text:
Interrupt Operation

All pending interrupts are indicated by their respective flag bit being equal to a ‘1’ in the PIRx register. All pending interrupts are resolved using the priority scheme explained in the Interrupt Priority section.

Once the interrupt source to be serviced is resolved, the program execution vectors to the resolved interrupt vector addresses, as explained in Interrupt Vector Table section. The vector number is also stored in the WREG register. Most of the flag bits are required to be cleared by the application software, but in some cases, device hardware clears the interrupt automatically. Some flag bits are read-only in the PIRx registers. These flags are a summary of the source interrupts, and the corresponding interrupt flags of the source must be cleared.

A valid interrupt can be either a high- or low-priority interrupt when in the main routine, or a high-priority interrupt when in a low-priority Interrupt Service Routine. Depending on the order of interrupt requests received and their relative timing, the CPU will be in a state of execution indicated by the STAT bit.

The state machine shown in Figure 1 and the subsequent sections detail the execution of interrupts when received in different orders.

Important: The state of GIEH/L is not changed by the hardware when servicing an interrupt. The internal state machine is used to keep track of execution states. These bits can be manipulated in the user code, resulting in transferring execution to the main routine and ignoring existing interrupts.
Figure 1. Vectored Interrupts State Transition Diagram


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# modec¶

## Purpose¶

Computes the mode of every column of a matrix.

## Format¶

m = modec(x)
Parameters

x (NxK matrix) – data

Returns

m (Kx1 matrix) – contains the mode of every column of x.

## Examples¶

### Basic example¶

// Create a 5x1 column vector
x = { 3,
4,
4,
3,
3 };

// Compute the mode of this column
m = modec(x);

After the code above, m is equal to:

3

### Multi-column mode from credit dataset¶

This example will load the variables, Cards, Age and Education from the dataset credit.dat and find the mode of each variable.

// Create the file name with full path
fname = getGAUSSHome() \$+ "examples/credit.dat";

// Load three variables by name
X = loadd(fname, "Cards + Age + Education");

// Compute the mode of each of the
// three columns in 'X'
m = modec(X);

After the code above, m will equal:

 2
44
16

Which tells us that for our sample, the most common:

• number of credit cards owned is 2.

• age is 44 years old.

• number of years of education is 16.

## Remarks¶

• If two more more numbers are tied for the most appearances, then modec() will select the smallest of them.

• modec() will not count missing values. However, if a column contains only missing values, a missing value will be returned, indicating that no valid numbers were found.

• To find the mode of the entire matrix, use the vecr() command to turn the matrix into a column vector before calling modec.

m = modec(vecr(x));

• To find out how many times each value is present, use counts().

• To compute the mode for subgroups of a variable, use aggregate().

format: plain


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Biochemical Reactions | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Biology Workbook Go to the latest version.

2.2: Biochemical Reactions

Created by: CK-12

Lesson 2.2: True or False

Name___________________ Class______________ Date________

Write true if the statement is true or false if the statement is false.

_____ 1. A substance that forms as a result of a chemical reaction is called a reactant.

_____ 2. Only some chemical reactions need energy to get started.

_____ 3. Biochemical reactions take place inside the cells.

_____ 4. A chemical reaction that releases heat is an exothermic reaction.

_____ 5. Most biochemical reactions need help to get started.

_____ 6. Anabolic reactions give off energy.

_____ 7. Metabolism is the sum of all the biochemical reactions in an organism.

_____ 8. In a chemical reaction, the quantity of an element may change.

_____ 9. During a chemical reaction, some bonds break and new bonds form.

_____ 10. Activation energy is the energy needed to start a chemical reaction.

_____ 11. An enzyme speeds up the reaction by lowering the activation energy.

_____ 12. In a chemical reaction, the number of atoms on one side of the arrow may differ from the number of atoms on the other side.

_____ 13. Matter is always conserved.

_____ 14. Understanding chemistry is needed to understand fully the processes within the cell.

_____ 15. In a chemical reaction, the quantity of each element does not change.

Name___________________ Class______________ Date________

Read these passages from the text and answer the questions that follow.

Biochemical Reactions and Enzymes

Biochemical reactions are chemical reactions that take place inside the cells of living things. Biochemistry is a relatively new field that emerged at the interface of biology and chemistry. Its emergence shows that knowledge of chemistry as well as biology is needed to understand fully the life processes of organisms at the level of the cell. The sum of all the biochemical reactions in an organism is called metabolism. It includes both exothermic and endothermic reactions.

Types of Biochemical Reactions

Exothermic reactions in organisms are called catabolic reactions. These reactions break down molecules into smaller units and release energy. An example of a catabolic reaction is the breakdown of glucose, which releases energy that cells need to carry out life processes. Endothermic reactions in organisms are called anabolic reactions. These reactions build up bigger molecules from smaller ones. An example of an anabolic reaction is the joining of amino acids to form a protein. Which type of reactions — catabolic or anabolic — do you think occur when your body digests food?

Enzymes

Most biochemical reactions in organisms need help in order to take place. Why is this the case? For one thing, temperatures are usually too low inside living things for biochemical reactions to occur quickly enough to maintain life. The concentrations of reactants may also be too low for them to come together and react. Where do the biochemical reactions get the help they need to proceed? The help comes from enzymes.

An enzyme is a protein that speeds up a biochemical reaction. An enzyme works by reducing the amount of activation energy needed to start the reaction. Less activation energy is needed when the correct enzyme is present than when it is not present.

Enzymes are involved in most biochemical reactions, and they do their job extremely well. A typical biochemical reaction could take several days to occur without an enzyme. With the proper enzyme, the same reaction can occur in just a split second! Without enzymes to speed up biochemical reactions, most organisms could not survive. The activities of enzymes depend on the temperature, ionic conditions, and the pH of the surroundings. Some enzymes work best at an acidic pH, while others work best in neutral environments.

Questions

1. What is an enzyme?

$\\\\\\\\\\$

2. How are biochemistry and metabolism related?

$\\\\\\\\\\\\$

3. Which type of reactions — catabolic or anabolic — do you think occur when your body digests food?

$\\\\\\\\\\$

4. How do enzymes work?

$\\\\\\\\\\$

5. What is activation energy?

$\\\\\\\\\\$

Lesson 2.2: Multiple Choice

Name___________________ Class______________ Date________

Circle the letter of the correct choice.

1. Reactants in the burning of methane include
1. CH4 and 2O2.
2. CO2 and 2H2O.
3. CH4 and CO2.
4. CO2 and 2O2.
2. Activities of enzymes depend on
1. pH.
2. temperature.
3. ionic conditions.
4. all of the above
3. An enzyme is a ____________.
1. carbohydrate
2. lipid
3. protein
4. nucleic acid
4. Reactions that take place inside cells are
1. cellular reactions.
2. enzyme reactions.
3. metabolic reactions.
4. biochemical reactions.
5. What is the main difference between an endothermic reaction and an exothermic reaction?
1. An endothermic reaction gives off energy and an exothermic reaction absorbs energy.
2. An exothermic reaction gives off energy and an endothermic reaction absorbs energy.
3. An endothermic reaction does not need activation energy.
4. Only endothermic reactions involve enzymes.
6. Another name for a “biological catalyst” could be a(n)
1. enzyme.
2. reactant.
3. activator.
4. metabolism.
7. The joining of amino acids to form a protein is a(n)
1. anabolic reaction.
2. catabolic reaction.
3. amino acid reaction.
4. polypeptide reaction.
8. The “push” needed to start a chemical reaction is the
1. enzymatic energy.
2. endothermic energy.
3. activation energy.
4. reactant energy.

Lesson 2.2: Vocabulary I

Name___________________ Class______________ Date________

Match the vocabulary word with the proper definition.

Definitions

_____ 1. represents a chemical reaction

_____ 2. a protein that speeds up a biochemical reaction

_____ 3. a substance that forms as a result of a chemical reaction

_____ 4. a substance that starts a chemical reaction

_____ 5. sum of all the biochemical reactions in an organism

_____ 6. a process that changes some chemical substances into others

_____ 7. exothermic reactions in organisms

_____ 8. endothermic reactions in organisms

_____ 9. chemical reactions that take place inside the cells of living things

_____ 10. a chemical reaction that releases energy

_____ 11. a chemical reaction that absorbs energy

_____ 12. the energy needed to start a chemical reaction

Terms

a. activation energy

b. anabolic reaction

c. biochemical reaction

d. catabolic reaction

e. chemical equation

f. chemical reaction

g. enzyme

h. endothermic

i. exothermic

j. metabolism

k. product

l. reactant

Lesson 2.2: Vocabulary II

Name___________________ Class______________ Date________

Fill in the blank with the appropriate term.

1. Biochemical reactions are chemical reactions that take place inside the ____________ of living things.

2. During a chemical reaction, the ____________ are used up to create the products.

3. All chemical reactions need ____________ to get started.

4. ____________ reactions in organisms are called catabolic reactions.

5. ____________ energy provides the push needed to start a chemical reaction.

6. Your ____________ includes both exothermic and endothermic reactions.

7. A chemical reaction involves the breaking and forming of ___________.

8. In a chemical reaction, all matter is ____________.

9. Energy can be released during a chemical reaction in the form of ____________ and light.

10. In a chemical reaction, there is the same amount of each ____________ in the products as there was in the reactants.

11. An ____________ reaction builds up bigger molecules from smaller ones.

12. An ____________ works by reducing the amount of activation energy needed to start the reaction.

Lesson 2.2: Critical Writing

Name___________________ Class______________ Date________

Thoroughly answer the question below. Use appropriate academic vocabulary and clear and complete sentences.

Describe the roles of enzymes in biochemical reactions. Use specifics in discussing how enzymes work. Provide an example of a biochemical reaction involving an enzyme.

Feb 23, 2012

Aug 26, 2014


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# Fundamentals of Transportation/Destination Choice/Background

Some additional background on Fundamentals of Transportation/Destination Choice

## HistoryEdit

Over the years, modelers have used several different formulations of trip distribution. The first was the Fratar or Growth model (which did not differentiate trips by purpose). This structure extrapolated a base year trip table to the future based on growth, but took no account of changing spatial accessibility due to increased supply or changes in travel patterns and congestion.

The next models developed were the gravity model and the intervening opportunities model. The most widely used formulation is still the gravity model.

While studying traffic in Baltimore, Maryland, Alan Voorhees developed a mathematical formula to predict traffic patterns based on land use. This formula has been instrumental in the design of numerous transportation and public works projects around the world. He wrote "A General Theory of Traffic Movement," (Voorhees, 1956) which applied the gravity model to trip distribution, which translates trips generated in an area to a matrix that identifies the number of trips from each origin to each destination, which can then be loaded onto the network.

Evaluation of several model forms in the 1960s concluded that "the gravity model and intervening opportunity model proved of about equal reliability and utility in simulating the 1948 and 1955 trip distribution for Washington, D.C." (Heanue and Pyers 1966). The Fratar model was shown to have weakness in areas experiencing land use changes. As comparisons between the models showed that either could be calibrated equally well to match observed conditions, because of computational ease, gravity models became more widely spread than intervening opportunities models. Some theoretical problems with the intervening opportunities model were discussed by Whitaker and West (1968) concerning its inability to account for all trips generated in a zone which makes it more difficult to calibrate, although techniques for dealing with the limitations have been developed by Ruiter (1967).

With the development of logit and other discrete choice techniques, new, demographically disaggregate approaches to travel demand were attempted. By including variables other than travel time in determining the probability of making a trip, it is expected to have a better prediction of travel behavior. The logit model and gravity model have been shown by Wilson (1967) to be of essentially the same form as used in statistical mechanics, the entropy maximization model. The application of these models differ in concept in that the gravity model uses impedance by travel time, perhaps stratified by socioeconomic variables, in determining the probability of trip making, while a discrete choice approach brings those variables inside the utility or impedance function. Discrete choice models require more information to estimate and more computational time.

Ben-Akiva and Lerman (1985) have developed combination destination choice and mode choice models using a logit formulation for work and non-work trips. Because of computational intensity, these formulations tended to aggregate traffic zones into larger districts or rings in estimation. In current application, some models, including for instance the transportation planning model used in Portland, Oregon use a logit formulation for destination choice. Allen (1984) used utilities from a logit based mode choice model in determining composite impedance for trip distribution. However, that approach, using mode choice log-sums implies that destination choice depends on the same variables as mode choice. Levinson and Kumar (1995) employ mode choice probabilities as a weighting factor and develops a specific impedance function or “f-curve” for each mode for work and non-work trip purposes.

## MathematicsEdit

At this point in the transportation planning process, the information for zonal interchange analysis is organized in an origin-destination table. On the left is listed trips produced in each zone. Along the top are listed the zones, and for each zone we list its attraction. The table is n x n, where n = the number of zones.

Each cell in our table is to contain the number of trips from zone i to zone j. We do not have these within cell numbers yet, although we have the row and column totals. With data organized this way, our task is to fill in the cells for tables headed t=1 through say t=n.

Actually, from home interview travel survey data and attraction analysis we have the cell information for t = 1. The data are a sample, so we generalize the sample to the universe. The techniques used for zonal interchange analysis explore the empirical rule that fits the t = 1 data. That rule is then used to generate cell data for t = 2, t = 3, t = 4, etc., to t = n.

The first technique developed to model zonal interchange involves a model such as this:

$T_{ij} = T_i\frac{{T_j f\left( {C_{ij} } \right)K_{ij} }} {{\sum_{j = 1}^n {T_j f\left( {C_{ij} } \right)K_{ij} } }}$

where:

• $T_{ij}$ : trips from i to j.
• $T_i$ : trips from i, as per our generation analysis
• $T_j$ : trips attracted to j, as per our generation analysis
• $f(C_{ij})$ : travel cost friction factor, say = $C_{ij}^b$
• $K_{ij}$ : Calibration parameter

Zone i generates $T_i$ trips; how many will go to zone j? That depends on the attractiveness of j compared to the attractiveness of all places; attractiveness is tempered by the distance a zone is from zone i. We compute the fraction comparing j to all places and multiply $T_i$ by it.

The rule is often of a gravity form:

$T_{ij} = a\frac{{P_i P_j }} {{C_{ij}^b }}$

where:

• $P_i; P_j$ : populations of i and j
• $a; b$ : parameters

But in the zonal interchange mode, we use numbers related to trip origins ($T_i$) and trip destinations ($T_j$) rather than populations.

There are lots of model forms because we may use weights and special calibration parameters, e.g., one could write say:

$T_{ij} = a\frac{{T_i^c T_j^d }} {{C_{ij}^b }}$

or $T_{ij} = \frac{{cT_i dT_j }} {{C_{ij}^b }}$

where:

• a, b, c, d are parameters
• $C_{ij}$ : travel cost (e.g. distance, money, time)
• $T_j$ : inbound trips, destinations
• $T_i$ : outbound trips, origin

## Entropy AnalysisEdit

Wilson (1970) gives us another way to think about zonal interchange problem. This section treats Wilson’s methodology to give a grasp of central ideas. To start, consider some trips where we have seven people in origin zones commuting to seven jobs in destination zones. One configuration of such trips will be:

Table: Configuration of Trips
zone 1 2 3
1 2 1 1
2 0 2 1

$w\left( {T_{ij} } \right) = \frac{{7!}} {{2!1!1!0!2!1!}} = 1260$

where $0!=1$

That configuration can appear in 1,260 ways. We have calculated the number of ways that configuration of trips might have occurred, and to explain the calculation, let’s recall those coin tossing experiments talked about so much in elementary statistics. The number of ways a two-sided coin can come up is 2n, where n is the number of times we toss the coin. If we toss the coin once, it can come up heads or tails, 2*1 = 2. If we toss it twice, it can come up HH, HT, TH, or TT, 4 ways, and 2*2 = 4. To ask the specific question about, say, four coins coming up all heads, we calculate 4!/4!0! =1 . Two heads and two tails would be 4!/2!2! = 6. We are solving the equation:

$w = \frac{{n!}} {{\prod_{i = 1}^n {n_i !} }}$

An important point is that as n gets larger, our distribution gets more and more peaked, and it is more and more reasonable to think of a most likely state.

However, the notion of most likely state comes not from this thinking; it comes from statistical mechanics, a field well known to Wilson and not so well known to transportation planners. The result from statistical mechanics is that a descending series is most likely. Think about the way the energy from lights in the classroom is affecting the air in the classroom. If the effect resulted in an ascending series, many of the atoms and molecules would be affected a lot and a few would be affected a little. The descending series would have a lot affected not at all or not much and only a few affected very much. We could take a given level of energy and compute excitation levels in ascending and descending series. Using the formula above, we would compute the ways particular series could occur, and we would concluded that descending series dominate.

That’s more or less Boltzmann’s Law,

$p_j = p_0 e^{\beta e_j }$

That is, the particles at any particular excitation level, j, will be a negative exponential function of the particles in the ground state, $p_0$, the excitation level, $e_j$, and a parameter $beta$, which is a function of the (average) energy available to the particles in the system.

The two paragraphs above have to do with ensemble methods of calculation developed by Gibbs, a topic well beyond the reach of these notes.

Returning to our O-D matrix, note that we have not used as much information as we would have from an O and D survey and from our earlier work on trip generation. For the same travel pattern in the O-D matrix used before, we would have row and column totals, i.e.:

Table: Illustrative O-D Matrix with row and column totals
zone 1 2 3
zone Ti \Tj 2 3 2
1 4 2 1 1
2 3 0 2 1

Consider the way the four folks might travel, 4!/2!1!1! = 12; consider three folks, 3!/0!2!1! = 3. All travel can be combined in 12*3 = 36 ways. The possible configuration of trips is, thus, seen to be much constrained by the column and row totals.

We put this point together with the earlier work with our matrix and the notion of most likely state to say that we want to

$\max w\left( {T_{ij} } \right) = \frac{{T!}} {{\prod_{ij} {Tij!} }}$

subject to

$\sum_j {T_{ij} = T_i } ; \sum_i {T_{ij} = T_j }$

where: $T = \sum_j {\sum_i {T_{ij} } } = \sum_i {T_i } = \sum_j {T_j }$

and this is the problem that we have solved above.

Wilson adds another consideration; he constrains the system to the amount of energy available (i.e., money), and we have the additional constraint,

$\sum_i {\sum_j {T_{ij} C_{ij} = C} }$

where C is the quantity of resources available and $C_{ij}$ is the travel cost from i to j.

The discussion thus far contains the central ideas in Wilson’s work, but we are not yet to the place where the reader will recognize the model as it is formulated by Wilson.

First, writing the function to be maximized using Lagrangian multipliers, we have:

$\frac{{T!}} {{\prod_{ij} {Tij!} }} + \sum_i {\lambda _i \left( {T_i - \sum_j {T_{ij} } } \right)} + \sum_j {\lambda _j \left( {T_j - \sum_i {T_{ij} } } \right) + \beta \left( {C - \sum_i {\sum_j {T_{ij} C_{ij} } } } \right)}$

where $\lambda_i, \lambda_j, and \beta$ are the Lagrange multipliers, $beta$ having an energy sense.

Second, it is convenient to maximize the natural log (ln) rather than w(Tij), for then we may use Stirling's approximation.

$\ln N! \approx N\ln N - N$

so $\frac{{\partial \ln N!}} {{\partial N}} \approx \ln N$

Third, evaluating the maximum, we have

$\frac{{\partial T}} {{\partial T_{ij} }} = - \ln T_{ij} - \lambda _i - \lambda _j - \beta C_{ij} = 0$

with solution

$\ln T_{ij} = - \lambda _i - \lambda _j - \beta C_{ij}$

$T_{ij} = e^{ - \lambda _i - \lambda _j - \beta C_{ij} }$

Finally, substituting this value of $T_ij$ back into our constraint equations, we have: $\sum_j {e^{ - \lambda _i - \lambda _j - \beta C_{ij} } } = 0 \sum_i {e^{ - \lambda _i - \lambda _j - \beta C_{ij} } } = 0$

and, taking the constant multiples outside of the summation sign

$e^{ - \lambda _j } = \frac{{T_i }} {{\sum_j {e^{ - \lambda _j - \beta C_{ij} } } }};e^{ - \lambda _i } = \frac{{T_j }} {{\sum_i {e^{ - \lambda _i - \beta C_{ij} } } }}$

let $\frac{{e^{ - \lambda _j } }} {{T_i }} = A_i ;\frac{{e^{ - \lambda _i } }} {{T_j }} = B_j$

we have

$T_{ij} = A_i B_j T_i T_j e^{ - \beta C_{ij} }$

which says that the most probable distribution of trips has a gravity model form, $T_{ij}$ is proportional to trip origins and destinations. Ai, Bj, and $\beta$ ensure constraints are met.

Turning now to computation, we have a large problem. First, we do not know the value of C, which earlier on we said had to do with the money available, it was a cost constraint. Consequently, we have to set $\beta$ to different values and then find the best set of values for $A_i$ and $B_j$. We know what $\beta$ means – the greater the value of $\beta$, the less the cost of average distance traveled. (Compare $\beta$ in Boltzmann’s Law noted earlier.) Second, the values of $beta_i$ and $\beta_j$ depend on each other. So for each value of $\beta$, we must use an iterative solution. There are computer programs to do this.

Wilson’s method has been applied to the Lowry model.

## ReferencesEdit

• Allen, B. 1984 Trip Distribution Using Composite Impedance Transportation Research Record 944 pp. 118-127
• Barnes, G. and Davis, G. 2000. Understanding Urban Travel Demand: Problems, Solutions, and the Role of Forecasting, University of Minnesota Center for Transportation Studies: Transportation and Regional Growth Study
• Ben-Akiva M. and Lerman S. 1985 Discrete Choice Analysis, MIT Press, Cambridge MA
• Boyce, D., Lupa, M. and Zhang, Y.F. 1994 Introducing “Feedback” into the Four-Step Travel Forecasting Procedure vs. the Equilibrium Solution of a Combined Model presented at 73rd Annual Meeting of Transportation Research Board
• Evans, Suzanne P. 1976 . Derivation and Analysis of Some Models for Combining Trip Distribution and Assignment. Transportation Research, Vol. 10, PP 37-57 1976
• Florian M., Nguyen S., and Ferland J. 1975 On the Combined Distribution-Assignment of Traffic", Transportation Science, Vol. 9, pp. 43-53, 1975
• Haney, D. 1972 Consistency in Transportation Demand and Evaluation Models, Highway Research Record 392, pp. 13-25 1972
• Hansen, W. G. 1959. How accessibility shapes land use. Journal of the American Institute of Planners, 25(2), 73-76.
• Heanue, Kevin E. and Pyers, Clyde E. 1966. A Comparative Evaluation of Trip Distribution Procedures,
• Levinson, D. and A. Kumar 1994 The Rational Locator: Why Travel Times Have Remained Stable, Journal of the American Planning Association, 60:3 319-332
• Levinson, D. and Kumar A. 1995. A Multi-modal Trip Distribution Model. Transportation Research Record #1466: 124-131.
• Portland MPO Report to Federal Transit Administration on Transit Modeling
• Reilly, W.J. 1929 “Methods for the Study of Retail Relationships” University of Texas Bulletin No 2944, Nov. 1929.
• Reilly, W.J., 1931 The Law of Retail Gravitation, New York.
• Ruiter, E. 1967 Improvements in Understanding, Calibrating, and Applying the Opportunity Model Highway Research Record No. 165 pp. 1-21
• Stewart, J.Q. 1948 “Demographic Gravitation: Evidence and Application” Sociometry Vol. XI Feb.-May 1948 pp 31-58.
• Stewart, J.Q., 1947. Empirical Mathematical Rules Concerning the Distribution and Equilibrium of Population, Geographical Review, Vol 37, 461-486.
• Stewart, J.Q., 1950. Potential of Population and its Relationship to Marketing. In: Theory in Marketing , R. Cox and W. Alderson (Eds) ( Richard D. Irwin, Inc., Homewood, Illinois).
• Stewart, J.Q., 1950. The Development of Social Physics, American Journal of Physics, Vol 18, 239-253
• Voorhees, Alan M., 1956, "A General Theory of Traffic Movement," 1955 Proceedings, Institute of Traffic Engineers, New Haven, Connecticut.
• Whitaker, R. and K. West 1968 The Intervening Opportunities Model: A Theoretical Consideration Highway Research Record 250 pp.1-7
• Wilson, A.G. A Statistical Theory of Spatial Distribution Models Transportation Research, Volume 1, pp. 253-269 1967
• Wohl, M. 1963 Demand, Cost, Price and Capacity Relationships Applied to Travel Forecasting. Highway Research Record 38:40-54
• Zipf, G.K., 1946. The Hypothesis on the Intercity Movement of Persons. American

Sociological Review, vol. 11, Oct

• Zipf, G.K., 1949. Human Behaviour and the Principle of Least Effort. Massachusetts


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# Tensor index question

I am looking at the solution in the book "Problem book in Relativity and Gravitation" for problem 10.6. I don't think I need to go into the details of the problem (I will do so if need be) because I am only confused about one step (the tensor stuff is still fairly new to me). The book can be found for free online here: http://apps.nrbook.com/relativity/index.html

So I start with the following equation

$$\xi_{\mu;\nu\lambda}-\xi_{\mu;\lambda\nu} = R_{\mu\sigma\lambda\nu}\xi^{\sigma}$$ where $R_{\mu\sigma\lambda\nu}$ is the Riemann curvature tensor, and the $\xi$ are killing vectors. The solutions manual says the following:

"Now we use the Killing equation $\xi_{\mu;\nu}=-\xi_{\nu;\mu}$ and contract $\mu$ and $\lambda$." $$\xi^{\nu;\lambda}_{\;\;\;;\lambda} + R^{\nu}_{\;\sigma}\xi^{\sigma}=-(\xi^{\mu}_{\;;\mu})^{;v}$$

I am having trouble getting this. So okay, I use the Killing equation, and I believe that gives me the following: $$-\xi_{\nu;\mu\lambda}-\xi_{\mu;\lambda\nu} = R_{\mu\sigma\lambda\nu}\xi^{\sigma}$$ So now I need to contract $\mu$ and $\lambda$. So to do that, I need to raise the appropriate indices. But what is tripping me up is dealing with the covariant derivative. Is the covariant derivative invariant through index contractions? Am I allowed to do something like this? $$\xi^{\nu}_{\;;\mu\lambda}=g^{\nu\delta}\xi_{\delta;\mu\lambda}$$

What does it mean to even raise or lower the covariant derivative? Since $\xi_{\nu;\mu}=\nabla_{\mu}\xi_\nu$. How does this differ for $\xi_{\nu}^{\;;\nu}$? Can I raise this in the same way as any other index???

• That is the beauty of tensor notation. One can raise or lower any of these indices with the aid of the metric tensor, regardless of whether they are associated with derivatives or not. – flippiefanus Sep 30 '16 at 4:32

To flesh out a bit more what @Erik Jörgenfelt answered (I think correctly):

I am not familiar with "raised"/(contravariant?) covariant derivatives but formally it is clear how to construct it: $$\xi_{\mu}^{~~;\nu}=g^{\nu\alpha}\xi_{\mu;\alpha}.$$ I would understand it as the normal covariant derivative with the differentiation index raised. Formally this is not a problem: the covariant derivative of a vector transforms like a tensor of rank two: so raising and lowering indices with the metric are well defined operations. So in that sense it is nothing new. If one accepts this it is straight forward to get from your first to your second equation. So lets first use the Killing equation: $$\xi_{\mu;\nu\lambda}-\xi_{\mu;\lambda\nu}=-\xi_{\nu;\mu\lambda}-\xi_{\mu;\lambda\nu}=R_{\mu\sigma\lambda\nu}\xi^\sigma.$$ Now before we contract something we need to raise something (lets raise $\mu$):

$$-{\xi_\nu^{~~;\mu}}_{;\lambda}-\xi^\mu_{~~;\lambda\nu}=R^\mu_{~~\sigma\lambda\nu}\xi^\sigma.$$

Now we sum over $\mu=\lambda$ and use $R^\lambda_{~~\sigma\lambda\nu}=R_{\sigma\nu}=R_{\nu\sigma}$ to get to:

$$-{\xi_\nu^{~~;\lambda}}_{;\lambda}-\xi^\lambda_{~~;\lambda\nu}=R_{\nu\sigma}\xi^\sigma.$$ So in the last two steps we contracted the indices $\mu$ and $\lambda$. To get to what we want out lets raise $\nu$:

\begin{align} -{\xi^{\nu;\lambda}}_{;\lambda}-{\xi^\lambda_{~~;\lambda}}^{;\nu}&=R^\nu_{~~\sigma}\xi^\sigma\\ {\xi^{\nu;\lambda}}_{;\lambda}+R^\nu_{~~\sigma}\xi^\sigma&=-{\xi^\lambda_{~~;\lambda}}^{;\nu}. \end{align}

This is equivalent to your second equation, it looks odd to me to use $\lambda$ and $\mu$ as summation indices after one contracted them. It is formally ok, since you can chose the labels for summation indices in different terms as you please, but I would stick with one of the two not both. We are left with a covariant derivative of a covariant divergence (RHS), the term @Erik Jörgenfelt explained (LHS 1. term) and the Ricci tensor contracted with $\xi$ (LHS 2. term).

So form a tensor calculus point of view this should be the way to get from your first to second equation.

In short: yes, you can manipulate the indices representing the covariant derivative just like you can any other indices. For example $$\xi^{\nu;\lambda}{}_{;\lambda} = g^{\lambda\mu}\xi^\nu{}_{;\mu\lambda} = \nabla^2\xi^\nu,$$ is just the covariant derivative taken twice in such a manner that the direction of the second covariant derivative is given by the gradient, i.e. the first covariant derivatve. Notice the similarity with the Laplacian of standard vector analysis.


text:
# Prove that If $x$ is an accumulation point of $S$, then every neighborhood of $x$ contains infinitely many points of $S$

Let $S \subseteq \mathbb{R}$.

To prove that every neighborhood of $x$ contains infinitely many points of $S$ whenever $x$ is an accumulation point of $S$, we will suppose to the contrary that there exists a neighborhood $N$ of $x$ such that

$$N \cap S = \{s_1, \cdots, s_n\}$$

Since $x$ is an accumulation point of $S$, then every deleted neighborhood $N^*_{\delta} \cap S \neq \varnothing$. We then select a particular $\delta = \min\{|x-s_i| \, : \, s_i \neq x\}$. Then $\delta > 0$, but we have

$$N^*_{\delta} \cap \, (N \cap S) = \varnothing$$

which is a contradiction, so every neighborhood of $x$ contains infinitely many points of $S$.

Are there any problems with this proof?

• The claim is not true in general. For example in a finite space with indiscrete topolgy, every point is an accumulation point of ecevery nonempty subset. Does the use of $|\cdot|$ mean that the problem is specifically about $S\subseteq \Bbb R$? – Hagen von Eitzen May 18 '16 at 20:03
• yes, $S \subseteq \mathbb{R}$ – Jeremiah Dunivin May 18 '16 at 20:05
• Looks good. You should point out that $N_{\delta}^* \cap S = \emptyset$ rather than $N_{\delta}^* \cap(N\cap S)$ though. – Alex G. May 18 '16 at 20:12
• But is it clear that $N^*_{\delta} \cap S = \varnothing$? – Jeremiah Dunivin May 18 '16 at 20:31
• @AlexG. I don't think we any information that $N^*_{\delta} \cap S = \varnothing$, since $S$ is an arbitrary subset of the reals. The only fact that we do have is that $N^*_{\delta} \cap (N\cap S)$ is empty by definition of $\delta$. Please correct if I'm wrong. – Jeremiah Dunivin May 19 '16 at 11:13

Let $B_\alpha$ be a neighborhood of $x$. That is:

$$B_\alpha = \{z : |x - z| < \alpha\},$$

for $\alpha > 0$.

Take the sequence $$a_n = x + \frac{1}{n}.$$ This converges to $x$ as $n$ approaches to $+\infty$. This means that $x$ is accumulation point for some sets that contain $s$. Moreover:

$$\forall \varepsilon > 0 \exists m \in \mathbb{N} : n > m \Rightarrow a_n \in B_\varepsilon.$$

Now, consider the sets $S = B_\varepsilon$, $N = B_\alpha$ and $S \cap N$. It's clear that you can always find a $m'$ such that

$$n > m' \Rightarrow a_n \in S \cap N,$$

and the $a_n$'s with $n > m'$ are, of course, infinite.

• Nice direct proof – Jeremiah Dunivin May 18 '16 at 21:36


text:
# text bbox, postscript formatting problem

It looks like there is a bug in the text bounding boxes. It causes some
problems in the postscript backend, where I noticed badly formatted logscale
ticks. It looks like the bug exists in the AGG backend as well, although it
is less noticeable there. I filed two bugs at the sf website, with images of
bboxes that dont fit their associated text. You can also run the script
below, but you will need a recent update of patches.py, John fixed a bug
there today.

I assigned the PS bug to myself. If anyone has some insight, I could
definitely use it.

Darren

http://sourceforge.net/tracker/index.php?func=detail&aid=1177396&group_id=80706&atid=560720

from pylab import *

text(0.5, 0.5, 'xx-small', props, fontsize='xx-small')
text(1.5, 0.5, 'x-small', props, fontsize='x-small')
text(2.5, 0.5, 'small', props, fontsize='small')
text(3.5, 0.5, 'medium', props, fontsize='medium')
text(4.5, 0.5, 'large', props, fontsize='large')
yticks([0,.5,1])

# the text bounding box

figure(1,figsize=(5,2))
axes()
xlim(0,5)
savefig('/home/darren/temp/bbox.eps')
savefig('/home/darren/temp/bbox.png')
show()

Hi Darren,

I assigned the PS bug to myself. If anyone has some insight, I could
definitely use it.

Sorry, I do not have much time at the moment, but I remember
something: the PostScript driver always got the horizontal text
extents slightly wrong, because PostScript seems to not do kerning by
itself. It is a small effect, but visible. This is especially
visible when comparing strings like "AAAAA" and "AVAVAV". Is this
what you are seeing?

To fix this one should get the kerning information from the font,
and do the kerning manually. Have a look at dvips output to get
inspiration.

All the best,
Jochen

···

On Mon, May 02, 2005 at 04:03:17PM -0400, Darren Dale wrote:
--
http://seehuhn.de/

I have a little more information. When I set ps.useafm =True in .matplotlibrc,
eps output looks good, there is no mismatch between the bboxes and the text.
I'm guessing the bug may be in either backend_ps.encodeTTFasPS, or somewhere
in ft2font.h. Unfortunately, I have zarro experience with C++, so I'm having
a hard time interpretting ft2font.

If anyone can offer advice, I am a little desperate. Half of my plots are
logscale, so I need to fix this in order to generate figures for my thesis.
Is there anywhere I can look to learn the basics of dealing with fonts at
this level? Any recommendations on C++ crash courses?

···

On Monday 02 May 2005 4:03 pm, Darren Dale wrote:

It looks like there is a bug in the text bounding boxes. It causes some
problems in the postscript backend, where I noticed badly formatted
logscale ticks. It looks like the bug exists in the AGG backend as well,
although it is less noticeable there. I filed two bugs at the sf website,
with images of bboxes that dont fit their associated text. You can also run
the script below, but you will need a recent update of patches.py, John
fixed a bug there today.

--
Darren S. Dale

Bard Hall
Department of Materials Science and Engineering
Cornell University
Ithaca, NY. 14850

dd55@...143...

Hi Jochen,

Thanks for responding. I'll continue looking into this tomorrow. For now, I
posted a new eps file in the sourceforge bugreport, which renders the
suggested strings. You are right, the effect is especially visible when
rendering 'AVAVAV'.

By the way, did you know MoonBuggy is in the gentoo package management tree?

···

On Tuesday 03 May 2005 9:51 am, Jochen Voss wrote:

Hi Darren,

On Mon, May 02, 2005 at 04:03:17PM -0400, Darren Dale wrote:
> I assigned the PS bug to myself. If anyone has some insight, I could
> definitely use it.

Sorry, I do not have much time at the moment, but I remember
something: the PostScript driver always got the horizontal text
extents slightly wrong, because PostScript seems to not do kerning by
itself. It is a small effect, but visible. This is especially
visible when comparing strings like "AAAAA" and "AVAVAV". Is this
what you are seeing?

To fix this one should get the kerning information from the font,
and do the kerning manually. Have a look at dvips output to get
inspiration.

--
Darren S. Dale

Bard Hall
Department of Materials Science and Engineering
Cornell University
Ithaca, NY. 14850

dd55@...143...

Hi Darren,

> I assigned the PS bug to myself. If anyone has some insight, I could
> definitely use it.

Sorry, I do not have much time at the moment, but I remember
something: the PostScript driver always got the horizontal text
extents slightly wrong, because PostScript seems to not do kerning by
itself. It is a small effect, but visible. This is especially
visible when comparing strings like "AAAAA" and "AVAVAV". Is this
what you are seeing?

To fix this one should get the kerning information from the font,
and do the kerning manually. Have a look at dvips output to get
inspiration.

I just finished reading Adobe's Technical Note #5012: The Type 42 Font Format
Specification. The only mention of kerning is on page 12, indicating that
performance can be improved by including only those type 42 tables that are
actually used by the TrueType rasterizer. It implies that kerning tables are
not used.

I also had a look at the dvips output, and get the general idea of how it
handles layout. So let me ask, why does mathtext break with afm fonts? Is it
inherent, or could mathtext.py (or backend_ps.py) be changed to cooperate
with afm? From looking at the dvips output, a naive individual like myself
might think it is pretty straight forward to render something like
r'My test string $A=A_0e^(i\theta)$'.
(I'm not saying it would be easy to implement.)

Dvips embeds four fonts to render my example string: CMMI7, CMR7, CMMI10,
CMR10. Here is the output, I clipped the start of the file up to the end of
the font defs:

%%EndFont
TeXDict begin 40258431 52099146 1000 600 600 (temp.dvi)
@start /Fa 150[23 86[32 18[{}2 58.1154 /CMMI7 rf /Fb
207[33 48[{}1 58.1154 /CMR7 rf /Fc 154[39 35[62 65[{}2
83.022 /CMMI10 rf /Fd 134[44 4[32 33 33 3[46 4[23 1[42
1[37 23[76 15[65 11[42 49[{}11 83.022 /CMR10 rf end
%%EndProlog
%%BeginSetup
%%Feature: *Resolution 600dpi
TeXDict begin
end
%%EndSetup
%%Page: 1 1

I'm not sure what that first block is doing (yet), but here is the important
part:

TeXDict begin 1 0 bop 639 523 a Fd(My)28 b(test)g(string)f
Fc(A)c Fd(=)g Fc(A)1420 535 y Fb(0)1457 523 y Fc(e)1496
493 y Fa(i\022)1926 5255 y Fd(1)p eop end

Its essentially a group of moveto statements and font changes. It looks like
the kerning support is sort of a hybrid, where you dont have to layout each
glyph independently, but layout is done manually between groups of glyphs.

I guess I havent considered rendering in AGG.

Darren

···

On Tuesday 03 May 2005 9:51 am, Jochen Voss wrote:

On Mon, May 02, 2005 at 04:03:17PM -0400, Darren Dale wrote:


text: Write the equilibrium constant expression for the overall reaction. Omitting the water molecules bound to Cu2+ for simplicity, we can write the equilibrium reactions as follows: The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four $$NH_3$$ ligands, not six. (8) This is NOT an exhaustive list of shapes, but these are the ones you are most likely to encounter in an introductory inorganic chemistry course. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0, That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2− complex ion. AP Chem students need to be able to write balanced net ionic equations for the formation of complex ions. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. The Roman numeral III indicates the oxidation state of Ruthenium. Subscribe to RSS headline updates from: Powered by FeedBurner. (7) If more than one type of ligand is present, the symbols are given in alphabetical order, eg, if Cl- and NH3 both occur as ligands in the same coordination compound, then, because C occurs before N in the alphabet, so we write [CrCl2(NH3)4]+ The formula of the complex is written [Co (H 2 O) 6] 3+ The hexaamminenickel (II) ion at the right has six ammonia molecules surrounding the central nickel 2+ ion in an octahedral arrangement. Write the formula and the name of each complex ion. Hence, the formula of pentaammineaquaruthenium (III) ion is [ Ru (NH 3) 5 H 2 O] 3 +. A complex is a substance in which a metal atom or ion is associated with a group of neutral molecules or anions called ligands.Coordination compounds are neutral substances (i.e. $Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{17.3.2}$. A negatively charged complex ion is called an anionic complex. The formation of a complex ion is a stepwise process and … No ads = no money for us = no free stuff for you! The shape, or geometry, of a number of complex ions is given in the table below: Footnotes: reference "Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005" (Red Book). Examples of complex ions are [Fe(H 2 O) 6] 2+, [CoCl 4] 2-, [Cu(NH 3) 4 (H 2 O) 2] 2+, [V(H 2 O) 6] 3+. Solve for the concentration of the complex ion. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation $$\ref{17.3.2}$$. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.). Note the change: previously we wrote the symbols for anionic ligands before the symbols for neutral ligands. X in the formula is equal the number of H 2 O ligands displaced by NH 3 ligands. Write the formula: The solution could also be called tetraquacopper (II) sulfate because there is a sulfate ion present. [Al(H 2 O) 6] 3+ is usually just called the hexaaquaaluminium ion rather than the hexaaquaaluminium(III) ion. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. Further additions of metal ion give solutions that contain insufficient salicylic acid to complex with all the metal, so absorbance due to the complex … Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M. If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water. Learn how in this brief webcast! The 2+ ion is almost the same colour as the hexaaquacopper(II) ion, and the 3+ ion is the hard-to-describe violet-blue-grey colour. Recent developments in chemistry written in language suitable for students. For negatively charged complex ions. Complex Ion Equilibria Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. In this case, the expression is as follows: $K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{17.3.3}$. Kf is called the formation constant. Missed the LibreFest? The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The reaction for the dissolution of silver bromide is as follows: $AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{17.3.4a}$, $K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{17.3.4b}$. The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Examples: The copper ammine ion, Cu (NH 3) 62+ is a complex ion. One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. # Objectives!! a. a complex ion with $\mathrm{Co}^{3+}$ as the central ion and three $\mathrm{NH}_{3}$ molecules and three CN- ions as ligands b. a complex ion with $\mathrm{Cr}^{3+}$ as the central ion and a coordination number of 6 with ethylenediamine ligands (4) This is true for the names of simple ligands. The central ion is the coordination center, while the molecules or ions bound to it are termed complexing agents or ligands. The key to using complex-ion equilibria to dissolve an insoluble salt is simple: Choose a complex for which K f is large enough that the concentration of the uncomplexed metal ion is too small for the ion product to … The aim of this investigation is to determine the formula of the complex ion formed from copper (II) ions and ammonia through colorimetry. The formation constants for some common complex ions are listed in Table $$\PageIndex{1}$$. To# explore# complex# ions# and# coordination numbers.# To#use#absorptionspectroscopy#and#Job’s#method# to#determine#the#formula#of#a#complex#ion.# # Introduction! Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $$\PageIndex{1}$$. Experiment!23:!Determinationof!theFormula!of!a!Complex!Ion! The equilibrium constant for the formation of the complex ion is the formation constant (Kf). Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. This form of additive nomenclature has been in use for a very long time, but it has problems, notably it may not accurately reflect the distribution of electrons within the complex ion. Curriculum Notes . Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions. (1) Historically, coordination compounds were considered to be formed by adding independently stable compounds to a simple central atom or ion. then the shape of the complex ion is linear, If coordination number (= number of ligands) = 4 Differing from zinc and mercury, cadmium can form the complex… We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Complex Formula with Wildcard Characters. A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. Technique:AbsorbanceSpectroscopy! The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. The development of phosphate substitutes is an area of intense research. For example, if the lookup value is “VIVO” and in the main table if it is “VIVO Mobile” we can still match using wildcard characters. Writing formula of complex ions To write the formula of a complex ion you must remember to: • Place the whole formula in square brackets [ ] • Place the charge on the outside of the square brackets [Ag(NH 3) 2]+ Working out the charge of complex ions OH‐has a charge of ‐1NH 3 has a charge of 0 A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M. The ferrocyanide ion {[Fe(CN)6]4−} is very stable, with a Kf of 1 × 1035. A coordination complex consists of a central atom or ion, which is usually metallic and is called the coordination centre, and a surrounding array of bound molecules or ions, that are in turn known as ligands or complexing agents. A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows: Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Concept!Under!Investigation:!Complex!Ions! Central Ion Name Complex Ion Name [Ag(NH 3)] 2 + ammine : 2 → di: silver(I) +1= x + 2(0) x = +1: diamminesilver(I) ion (complex is a cation) THEORY The Ni 2+ cation forms a complex ion with H 2 O ligands in aqueous solutions. Coordination compounds are neutral substances (i.e. This is the traditional slogan but we can still fetch the data by using particle lookup value. Send Gift Now Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Empirical formula determination is difficult with a mixture of the two complex ions. Identify the oxidation state of the central metal ion (shown in parantheses): Calculate the total charge on the ligands: Write the formula giving the central metal ion first followed by the ligands, everything enclosed in square brackets followed by the charge on the ion as a superscript: If coordination number (= number of ligands) = 2 The numbers of ligands in a complex are specified using a Greek prefix: Oxidation state of the central metal atom is shown with a Roman numeral in parantheses at the end of the metal's name: The complex ion is an anion, therefore the name will end in ferrate(III), Ligands are named before central metal ion: hexacyanidoferrate(III). They are a type of coordination complex. As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Also note that hydride becomes hydrido when coordinating to all elements EXCEPT boron. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled. The formula of complex ion formed when aluminum hydroxide dissolves in sodium hydroxide will be: Select the correct answer below: [AlOH]2+ [Al(OH)2]+ If the oxidation state is zero, then an arabic 0 is placed in parentheses. Anionic ligands have names ending in 'o'. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. Six is the most common coordination number, although 4 and 2 are also known. The hydroxyaluminate or hydrated aluminate can be precipitated and then calcined to produce anhydrous aluminates. eg, [CoCl(NH3)5]Cl2 is pentaaminechloridocobalt(2+) chloride. If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts. To determine: The formula of the complex ion pentaammineaquaruthenium (III) ion. The replacement of water molecules from [Cu(H2O)6]2+ by ammonia occurs in sequential steps. Pay for 5 months, gift an ENTIRE YEAR to someone special! So the equilibrium lies to the right, and this a stable complex ion. Note the change of name when water is a ligand from the previous "aquo" to the new "aqua". Legal. ? A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion. Naming Coordination Compounds. The value of x indicates that our assumption was justified. For a more complete description, see https://www.youtube.com/watch?v=IQNcLH6OZK0. As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. There are other, arguably better, systematic IUPAC ways to name a complex ion, for example, you could name the complex ion as an ion (see Naming cations and Naming Anions) in which case you enclose the charge on the ion in round brackets after the additive name (no need to try to determine the oxidation state of the metal and hence removes the problem of electron distribution). Given: Ksp of AgCl, Kf of AgCl2−, and KCl concentration, Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions. VLOOKUP requires an exact lookup value to be matched to fetch the data. Observe the peacock blue solution of copper (II) sulfate. For complex ligand names, the prefixes bis, tris, tetrakis etc are used with enclosing marks around the multiplicand in order to aviod ambiguity. (3) The names of cationic ligands are also used without modification, even if the name ends in "ide", "ate" or "ite". In the given ion, metal ion Ruthenium (Ru 3 +) contains five Ammonia (NH 3) and one Water (aqua) ligand. Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Their order in the formula is the opposite of that in the complex's name since one uses their chemical symbols and the other uses the names of the ligands. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. You will learn more about coordination compounds in the lab lectures of experiment 4 in this course. Each blog post includes links to relevant AUS-e-TUTE tutorials and problems to solve. Have questions or comments? Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. then the shape of the complex ion is tetrahedral or square-planar, If coordination number (= number of ligands) = 6 then the shape of the complex ion is octahedral. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K4[Fe(CN)6]. (6) If more than one type of ligand is present, the ligands are named in alphabetical order (disregard any multiplicative prefixes when determining that order), ⚛ iron(II) → ferrate(II) The formation of a complex ion by adding a complexing agent increases the solubility of a compound. And notice the equilibrium constant for this. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together: Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. Splitting is greater if the ion is octahedral than if it is tetrahedral, and therefore the colour will change with a change of co-ordination. Identify the oxidation state on the central metal ion (shown in Roman numerals parantheses), Calculate the total charge on the ligands. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. uncharged) in which at least one ion is present as a complex. ⚛ tin(IV) → stannate(IV). When mixed with aqueous NH 3, some of the 6 H 2 O ligands are displaced with NH 3 ligands forming a complex ion Ni(H 2 O) 6-X (NH 3) X 2+. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. (a) What is the formula of a complex ion composed of one Co 3+ ion, three ammonia molecules, and three Cl − ions? An aqueous solution of CuSO4 consists of hydrated Cu2+ ions in the form of pale blue [Cu(H2O)6]2+ (left). Figure $$\PageIndex{2}$$: An MRI Image of the Heart, Arteries, and Veins. Correct formula for the complex ion is [Ni(en) 3 ]2+ 16 Determining the Coordination Number of Ni 2+ and Cu by Enthalpy Sample Data Calculations for the [Ni(OH 26) ]2+ series: Trial 1 Q = [55 ml soln x 1.00 g soln x 0.80 °C x 4.18 J + 0.80 °C x 10 J] -1 When dissolved in bases it forms hydroxyaluminate ions in the same way as aluminium hydroxide or aluminium salts. Given: mass of Cu2+ salt and volume and concentration of ammonia solution, Asked for: equilibrium concentration of Cu2+(aq). ⚛ gold(I) → aurate(I) We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible. ⚛ silver(I) → argentate(I) Write the formula of a complex ion with four water molecules and two ONO- ions connected to an Fe (III) ion. Aluminate oxyanions. Its formula is Ni(H 2 O) 6 2+. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The sulfate ion is not part of the complex ion. Watch the recordings here on Youtube! Please enable javascript and pop-ups to view all page content. The mixture becomes clear due to the formation of a complex ion that is soluble in water. The complexing agents used for gadolinium are ligands such as DTPA5− (diethylene triamine pentaacetic acid), whose fully protonated form is shown here. 1.6 times 10 to the seventh. To be introduced to complex ions, including ligands. Complex ions that form with the Fe3+ have a different number of oxalate groups than those that form with Fe2+. (2) Previously, anionic ligands were named by removal of "ide" and replacement by "o", for example chloride became chloro, and, removal of "ate" and replacement with "o" for example sulfate became sulfo. Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. Write the formula giving the central metal ion first followed by the ligands, and enclose everything except the charge in square brackets: The coordination number of the complex ion is the number of bonds formed between the central metal ion and its ligands. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. A complex ion is a polyatomic species consisting of a central metal ion surrounded by several ligands. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2−). As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br− ions are very low (7.31 × 10−7 M). A coordination complex is a substance in which a metal atom or ion accepts electrons from (and thus associates with) a group of neutral molecules or anions called ligands.A complex can be an anion, a cation ion, or a neutral molecule. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Adding a complexing agent such as pyrophosphate (O3POPO34−, or P2O74−) or triphosphate (P3O105−) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large: However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. The nickel ion has an oxidation state of 2+ and each of the ammonia molecules is neutral, so the overall charge on the complex ion is: 2+ + 0 = 2+ ]^2+ ion the four ammine ligands are named first. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $$\PageIndex{2}$$). When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. ⚛ copper(I) → cuprate(I) Calculate the solubility of AgCl in each situation: At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2−. The table below lists some common polar molecules which can be present as ligands: The table below lists some anions which can be present as ligands: Example: Name the complex ion with the forumla [Fe(CN)6]3-, Example: write the formula for the complex ion tetraamminecopper(II). A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H 2 O) 6] 3+ ion. D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2−] = x and [Cl−] = 1.0 − x. Naming of these coordination compounds was therefore based on an additive principle whereby the names of added compounds and the central atom were combined. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt. Hence, the name of the complex ion is tetra ammine copper (2) 3) A complex anion named by writting ligands followed by name of the metal ending with suffix- ate. Aluminium oxide (alumina) is amphoteric: it dissolves in both bases and acids. It is followed by copper metal along with its oxidation state. Note the changes: final "e" is removed and replaced by "o", chloride becomes chlorido and sulfate becomes sulfato. Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. One such example occurs in conventional black-and-white photography. Amminetetraaquachromium(II) ion would be written as [Cr(H 2 O) 4 (NH 3)] +2. uncharged) in which at least one ion is present as a complex. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ion. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant (Kf). I. In this tutorial we will name the complex ion using the oxidation state for the central metal atom. (1999). Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32−). As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ion {[Cu(H2O)6]2+}. Most transition metal cations can do something interesting in solution, they can interact with specific ligands to form complex ions. Using Complex Ion Equilibria to Dissolve an Insoluble Salt . Ligands are named before the central metal atom. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations. Some content on this page could not be displayed. Write the formula and name of (a) a complex ion having \mathrm{Cr}^{3+} as the central ion and two \mathrm{NH}_{3} molecules and four \mathrm{Cl}^{-} ions as l… Give the gift of Numerade. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. Please do not block ads on this website. Neutral ligands are named as the molecule without modification, with these notable exceptions: The numbers of ligands in a complex are specified using the Greek prefixes: The name of a cationic complex ion ends in the name of the central metal ion with the, The name of an anionic complex ion ends in '. The reaction of Ag+ with thiosulfate is as follows: $Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{17.3.5a}$, $K_f = 2.9 \times 10^{13} \label{17.3.5b}$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion. (adsbygoogle = window.adsbygoogle || []).push({}); Want chemistry games, drills, tests and more? If 12.5 g of Cu(NO3)2•6H2O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)? The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. Many metal-containing compounds, especially those of transition metals, are coordination complexes. [ "article:topic", "showtoc:no", "license:ccbyncsa", "program:hidden" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FGeneral_Chemistry%2FBook%253A_Chemistry_(Averill_and_Eldredge)%2F17%253A_Solubility_and_Complexation_Equilibria%2F17.3%253A_The_Formation_of_Complex_Ions, \begin{align}\mathrm{Cu^{2+}(aq)}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)]^{2+}_{(aq)}}\hspace{5mm}K_1, \[12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M}, \begin{align}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13}, \begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13}, \(\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10}, \(K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0-x}\approx1.9\times10^{-5}=x, \[Ca^{2+}_{(aq)} + O_3POPO^{4−}_{4(aq)} \rightleftharpoons [Ca(O_3POPO_3)]^{2−}_{(aq)} \label{17.3.7a}, 17.0: Prelude to Solubility and Complexation Equilibria, The Effect of the Formation of Complex Ions on Solubility, https://www.youtube.com/watch?v=IQNcLH6OZK0, information contact us at info@libretexts.org, status page at https://status.libretexts.org. 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# Conditional distribution and correlation in multivariate normal distribution.

Consider the random vector $(X_1, X_2, X_3)^T \sim \mathcal{N}_3 (\underline{\mu}, C)$, where $\underline{\mu}=(\mu_1, \mu_2, \mu_3)^T$ and

$$C=\begin{bmatrix} 1 & \rho & \rho^2 \\ \rho & 1\ & 0\\ \rho^2 & 0 & 1 \end{bmatrix}$$

$(0 < \rho < 1)$ is a given parameter. Determine the conditional distribution of $(X_1, X_2)^T$ conditioned on $X_3=x_3$.

Now suppose, that $\mu_1=\mu_2=\mu_3=0$, so $(X_1, X_2, X_3)^T \sim \mathcal{N}_3 (\underline{0}, C)$ and $C$ remains the same.

Find the multiple correlation between $(X_1, X_2)^T$ and $X_3$.

Find the partial correlation between $X_1$ and $X_2$, after eliminating the effect of $X_3$.

I am familiar with the definition of conditional distribution and correlation, but I haven't really solved any exercises in multivariate cases. Anything which can help me to get started is appreciated.

• I'm assuming all instances of ! are indeed exclamation marks rather than factorial notation. Please remove these to avoid ambiguity. – Jason Oct 3 '17 at 18:25
• Hopefully it won't cause confusion now :) – Atvin Oct 3 '17 at 18:26