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The radioactive nuclei Al26 and Fe60 in the Cosmos and in the solar system - Archive ouverte HAL Access content directly
Journal Articles Publications of the Astronomical Society of Australia Year : 2021

## The radioactive nuclei Al26 and Fe60 in the Cosmos and in the solar system

(1) , , , , , , , , , , , , , , , , ,
1
Roland Diehl
• Function : Author
M. Lugaro
A. Heger
A. Sieverding
X. Tang
K. Li
E. Li
• Function : Author
C. Doherty
M. Krause
A. Wallner
N. Prantzos
H. Brinkman
J. den Hartogh
B. Wehmeyer
• Function : Author
A. Yagüe López
M. Pleintinger
• Function : Author
P. Banerjee
W. Wang

#### Abstract

Abstract The cosmic evolution of the chemical elements from the Big Bang to the present time is driven by nuclear fusion reactions inside stars and stellar explosions. A cycle of matter recurrently re-processes metal-enriched stellar ejecta into the next generation of stars. The study of cosmic nucleosynthesis and this matter cycle requires the understanding of the physics of nuclear reactions, of the conditions at which the nuclear reactions are activated inside the stars and stellar explosions, of the stellar ejection mechanisms through winds and explosions, and of the transport of the ejecta towards the next cycle, from hot plasma to cold, star-forming gas. Due to the long timescales of stellar evolution, and because of the infrequent occurrence of stellar explosions, observational studies are challenging, as they have biases in time and space as well as different sensitivities related to the various astronomical methods. Here, we describe in detail the astrophysical and nuclear-physical processes involved in creating two radioactive isotopes useful in such studies, $^{26}\mathrm{Al}$ and $^{60}\mathrm{Fe}$ . Due to their radioactive lifetime of the order of a million years, these isotopes are suitable to characterise simultaneously the processes of nuclear fusion reactions and of interstellar transport. We describe and discuss the nuclear reactions involved in the production and destruction of $^{26}\mathrm{Al}$ and $^{60}\mathrm{Fe}$ , the key characteristics of the stellar sites of their nucleosynthesis and their interstellar journey after ejection from the nucleosynthesis sites. This allows us to connect the theoretical astrophysical aspects to the variety of astronomical messengers presented here, from stardust and cosmic-ray composition measurements, through observation of $\gamma$ rays produced by radioactivity, to material deposited in deep-sea ocean crusts and to the inferred composition of the first solids that have formed in the Solar System. We show that considering measurements of the isotopic ratio of $^{26}\mathrm{Al}$ to $^{60}\mathrm{Fe}$ eliminate some of the unknowns when interpreting astronomical results, and discuss the lessons learned from these two isotopes on cosmic chemical evolution. This review paper has emerged from an ISSI-BJ Team project in 2017–2019, bringing together nuclear physicists, astronomers, and astrophysicists in this inter-disciplinary discussion.

#### Domains

Physics [physics]

### Dates and versions

hal-03876336 , version 1 (28-11-2022)

### Identifiers

• HAL Id : hal-03876336 , version 1
• ARXIV :
• DOI :

### Cite

Roland Diehl, M. Lugaro, A. Heger, A. Sieverding, X. Tang, et al.. The radioactive nuclei Al26 and Fe60 in the Cosmos and in the solar system. Publications of the Astronomical Society of Australia, 2021, 38, pp.e062. ⟨10.1017/pasa.2021.48⟩. ⟨hal-03876336⟩

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Question

# A clock, with a brass pendulum, keeps correct time at $$20^\circ C$$, but losses $$8.212 s$$ per day, when the temperature rises to $$30^\circ C$$. The coefficient of linear expansion of brass is:-

A
25×106/C
B
19×106/C
C
20×206/C
D
11×106/C

Solution

## The correct option is B $$19 \times 10^{-6}/^\circ C$$Loss of time in a day= $$\cfrac {1}{2}\times \alpha\times (t_2-t_1)\times 86400$$So, $$8.212= 1/2\times \alpha \times 10 \times 86400$$Answer is $$19 \times 10^{-16}/^oC$$Physics

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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

In survey research, our datasets nearly always comprise variables with mixed measurement levels – in particular, nominal, ordinal and continuous, or in R-speak, unordered factors, ordered factors and numeric variables. Sometimes it is useful to be able to do blanket tests of one set of variables (possibly of mixed level) against another without having to worry about which test to use.

For this we have developed an omni function which can do binary tests of significance between pairs of variables, either of which can be any of the three aforementioned levels. We have also generalised the function to include other kinds of variables such as lat/lon for GIS applications, and to distinguish between integer and continuous variables, but the version I am posting below sticks to just those three levels. Certainly one can argue about which tests are applicable in which precise case, but at least the principle might be interesting to my dear readeRs.

I will write another post soon about using this function in order to display heatmaps of significance levels.

The function returns the p value, together with attributes for the sample size and test used. It is also convenient to test if the two variables are literally the same variable. You can do this by providing your variables with an attribute “varnames”. So if attr(x,”varnames”) is the same as attr(y,”varnames”) then the function returns 1 (instead of 0, which would be the result if you hadn’t provided those attributes).

 {r} #some helper functions classer=function(x){ y=class(x)[1] s=switch(EXPR=y,"integer"="con","factor"="nom","character"="str","numeric"="con","ordered"="ord","logical"="log") s } xc=function(stri,sepp=" ") (strsplit(stri, sepp)[[1]]) #so you can type xc("red blue green") instead of c("red","blue","green") #now comes the main function xtabstat=function(v1=xy,v2,level1="nom",level2="nom",spv=F,...){ p=1 if(length(unique(v1))`


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# Union of two proper subgroups [closed]

When can a group be written as the set-theoretic union of its proper subgroups?

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Whether this is an actual homework question or not, it's not the right level for MO. Voting to close. –  Victor Protsak Nov 16 '10 at 5:56
add comment

## closed as too localized by Gjergji Zaimi, Victor Protsak, Gerry Myerson, Denis Serre, Andreas ThomNov 16 '10 at 6:45

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

## 3 Answers

The question does not match the title... As has been noted, a group $G$ is the union of proper subgroups if and only if $G$ is not cyclic. No group is the union of two proper subgroups (simple exercise often assigned as homework).

A more interesting question is: when is a group a union of $n$ proper subgroups, $n\gt 2$, but no fewer?

Theorem (Scorza) A group $G$ is the union of three proper subgroups if and only if $G$ has a quotient isomorphic to $C_2\times C_2$.

Theorem (Cohn) A group $G$ is the union of four proper subgroups and no fewer if and only if $G$ has a quotient isomorphic to $S_3$, or a quotient isomorphic to $C_3\times C_3$. A group $G$ is a union of five proper subgroups but no fewer if and only if it has a quotient isomorphic to $A_5$. A group is a union of six proper subgroups but no fewer if and only if it has a quotient isomorphic to the dihedral group of order $10$, a quotient isomorphic to $C_5\times C_5$, or a quotient isomorphic to $\langle x,y\mid x^5, y^4, x^2yx^{-1}y^{-1}\rangle$.

Theorem (Tomkinson) There are no groups that are the union of seven proper subgroups but no fewer.

I seem to remember it has been shown that for any $n\gt 2$, there is a finite set of groups $S(n)$ so that $G$ is the union of $n$ proper subgroups but no fewer if and only if $G$ has a quotient isomorphic to a group in $S(n)$. The minimal number of subgroups that cover the symmetric and alternating groups $S_k$ and $A_k$ have only been found for smallish values of $k$, though upper and lower bounds are known.

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i am confused by your notation in the last line of your theorem due to cohn: what are a and b? should they be x and y? –  Sean Tilson Nov 16 '10 at 6:44
@Sean Tilson: Oops; yes. Sorry about that. –  Arturo Magidin Nov 16 '10 at 14:38
add comment

If and only if $G$ is not cyclic. If $G$ is a non-cyclic group, you can write $G = \bigcup_{H \le G} H$. The LHS clearly contains the RHS. If $g \in G$, then $g$ generates some proper subgroup $H$ of $G$, hence $g$ is in the RHS. If $G$ is cyclic, then some $g \in G$ generates the whole group, so it can't lie in a proper subgroup.

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add comment

Almost* always - simply write $\langle g \rangle$ for the cyclic subgroup generated by $g\in G$. Then: $$G \subseteq \bigcup_{g\in G}\langle g\rangle \subseteq \bigcup_{H<G} H.$$

I'm not sure this is a research level question, which makes me sad, since it's the first time I've seen a question on MO I felt like I could honest-to-goodness provide a complete answer.

*Edit: I neglected to say, as I was "shooting from the hip" that this requires $\langle g \rangle <G$ (as the notation above should indicate) for all $g\in G$. This is true iff $G$ is not cyclic, as others have now pointed out. Thanks to Sándor for pointing this out courteously. I will note that the distinction between < and ≤ is somewhat important, and has been elsewhere neglected. I also think I would like to make it clear that I edited this response. This has also been elsewhere neglected.

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There are other places as well as other questions; you have to start somewhere. Gerhard "Ask Me About System Design" Paseman, 2010.11.15 –  Gerhard Paseman Nov 16 '10 at 5:34
Kellen, unfortunately, your answer is not entirely correct. If $G$ is cyclic and $g\in G$ is a generator, then $g$ is not contained in a proper subgroup. Otherwise what you wrote is the solution. –  Sándor Kovács Nov 16 '10 at 5:45
@Gerhard: Thank you for your comment, but I believe I can handle where and when "to start." I am quite capable of being active here and "other places" at my own discretion - even if I'm new to MO and this is an "easy" question. If I didn't assume otherwise, I might have taken this comment as "go elsewhere for easier questions to answer." If you mean well, as I do assume, I would suggest more careful word choice. What sounds like vague, sage, or cryptic advice may - in fact - also sound like a discouraging comment depending on how much the reader intends on exercising the principle of charity. –  Kellen Myers Nov 16 '10 at 16:49
It is meant to be a self-discriminatory comment: if your character is such that you take from it the message "go away", then react to it that way; if you take from it "keep trying", then react to it that way. I am willing for your character to take either or both (or other) interpretations, but what you do is your responsibility. Thank you for your acknowledgment and your observation on my choice of words. Gerhard "Ask Me About System Design" Paseman, 2010.11.16 –  Gerhard Paseman Nov 16 '10 at 19:22
Thanks Gerhard, but in the future, I would ask you not make comments about my character, or worse, comments intended to test or build my character. I do not believe my character is an appropriate topic of conversation - implicitly or explicitly - in a public form for professional collaboration & discussion. –  Kellen Myers Nov 16 '10 at 20:21
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# Yogurt

The board is 35 yogurts. How many boards of yogurts must be ordered for 886 people in the canteen?

Result

x =  26

#### Solution:

$x= 886/35 = 25.314 \doteq 26$

Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Be the first to comment!

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2. Crates 2
One crate will hold 50 oranges. If Bob needs to ship 932 oranges, how many crates will he need?
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Rudi has 4 cups of almonds. His trail mix recipe calls for 2/3 cup of almonds. How many batches of trail mix can he make?
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# Langlands dual and integrable representations

Assume I successfully classified the integrable representations of a certain semi-simple Lie group $G$. Given this information, what do I know about the integrable representations of $G^\vee$, the Langlans dual of $G$? More generally, is there any direct information I can learn about $G^\vee$ given the integrable reps. of $G$? Or is there no straightforward relation between these objects?

• I'm new to these concepts, so I hope the question is not too vague or naïve. I apologise in advance if it is. Also, I'm a physicist, so feel free to use physicists language if convenient. – AccidentalFourierTransform May 25 '18 at 23:40


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# Finding the acceleration at an angle [closed]

"What's the maximum acceleration you can achieve in a a water-slide at a 34 degree angle (If you can't use your arms and legs)"?

This is the free-body-diagram that I drew, assuming $g = 10m/s^2$:

The maximum acceleration is when there is no friction. So when I set friction to zero, the maximum acceleration is just the value of a:

$sin(34^o) = 10/a$

$a = 10 / sin(34^o) = 17.9m/s^2$

However, the book says it's $5.5m/s^2$. What am i doing wrong?

## closed as off-topic by ACuriousMind♦Sep 28 '17 at 22:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

You should start with a = g cos(90-34), the (90-34) is the angle between g and a.

Then a = 10 cos (56) = 5.5 m/s^2

so you could also go with, a = g sin (34).

I would like to point out that if you used the a as the hypotenuse this would mean that a is the resultant, which is always greater in value than its horizontal and vertical components, which is not the case here. Use g as a hypotenuse (g is the resultant) and you will get correct results.

• That doesn't seem correct. cos(56) = g / a, so a = g / cos 56. – Peatherfed Apr 7 '13 at 21:06
• Draw a line starting from the tip of arrow g on your figure perpendicular on the line a, then you will have your correct triangle. I edited my answer to give you the cause of choosing this triangle over the other. Always check which is the resultant and which is the component. Good Luck – Force Apr 7 '13 at 21:21
• Always use the resultant as the hypotenuse. An easier way to calculate this is, when there's a greater acceleration (g), reduced by being inclined (a), then (a) will always equal to g cos (x) where x is the angle between them. So all you have to do is figure out the angle between the two forces, which is causing which, and go with it. – Force Apr 7 '13 at 21:35

you did 10/sin(34) when its supposed to be multiplication so it would be 10*sin(34) which equals 5.5m/s^2


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# 2.15 Solving Circuit Systems

A six-step process for solving any linear DC circuit, with examples. 14 min read

In the previous sections building up to Kirchhoff’s Voltage Law and Kirchhoff’s Current Law we examined how to turn the structure of a circuit network into a number of equations. In this section, we’ll finally combine the structure of the network with the individual component behavior to fully setup and solve the system of equations describing a circuit.

## Component Constituent Equations

The KCL and KVL equations tell us how various currents and voltage relate to each other throughout the circuit, but they do not tell us anything about the behavior of each component itself.

For that, we need the equation for each component that describes the voltage-current relationship for that particular component.

In general, these can be just about anything: for example, they can be nonlinear equations (as we’ll see for example in diodes), they can have $\frac{d}{dt}$ in them (as we’ll see for example in capacitors and inductors), they can even reference other variables elsewhere in the circuit (as we’ll see for example in controlled sources, used in modeling transistors).

However, so far we’ve only looked at three types of simple, linear elements: ideal voltage sources, ideal current sources, and ideal resistors. All three of these have very simple constitutent equations:

\begin{align} v_7 & = 5 \ \text{Volts} &&\text{Ideal Voltage Source}\\ i_8 & = 1 \ \text{Ampere} &&\text{Ideal Current Source}\\ v_9 & = (200 \ \text{Ohms})*i_9 &&\text{Resistor} \end{align}

These are all simple linear relationships created by each type of element. The ideal voltage source is setting a particular voltage difference equal to a constant. The ideal current source is setting a branch current equal to a constant. And the resistor is a simple line through the origin.

Each two-terminal component generates a single equation. Next, we’ll combine these component constituent equations with the structural equations to solve a full circuit.

## Solving Circuits

Here is a general process that applies to any network that follows the Lumped Element Model. This process is closely related to Modified Nodal Analysis, which merges a few algebraic reduction steps into the setup process at the cost of added complexity to understand. It is essential to practice and remember this general process:

1. Label Node Voltages. Choose a ground node. Label all non-ground voltages. (For $N$ nodes we’ll have $N-1$ labeled voltages.)
2. Label Branch Currents. Indicate flow direction with an arrow. Where two or more components are strictly in series, use a single branch current variable. (For review on steps 1 and 2, see Labeling Voltages, Currents, and Nodes.)
3. Write Kirchhoff’s Current Law Equations. For each (non-ground) node, write the KCL equation from branch currents. When only two components meet at a given node and you’ve labeled only a single branch current variable for the connected components, you may simply omit this KCL equation because you’ve already included it when considering the two paths to have the same branch current. (For review, including a discussion of why we don’t write the KCL equation at the ground node, see Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL).)
4. Write Component Constituent Equations. For each (two-terminal) component, write the component-specific constituent equation for that component, which relates its voltage difference (expressed as a difference of the corresponding nodal voltages) and its branch current. Be careful to keep track of signs. Note that this step implicitly uses KVL’s definition of voltage difference. (For review, see Ideal Sources and Resistance and Ohm’s Law.)
5. Setup the System of Equations. Combine the unknown variables from steps 1 and 2 with the equations from steps 3 and 4. If you’ve done the steps properly, you should have the same number of unknowns as equations.
6. Solve the System of Equations. Find the resulting branch currents and node voltages. Use these to answer your questions of interest. (For review: apply the techniques from the Systems of Equations section to solve the simultaneous equations.)

As we’ll discuss in greater detail in future chapters, nonlinear elements are included either by solving them algebraically where possible, or via Newton’s Method of iterated relinearization as demonstrated in Linear & Nonlinear.

Note that 3+-terminal components are generally included in this formulation by modeling them as a collection of two-terminal elements. In doing so, these higher-order components may each generate several equations in step 4, and may also have multiple internal branch currents in step 2, as well as internal nodes in step 1!

## A Simple Example

First, let’s walk through all six steps on a very simple circuit with just three components in a single series loop:

Exercise Click to open and simulate the circuit above.

Step 1 is already done by labeling nodes X and Y as shown, as well as indicating the third node as a ground. This gives us two variables: $V_X$ and $V_Y$ .

Step 2 is already done by labeling current $i_1$ along with a directional arrow. As there’s only one branch current, there’s only one current variable: $i_1$ .

Step 3 is trivial for this circuit. At node X, only two devices connect, and the two current flows are labeled with the same branch current variable. This means we can omit the KCL equation. The same goes for node Y, so zero equations are generated by step 3.

Step 4 is what we’ll practice in more detail now by going through each component and writing its constitutent equation in terms of the variables we’ve already defined in steps 1 and 2.

For the voltage source V1: we have a relationship for the voltage difference at its two terminals. To use the component terminal notation we introduced in the Labeling Voltages, Currents, and Nodes section:

$$\text{V(V1.nA)} - \text{V(V1.nB)} = 12$$

In this case, the terminal V1.nA is connected to node X, and the terminal V1.nB is connected to a ground node:

$$\text{V(X)} - \text{V(ground)} = 12$$

By definition, $\text{V(ground)} = 0$ , so we have:

$$\text{V(X)} = 12$$

or in the node subscript notation we used in step 1:

$$V_X = 12$$

This is our first equation for step 4.

For the resistor R1: regardless of which terminal we choose to call terminals “nA” and “nB” of the resistor, the same relationship applies:

$$\text{V(R1.nA)} - \text{V(R1.nB)} = (100 \ \Omega) \ \text{I(R1.nA)}$$

where I(R1.nA) is defined as the current into the nA terminal.

Resistor R1 is connected between nodes X and Y. We can choose either orientation, but let’s suppose that we pick node X as R1.nA, and node Y as R1.nB. In this case, our equation becomes:

$$\text{V(X)} - \text{V(Y)} = (100 \ \Omega) \ \text{I(R1.nA)}$$

We now have to define I(R1.nA) in terms of the branch current $i_1$ . As the arrow of branch current flow is flowing into R1 from node X, we simply have $\text{I(R1.nA)} = i_1$ , so:

$$\text{V(X)} - \text{V(Y)} = (100 \ \Omega) \ i_1$$

or in the node subscript notation we used in step 1:

$$V_X - V_Y = (100 \ \Omega) \ i_1$$

This is our second equation for step 4.

For the resistor R2: again, the same relationship applies:

$$\text{V(R2.nA)} - \text{V(R2.nB)} = (100 \ \Omega) \ \text{I(R2.nA)}$$

Let’s choose node Y as R2.nA, and the ground node as R2.nB (so $\text{V(R2.nB)} = \text{V(ground)} = 0$ ). As the branch current $i_1$ flows into R2.nA, we have $\text{I(R2.nA)} = i_1$ , so:

$$\text{V(Y)} = (100 \ \Omega) \ i_1$$

or in the node subscript notation we used in step 1:

$$V_Y = (100 \ \Omega) \ i_1$$

This is our third and final equation for step 4.

Step 5 involves simply gathering and writing out all the equations from steps 3 and 4:

\begin{align} V_X & = 12 \\ V_X - V_Y & = (100 \ \Omega) \ i_1 \\ V_Y & = (100 \ \Omega) \ i_1 \end{align}

This is a fairly simple system of equations to solve directly, but in general, we might want to rearrange them so all variable terms are on the left-hand side, and all constants on the right:

\begin{align} V_X & = 12 \\ V_X - V_Y - (100 \ \Omega) \ i_1 & = 0\\ V_Y - (100 \ \Omega) \ i_1 & = 0 \end{align}

Equivalently as a (unitless) matrix equation:

$$$$\begin{bmatrix} 1 & 0 & 0 \\ 1 & -1 & -100 \\ 0 & 1 & -100 \end{bmatrix} \begin{bmatrix} V_X \\ V_Y \\ i_1 \end{bmatrix} = \begin{bmatrix} 12 \\ 0 \\ 0 \end{bmatrix}$$$$

We could equivalently write this matrix equation even more compactly as an augmented matrix as described in the Systems of Equations section:

$$$$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 12 \\ 1 & -1 & -100 & 0 \\ 0 & 1 & -100 & 0 \end{array} \right] \ \text{for} \ (V_X, V_Y, i_1)$$$$

Step 6 is to solve this system of simultaneous equations now that we’ve set it up. We’ll follow the process outlined in the Systems of Equations section for solving by hand.

First, we’ll see the first equation sets $\boxed{V_X = 12}$ . We can use that to create a smaller augmented matrix:

$$$$\left[ \begin{array}{cc|c} -1 & -100 & -12 \\ 1 & -100 & 0 \end{array} \right] \ \text{for} \ (V_Y, i_1)$$$$

Now we can observe that the second row sets $V_Y = 100 i_1$ , from which we can decide to replace the variable $V_Y$ in the remaining equation by $100 i_1$ :

$$$$\left[ \begin{array}{c|c} -200 & -12 \\ \end{array} \right] \ \text{for} \ (i_1)$$$$

From this we find that $\boxed{i_1 = \frac {-12} {-200} = 0.06}$ , or 60 mA.

Finally, we can expand our earlier simplification for $\boxed{V_Y = 100 \cdot 0.06 = 6}$ , or 6 volts.

We’ve now successfully solved for all our unknowns:

$$$$\begin{bmatrix} V_X \\ V_Y \\ i_1 \end{bmatrix} = \begin{bmatrix} 12 \\ 6 \\ 0.06 \end{bmatrix}$$$$

As discussed in Labeling Voltages, Currents, and Nodes, we can now use these three answers to find the answers to a much larger space of possible voltage difference and terminal current questions we may wish to ask about the circuit.

You can confirm these values by using the CircuitLab simulation software. To do so, simply click the schematic above to open it in a new browser tab, then click “Simulate,” click “Run DC Solver,” and read off the values listed.

## A Closer Look at Resistors

While a voltage source or current source has a defined orientation or direction, a resistor can have its two terminals swapped with no change in behavior in the surrounding circuit. (Sometimes, this property is called “unpolarized.”) Regardless of orientation, we can arbitrarily call one of the resistor’s terminals “nA” and the other “nB”. Now, whichever way we choose those, the same equation holds:

$$v_{A B} = i_A R$$

where $i_A$ is the current into terminal A. In the more verbose style as used within the CircuitLab simulation software:

$$\text{V(R1.nA)} - \text{V(R1.nB)} = \text{I(R1.nA)} \cdot \text{R1.R}$$

This equation says that the voltage difference between the two terminals of the resistor R1 (specifically the voltage at the nA terminal with respect to the nB terminal) is equal to the current into terminal R1.nA multiplied by the resistance R1.R. For resistors, this equation is true regardless of which way we label the two terminals.

Let’s review the simple example circuit above:

For this part, let’s take a closer look only at just resistor R1, which is connected between nodes X and Y. Let’s cut it out of the circuit and show the node labels X and Y, as well as the component terminals:

We’ve drawn two copies of R1, called R1_alpha and R1_beta. Note that the “beta” copy has the resistor rotated (or flipped) around, so that the designated nA and nB terminals are in opposite positions. Regardless, in both copies, the node names X and Y are unchanged, and the branch current $i_1$ is unchanged.

In the alpha copy:

\begin{align} \text{V(R1.nA)} - \text{V(R1.nB)} & = \text{I(R1.nA)} \cdot \text{R1.R} \\ \text{V(R1.nA)} & = \text{V(X)} \\ \text{V(R1.nB)} & = \text{V(Y)} \\ \text{I(R1.nA)} & = i_1 \\ \text{V(X)} - \text{V(Y)} & = i_1 \cdot \text{R1.R} \end{align}

In the beta copy: we’ve now switched the mapping between terminals and nodes. Additionally, now the branch current direction of $i_1$ is in the opposite direction relative to our newly selected terminal A because if we follow the direction of the branch current arrow we find that it flows out of R1.nA. We’ve marked the three changes resulting from flipping the terminals of the resistor in red:

\begin{align} \text{V(R1.nA)} - \text{V(R1.nB)} & = \text{I(R1.nA)} \cdot \text{R1.R} \\ \text{V(R1.nA)} & = \text{V(} \color{red}{\text{Y}} \text{)} \\ \text{V(R1.nB)} & = \text{V(} \color{red}{\text{X}} \text{)} \\ \text{I(R1.nA)} & = \color{red}{-} i_1 \\ \text{V(} \color{red}{\text{Y}} \text{)} - \text{V(} \color{red}{\text{X}} \text{)} & = \color{red}{-} i_1 \cdot \text{R1.R} \end{align}

With respect to the node voltages and branch currents: the resulting equations in both cases are completely identical! Multiply the final beta-case equation by -1 to get the alpha-case equation:

\begin{align} \color{green}{-1} \cdot \big( \text{V(} \color{red}{\text{Y}} \text{)} - \text{V(} \color{red}{\text{X}} \text{)} \big) & = \color{green}{-1} \cdot \big( \color{red}{-} i_1 \cdot \text{R1.R} \big) \\ - \text{V(} \color{red}{\text{Y}} \text{)} + \text{V(} \color{red}{\text{X}} \text{)} & = \color{green}{+} i_1 \cdot \text{R1.R} \\ \text{V(} \color{green}{\text{X}} \text{)} - \text{V(} \color{green}{\text{Y}} \text{)} & = \color{green}{+} i_1 \cdot \text{R1.R} \\ \end{align}

Note that this reversability is only the case for unpolarized components such as resistors. In future sections we will see that inductors and (unpolarized) capacitors also have this property. Because it’s possible to define the resistor’s direction in two ways, this is a common source of confusion when setting up resistive circuits to solve.

With respect to resistors in particular, it may help to remember that their instantaneous power dissipation is always positive. Resistors can only consume power from the circuit and subsequently heat up due to current flow; they can’t turn thermal energy back into electrical energy. (See also: Thermodynamics, Energy & Equilibrium.)

## A Larger Example

Let’s walk through all six steps with a more complicated example circuit:

Exercise Click to open and simulate the circuit above.

Step 1 is already done by labeling four non-ground nodes W, X, Y, Z as shown; the fifth node is marked as ground. This gives us four variables: $V_W, V_X, V_Y, V_Z$ .

Step 2 is already done by labeling five branch current variables $i_1, i_2, i_3, i_4, i_5$ along with directional arrows for each.

Step 3 is to write Kirchhoff’s Current Law at each non-ground node. For node W, no equation is necessary, because we have only the single branch current $i_2$ flowing in and out of that node. A similar argument goes for node Y and branch current $i_3$ . For nodes X and Z:

\begin{align} i_1 + i_2 - i_3 & = 0 && \text{(node X)} \\ i_3 - i_4 - i_5 & = 0 && \text{(node Z)} \end{align}

Step 4 is to write the constitutent equation for each of the 7 components as a function of the node voltage and branch current variables we’ve defined in steps 1 and 2.

For voltage source V1:

The positive terminal is connected to node X, and the negative to ground, so:

\begin{align} \text{V(V1.nA)} - \text{V(V1.nB)} & = 48 \\ \text{V(X)} - \text{V(ground)} & = 48 \\ V_X - 0 & = 48 \\ V_X & = 48 \end{align}

For resistor R1:

\begin{align} \text{V(R1.nA)} - \text{V(R1.nB)} & = 8 \ \text{I(R1.nA)} \\ \text{V(X)} - \text{V(Y)} & = 8 \ \text{I(R1.nA)} \\ V_X - V_Y & = 8 i_3 \end{align}

For voltage source V2:

\begin{align} \text{V(V2.nA)} - \text{V(V2.nB)} & = 19 \\ V_Y - V_Z & = 19 \end{align}

For resistor R2:

\begin{align} \text{V(R2.nA)} - \text{V(R2.nB)} & = 2.5 \ \text{I(R2.nA)} \\ \text{V(Z)} - \text{V(ground)} & = 2.5 \ \text{I(R2.nA)} \\ V_Z & = 2.5 i_4 \end{align}

For resistor R3:

\begin{align} \text{V(R3.nA)} - \text{V(R3.nB)} & = 5 \ \text{I(R3.nA)} \\ \text{V(Z)} - \text{V(ground)} & = 5 \ \text{I(R3.nA)} \\ V_Z & = 5 i_5 \end{align}

For resistor R4:

\begin{align} \text{V(R4.nA)} - \text{V(R4.nB)} & = 1 \ \text{I(R4.nA)} \\ \text{V(W)} - \text{V(X)} & = 1 \ \text{I(R4.nA)} \\ V_W - V_X & = i_2 \end{align}

For current source I1:

$$i_2 = 2$$

Step 5 is to gather all the equations from steps 3 and 4:

\begin{align} i_1 + i_2 - i_3 & = 0 \\ i_3 - i_4 - i_5 & = 0 \\ V_X & = 48 \\ V_X - V_Y & = 8 i_3 \\ V_Y - V_Z & = 19 \\ V_Z & = 2.5 i_4 \\ V_Z & = 5 i_5 \\ V_W - V_X & = i_2 \\ i_2 = 2 \end{align}

This is a system of 9 equations and 9 unknowns. While it may appear daunting, notice that each individual equation is quite simple. We’ll be able to take advantage of this to solve the system fairly quickly. First, let’s represent the system with all variables on one side:

\begin{align} i_1 + i_2 - i_3 & = 0 \\ i_3 - i_4 - i_5 & = 0 \\ V_X & = 48 \\ V_X - V_Y - 8 i_3 & = 0 \\ V_Y - V_Z & = 19 \\ V_Z - 2.5 i_4 & = 0 \\ V_Z - 5 i_5 & = 0 \\ V_W - V_X - i_2 & = 0 \\ i_2 & = 2 \end{align}

Then as a matrix equation:

$$$$\begin{bmatrix} 1 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -8 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & -2.5 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & -5 & 0 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ i_3 \\ i_4 \\ i_5 \\ V_W \\ V_X \\ V_Y \\ V_Z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 48 \\ 0 \\ 19 \\ 0 \\ 0 \\ 0 \\ 2 \end{bmatrix}$$$$

And as an augmented matrix:

$$$$\left[ \begin{array}{ccccccccc|c} 1 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 48 \\ 0 & 0 & -8 & 0 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 19 \\ 0 & 0 & 0 & -2.5 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & -5 & 0 & 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \end{array} \right] \ \text{for} \ (i_1, i_2, i_3, i_4, i_5, V_W, V_X, V_Y, V_Z)$$$$

While this is a fairly large system to approach by hand, notice that most cells of the matrix are 0. This is known as a “sparse” matrix. The process of solving a sparse matrix is considerably faster than solving a dense matrix of the same size.

Step 6 is to solve the system of equations. Again, we’ll follow the process outlined in the Systems of Equations section for solving by hand. In practice, with some experience solving circuits you will not have to write out the entire system, but instead may be able to solve parts in place as you go by applying parts of the same process in your head without as much written bookkeeping. For this example, though, we’ll work through writing our work at every step.

First, we’ll notice that the 3rd and 9th rows have simply one variable equal to a constant: $\boxed{V_X = 48}$ and $\boxed{i_2 = 2}$ . We can remove these two equations and these two variables from the system by substituting their values everywhere else they’re referenced:

$$$$\left[ \begin{array}{ccccccc|c} 1 & -1 & 0 & 0 & 0 & 0 & 0 & -2 \\ 0 & 1 & -1 & -1 & 0 & 0 & 0 & 0 \\ 0 & -8 & 0 & 0 & 0 & -1 & 0 & -48 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 & 19 \\ 0 & 0 & -2.5 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -5 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 50 \end{array} \right] \ \text{for} \ (i_1, i_3, i_4, i_5, V_W, V_Y, V_Z)$$$$

In this 7x7 system, the final equation says $\boxed{V_W = 50}$ , which is easy to eliminate as the variable is not referenced anywhere else in the system.

$$$$\left[ \begin{array}{cccccc|c} 1 & -1 & 0 & 0 & 0 & 0 & -2 \\ 0 & 1 & -1 & -1 & 0 & 0 & 0 \\ 0 & -8 & 0 & 0 & -1 & 0 & -48 \\ 0 & 0 & 0 & 0 & 1 & -1 & 19 \\ 0 & 0 & -2.5 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -5 & 0 & 1 & 0 \end{array} \right] \ \text{for} \ (i_1, i_3, i_4, i_5, V_Y, V_Z)$$$$

In the 6x6 system, the final two lines have two variables and zero constant term, so they say simply $V_Z = 2.5 i_4$ and $V_Z = 5 i_5$ . Out of these three variables, we can choose to keep any one and eliminate the other two. Let’s keep $V_Z$ and replace $i_4 = 0.4 V_Z$ and $i_5 = 0.2 V_Z$ :

$$$$\left[ \begin{array}{cccc|c} 1 & -1 & 0 & 0 & -2 \\ 0 & 1 & 0 & -0.6 & 0 \\ 0 & -8 & -1 & 0 & -48 \\ 0 & 0 & 1 & -1 & 19 \\ \end{array} \right] \ \text{for} \ (i_1, i_3, V_Y, V_Z)$$$$

These substitutions modified the 2nd remaining equation, leaving us again with an easily reducible $i_3 - 0.6 V_Z = 0$ , or $i_3 = 0.6 V_Z$ . We can remove $i_3$ everywhere and replace it with $0.6 V_Z$ :

$$$$\left[ \begin{array}{ccc|c} 1 & 0 & -0.6 & -2 \\ 0 & -1 & -4.8 & -48 \\ 0 & 1 & -1 & 19 \\ \end{array} \right] \ \text{for} \ (i_1, V_Y, V_Z)$$$$

We’re now down to a 3x3 system, but have no easy substitutions left. Now we can use row operations to perform Gaussian elimination. Add the second row to the third row, and store the result in place of the 3rd row ($R_2 + R_3 \rightarrow R_3$ ):

$$$$\left[ \begin{array}{ccc|c} 1 & 0 & -0.6 & -2 \\ 0 & -1 & -4.8 & -48 \\ 0 & 0 & -5.8 & -29 \\ \end{array} \right] \ \text{for} \ (i_1, V_Y, V_Z)$$$$

At this point, we have an upper triangular matrix, so the system is easily solved from the bottom row up: $-5.8 V_Z = -29$ , or $\boxed{V_Z} = 5$ . Next, $-V_Y -4.8 (5) = -48$ , or $\boxed{V_Y = 24}$ . And finally, $i_1 - 0.6 (5) = -2$ , or $\boxed{i_1 = 1}$ .

From here, we can expand the proportional substitutions we made earlier:

\begin{align} i_4 & = 0.4 V_Z = 2 \\ i_5 & = 0.2 V_Z = 1 \\ i_3 & = 0.6 V_Z = 3 \end{align}

We now have solved all nine variables:

$$$$\begin{bmatrix} i_1 \\ i_2 \\ i_3 \\ i_4 \\ i_5 \\ V_W \\ V_X \\ V_Y \\ V_Z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 2 \\ 1 \\ 50 \\ 48 \\ 24 \\ 5 \end{bmatrix}$$$$

You can confirm that this solution is correct by clicking the schematic above, then clicking “Simulate,” and “Run DC Solver” to print the voltage and current values as shown. You can also substitute into any of the equations we wrote earlier to double-check that these satisfy all constraints.

With practice and experience, you will be able to make many of the simplifications we described mathematically as you’re working with the schematic, saving you the trouble of setting up the entire system of equations.

(In fact, other circuit analysis methods with fancy-sounding names like “Modified Nodal Analysis” are mathematically identical to the six-step method presented here, and essentially just do some of these simplifications at the same time as setting up the equations.)

An experienced electrical engineer would be able to look at the schematic and instantly label, for example, $V_W=50$ – because the first two variables we solved ($V_X=48$ and $i_2 = 2$ ) are so obvious that they wouldn’t be worth labeling! They would then start to look at R1, V2, R2, and R3 as if they were a branch composed of a single voltage source and a single equivalent resistance $R1 + (R2 // R3)$ , and use that to solve for the total current $i_3$ . All in all, the experienced engineer would effectively follow the same mathematical procedure done here but with more “intuition” about what to solve for next. However, it’s powerful to remember that any circuit can always be solved using the technique presented here.

## What’s Next

In the next section, Superposition, we’ll look at one special method of solving circuits with multiple sources.


text:
# The similarity between an ordinal level and an interval level is that.

Question
Similarity
The similarity between an ordinal level and an interval level is that.

2020-10-24
Step 1
Ordinal level of measurement: When the possible observations are tags or categories or names, with a natural ordering, the level of measurement is ordinal, arithmetic operations cannot be performed even if the values are numerical.
Interval level of measurement: When the possible observations are numerical values that can be added or subtracted, but not divided or multiplied, the level of measurement is interval, zero is not absolute, but relative. Step 2
In both ordinal and interval data, there is natural order of the data. The sequence in which the values of the variable occur is meaningful in both cases.
The mode and median exist for both of the scales. This is because, the mode is the most frequently occurring in a data set, which can be obtained for both ordinal and interval scale variable, again, the median is the value that lies in the middle most position of an ordered arrangement of all value in the data set, which can also be obtained for both the scales.
Moreover, the division and multiplication operations cannot be done for both of the variables.

### Relevant Questions

Describe one similarity and one difference between the definitions of $$\displaystyle{\sin{{0}}}{\quad\text{and}\quad}{\cos{{0}}}$$, where 0 is an acute angle of a right triangle.
Describe one similarity and one difference between the definitions of $$\displaystyle{\sin{{0}}},{\quad\text{and}\quad}{\cos{{0}}}$$, where 0 is an acute angle of a right triangle.
Give the explanation for the relation between the scale factor and the similarity ratio.
Given:
The given statement is that the MNOP is a dilation of ABCD.
Prove that the similarity of polygons is an equivalence relation.
To determine:The difference and similarity between the graphs of the equations $$\displaystyle\frac{{{x}^{{2}}}}{{9}}-\frac{{{y}^{{2}}}}{{1}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{3}\right)}^{{2}}}}{{9}}-\frac{{{\left({y}+{3}\right)}^{{2}}}}{{1}}={1}$$
$$\displaystyle{y}^{{2}}={4}{x}{\quad\text{and}\quad}{\left({y}-{1}\right)}^{{2}}={4}{\left({x}-{1}\right)}$$
To determine:One similarity and one difference between the graphs of $$\displaystyle\frac{{{x}^{{2}}}}{{{25}}}+\frac{{{y}^{{2}}}}{{{16}}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{16}}}={1}$$
Describe one similarity and one difference between the graphs of $$\displaystyle{x}\frac{{2}}{{25}}+{y}\frac{{2}}{{16}}={1}{\quad\text{and}\quad}{\left({x}-{1}\right)}\frac{{2}}{{25}}+{\left({y}-{1}\right)}\frac{{2}}{{16}}={1}$$.
Describe one similarity and one difference between the graphs of $$\displaystyle\frac{{x}^{{2}}}{{9}}-\frac{{y}^{{2}}}{{1}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{3}\right)}^{{2}}}}{{9}}-\frac{{{\left({y}+{3}\right)}^{{2}}}}{{1}}={1}$$.


text:
# SparsePak IFU / Performance

## Overview

A full description of the performance can be found in

SparsePak: A Formatted Fiber Field-Unit for The WIYN Telescope Bench Spectrograph. I. Design, Construction, and Calibration , Bershady, M. A., Andersen, Harker, J., Ramsey, L. W., Verheijen, M. A. W. 2004, PASP, 116, 565 (preprint: [local pdf] [astro-ph/0403456])

and in the forthcoming paper:

SparsePak: A Formatted Fiber Field-Unit for The WIYN Telescope Bench Spectrograph. II. On-Sky Performance, Bershady, M. A., Andersen, D. R., Verheijen, M. A. W., Westfall, K. B., Crawford, S. M., Swaters, R. A. 2003, submitted to ApJSupp

The above results were based on the following measurements made during instrument commissioning:

• Lab test reults on FRD and throughput from 450 nm to 800 nm;
• On-telescope comparisons to Densepak and Red Hydra cables at 515 nm, 660 nm, and 860 nm;
• On-telescope performance at 515 nm, 660 nm, and 860 nm: resolution, vignetting, and sky levels using the (echelle); and
• On-telescope scattered light tests at 5150 and 8650 A (O11 & O6 of the echelle).

A summary of the primary results are presented here.

## Transmission

Fiber transmission (throughput) of 13 of the 82 SparsePak fibers and the SparsePak Reference Cable, measured using the test-bench in our lab, as described in the text and illustrated in Figure 13. Symbols denote different fibers, as identified in the key, and are offset in wavelength within groups for presentation purposes. Light symbols (gray instead of black) are fibers for which the throughput variance across bands was larger than 15%. These fibers either had measurement-log notes indicating unsatisfactory setup of the pin-hole image on the fiber face, or beam-mode measurement variances indicating the lamp stability was poor. As such, these measurements should be viewed as suspect, consistent with the unusually high or low measured transmittance. The large, grey circles are the median of the 8 good fiber measurements. The solid curve is the expected transmittance of 25.4m of Polymicro ultra-low OH- fused-silica fiber, based on Polymicro's figures, combined with two silica-air interfaces (3.43% per interface). The dotted curves represent the normalized broad-band filter transmission, convolved with the CCD response function. From left to right: B, V, R, I. For V,R,I, note the excellent agreement between the measured and expected values, the measurable difference between different fibers. The B-band measurements appear high relative to expectations. The dashed curve curve takes into account the effective band-pass given the expected fiber transmittance. Not taken into account is the spectrum of the light source. The likely effective wavelength of the B-band measurements is 4900A. (Ref: Fig. 5, Bershady et al. 2004, PASP, 116, 565.)

[LEFT] Relative encircled energy (EE) at an output f-ratio of f/6.3 as a function of wavelength for 4 of the SparsePak fibers (labeled) fed with a f/6.3 input beam. These represent four different measures of FRD wavelength dependence for the 500 micron core Polymicro ultra-low OH- fibers. Error-bars represent estimated observational uncertainties. Wavelength offsets of points are for presentation purposes. (Ref: Fig. 8, Bershady et al. 2004, PASP, 116, 565.)

[RIGHT] Fractional output encircled energy (EE) from fibers as a function of WIYN Bench Spectrograph collimator focal ratio (bottom axis) or collimator focal length (top axis). This assumes a 152mm collimated beam diameter and that all fibers are fed with a f/6.3 beam. The mean SparsePak beam profile (thick, solid line) the range for the 13 measured SparsePak fibers (grey shaded area), and reference cable (thin solid curve) are based on laboratory measurements (cf. Figure 15) using the f/6.3 input beam shown as a thin solid curve with solid circles. Comparable curves, as measured {\it on the telescope} for Densepak (300 micron fibers), and the two Hydra cables (with blue,'' 310 micron and red,'' 200 micron fibers), are shown for comparison (private communication, P. Smith & C. Conselice; see text for further details). These measurements use the WIYN f/6.3 beam (accounting for the central obstruction -- see text), shown as the thin solid curve with open triangles. The very bottom scale (relative spectral degradation) indicates how the spectral resolution of the Bench would alter (worst case) due to changes in system demagnification as a function of changes in the collimator focal length at fixed camera focal length. The current Bench has a f/6.7 collimator for a 152mm collimated beam. This figure illustrates the effects of FRD on light losses for the Bench Spectrograph, and how optimization trades might be made between throughput and spectral resolution for redesign of the Bench Spectrograph collimator. (Ref: Fig. 10, Bershady et al. 2004, PASP, 116, 565.)

## Vignetting

Information will be forthcoming with publication of second SparsePak commissioning paper.

## Scattered Light

Information will be forthcoming with publication of second SparsePak commissioning paper.

## Spectral Resolution

Detailed table is forthcoming with publication of second SparsePak commissioning paper. For now see table under setups.

## Sky Substraction

Information will be forthcoming with publication of second SparsePak commissioning paper.

last updated: Jun 08, 2004 (mab@astro.wisc.edu)


text:
# getFileLocation

## Usage

getFileLocation(fileObject)

## Description

function to get the location of the attachment to a File

## Details

Returns a string which is the path to the file attached to a File object. If the File 'links' to an external URL (not downloaded to the local file system), returns the URL itself.

File synGet synStore


text:
Web: http://arxiv.org/abs/2101.01350

May 12, 2022, 1:11 a.m. | Xiangyu Yang, Jiashan Wang, Hao Wang

This paper primarily focuses on computing the Euclidean projection of a
vector onto the $\ell_{p}$ ball in which $p\in(0,1)$. Such a problem emerges as
the core building block in statistical machine learning and signal processing
tasks because of its ability to promote the sparsity of the desired solution.
However, efficient numerical algorithms for finding the projections are still
not available, particularly in large-scale optimization. To meet this
challenge, we first derive the first-order necessary optimality conditions of
this problem. Based …

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text:
## sigma and pi bonds

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

Elyssa Smith 4B
Posts: 58
Joined: Wed Sep 18, 2019 12:22 am

### sigma and pi bonds

How can you figure out the hybridization of sigma and pi bonds? How do the hybridization's differ if something is double bonded and has sigma and pi bonds?

905416023
Posts: 54
Joined: Thu Jul 25, 2019 12:17 am

### Re: sigma and pi bonds

Hybridization depends on the # of regions of electron density. 2 regions of electron density = sp hybridization. 3 regions of electron density = sp2 hybridization. 4 regions of electron density = sp3 hybridization. 5 regions of electron density = sp3d hybridization. 6 regions of electron density = sp3d2 hybridization. And you just put the row # in front of the s. So if it's in row 2 it would be 2sp.

Amina Durrani 3G
Posts: 52
Joined: Sat Jul 20, 2019 12:16 am

### Re: sigma and pi bonds

If something is double bonded, then the sigma bond has sp^(number of electron density) hybridization, and the pi bond has p hybridization (ex 2p, with 2 being the row number), since in pi bonds, p orbitals are overlapping.

DMuth_1J
Posts: 63
Joined: Thu Jul 11, 2019 12:15 am

### Re: sigma and pi bonds

I thought hybridization was reliant upon the number of bonds, not just the electron density. Do lone pairs increase the hybridization as well?


text:
Contest Duration: - (local time) (100 minutes) Back to Home

## G - Teleporting Takahashi Editorial by en_translator

This is a problem on a three-dimensional grid, but we will first consider the simplified version on a one-dimensional grid, then extend it to two-dimensional, and the original three-dimensional version.

## [1] One-dimensional version

We consider the following problem.

At first he is in square $$0$$ on an infinite one-dimensional grid. When he is in square $$x$$, he can move to square $$x-1$$ or square $$x+1$$. How many ways are there to reach square $$X$$ after $$N$$ moves?

If $$N < |X|$$, then it is impossible to reach square $$X$$ after $$N$$ moves, so the answer is $$0$$.
Also, after moving odd number of times, the resulting position $$x$$ is always odd, and after moving even number of times, $$x$$ is always even, so if $$N \bmod 2 \neq X \bmod 2$$, the answer is $$0$$ too.

We consider other cases. If $$X \geq 0$$ (if $$X < 0$$), “he is in square $$X$$ after exactly $$N$$ moves” if and only if “in the $$N$$ moves, he moves $$(N+|X|)/2$$ times in the positive (negative) direction and $$(N-|X|)/2$$ times in the negative (positive) direction of $$x$$ coordinate.” Therefore, the desired number of ways can be expressed with a binomial coefficient as $$\binom{N}{(N+|X|)/2}$$.

Therefore, the number of ways $$f_1(N, X)$$ to reach square $$X$$ after exactly $$N$$ moves is

$f_1(N, X) = \begin{cases} \displaystyle\binom{N}{(N+|X|)/2} &(\text{if }N \geq |X|\text{ and }N \bmod 2 = X \bmod 2)\\ 0&(\text{otherwise}). \end{cases}$

By precalculating the factorials, $$f_1(N, X)$$ can be calculated for given $$N$$ and $$X$$ in an $$\mathrm{O}(1)$$ time.

## [2] Two-dimensional version

Consider the following problem.

At first he is in square $$(0, 0)$$ on an infinite two-dimensional grid. When he is in square $$(x, y)$$, he can move to square $$(x-1, y), (x+1, y), (x, y-1)$$ or square $$(x, y+1)$$. How many ways are there to reach square $$X$$ after $$N$$ moves?

In order to solve this problem, we introduce the following coordinate transformation on $$xy$$-coordinate:

$\begin{cases} u = x+y\\ v = x-y\\ \end{cases}$

and we will rephrase the problem as a problem on $$uv$$-coordinate. Also, let $$(U, V) := (X+Y, X-Y)$$.

Then, “the number of ways to reach square $$(X, Y)$$ after $$N$$ moves” is equal to “the number of ways to reach $$(u, v) = (U, V)$$ after $$N$$ moves.”

Also, the counterpart of “when he is in square $$(x, y)$$, he can move to square $$(x-1, y), (x+1, y), (x, y-1)$$ or square $$(x, y+1)$$” on $$uv$$-coordinate system is “when he is at $$(u, v)$$, he can move to $$(u-1, v-1), (u-1, v+1), (u+1, v-1)$$, or $$(u+1, v+1)$$. Therefore, on each move, “whether to increment or decrement $$u$$” and “whether to increment or decrement $$v$$” can be determined independently.

Hence, in order to determine a way to reach $$(u, v) = (U, V)$$ after exactly $$N$$ moves, we can choose the following two routes independently:

• the route along the $$u$$ axis to reach $$u = U$$ after $$N$$ moves ( $$f_1(N, U)$$ choices) and
• the route along the $$v$$ axis to reach $$v = V$$ after $$N$$ moves ( $$f_1(N, V)$$ choices),

so the number of ways $$f_2(N, X, Y)$$ to reach square $$(X, Y)$$ after exactly $$N$$ moves can be expressed as

$f_2(N, X, Y) = f_1(N, U) f_1(N, V) = f_1(N, X+Y) f_1(N, X-Y).$

## [3] Three-dimensional problem

We now consider the original three-dimensional problem.

We will count the number of ways to reach square $$(X, Y, Z)$$ such that he moves in the direction of $$z$$ axis $$k$$ times. Such a way is determined by the following three factors:

• there are $$\binom{N}{k}$$ ways of choosing which $$k$$ moves of the $$N$$ moves are in the direction of $$z$$ axis;
• the $$k$$ moves along $$z$$ axis are chosen so that he reaches $$z = Z$$ after $$k$$ moves along $$z$$ axis, so there are $$f_1(k, Z)$$ ways to do so;
• the other $$(N-k)$$ moves in the direction other than $$x$$ axis are chosen so that he reaches $$(x, y) = (X, Y)$$ after $$(N-k)$$ moves on the two-dimensional $$xy$$-grid, so there are $$f_2(N-k, X, Y)$$ ways to do so.

Therefore, the number of ways to reach square $$(X, Y, Z)$$ such that he moves in the direction of $$z$$ axis $$k$$ times is

$\binom{N}{k} f_1(k, Z) f_2(N-k, X, Y).$

By finding the sum over $$k = 0, 1, \ldots, N$$, the answer for this problem $$f_3(N, X, Y, Z)$$ is obtained as

$f_3(N, X, Y, Z) = \sum_{k = 0}^N \binom{N}{k} f_1(k, Z) f_2(N-k, X, Y) = \sum_{k = 0}^N \binom{N}{k} f_1(k, Z) f_1(N-k, X+Y) f_1(N-k, X-Y).$

With proper precalculations, necessary binomial coefficients can be computed in an $$\mathrm{O}(1)$$ time each, so $$f_3(N, X, Y, Z)$$ can be computed in a total of $$\mathrm{O}(N)$$ time.

Hence, the problem can be solved in a total of $$\mathrm{O}(N)$$ time.

posted:
last update:


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# Complex analysis - maximum modulus & analytic function

1. May 9, 2008

### romiet3625

[SOLVED] complex analysis - maximum modulus & analytic function

Hi all, I'm having difficulty figuring out how to do the following two problems in complex analysis. I need help!

1. Consider the infinite strip -$$\pi$$< Im z < $$\pi$$. Does maximum modulus principle apply to this strip? Why or why not? (Hint: e$$^{i\pi}$$ = e$$^{-i\pi}$$ = 1)

2. Show that if f(z) is analytic and Re f(z) is bounded in the complex plane, then f(z) is constant. What if Im f(z) is bounded? (Hint: Consider e$$^{f(z)}$$ and its absolute value)

Thank you for your time and thank you for any help.

2. May 9, 2008

### HallsofIvy

Staff Emeritus
You are given hints and you haven't used them? What have YOU done on this? No one can help you if you don't show us what YOU know about these problems and where you are having difficulties.


text:
# I Ladder operators and SU(2) representation

#### kelly0303

Hello! I read in many places the derivation of the representation for SU(2) using ladder operators and in all of the places they say that, due to the fact that we are looking for a finite dimensional representation, the ladder must end at a point, hence why we have an eigenvector of $L_3$ (usually) such that, when acted on with the raising operator gives zero, and the same for the lowering operator. I didn't find an explanation as to why this must be unique. From a physics point of view it makes sense, as you have an irreducible representation of a spin J particle, so in order for the representation to allow for all the spin-z component, all the eigenvectors must have a different eigenvalue, hence why you have just one eigenvector at the top and at the bottom, and in general for any eigenvalue of $L_3$, but I am not sure I understand mathematically, why can't you have more than one vector with the same eigenvalue in an irrep of SU(2). Can someone help me? Thank you!

Related Linear and Abstract Algebra News on Phys.org

#### fresh_42

Mentor
2018 Award
Hello! I read in many places the derivation of the representation for SU(2) using ladder operators and in all of the places they say that, due to the fact that we are looking for a finite dimensional representation, the ladder must end at a point, hence why we have an eigenvector of $L_3$ (usually) such that, when acted on with the raising operator gives zero, and the same for the lowering operator. I didn't find an explanation as to why this must be unique. From a physics point of view it makes sense, as you have an irreducible representation of a spin J particle, so in order for the representation to allow for all the spin-z component, all the eigenvectors must have a different eigenvalue, hence why you have just one eigenvector at the top and at the bottom, and in general for any eigenvalue of $L_3$, but I am not sure I understand mathematically, why can't you have more than one vector with the same eigenvalue in an irrep of SU(2). Can someone help me? Thank you!
You can find the underlying theorem (7.1.) here: https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
For the mathematical derivation I recommend [6] in the sources. The main reason is, that all invariant subspaces are one-dimensional, and the upper triangular matrix shifts those subspaces one place higher, the lower triangular one place less.

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text:
# American Institute of Mathematical Sciences

2015, 9: 365-405. doi: 10.3934/jmd.2015.9.365

## Dynamical cubes and a criteria for systems having product extensions

 1 Centro de Modelamiento Matemático and Departamento de Ingeniería Matemática, Universidad de Chile, Av. Blanco Encalada 2120, Santiago 2 Department of Mathematics, Northwestern University, 2033 Sheridan Road Evanston, IL 60208-2730, United States

Received  March 2015 Revised  November 2015 Published  December 2015

For minimal $\mathbb{Z}^{2}$-topological dynamical systems, we introduce a cube structure and a variation of the usual regional proximality relation for $\mathbb{Z}^2$ actions, which allow us to characterize product systems and their factors. We also introduce the concept of topological magic systems, which is the topological counterpart of measure theoretic magic systems introduced by Host in his study of multiple averages for commuting transformations. Roughly speaking, magic systems have less intricate dynamics, and we show that every minimal $\mathbb{Z}^2$ dynamical system has a magic extension. We give various applications of these structures, including the construction of some special factors in topological dynamics of $\mathbb{Z}^2$ actions and a computation of the automorphism group of the minimal Robinson tiling.
Citation: Sebastián Donoso, Wenbo Sun. Dynamical cubes and a criteria for systems having product extensions. Journal of Modern Dynamics, 2015, 9: 365-405. doi: 10.3934/jmd.2015.9.365
##### References:
 [1] J. Auslander, Minimal Flows and Their Extensions,, North-Holland Mathematics Studies, (1988). Google Scholar [2] L. Auslander, L. Green and F. Hahn, Flows on Homogeneous Spaces,, Ann. Math. Studies, (1963). Google Scholar [3] T. Austin, On the norm convergence of non-conventional ergodic averages,, Ergodic Theory Dynam. Systems, 30 (2010), 321. doi: 10.1017/S014338570900011X. Google Scholar [4] F. Blanchard, B. Host and A. Maass, Topological complexity,, Ergodic Theory Dynam. Systems, 20 (2000), 641. doi: 10.1017/S0143385700000341. Google Scholar [5] Q. Chu, Multiple recurrence for two commuting transformations,, Ergodic Theory Dynam. Systems, 31 (2011), 771. doi: 10.1017/S0143385710000258. Google Scholar [6] S. Donoso, Enveloping semigroups of systems of order d,, Discrete Contin. Dyn. Sys., 34 (2014), 2729. doi: 10.3934/dcds.2014.34.2729. Google Scholar [7] R. Ellis, Lectures on Topological Dynamics,, W. A. Benjamin, (1969). Google Scholar [8] F. Gälher, A. Julien and J. Savinien, Combinatorics and topology of the Robinson tiling,, C. R. Math. Acad. Sci. Paris, 350 (2012), 627. doi: 10.1016/j.crma.2012.06.007. Google Scholar [9] E. Glasner, Ergodic Theory via Joinings,, Mathematical Surveys and Monographs, (2003). doi: 10.1090/surv/101. Google Scholar [10] E. Glasner, Topological ergodic decompositions and applications to products of powers of a minimal transformation,, J. Anal. Math., 64 (1994), 241. doi: 10.1007/BF03008411. Google Scholar [11] B. Host, Ergodic seminorms for commuting transformations and applications,, Studia Math., 195 (2009), 31. doi: 10.4064/sm195-1-3. Google Scholar [12] B. Host and B. Kra, Nonconventional averages and nilmanifolds,, Ann. of Math. (2), 161 (2005), 397. doi: 10.4007/annals.2005.161.397. Google Scholar [13] B. Host, B. Kra and A. Maass, Nilsequences and a structure theorem for topological dynamical systems,, Adv. Math., 224 (2010), 103. doi: 10.1016/j.aim.2009.11.009. Google Scholar [14] W. Huang, S. Shao and X. Ye, Nil Bohr-sets and almost automorphy of higher order,, Memoirs of Amer. Math. Soc., 241 (). doi: 10.1090/memo/1143. Google Scholar [15] A. Leibman, Pointwise convergence of ergodic averages for polynomial sequences of rotations of a nilmanifold,, Ergodic Theory Dynam. Systems, 25 (2005), 201. doi: 10.1017/S0143385704000215. Google Scholar [16] S. Mozes, Tilings, substitution systems and dynamical systems generated by them,, J. Anal. Math., 53 (1989), 139. doi: 10.1007/BF02793412. Google Scholar [17] J. Olli, Endomorphisms of Sturmian systems and the discrete chair substitution tiling system,, Discrete Contin. Dyn. Syst., 33 (2013), 4173. doi: 10.3934/dcds.2013.33.4173. Google Scholar [18] K. E. Petersen, Disjointness and weak mixing of minimal sets,, Proc. Amer. Math. Soc., 24 (1970), 278. doi: 10.1090/S0002-9939-1970-0250283-7. Google Scholar [19] M. Queffélec, Substitution Dynamical Systems-Spectral Analysis,, Second edition, (1294). doi: 10.1007/978-3-642-11212-6. Google Scholar [20] C. Radin, Miles of Tiles,, Student Mathematical Library, (1999). doi: 10.1090/stml/001. Google Scholar [21] S. Shao and X. Ye, Regionally proximal relation of order $d$ is an equivalence one for minimal systems and a combinatorial consequence,, Adv. Math., 231 (2012), 1786. doi: 10.1016/j.aim.2012.07.012. Google Scholar [22] T. Tao, Norm convergence of multiple ergodic averages for commuting transformations,, Ergodic Theory Dynam. Systems, 28 (2008), 657. doi: 10.1017/S0143385708000011. Google Scholar [23] H. Towsner, Convergence of diagonal ergodic averages,, Ergodic Theory Dynam. Systems, 29 (2009), 1309. doi: 10.1017/S0143385708000722. Google Scholar [24] S. Tu and X. Ye, Dynamical parallelepipeds in minimal systems,, J. Dynam. Differential Equations, 25 (2013), 765. doi: 10.1007/s10884-013-9313-6. Google Scholar

show all references

##### References:
 [1] J. Auslander, Minimal Flows and Their Extensions,, North-Holland Mathematics Studies, (1988). Google Scholar [2] L. Auslander, L. Green and F. Hahn, Flows on Homogeneous Spaces,, Ann. Math. Studies, (1963). Google Scholar [3] T. Austin, On the norm convergence of non-conventional ergodic averages,, Ergodic Theory Dynam. Systems, 30 (2010), 321. doi: 10.1017/S014338570900011X. Google Scholar [4] F. Blanchard, B. Host and A. Maass, Topological complexity,, Ergodic Theory Dynam. Systems, 20 (2000), 641. doi: 10.1017/S0143385700000341. Google Scholar [5] Q. Chu, Multiple recurrence for two commuting transformations,, Ergodic Theory Dynam. Systems, 31 (2011), 771. doi: 10.1017/S0143385710000258. Google Scholar [6] S. Donoso, Enveloping semigroups of systems of order d,, Discrete Contin. Dyn. Sys., 34 (2014), 2729. doi: 10.3934/dcds.2014.34.2729. Google Scholar [7] R. Ellis, Lectures on Topological Dynamics,, W. A. Benjamin, (1969). Google Scholar [8] F. Gälher, A. Julien and J. Savinien, Combinatorics and topology of the Robinson tiling,, C. R. Math. Acad. Sci. Paris, 350 (2012), 627. doi: 10.1016/j.crma.2012.06.007. Google Scholar [9] E. Glasner, Ergodic Theory via Joinings,, Mathematical Surveys and Monographs, (2003). doi: 10.1090/surv/101. Google Scholar [10] E. Glasner, Topological ergodic decompositions and applications to products of powers of a minimal transformation,, J. Anal. Math., 64 (1994), 241. doi: 10.1007/BF03008411. Google Scholar [11] B. Host, Ergodic seminorms for commuting transformations and applications,, Studia Math., 195 (2009), 31. doi: 10.4064/sm195-1-3. Google Scholar [12] B. Host and B. Kra, Nonconventional averages and nilmanifolds,, Ann. of Math. (2), 161 (2005), 397. doi: 10.4007/annals.2005.161.397. Google Scholar [13] B. Host, B. Kra and A. Maass, Nilsequences and a structure theorem for topological dynamical systems,, Adv. Math., 224 (2010), 103. doi: 10.1016/j.aim.2009.11.009. Google Scholar [14] W. Huang, S. Shao and X. Ye, Nil Bohr-sets and almost automorphy of higher order,, Memoirs of Amer. Math. Soc., 241 (). doi: 10.1090/memo/1143. Google Scholar [15] A. Leibman, Pointwise convergence of ergodic averages for polynomial sequences of rotations of a nilmanifold,, Ergodic Theory Dynam. Systems, 25 (2005), 201. doi: 10.1017/S0143385704000215. Google Scholar [16] S. Mozes, Tilings, substitution systems and dynamical systems generated by them,, J. Anal. Math., 53 (1989), 139. doi: 10.1007/BF02793412. Google Scholar [17] J. Olli, Endomorphisms of Sturmian systems and the discrete chair substitution tiling system,, Discrete Contin. Dyn. Syst., 33 (2013), 4173. doi: 10.3934/dcds.2013.33.4173. Google Scholar [18] K. E. Petersen, Disjointness and weak mixing of minimal sets,, Proc. Amer. Math. Soc., 24 (1970), 278. doi: 10.1090/S0002-9939-1970-0250283-7. Google Scholar [19] M. Queffélec, Substitution Dynamical Systems-Spectral Analysis,, Second edition, (1294). doi: 10.1007/978-3-642-11212-6. Google Scholar [20] C. Radin, Miles of Tiles,, Student Mathematical Library, (1999). doi: 10.1090/stml/001. Google Scholar [21] S. Shao and X. Ye, Regionally proximal relation of order $d$ is an equivalence one for minimal systems and a combinatorial consequence,, Adv. Math., 231 (2012), 1786. doi: 10.1016/j.aim.2012.07.012. Google Scholar [22] T. Tao, Norm convergence of multiple ergodic averages for commuting transformations,, Ergodic Theory Dynam. Systems, 28 (2008), 657. doi: 10.1017/S0143385708000011. Google Scholar [23] H. Towsner, Convergence of diagonal ergodic averages,, Ergodic Theory Dynam. Systems, 29 (2009), 1309. doi: 10.1017/S0143385708000722. Google Scholar [24] S. Tu and X. Ye, Dynamical parallelepipeds in minimal systems,, J. Dynam. Differential Equations, 25 (2013), 765. doi: 10.1007/s10884-013-9313-6. Google Scholar
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## Essential University Physics: Volume 1 (4th Edition) Clone

Recall that $H_2$ weights $3.3 \times 10^{-27}$ kg, and Sulfur Dioxide weighs $1.1 \times 10^{-25}$ kg. Thus: $v_ = \sqrt{\frac{3(T)}{m}}$ For Hydrogen: $v = \sqrt{\frac{3(1.4 \times 10^{-23})(75)}{3.3 \times 10^{-27}}} = 967 m/s$ For Sulfur Dioxode: $v = \sqrt{\frac{3(1.4 \times 10^{-23})(350)}{1.1 \times 10^{-25}}} = 370 m/s$


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My Math Forum 2 Cylinders Attached Perpendicular, Find Length of Second Cylinder

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 August 8th, 2014, 11:27 AM #1 Newbie   Joined: Aug 2014 From: canada Posts: 2 Thanks: 0 2 Cylinders Attached Perpendicular, Find Length of Second Cylinder I have this math problem I am trying to solve. So if I have two cylinders perpendicular attached and I have both radius' (R1 and R2) and length from first cylinders radius to top of second cylinder (L1), how can I find the length of the second cylinder (L2)? Last edited by zipaway; August 8th, 2014 at 11:56 AM.
 August 8th, 2014, 11:36 AM #2 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Well, assuming $L_1$ goes from the center of the horizontal cylinder to the top of the vertical cylinder, then it is easy to see that $L_1-L_2=R_1$, so $L_2=L_1-R_1$. To minimize $L_2$, you would want to minimize $L_1$ and maximize $R_1$.
 August 8th, 2014, 11:56 AM #3 Newbie   Joined: Aug 2014 From: canada Posts: 2 Thanks: 0 But we are given R1, R2 and L1. How would we go about finding L2?
 August 8th, 2014, 11:57 AM #4 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 I just told you. $L_2=L_1-R_1$.
 August 8th, 2014, 07:09 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Making the same assumption about $L_1$ we have $\displaystyle L_2=L_1-\sqrt{R_1^2-R_2^2}$ if the largest external length of the upright cylinder is to be considered. Last edited by greg1313; August 9th, 2014 at 01:49 PM.

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# How to rationalise with MO theory that CO is a two-electron donor through carbon?

A question I am looking at is as follows:

$\ce{CO}$ is isoelectronic with $\ce{N2}$. Sketch MO diagrams for $\ce{CO}$ and $\ce{N2}$. Point out key differences between the diagrams and use the diagram to explain why $\ce{CO}$ acts as a two-electron donor through carbon rather than through oxygen.

Understandably, the key difference between these molecules is that $\ce{CO}$ is heteronuclear, and thus will have differences in energy between the molecular orbital and the atoms.

But I can't explain why $\ce{CO}$ is a two-electron donor using MO theory, even though I can with Lewis?

Does anyone have any thoughts?

In the above MO diagram, the 5σ is the HOMO. But it is closest in energy to oxygen's 2p orbitals, so why is it centered on carbon?

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one. I recommend you have a look at: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory? My answer there contains a full MO diagram of CO which should make it easier to comprehend. – Martin - マーチン May 16 '16 at 11:08
• Hi Martin. Thank for your reply! I'm struggling to understand the explanation given in the linked thread. This question is actually an exam question, so I wouldn't have access to the data you've given there. Is there a way to see from the MO diagram alone, why CO is a two-electron donor through C? Also, how do I upvote you? – Adam May 16 '16 at 11:26
• Upvoting is a privilege you have to earn on this site. But it is not hard. I recommend you include your MO sketches into the post and you'll soon gather a few upvotes that get you there. When you consider the donor ability you usually look at the HOMO. From your graph (or the MO diagram linked) you should be able to tell which atom has the largest contribution to it. If you figure the solution out by yourself, you are more than welcome to answer your own question. This way you can also earn some reputation points. – Martin - マーチン May 16 '16 at 11:36
• I see. I have provided an image of the MO diagram and a suggestion! – Adam May 16 '16 at 11:49
• So overall, how could I tell exclusively from the MO diagram that HOMO electrons are centered on C and not on O? – Adam May 16 '16 at 23:24

The electrons in the frontier orbital(s) play a special role for the chemical reactivity. In $\ce{CO}$, the HOMO is the $5 \sigma$ orbital (ref: your diagram), and has mainly $\ce{C}$ character. The two electrons in it act like a lone pair on the carbon.

The predominant $\ce{C}$ character of the HOMO accounts for the reactivity via carbon ($\sigma$ donor

Side note: $\ce{CO}$ is also a $\pi ^*$ acceptor due to the LUMO (which is also predominantly $\ce{C}$ in character.

So overall, how could I tell exclusively from the MO diagram that HOMO electrons are centered on C and not on O?

Don't think you can. The contribution to MOs is determind by the coefficients in the linear combination. You should see Martin's calculated values in the answer link he provided.

• Well, there is a hint that the lower orbitals will generally be centred on the electronegative partner and the higher ones generally on the electropositive partner, but that’s a weak indication at best. – Jan Sep 21 '16 at 9:44
• Agreed. It has its uses, say for something like HF where there are very few orbitals overlapping, but in CO it's really difficult (more precisely, I found it difficult) to use "intuition" because the C2s, C2p, O2p all interact significantly. Even the usual adage of "AOs close in energy to the MO contribute more" fails here because the HOMO is very close to O2p energy level. – orthocresol Sep 21 '16 at 10:06
• Not only that, but there are still sources out there stating that the HOMO of CO is antibonding in character. If we can't even get that right by intuition, it is not hard to imagine that we will fail at predicting more complicated features. – Martin - マーチン Sep 21 '16 at 10:19

I had to give this a good, long thought, after I initially thought this is a question which is hardly answerable with a molecular orbitals diagram by itself. Now that I have spent quite some time thinking about it, I must admit I enjoy this exercise a lot.

The key to explaining that carbon monoxide is a two electron donor through carbon is to start with with a very fundamental MO scheme. This is basically to get you dimensions right and not to mix things up later. It is also one of the things I would consider a prerequisite for these kinds of exercises.

The above diagram is for a symmetric diatomic molecule with no s-p mixing. This is true for the heavier main group elements; have a look at this explanation by Wildcat.

For dinitrogen we will have to adapt this scheme to include s-p mixing. Here it is important to know, that the interaction is still strong enough to raise the 2σ valence orbital above the 1π orbitals.

With pen and paper it is a bit too difficult to draw the MO, so it is probably easier to exaggerate the respective contributions to the orbitals. For example the 1σ orbital in the non mixing case is comprised of pure s-orbitals. When it is mixing, there has to be some s-p character which is difficult to draw. As a general guideline electron density concentrates along the bonding axis for bonding orbitals, while for anti-bonding orbitals in the lone-pair region. The π orbitals largely remain untouched by this.

From there we can use the fact, that carbon monoxide is isoelectronic with dinitrogen. We have to skew the energies of the atomic orbitals. The energy of the atomic orbitals is decreasing from left to right in the period; therefore carbon will have slightly elevated levels and nitrogen's are lowered.
In this case it is easier to not tinker with the ordering of the MO as this is just a pen and paper exercise. However, one should be aware, that the levels will overall change in this molecule.

The atomic orbitals that are closer to the energy of the molecular orbital will have a larger coefficient. Therefore 1σ and 1π will be polarised towards oxygen. On the other hand 2σ and 3σ will be polarised towards carbon.
One can also argue this with Bent's rule (gold book), emphasis mine:

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

The rule itself (is an observation, which) applies to more complex molecules, but can be applied to this situation, too. Oxygen is more electronegative than carbon therefore the bonding hybrid orbitals will have more p-character directed towards oxygen. At carbon itself will be more s-character for the lone pair, which is what we observe for the HOMO, when we apply the polarisation, too.

From this deductive chain one can reasonably explain, that the HOMO of carbon monoxide must be of σ symmetry (two electron donor) and have the largest coefficient at the carbon atom and will therefore rather coordinate there than through the oxygen.

• It seems reasonable but slightly awkward to apply Bent's rule here, considering that CO is a diatomic and thus, there are no "ligands" and "central atom". – Tan Yong Boon Jun 7 '19 at 1:35