law of Heating . Re: Converting Between keV and Joules. Understanding how electrons are arranged in an atom is a building block of Chem I. The dimensional formula of Joule is given by, [M 1 L 2 T-2] Where, M = Mass; L = Length; T = Time; Derivation. But it was Neil Bohr, who asserted that electrons … 30 Oct. joule formula chemistry. How to convert energy in joules (J) to calories (cal). This is the currently selected item. s = x . Energy Level Formula. Heat and temperature. Specific heat and latent heat of fusion and vaporization. Key Takeaways: Heat of Fusion for Melting Ice. 1.6022 x 10^-19 J/eV is a ratio that is used to convert between the two units since most equations use Energy in Joules. P) is simply the product of current times voltage. TL;DR (Too Long; Didn't Read) To calculate the amount of heat released in a chemical reaction, use the equation Q = mc ΔT, where Q is the heat energy transferred (in joules), m is the mass of the liquid being heated (in kilograms), c is the specific heat capacity of the liquid (joule per kilogram degrees Celsius), and ΔT is the change in temperature of the liquid (degrees Celsius). References/ Further Reading. Its units are usually Joules per gram (J/g) or calories per gram (cal/g). How to convert joules to calories. A chemistry tutorial on converting between joules and calories formulas thermodynamic data vapor pressure of water Calories and Joules . Not all gases undergo a cooling effect upon expansion. More on internal energy. The larger the number of the energy level, the farther it is from the nucleus.

joules chemistry formula

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# Is there a ring so that any two distinct non-zero elements do not commute?

I was wondering if there was a ring so that any two distinct non-zero elements do not commute.

Formally, is there a ring $R\not=\{0\}$ so that

$$\forall x,y\in R\setminus\{0\}, x\not= y\implies xy\not=yx$$

Obviously, $\{0\}$ and $\Bbb Z/2\Bbb Z$ work but I'm searching for a non-trivial example (or a proof that there is none).

If $R$ has a $1$, then if we have $x\in R$ so that $x\not=0$ and $x\not=1$ (that is, $R\not=\Bbb Z/2\Bbb Z$), $x$ and $1$ commute.

$$\boxed{R\text{ doesn't have }1}$$

Let $x\in R$

$(x+x)x=xx+xx=x(x+x)$

We have $x=0$ or $x+x=0$ or $x+x=x$

So $x+x=0$

$$\boxed{R\text{ has characteristic }2}$$

Suppose we have a nilpotent element $x\in R$

$\exists n\in\Bbb N^*,x^{n}=0$

If $n=1$, $x=0$

If $n=2$, $x^2=0$

If $n>2$, $x^{n-1}\not=x$ but they commute

$$\boxed{\text{The square of a nilpotent element is }0}$$

(Given by Arthur in the comments)

Let $x\in R$

$x$ and $x^2$ commute so $x=0$ or $x^2=0$ or $x=x^2$

$$\boxed{\text{Every element is either nilpotent or idempotent.}}$$

• You can add that all elements are either nilpotent or idempotent by considering $x$ and $x^2$. – Arthur Mar 14 '14 at 15:44
• Perhaps the Klein 4-group with $ab=a,ba=b,a(a+b)=a^2+a=0,(a+b)a=a+b,b(a+b)=0,(a+b)b=a+b,(a+b)^2=a+b+a+b=0$? Distibutivity is automatic, and every three-way product I've checked is associative, but there are 24 or so to check and I don't see how to do it efficiently. – Kevin Carlson Mar 14 '14 at 16:17
• @KevinCarlson One way to see it quickly is that the semigroup $S=\langle a,b\mid a^2=ab=a,b^2=ba=b\rangle$, which has two elements, is certainly associative (every product is just the letter on the farthest left, and that doesn't change with association.) Then the semigroup ring $F_2[S]$ is automatically associative. – rschwieb Mar 14 '14 at 19:42

Kevin Carlson's suggestion in the comments does the job. Namely, take the $\mathbb Z/2\mathbb Z$-vector space $$R=\mathbb Z/2\mathbb Z v\oplus\mathbb Z/2\mathbb Z w$$ and define multiplication by setting $v^2=v=vw$, $w^2=w=wv$ and extending linearly to the sum. In particular, $(av+bw)(cv+dw)=a(c+d)v+b(c+d)w=(c+d)(av+bw)$ for $a,b,c,d\in\mathbb Z/ 2\mathbb Z$. This multiplication is easily checked to be associative by verifying that $$(av+bw)(cv+dw)(ev+fw)=(c+d)(e+f)(av+bw)$$ in either order of multiplication. (In fact, this is associative over any ring of coefficients.)
Though Kevin already demonstrated that $R$ satisfies: $$\forall x,y\in R\setminus\{0\}, x\not= y\implies xy\not=yx$$ but for completeness I will include that argument here. Suppose $x=av+bw\neq 0$, $y=cv+dw\neq 0$, and $xy=yx$. Then $$a(c+d)v+b(c+d)w=c(a+b)v+d(a+b)w$$ so in particular, $a(c+d)=c(a+b)$ and $d(a+b)=b(c+d)$ and thus $ad=cb$. Now if $a=0$, then since $x\neq 0$ we must have $b=1$, hence $c=0$ and $d=1$, so $x=y$. An identical argument applies to the case $b=0$, so we can assume $a=b=1$. But then $c=d=1$ so again $x=y$. Therefore, non-equal non-zero elements do not commute. (I should also note that here, we need that the coefficients are all in $\mathbb Z/2\mathbb Z$. Since $v,w$ are idempotent, it is easy to see that an arbitrary coefficient ring of character $2$ would not work from the properties listed in the OP.)
Consider an abelian (additive) group $$G =\oplus_{\alpha\in A} {\mathbb Z}/2 {\mathbb Z},$$ where $A$ is any set of cardinality $\ge 2$. The useful property that $G$ satisfies is that $x+x=0$ for every $x\in G$. Define the following product operation on $G$: For all $x, y\in G$, $$xy=y$$ unless $x=0$, in which case, of course, $xy=0$. It is then straightforward to verify that $G$ with these two binary operations is a ring where $xy=yx$ if and only if either $x=y$, or $x=0$, or $y=0$.