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In this project, you will employ several supervised algorithms of your choice to accurately model individuals’ income using data collected from the 1994 U.S. Census. You will then choose the best candidate algorithm from preliminary results and further optimize this algorithm to best model the data. Your goal with this implementation is to construct a model that accurately predicts whether an individual makes more than $50,000. This sort of task can arise in a non-profit setting, where organizations survive on donations. Understanding an individual’s income can help a non-profit better understand how large of a donation to request, or whether or not they should reach out to begin with. While it can be difficult to determine an individual’s general income bracket directly from public sources, we can (as we will see) infer this value from other publically available features. The dataset for this project originates from the UCI Machine Learning Repository. The datset was donated by Ron Kohavi and Barry Becker, after being published in the article “Scaling Up the Accuracy of Naive-Bayes Classifiers: A Decision-Tree Hybrid”. You can find the article by Ron Kohavi online. The data we investigate here consists of small changes to the original dataset, such as removing the 'fnlwgt' feature and records with missing or ill-formatted entries. ## Exploring the Data Run the code cell below to load necessary Python libraries and load the census data. Note that the last column from this dataset, 'income', will be our target label (whether an individual makes more than, or at most,$50,000 annually). All other columns are features about each individual in the census database.

age workclass education_level education-num marital-status occupation relationship race sex capital-gain capital-loss hours-per-week native-country income
0 39 State-gov Bachelors 13.0 Never-married Adm-clerical Not-in-family White Male 2174.0 0.0 40.0 United-States <=50K

### Implementation: Data Exploration

A cursory investigation of the dataset will determine how many individuals fit into either group, and will tell us about the percentage of these individuals making more than $50,000. In the code cell below, you will need to compute the following: • The total number of records, 'n_records' • The number of individuals making more than$50,000 annually, 'n_greater_50k'.
• The number of individuals making at most $50,000 annually, 'n_at_most_50k'. • The percentage of individuals making more than$50,000 annually, 'greater_percent'.

** HINT: ** You may need to look at the table above to understand how the 'income' entries are formatted.

Total number of records: 45222
Individuals making more than $50,000: 11208 Individuals making at most$50,000: 34014
Percentage of individuals making more than $50,000: 24.78439697492371% ** Featureset Exploration ** • age: continuous. • workclass: Private, Self-emp-not-inc, Self-emp-inc, Federal-gov, Local-gov, State-gov, Without-pay, Never-worked. • education: Bachelors, Some-college, 11th, HS-grad, Prof-school, Assoc-acdm, Assoc-voc, 9th, 7th-8th, 12th, Masters, 1st-4th, 10th, Doctorate, 5th-6th, Preschool. • education-num: continuous. • marital-status: Married-civ-spouse, Divorced, Never-married, Separated, Widowed, Married-spouse-absent, Married-AF-spouse. • occupation: Tech-support, Craft-repair, Other-service, Sales, Exec-managerial, Prof-specialty, Handlers-cleaners, Machine-op-inspct, Adm-clerical, Farming-fishing, Transport-moving, Priv-house-serv, Protective-serv, Armed-Forces. • relationship: Wife, Own-child, Husband, Not-in-family, Other-relative, Unmarried. • race: Black, White, Asian-Pac-Islander, Amer-Indian-Eskimo, Other. • sex: Female, Male. • capital-gain: continuous. • capital-loss: continuous. • hours-per-week: continuous. • native-country: United-States, Cambodia, England, Puerto-Rico, Canada, Germany, Outlying-US(Guam-USVI-etc), India, Japan, Greece, South, China, Cuba, Iran, Honduras, Philippines, Italy, Poland, Jamaica, Vietnam, Mexico, Portugal, Ireland, France, Dominican-Republic, Laos, Ecuador, Taiwan, Haiti, Columbia, Hungary, Guatemala, Nicaragua, Scotland, Thailand, Yugoslavia, El-Salvador, Trinadad&Tobago, Peru, Hong, Holand-Netherlands. ## Preparing the Data Before data can be used as input for machine learning algorithms, it often must be cleaned, formatted, and restructured — this is typically known as preprocessing. Fortunately, for this dataset, there are no invalid or missing entries we must deal with, however, there are some qualities about certain features that must be adjusted. This preprocessing can help tremendously with the outcome and predictive power of nearly all learning algorithms. ### Transforming Skewed Continuous Features A dataset may sometimes contain at least one feature whose values tend to lie near a single number, but will also have a non-trivial number of vastly larger or smaller values than that single number. Algorithms can be sensitive to such distributions of values and can underperform if the range is not properly normalized. With the census dataset two features fit this description: ‘capital-gain' and 'capital-loss'. Run the code cell below to plot a histogram of these two features. Note the range of the values present and how they are distributed. For highly-skewed feature distributions such as 'capital-gain' and 'capital-loss', it is common practice to apply a logarithmic transformation on the data so that the very large and very small values do not negatively affect the performance of a learning algorithm. Using a logarithmic transformation significantly reduces the range of values caused by outliers. Care must be taken when applying this transformation however: The logarithm of 0 is undefined, so we must translate the values by a small amount above 0 to apply the the logarithm successfully. Run the code cell below to perform a transformation on the data and visualize the results. Again, note the range of values and how they are distributed. ### Normalizing Numerical Features In addition to performing transformations on features that are highly skewed, it is often good practice to perform some type of scaling on numerical features. Applying a scaling to the data does not change the shape of each feature’s distribution (such as 'capital-gain' or 'capital-loss' above); however, normalization ensures that each feature is treated equally when applying supervised learners. Note that once scaling is applied, observing the data in its raw form will no longer have the same original meaning, as exampled below. Run the code cell below to normalize each numerical feature. We will use sklearn.preprocessing.MinMaxScaler for this. /home/jason/anaconda3/lib/python3.7/site-packages/sklearn/preprocessing/data.py:334: DataConversionWarning: Data with input dtype int64, float64 were all converted to float64 by MinMaxScaler. return self.partial_fit(X, y) age workclass education_level education-num marital-status occupation relationship race sex capital-gain capital-loss hours-per-week native-country 0 0.301370 State-gov Bachelors 0.800000 Never-married Adm-clerical Not-in-family White Male 0.667492 0.0 0.397959 United-States 1 0.452055 Self-emp-not-inc Bachelors 0.800000 Married-civ-spouse Exec-managerial Husband White Male 0.000000 0.0 0.122449 United-States 2 0.287671 Private HS-grad 0.533333 Divorced Handlers-cleaners Not-in-family White Male 0.000000 0.0 0.397959 United-States 3 0.493151 Private 11th 0.400000 Married-civ-spouse Handlers-cleaners Husband Black Male 0.000000 0.0 0.397959 United-States 4 0.150685 Private Bachelors 0.800000 Married-civ-spouse Prof-specialty Wife Black Female 0.000000 0.0 0.397959 Cuba ### Implementation: Data Preprocessing From the table in Exploring the Data above, we can see there are several features for each record that are non-numeric. Typically, learning algorithms expect input to be numeric, which requires that non-numeric features (called categorical variables) be converted. One popular way to convert categorical variables is by using the one-hot encoding scheme. One-hot encoding creates a “dummy” variable for each possible category of each non-numeric feature. For example, assume someFeature has three possible entries: A, B, or C. We then encode this feature into someFeature_A, someFeature_B and someFeature_C. someFeature someFeature_A someFeature_B someFeature_C 0 B 0 1 0 1 C —-> one-hot encode —-> 0 0 1 2 A 1 0 0 Additionally, as with the non-numeric features, we need to convert the non-numeric target label, 'income' to numerical values for the learning algorithm to work. Since there are only two possible categories for this label (“<=50K” and “>50K”), we can avoid using one-hot encoding and simply encode these two categories as 0 and 1, respectively. In code cell below, you will need to implement the following: • Use pandas.get_dummies() to perform one-hot encoding on the 'features_log_minmax_transform' data. • Convert the target label 'income_raw' to numerical entries. • Set records with “<=50K” to 0 and records with “>50K” to 1. 103 total features after one-hot encoding. ### Shuffle and Split Data Now all categorical variables have been converted into numerical features, and all numerical features have been normalized. As always, we will now split the data (both features and their labels) into training and test sets. 80% of the data will be used for training and 20% for testing. Run the code cell below to perform this split. Training set has 36177 samples. Testing set has 9045 samples. ## Evaluating Model Performance In this section, we will investigate four different algorithms, and determine which is best at modeling the data. Three of these algorithms will be supervised learners of your choice, and the fourth algorithm is known as a naive predictor. ### Metrics and the Naive Predictor CharityML, equipped with their research, knows individuals that make more than$50,000 are most likely to donate to their charity. Because of this, CharityML is particularly interested in predicting who makes more than $50,000 accurately. It would seem that using accuracy as a metric for evaluating a particular model’s performace would be appropriate. Additionally, identifying someone that does not make more than$50,000 as someone who does would be detrimental to CharityML, since they are looking to find individuals willing to donate. Therefore, a model’s ability to precisely predict those that make more than $50,000 is more important than the model’s ability to recall those individuals. We can use F-beta score as a metric that considers both precision and recall: $$F_{\beta} = (1 + \beta^2) \cdot \frac{precision \cdot recall}{\left( \beta^2 \cdot precision \right) + recall}$$ In particular, when$\beta = 0.5$, more emphasis is placed on precision. This is called the F$_{0.5}$score (or F-score for simplicity). Looking at the distribution of classes (those who make at most$50,000, and those who make more), it’s clear most individuals do not make more than $50,000. This can greatly affect accuracy, since we could simply say *”this person does not make more than$50,000”* and generally be right, without ever looking at the data! Making such a statement would be called naive, since we have not considered any information to substantiate the claim. It is always important to consider the naive prediction for your data, to help establish a benchmark for whether a model is performing well. That been said, using that prediction would be pointless: If we predicted all people made less than $50,000, CharityML would identify no one as donors. #### Note: Recap of accuracy, precision, recall ** Accuracy ** measures how often the classifier makes the correct prediction. It’s the ratio of the number of correct predictions to the total number of predictions (the number of test data points). ** Precision ** tells us what proportion of messages we classified as spam, actually were spam. It is a ratio of true positives(words classified as spam, and which are actually spam) to all positives(all words classified as spam, irrespective of whether that was the correct classificatio), in other words it is the ratio of [True Positives/(True Positives + False Positives)] ** Recall(sensitivity)** tells us what proportion of messages that actually were spam were classified by us as spam. It is a ratio of true positives(words classified as spam, and which are actually spam) to all the words that were actually spam, in other words it is the ratio of [True Positives/(True Positives + False Negatives)] For classification problems that are skewed in their classification distributions like in our case, for example if we had a 100 text messages and only 2 were spam and the rest 98 weren’t, accuracy by itself is not a very good metric. We could classify 90 messages as not spam(including the 2 that were spam but we classify them as not spam, hence they would be false negatives) and 10 as spam(all 10 false positives) and still get a reasonably good accuracy score. For such cases, precision and recall come in very handy. These two metrics can be combined to get the F1 score, which is weighted average(harmonic mean) of the precision and recall scores. This score can range from 0 to 1, with 1 being the best possible F1 score(we take the harmonic mean as we are dealing with ratios). ### Question 1 - Naive Predictor Performace • If we chose a model that always predicted an individual made more than$50,000, what would that model’s accuracy and F-score be on this dataset? You must use the code cell below and assign your results to 'accuracy' and 'fscore' to be used later.

** Please note ** that the the purpose of generating a naive predictor is simply to show what a base model without any intelligence would look like. In the real world, ideally your base model would be either the results of a previous model or could be based on a research paper upon which you are looking to improve. When there is no benchmark model set, getting a result better than random choice is a place you could start from.

** HINT: **

• When we have a model that always predicts ‘1’ (i.e. the individual makes more than 50k) then our model will have no True Negatives(TN) or False Negatives(FN) as we are not making any negative(‘0’ value) predictions. Therefore our Accuracy in this case becomes the same as our Precision(True Positives/(True Positives + False Positives)) as every prediction that we have made with value ‘1’ that should have ‘0’ becomes a False Positive; therefore our denominator in this case is the total number of records we have in total.
• Our Recall score(True Positives/(True Positives + False Negatives)) in this setting becomes 1 as we have no False Negatives.
Naive Predictor: [Accuracy score: 0.2478, F-score: 0.2917]

### Supervised Learning Models

The following are some of the supervised learning models that are currently available in scikit-learn that you may choose from:

• Gaussian Naive Bayes (GaussianNB)
• Decision Trees
• K-Nearest Neighbors (KNeighbors)
• Stochastic Gradient Descent Classifier (SGDC)
• Support Vector Machines (SVM)
• Logistic Regression

### Question 2 - Model Application

List three of the supervised learning models above that are appropriate for this problem that you will test on the census data. For each model chosen

• Describe one real-world application in industry where the model can be applied.
• What are the strengths of the model; when does it perform well?
• What are the weaknesses of the model; when does it perform poorly?
• What makes this model a good candidate for the problem, given what you know about the data?

** HINT: **

### 1. Logistic Regression

1) It can be used in banking system to determine whether a customer will default or not.

2) When the targeted is a dummy variable this model can perform well. It’s highly interpretable and very efficient.

3) It is not suitable for non-linear problems and it is relatively perform worse than other complex methods.

4) In this problem we want to estimate the dummy which are people who have salary over or below 50k. This model is exactly suitable for this situation.

### 2. Ensemble Methods (Random Forest)

1) It can be used to determine span e-mails or classify user’s profile.

2) It performs better than many other simple models and not likely to be overfitting. It suits non-linear problems.

3) It is not suitable for linear problems, it is hard to interpret, and it requires high computational power.

4) In this problem we have a classification problem and ensemble methods can give us a good model.

### 3.Support Vector Machines (SVM)

1) It can also be used for classification problems like whether a customer will default or not.

2) SVM is effective in high dimensional spaces and fast.

3) Not suitable for large dataset.

4) In our problem, we have many dimensions, SVM might perform good.

### Implementation - Creating a Training and Predicting Pipeline

To properly evaluate the performance of each model you’ve chosen, it’s important that you create a training and predicting pipeline that allows you to quickly and effectively train models using various sizes of training data and perform predictions on the testing data. Your implementation here will be used in the following section.
In the code block below, you will need to implement the following:

• Import fbeta_score and accuracy_score from sklearn.metrics.
• Fit the learner to the sampled training data and record the training time.
• Perform predictions on the test data X_test, and also on the first 300 training points X_train[:300].
• Record the total prediction time.
• Calculate the accuracy score for both the training subset and testing set.
• Calculate the F-score for both the training subset and testing set.
• Make sure that you set the beta parameter!

### Implementation: Initial Model Evaluation

In the code cell, you will need to implement the following:

• Import the three supervised learning models you’ve discussed in the previous section.
• Initialize the three models and store them in 'clf_A', 'clf_B', and 'clf_C'.
• Use a 'random_state' for each model you use, if provided.
• Note: Use the default settings for each model — you will tune one specific model in a later section.
• Calculate the number of records equal to 1%, 10%, and 100% of the training data.
• Store those values in 'samples_1', 'samples_10', and 'samples_100' respectively.

Note: Depending on which algorithms you chose, the following implementation may take some time to run!

LogisticRegression trained on 361 samples.
LogisticRegression trained on 3617 samples.
LogisticRegression trained on 36177 samples.
RandomForestClassifier trained on 361 samples.
RandomForestClassifier trained on 3617 samples.
RandomForestClassifier trained on 36177 samples.
SVC trained on 361 samples.
SVC trained on 3617 samples.
SVC trained on 36177 samples.

## Improving Results

In this final section, you will choose from the three supervised learning models the best model to use on the student data. You will then perform a grid search optimization for the model over the entire training set (X_train and y_train) by tuning at least one parameter to improve upon the untuned model’s F-score.

### Question 3 - Choosing the Best Model

• Based on the evaluation you performed earlier, in one to two paragraphs, explain to CharityML which of the three models you believe to be most appropriate for the task of identifying individuals that make more than $50,000. ** HINT: ** Look at the graph at the bottom left from the cell above(the visualization created by vs.evaluate(results, accuracy, fscore)) and check the F score for the testing set when 100% of the training set is used. Which model has the highest score? Your answer should include discussion of the: • metrics - F score on the testing when 100% of the training data is used, • prediction/training time • the algorithm’s suitability for the data. *Answer: * The Logistic Regression model is the best model in this situation. When 100% of the training data is used, the Logistic Regression model has the second highest F-score on testing set with only very slightly difference compared with the best model (SVM). However, it has the lowest training and prediction times, which is faster than the third(best) model quite a lot. Therefore it is wise to choose Logistic Regression model. Actually, Logistic Regression model is very suitable for this data because the targeted variable is a dummy. ### Question 4 - Describing the Model in Layman’s Terms • In one to two paragraphs, explain to CharityML, in layman’s terms, how the final model chosen is supposed to work. Be sure that you are describing the major qualities of the model, such as how the model is trained and how the model makes a prediction. Avoid using advanced mathematical jargon, such as describing equations. ** HINT: ** When explaining your model, if using external resources please include all citations. *Answer: * The logistic regression model will create a linear boundary cutting the data points, which can classify whether people have high income or not. It estimates the probability of a person belongs to high income group or low income group based on the change of explaining variables. Logistic regression uses the natural logarithm function to find the relationship between the variables and uses test data to find the coefficients. The function can then predict the future results using these coefficients in the logistic equation.[1] Therefore, this model can split our dataset successfully. It can be seen in th graph, different from linear model, logistic model is a non-linear function with probability between 0 to 1. [1] Wikipedia. (2020). Logistic Regression. [online] Available at: https://simple.wikipedia.org/wiki/Logistic_Regression [Accessed 13 Jan. 2020]. ### Implementation: Model Tuning Fine tune the chosen model. Use grid search (GridSearchCV) with at least one important parameter tuned with at least 3 different values. You will need to use the entire training set for this. In the code cell below, you will need to implement the following: • Import sklearn.grid_search.GridSearchCV and sklearn.metrics.make_scorer. • Initialize the classifier you’ve chosen and store it in clf. • Set a random_state if one is available to the same state you set before. • Create a dictionary of parameters you wish to tune for the chosen model. • Example: parameters = {'parameter' : [list of values]}. • Note: Avoid tuning the max_features parameter of your learner if that parameter is available! • Use make_scorer to create an fbeta_score scoring object (with$\beta = 0.5$). • Perform grid search on the classifier clf using the 'scorer', and store it in grid_obj. • Fit the grid search object to the training data (X_train, y_train), and store it in grid_fit. Note: Depending on the algorithm chosen and the parameter list, the following implementation may take some time to run! Unoptimized model ------ Accuracy score on testing data: 0.8419 F-score on testing data: 0.6832 Optimized Model ------ Final accuracy score on the testing data: 0.8418 Final F-score on the testing data: 0.6829 ### Question 5 - Final Model Evaluation • What is your optimized model’s accuracy and F-score on the testing data? • Are these scores better or worse than the unoptimized model? • How do the results from your optimized model compare to the naive predictor benchmarks you found earlier in Question 1?_ Note: Fill in the table below with your results, and then provide discussion in the Answer box. #### Results: Metric Unoptimized Model Optimized Model Accuracy Score 0.8419 0.8418 F-score 0.6832 0.6829 *Answer: * The scores for the optimized model is worse than the unoptimized model. The scores for the optimized model improves a lot compared with the naive model. ## Feature Importance An important task when performing supervised learning on a dataset like the census data we study here is determining which features provide the most predictive power. By focusing on the relationship between only a few crucial features and the target label we simplify our understanding of the phenomenon, which is most always a useful thing to do. In the case of this project, that means we wish to identify a small number of features that most strongly predict whether an individual makes at most or more than$50,000.

Choose a scikit-learn classifier (e.g., adaboost, random forests) that has a feature_importance_ attribute, which is a function that ranks the importance of features according to the chosen classifier. In the next python cell fit this classifier to training set and use this attribute to determine the top 5 most important features for the census dataset.

### Question 6 - Feature Relevance Observation

When Exploring the Data, it was shown there are thirteen available features for each individual on record in the census data. Of these thirteen records, which five features do you believe to be most important for prediction, and in what order would you rank them and why?

1. education_num: more education might get better jobs

2. hours-per-week: more working hours more money

3. age: higher age implies more experience thus more income

4. marital status: married people might have more incentive to earn more money

5. sex: it might exists sex discrimination.

### Implementation - Extracting Feature Importance

Choose a scikit-learn supervised learning algorithm that has a feature_importance_ attribute availble for it. This attribute is a function that ranks the importance of each feature when making predictions based on the chosen algorithm.

In the code cell below, you will need to implement the following:

• Import a supervised learning model from sklearn if it is different from the three used earlier.
• Train the supervised model on the entire training set.
• Extract the feature importances using '.feature_importances_'.
/home/jason/anaconda3/lib/python3.7/site-packages/sklearn/model_selection/_split.py:2053: FutureWarning: You should specify a value for 'cv' instead of relying on the default value. The default value will change from 3 to 5 in version 0.22.
warnings.warn(CV_WARNING, FutureWarning)

### Question 7 - Extracting Feature Importance

Observe the visualization created above which displays the five most relevant features for predicting if an individual makes at most or above \$50,000.

• How do these five features compare to the five features you discussed in Question 6?
• If you were close to the same answer, how does this visualization confirm your thoughts?
• If you were not close, why do you think these features are more relevant?

Compared with what I thought before, the importance of marital status, ages and years of education are comfirmed by the graph because their weights are high. However, capital-gain and relationship_Husband is not what I predicted.

For capital-gain it might be only whealthy people can have more gain in capital since they have more free money to invest. For relationship, it is kind of overlap with the marital status so the reason is similar as what I explained before that married people are more likely try to earn more money to support the family especially for males.

### Feature Selection

How does a model perform if we only use a subset of all the available features in the data? With less features required to train, the expectation is that training and prediction time is much lower — at the cost of performance metrics. From the visualization above, we see that the top five most important features contribute more than half of the importance of all features present in the data. This hints that we can attempt to reduce the feature space and simplify the information required for the model to learn. The code cell below will use the same optimized model you found earlier, and train it on the same training set with only the top five important features.

Final Model trained on full data
------
Accuracy on testing data: 0.8418
F-score on testing data: 0.6829

Final Model trained on reduced data
------
Accuracy on testing data: 0.8258
F-score on testing data: 0.6462

/home/jason/anaconda3/lib/python3.7/site-packages/sklearn/utils/validation.py:761: DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel().
y = column_or_1d(y, warn=True)

### Question 8 - Effects of Feature Selection

• How does the final model’s F-score and accuracy score on the reduced data using only five features compare to those same scores when all features are used?
• If training time was a factor, would you consider using the reduced data as your training set?


text:
# Proving Pick's Formula

Pick's formula is a great topic for student exploration, but why does it work?

Henri Picciotto

Part of this page is based on a meeting of a group of Bay Area math teachers (Escape from the Textbook!) in 2010.
In particular, I would like to thank Kim Seashore and Avery Pickford, who contributed key ideas, and Dan Bennett,
who took detailed notes, some of which I paraphrase below. Dan also created some of the figures I use here.

We will call simple polygons (no crossing, no holes) with vertices on lattice points geoboard polygons. Pick's formula for the area of a geoboard polygon is A = I + B/2 – 1, where A = area, I = interior lattice points, and B = boundary lattice points.

For example, in the figure above, the quadrilateral has I = 7; and B = 5, so the area should be 8.5.
You can confirm this is correct by subtracting the area of three right triangles from the area of the enclosing rectangle.

## In the Classroom

Discovering this formula is within the reach of middle school or high school students, given a bit of guidance from the teacher. It is a beautiful and surprising result, which provides a wonderful connection between algebra and geometry, and it is both accessible and challenging in just the right way.

Students can make the polygons on dot paper, or using rubber bands on geoboards. Good hints to support student exploration:
- Keep B or I constant, and see how changing the other affects the area.
See also Lab 8.6 in Geometry Labs for a more guided discovery. In fact, that book contains much else for students to explore on geoboards and dot paper.

## The Proof

The beginning of a full proof of Pick's formula may be appropriate for some students, but the rest of this page is mostly intended for teachers. All polygons mentioned are geoboard polygons.

The outline of the proof is as follows:

1. The formula works for rectangles (horizontal-vertical sides.)
2. The formula works for right triangles (horizontal-vertical legs.)
3. If the formula works for two adjacent polygons, it works for the polygon they jointly create. And likewise if a polygon is removed from another, and both satisfy Pick's formula, then the remaining figure, if it is a polygon, will satisfy the formula.
4. The formula works for any triangle.
5. Finally, since the area of any geoboard polygon is the sum or difference of areas of triangles, the formula always works.

## Rectangles

Here are two visual arguments to explain the formula in the case of a (non-tilted) rectangle.

In this figure, each interior point and half the boundary points are matched with one square of area 1, the square immediately above and to the left of the point. It's almost a one-to-one pairing, but one of the squares lies outside the rectangle. Thus A = I + B/2 – 1.

In this figure, each interior point is surrounded by a square of area 1. Each non-vertex boundary point contributes half a unit square. Each vertex contributes a quarter of a unit square. Adding it all up, we get:

$A=I+ B−4 2 +4( 1 4 ) MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2da9iaadMeacqGHRaWkdaWcaaqaaiaadkeacqGHsislcaaI0aaabaGaaGOmaaaacqGHRaWkcaaI0aWaaeWaaeaadaWcaaqaaiaaigdaaeaacaaI0aaaaaGaayjkaiaawMcaaaaa@3983@$

which is equivalent to the desired formula.

## Right Triangles

We will now prove that Pick's formula holds for right triangles whose legs are horizontal and vertical, like the ones in the figure above. The area of such a triangle is half the area of the enclosing rectangle.

Say that the rectangle's area is A, and it has I interior points, and B boundary points. Say that the triangle's hypotenuse passes through d of the rectangle's interior points, and that the triangle has i interior points, b boundary points, and area a. In the above examples, d=0 and d=2.

We have $i= I−d 2 MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamyAaiabg2da9maalaaabaGaamysaiabgkHiTiaadsgaaeaacaaIYaaaaaaa@337B@$

$b= B 2 +1+d MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOyaiabg2da9maalaaabaGaamOqaaqaaiaaikdaaaGaey4kaSIaaGymaiabgUcaRiaadsgaaaa@34FF@$

and therefore:

$I=2i+d B=2b−2d−2 MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaqabeaacaWGjbGaeyypa0JaaGOmaiaadMgacqGHRaWkcaWGKbaabaGaamOqaiabg2da9iaaikdacaWGIbGaeyOeI0IaaGOmaiaadsgacqGHsislcaaIYaaaaaa@3B12@$ But: $a= A 2 = I 2 + B 4 − 1 2 MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2da9maalaaabaGaamyqaaqaaiaaikdaaaGaeyypa0ZaaSaaaeaacaWGjbaabaGaaGOmaaaacqGHRaWkdaWcaaqaaiaadkeaaeaacaaI0aaaaiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaaaaa@3920@$

Substituting the above values for I and B yields:

$a=i+ b 2 −1 MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2da9iaadMgacqGHRaWkdaWcaaqaaiaadkgaaeaacaaIYaaaaiabgkHiTiaaigdaaaa@352E@$

...which is what we were trying to prove.

## Combining Areas

Next, we will prove that if two adjacent geoboard polygons satisfy Pick's formula, then the combined polygon, whose area is the sum of the two areas, also does.

Let us call the variables A, I, and B for the combined polygon, and a1, i1, b1, a2, i2, b2, for the two components. Finally, let us say the two polygons share S boundary points. (Note that to be able to combine the polygons into one, S must be at least 2. In the above examples, S=4 and S=3.)

We have:

$I= i 1 + i 2 +S−2 B= b 1 + b 2 −2S+2$

But we are assuming the component pieces satisfy Pick's formula, so:

$A= a 1 + a 2 =( i 1 + b 1 2 −1 )+( i 2 + b 2 2 −1 )+S−S+1−1$

which can be rewritten as:

$A= i 1 + i 2 +S−2+ b 1 + b 2 −2S+2 2 −1 A=I+ B 2 −1 MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaabbeaacaWGbbGaeyypa0JaamyAamaaBaaaleaacaaIXaaabeaakiabgUcaRiaadMgadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkcaWGtbGaeyOeI0IaaGOmaiabgUcaRmaalaaabaGaamOyamaaBaaaleaacaaIXaaabeaakiabgUcaRiaadkgadaWgaaWcbaGaaGOmaaqabaGccqGHsislcaaIYaGaam4uaiabgUcaRiaaikdaaeaacaaIYaaaaiabgkHiTiaaigdaaeaacaWGbbGaeyypa0JaamysaiabgUcaRmaalaaabaGaamOqaaqaaiaaikdaaaGaeyOeI0IaaGymaaaaaa@4AB0@$

Just what we wanted! A similar argument can be used to conclude that if a compound geoboard polygon and one of the component polygons satisfy Pick's formula, then the other component polygon also does. (In other words, subtracting areas instead of adding them.)

It follows that any geoboard triangle will satisfy the formula, since any triangle can be obtained by subtracting right triangles from an enclosing rectangle:

And since any geoboard polygon can be decomposed into geoboard triangles, we conclude that Pick's formula applies to all geoboard polygons. QED.

## Reflection

Proof serves different purposes. One of them is to confirm something is true. Another is to help deepen one's understanding of the question under discussion.

I have been familiar with Pick's formula for decades, and have introduced it to many students and teachers. Still, I didn't feel I had a solid grasp of it until I saw the visual explanations of the rectangle case. Seeing those makes it crystal clear why the B term needs to be divided by 2, and what the -1 is about. So even though the rectangle case does not settle the matter, it is in some ways the most important part of the proof. It is the one part that seems worth thinking about presenting in the classroom, once the formula has been discovered by the students.

The rest of the proof is necessary to confirm the formula will always work, but it only adds a little to the gut-level understanding. Perhaps that is because those are more algebraic than visual? Still, it is satisfying to see how the algebraic manipulations work out.

That said, even though I am pleased with the proof, I don't think I'm done thinking about Pick's formula. Here is an interesting question that came up at our 2010 meeting.

Consider a triangle with a given base, and move the third vertex parallel to the base, as in this figure:

The area does not change, of course, but the number of interior and boundary points does. In the dashed triangle, I = 5 and B = 7. In the other two, I = 4 and B = 9. When we lost an interior point, we gained two boundary points. When we lost two boundary points in one place, we regained them elsewhere. But why? Pick's formula guarantees that will always happen, but that does not feel satisfying as an explanation. This is worthy of further exploration, reflection, and discussion with colleagues, and perhaps with students.

Related pages on this site:
Geoboard Activities
Geometry Labs (Sections 6, 8-11)
Algebra: Themes, Tools, Concepts (index of geoboard lessons)
Connecting the Dots (presentation)
The Pythagorean Geoboard (debate)
Teachers' Mathematics

format: plain


text:
## The number of powers of 2 in a representation of large even integers. II.(English)Zbl 0924.11086

It was shown by Yu. V. Linnik [Mat. Sb. Nov. Ser. 32(74), 3-60 (1953; Zbl 0051.03402)] that every sufficiently large even integer $$N$$ can be written as a sum of two primes and a bounded number of powers is 2 $N-p_1+ p_2+ 2^{\nu_1}+\cdots+ 2^{\nu_k}, \quad k\leq k_0.$ The method was simplified and improved by P. X. Gallagher [Invent. Math. 29, 125-142 (1975; Zbl 0305.10044)]. In the first part [Sci. China, Ser. A 41, 386-398 (1998)] the authors showed that on the generalized Riemann hypothesis 770 powers of 2 suffice. They now prove unconditionally that $$k_0= 54 000$$ is a possible numerical value for $$k_0$$. The proof follows Gallagher’s method, combined with explicit numerical bounds for zero free regions and density results for Dirichlet’s $$L$$-functions.

### MSC:

 11P32 Goldbach-type theorems; other additive questions involving primes 11M26 Nonreal zeros of $$\zeta (s)$$ and $$L(s, \chi)$$; Riemann and other hypotheses 11P55 Applications of the Hardy-Littlewood method 11M06 $$\zeta (s)$$ and $$L(s, \chi)$$

### Citations:

Zbl 0990.46081; Zbl 0051.03402; Zbl 0305.10044
Full Text:

### References:

 [1] Linnik, Yu. V., Prime numbers and powers of two,Trudy Mat. Inst. Steklov, 1951, 38: 151. [2] Linnik, Yu. V., Addition of prime numbers and powers of one and the same number,Mat. Sb. (N.S.), 1953, 32: 3. [3] Gallagher, P. X., Primes and powers of 2,Invent. Math., 1975, 29: 125. · Zbl 0305.10044 [4] Liu, J. Y., Liu, M. C., Wang, T. Z., The number of powers of 2 in a representation of large even integers (I),Science in China, Ser. A, 1998, 41: 386. · Zbl 1029.11049 [5] McCurley, K. S., Explicit zero-free regions for DirichletL-functions,J. Number Theory, 1984, 19:7. · Zbl 0536.10035 [6] Chen, J. R., Liu, J. M., On the least prime in an arithmetical progression and theorems concerning the zeros of Dirichlet’sL-function (III),Science in China, Ser. A, 1989, 32: 654. [7] Graham, S., On Linnik’s constant,Acta Arith., 1981, 39: 163. · Zbl 0379.10028 [8] Davenport, H.,Multiplicative Number Theory, 2nd ed., New York: Springer-Verlag, 1980, 87. · Zbl 0453.10002 [9] Graham, S., Applications of sieve methods,Ph D Thesis, University of Michigan, 1977. [10] Prachar, K.,Primzahlverteilung, Berlin: Springer-Verlag, 1957. [11] Chen, J. R., On the least prime in an arithmetical progression and theorems concerning the zeros of Dirichlet’sL-function (II),Sci. Sin., Ser. A, 1979, 22: 859. [12] Heath-Brown, D. R., Hybrid bounds for DirichletL-functions (II),Quart. J. Math. Oxford, Ser (2), 1980, 31: 157. · Zbl 0427.10025 [13] Jutila, M., On Linnik’s constant,Math. Scand., 1977, 41: 45. · Zbl 0363.10026 [14] Montgomery, H. L.,Topics in Multiplicative Number Theory, Lecture Notes in Mathematics, Vol. 227, Berlin-New York: Springer-Verlag, 1975. · Zbl 0216.03501 [15] Pan, C. D., Pan, C. B.,Goldbach Conjecture, Beijing: Science Press, 1992. [16] Pan, C. D., Pan, C. B.,Fundamentals of Analytic Number Theory (in Chinese), Beijing: Science Press, 1991. · Zbl 0738.55007 [17] Chen, J. R., The exceptional set of Goldbach numbers (II),Sci.Sin., Ser. A., 1983, 26: 714. · Zbl 0513.10045
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Location via proxy:   [ UP ]   [Report a bug]   [Manage cookies]    No cookies    No scripts    No ads    No referrer    Show this form

# CGIProxy Installation

CGIProxy Home | News | Installation | Usage | Options | Limits | Disclaimer |

Subscribe to announcement list:

As of version 2.2.1, CGIProxy now uses an installation wizard, so this page has been completely rewritten. Please tell me about any installation problems you have on any platform, so I can fix them!

You need to install CGIProxy on a machine that's outside of any censoring firewall, but that is still accessible to people behind the firewall. You do NOT need to install anything on the browsing machine(s). Once CGIProxy is installed on a machine, any number of people can use it, if they know its URL.

## Prerequisites

CGIProxy requires Perl and OpenSSL, but those are already installed on almost all Unix servers. If you're using a Windows server, install both of these.

There are four ways CGIProxy can run: as a CGI script, as a mod_perl script, as a FastCGI script, or with its own embedded server. As a CGI script is the slowest to run, and as a mod_perl or FastCGI script are the fastest to run. Depending on which of these ways you choose, your Web server needs to be configured to support it, as detailed in the following sections.

If you're installing on a CentOS server: CentOS doesn't include all standard Perl modules by default, so install those first by running "yum install epel-release" as root. You can't even run the installation wizard without first doing this. In addition, the CPAN modules required by CGIProxy must be installed as root on CentOS, so if the installation wizard isn't run as root, you'll need to install those modules separately from the wizard. The wizard will tell you how to do this.

If you're installing on a Windows server: The commands below that run nph-proxy.cgi are for Unix, but are almost the same for the Windows command line. In Windows, make two changes: start a command with "perl", and replace "/" with "\". You can also remove an initial "./". For example, instead of running "./nph-proxy.cgi command", run "perl nph-proxy.cgi command".

### As a CGI script

Your Web server has to be configured to support CGI scripts such that any executable file ending in ".cgi" is treated as a CGI script. Your webmaster normally does this. For Apache, you need something like this in the section of httpd.conf that controls the directory where nph-proxy.cgi will be (normally a <Directory> or <Files> block):

    AddHandler cgi-script .cgi
Options +ExecCGI

### As a mod_perl script

mod_perl is an optional feature of Apache servers. With mod_perl, CGIProxy can run much faster than as a CGI script. Your webmaster normally configures mod_perl. If you're configuring your own mod_perl, you can use this block in your httpd.conf to make all scripts ending in ".pl" use mod_perl:

<Files *.pl>
Options +ExecCGI
SetHandler perl-script
PerlResponseHandler ModPerl::Registry
<Files nph-*>
</Files>
</Files>

### As a FastCGI script

FastCGI scripts are similar to CGI scripts in some ways, but run faster than CGI scripts. FastCGI can be used with most Web servers, including those that don't support CGI scripts, such as nginx. A FastCGI script runs as multiple managed processes on a server, and is called by the Web server process.

After installing CGIProxy, configure your Web server to use CGIProxy as a FastCGI script, using your settings of $SECRET_PATH and$FCGI_SOCKET; you can view these settings in either the configuration menu or the file cgiproxy.conf . Examples of how to configure the nginx and Apache servers are just below.

To start the FastCGI process, run the command "cgiproxy/bin/nph-proxy.cgi start-fcgi" from your home directory.

To stop the FastCGI process in Unix, run the command "killall nph-proxy.cgi".

#### Configuring nginx with FastCGI

Include a section like this in your nginx.conf, within the secure server section:

location /secret/ {
fastcgi_pass             localhost:8002;
fastcgi_split_path_info  ^(/secret)(/?.*)$; fastcgi_param SCRIPT_NAME$fastcgi_script_name;
fastcgi_param            PATH_INFO $fastcgi_path_info; include fastcgi.conf; # if not included elsewhere }  The first three lines need to match your configuration in nph-proxy.cgi : replace "secret" (in two places!) with your setting of$SECRET_PATH, and replace "8002" with your setting of $FCGI_SOCKET if you changed it from the default. #### Configuring Apache with FastCGI Include a line like this in your httpd.conf: FastCgiExternalServer /var/www/html/secret -host localhost:8002  Replace "/var/www/html/secret" with your setting of$SECRET_PATH appended to your DocumentRoot, and replace "8002" with your setting of $FCGI_SOCKET if you changed it from the default. ### Using CGIProxy's embedded server CGIProxy includes an embedded Web server, so you don't need an external Web server to run it. However, you will need a key pair. (All secure servers need this, but external servers usually have a key pair already installed by your hosting provider.) A key pair consists of a certificate (visible to the public) and a private key (which you must keep secret). You can get a key pair for your server from a certificate authority (CA). Most certificate authorities charge an annual fee for a key pair, but Let's Encrypt provides them for free. They also provide the Certbot tool to automate the process of managing your certificate. After you install CGIProxy, you'll need to copy the certificate and private key files into the "cgiproxy" directory under your home directory. After installing CGIProxy, start the embedded server by running the command "cgiproxy/bin/nph-proxy.cgi start-server" from your home directory. After it starts, it will tell you the URL to access the proxy with (including the actual port number), and the process ID of the running server. To stop the embedded server in Unix, run the command "killall nph-proxy.cgi". ## Installing CGIProxy Once the prerequisites are in place, then to install CGIProxy: 1. Upload the distribution file (the file ending in ".tar.gz") to your Web server. 2. Log in to a shell account on the server. 3. Unpack the distribution (on Unix, run the command "tar xzvf cgiproxy.*.tar.gz"). In it are two files, nph-proxy.cgi and README . nph-proxy.cgi is the program you'll be installing and running. That's it-- just one file. 4. If you want to rename nph-proxy.cgi (see below), do it now. 5. Run the command "./nph-proxy.cgi install". This does almost everything required to install CGIProxy. It runs a simple installation wizard that asks you a few questions. Ideally you can run this as root, but it should work even if you don't have root access. 6. If using the embedded server: Copy both files of your key pair (described in "Prerequisites" above) into the "cgiproxy" directory under your home directory. The filenames should match the configuration variables$CERTIFICATE_FILE and $PRIVATE_KEY_FILE, which are "plain-cert.pem" and "plain-key.pem" unless you changed them. 7. If installing on Windows: Add a daily or hourly task to purge the database in the Task Scheduler, using the command "\path\to\script\nph-proxy.cgi" (replace with the correct path) and the parameter "purge-db". As part of the wizard, you will see a simple menu that lets you modify any configuration variables in CGIProxy. (You probably don't need to change anything.) When you're finished, enter "0" to save your settings and exit the configuration menu. After installing CGIProxy, you can always run the configuration menu again by running "./nph-proxy.cgi config". After the configuration, the wizard will install all the Perl CPAN modules needed by CGIProxy. This may generate a lot of scrolling text, which you can ignore. If it asks you any questions, you can just hit <enter>. To only install these modules, and not do any other installation tasks, run "./nph-proxy.cgi install-modules". You can rename nph-proxy.cgi if you want. However, if you're installing it as a CGI script or under mod_perl, be sure the name still starts with "nph-". (You can change that requirement for mod_perl, by reconfiguring mod_perl.) ## After you install Please review the security guidelines before doing anything further. If you want your proxy to be usable by other people, you need to communicate the proxy URL to them. Try to use a secure method to do so, or else the proxy could easily be discovered by censors and blocked, or worse. In particular, don't be tempted to post your proxy URL to any public sites that list available proxies. If you do, the censors will quickly see your proxy and block it, or there may be even worse consequences. These proxy-listing sites may be useful for anonymity, but for getting around censorship they are dangerous! They make the censors' job easier. In addition, people should only be using proxies installed by people or organizations that they trust. If heavy use of this proxy puts too much load on your server, see "NOTES ON PERFORMANCE" near the top of the source code. The biggest improvement comes from running this under mod_perl or FastCGI. ## Troubleshooting ### "Bad Request" errors These happen when you accumulate too many cookies for the Web server to handle. To fix it, use a database with CGIProxy. The easiest database to use is SQLite-- to use it, just set$DB_DRIVER to "SQLite". To use a running database engine like MariaDB/MySQL or Oracle, you must also create a database account for CGIProxy to use, and set $DB_USER and$DB_PASS.

IMPORTANT: Whenever you use a database, it will accumulate old data that you need to purge periodically for both performance and security reasons. To purge the database, run "./nph-proxy.cgi purge-db" regularly. On a Unix server, do this automatically with a cron job . The CGIProxy installation wizard normally adds this cron job for you, but if it doesn't, then you can do it yourself: run "crontab -e" to edit your list of cron jobs, and add a line like this to purge the database every morning at e.g. 2:13am:

 13 2 * * * /path/to/script/nph-proxy.cgi purge-db

Replace "/path/to/script/" with the full path to nph-proxy.cgi . If you'd rather purge the database every hour, change the "2" above to "*".

If you ran "./nph-proxy.cgi install" as root, then do this as root too, but instead of running "crontab -e", run "crontab -e -u $RUN_AS_USER", where$RUN_AS_USER is that username from the CGIProxy configuration. This modifies the list of cron jobs belonging to \$RUN_AS_USER .

On a Windows server, use the Task Scheduler to do this instead of a cron job.


text:
Le contenu de cette page n’est pas disponible en français. Veuillez nous en excuser.

# Cosmology & Gravitation

This series consists of talks in the areas of Cosmology, Gravitation and Particle Physics.

## Seminar Series Events/Videos

Currently there are no upcoming talks in this series.

## Summer Undergraduate Student Talk

Vendredi aoû 14, 2009
Speaker(s):
Collection/Series:
Scientific Areas:

## Undergrad Research Project Talk

Jeudi aoû 06, 2009
Speaker(s):
Collection/Series:
Scientific Areas:

## Undergrad Research Project Talk

Jeudi aoû 06, 2009
Speaker(s):
Collection/Series:
Scientific Areas:

## The Beauty and Basics of BAO

Vendredi juil 31, 2009
Speaker(s):

The Baryon Acoustic Oscillations (BAO) are the latest weapon in the quest for precision cosmology and dark energy. Many presentations on BAO are complicated and unclear and I will therefore present BAO with particular emphasis on trying to give the simplest theoretical description, both at the linear and nonlinear level, and will describe some of the observational challenges to measuring BAO.

Collection/Series:
Scientific Areas:

## Galactic Dynamics: an overview for physicists

Mardi juil 14, 2009
Speaker(s):

The standard theorist's model for the dynamics of galaxies is the limit of a Newtonian N-body system at fixed mass as the number of particles goes to infinity - i.e a phase space fluid After going over conventional wisdom, some interesting open issues which remain will be highlighted, and their relation to real galaxies explored.

Collection/Series:
Scientific Areas:

## Plebanski formulation / Ashtekar variables

Mercredi juil 08, 2009
Speaker(s):
Collection/Series:
Scientific Areas:

## Energy transfer in warped string compactifications and warped Kaluza-Klein modes as a dark matter candidate

Mardi juin 16, 2009

Strongly warped regions, or throats, are a common feature of string theory compactifications. In the early, hot universe, energy will be transferred between these throats and between throats and the standard model. Using the gauge-gravity duality, we calculate the rate of this energy transfer. Due to the warping, the resulting decay rate of throat-localized Kaluza-Klein states to other throats or the standard model is strongly suppressed. If their lifetime is longer than the current age of the universe, these states are an interesting dark matter candidate.

Collection/Series:
Scientific Areas:

## Cartan GR / differential forms

Mercredi juin 10, 2009
Speaker(s):
Collection/Series:
Scientific Areas:

## A natural origin of primordial density perturbations

Mardi juin 09, 2009
Speaker(s):

We suggest here a mechanism for the seeding of the primordial density fluctuations. We point out that a process like reheating at the end of inflation will inevitably generate perturbations, even on superhorizon scales, by the local diffusion of energy. Provided that the final temperature is of order the GUT scale, the density contrast $\delta_R$ for spheres of radius $R$ will be of order $10^{-5}$ at horizon entry, consistent with the values measured by \texttt{WMAP}.

Collection/Series:
Scientific Areas:

## Spectral variation of the WMAP 5-year degree scale anisotropy

Lundi juin 08, 2009
Speaker(s):

The black body nature of the first acoustic peak of the cosmic microwave background (CMB) was tested using foreground reduced WMAP 5-year data, by producing subtraction maps between pairs of cosmological bands, viz. the Q, V, and W bands, for masked sky areas that avoid the Galactic disk. The resulting maps revealed a non black body signal that has three main properties.

Collection/Series:
Scientific Areas:

## RECENT PUBLIC LECTURE

### Janna Levin: Black Hole Blues and Other Songs from Outer Space

Speaker: Janna Levin


text:
Astronomy 2e

# 16.2Mass, Energy, and the Theory of Relativity

Astronomy 2e16.2 Mass, Energy, and the Theory of Relativity

### Learning Objectives

By the end of this section, you will be able to:

• Explain how matter can be converted into energy
• Describe the particles that make up atoms
• Describe the nucleus of an atom
• Understand the nuclear forces that hold atoms together
• Trace the nuclear reactions in the solar interior

As we have seen, energy cannot be created or destroyed, but only converted from one form to another. One of the remarkable conclusions derived by Albert Einstein (see Albert Einstein) when he developed his theory of relativity is that matter can be considered a form of energy too and can be converted into energy. Furthermore, energy can also be converted into matter. This seemed to contradict what humans had learned over thousands of years by studying nature. Matter is something we can see and touch, whereas energy is something objects have when they do things like move or heat up. The idea that matter or energy can be converted into each other seemed as outrageous as saying you could accelerate a car by turning the bumper into more speed, or that you could create a bigger front seat by slowing down your car. That would be pretty difficult to believe; yet, the universe actually works somewhat like that.

### Converting Matter into Energy

The remarkable equivalence between matter and energy is given in one of the most famous equations:

$E=mc2E=mc2$

In this equation, E stands for energy, m stands for mass, and c, the constant that relates the two, is the speed of light . Note that mass is a measure of the quantity of matter, so the significance of this equation is that matter can be converted into energy and energy can be converted into matter. Let’s compare this equation of converting matter and energy to some common conversion equations that have the same form:

$inches=feet×12,or cents=dollars×100inches=feet×12,or cents=dollars×100$

Just as each conversion formula allows you to calculate the conversion of one thing into another, when we convert matter into energy, we consider how much mass the matter has. The conversion factor in this case turns out not to be either 12 or 100, as in our examples, but another constant quantity: the speed of light squared. Note that matter does not have to travel at the speed of light (or the speed of light squared) for this conversion to occur. The factor of c2 is just the number that Einstein showed must be used to relate mass and energy.

Notice that this formula does not tell us how to convert mass into energy, just as the formula for cents does not tell us where to exchange coins for a dollar bill. The formulas merely tell us what the equivalent values are if we succeed in making the conversion. When Einstein first derived his formula in 1905, no one had the faintest idea how to convert mass into energy in any practical way. Einstein himself tried to discourage speculation that the large-scale conversion of atomic mass into energy would be feasible in the near future. Today, as a result of developments in nuclear physics, we regularly convert mass into energy in power plants, nuclear weapons, and high-energy physics experiments in particle accelerators.

Because the speed of light squared (c2) is a very large quantity, the conversion of even a small amount of mass results in a very large amount of energy. For example, the complete conversion of 1 gram of matter (about 1/28 ounce, or approximately 1 paperclip) would produce as much energy as the burning of 15,000 barrels of oil.

Scientists soon realized that the conversion of mass into energy is the source of the Sun’s heat and light. With Einstein’s equation of E = mc2, we can calculate that the amount of energy radiated by the Sun could be produced by the complete conversion of about 4 million tons of matter into energy inside the Sun each second. Destroying 4 million tons per second sounds like a lot when compared to earthly things, but bear in mind that the Sun is a very big reservoir of matter. In fact, we will see that the Sun contains more than enough mass to destroy such huge amounts of matter and still continue shining at its present rate for billions of years.

But knowing all that still does not tell us how mass can be converted into energy. To understand the process that actually occurs in the Sun, we need to explore the structure of the atom a bit further.

### Voyagers in Astronomy

#### Albert Einstein

For a large part of his life, Albert Einstein (Figure 16.3) was one of the most recognized celebrities of his day. Strangers stopped him on the street, and people all over the world asked him for endorsements, advice, and assistance. In fact, when Einstein and the great film star Charlie Chaplin met in California, they found they shared similar feelings about the loss of privacy that came with fame. Einstein’s name was a household word despite the fact that most people did not understand the ideas that had made him famous.

Einstein was born in 1879 in Ulm, Germany. Legend has it that he did not do well in school (even in arithmetic), and thousands of students have since attempted to justify a bad grade by referring to this story. Alas, like many legends, this one is not true. Records indicate that although he tended to rebel against the authoritarian teaching style in vogue in Germany at that time, Einstein was a good student.

After graduating from the Federal Polytechnic Institute in Zurich, Switzerland, Einstein at first had trouble getting a job (even as a high school teacher), but he eventually became an examiner in the Swiss Patent Office. Working in his spare time, without the benefit of a university environment but using his superb physical intuition, he wrote four papers in 1905 that would ultimately transform the way physicists looked at the world.

One of these, which earned Einstein the Nobel Prize in 1921, set part of the foundation of quantum mechanics—the rich, puzzling, and remarkable theory of the subatomic realm. But his most important paper presented the special theory of relativity, a reexamination of space, time, and motion that added a whole new level of sophistication to our understanding of those concepts. The famed equation E = mc2 was actually a relatively minor part of this theory, added in a later paper.

In 1916, Einstein published his general theory of relativity, which was, among other things, a fundamentally new description of gravity (see Black Holes and Curved Spacetime). When this theory was confirmed by measurements of the “bending of starlight” during a 1919 eclipse (The New York Times headline read, “Lights All Askew in the Heavens”), Einstein became world famous.

In 1933, to escape Nazi persecution, Einstein left his professorship in Berlin and settled in the United States at the newly created Institute for Advanced Studies at Princeton. He remained there until his death in 1955, writing, lecturing, and espousing a variety of intellectual and political causes. For example, he agreed to sign a letter written by Leo Szilard and other scientists in 1939, alerting President Roosevelt to the dangers of allowing Nazi Germany to develop the atomic bomb first. And in 1952, Einstein was offered the second presidency of Israel. In declining the position, he said, “I know a little about nature and hardly anything about men.”

Figure 16.3 Albert Einstein (1879–1955). This portrait of Einstein was taken in 1912. (credit: modification of work by J. F. Langhans)

### Elementary Particles

The fundamental components of atoms are the proton, neutron, and electron (see The Structure of the Atom).

Protons, neutrons, and electrons are by no means all the particles that exist. First, for each kind of particle, there is a corresponding but opposite antiparticle. If the particle carries a charge, its antiparticle has the opposite charge. The antielectron is the positron, which has the same mass as the electron but is positively charged. Similarly, the antiproton has a negative charge. The remarkable thing about such antimatter is that when a particle comes into contact with its antiparticle, the original particles are annihilated, and substantial amounts of energy in the form of photons are produced.

Since our world is made exclusively of ordinary particles of matter, antimatter cannot survive for very long. But individual antiparticles are found in cosmic rays (particles that arrive at the top of Earth’s atmosphere from space) and can be created in particle accelerators. And, as we will see in a moment, antimatter is created in the core of the Sun and other stars.

Science fiction fans may be familiar with antimatter from the Star Trek television series and films. The Starship Enterprise is propelled by the careful combining of matter and antimatter in the ship’s engine room. According to E = mc2, the annihilation of matter and antimatter can produce a huge amount of energy, but keeping the antimatter fuel from touching the ship before it is needed must be a big problem. No wonder Scotty, the chief engineer in the original TV show, always looked worried!

In 1933, physicist Wolfgang Pauli (Figure 16.4) suggested that there might be another type of elementary particle. Energy seemed to disappear when certain types of nuclear reactions took place, violating the law of conservation of energy. Pauli was reluctant to accept the idea that one of the basic laws of physics was wrong, and he suggested a “desperate remedy.” Perhaps a so-far-undetected particle, which was given the name neutrino (“little neutral one”), carried away the “missing” energy. He suggested that neutrinos were particles with zero mass, and that like photons, they moved with the speed of light.

Figure 16.4 Wolfgang Pauli in 1945. Pauli is considered the “father” of the neutrino, having conceived of it in 1933.

The elusive neutrino was not detected until 1956. The reason it was so hard to find is that neutrinos interact very weakly with other matter and therefore are very difficult to detect. Earth is more transparent to a neutrino than the thinnest and cleanest pane of glass is to a photon of light. In fact, most neutrinos can pass completely through a star or planet without being absorbed. As we shall see, this behavior of neutrinos makes them a very important tool for studying the Sun. Since Pauli’s prediction, scientists have learned a lot more about the neutrino. We now know that there are three different types of neutrinos, and in 1998, neutrinos were discovered to have a tiny amount of mass. Indeed, it is so tiny that electrons are at least 500,000 times more massive. Ongoing research is focused on determining the mass of neutrinos more precisely, and it may still turn out that one of the three types is massless. We will return to the subject of neutrinos later in this chapter.

Some of the properties of the proton, electron, neutron, and neutrino are summarized in Table 16.1. (Other subatomic particles have been produced by experiments with particle accelerators, but they do not play a role in the generation of solar energy.)

Properties of Some Common Particles
Particle Mass (kg) Charge
Proton $1.67265×10-271.67265×10-27$ +1
Neutron $1.67495×10-271.67495×10-27$ 0
Electron $9.11×10-319.11×10-31$ −1
Neutrino $<2×10-36<2×10-36$ (uncertain) 0
Table 16.1

### The Atomic Nucleus

The nucleus of an atom is not just a loose collection of elementary particles. Inside the nucleus, particles are held together by a very powerful force called the strong nuclear force. This is short-range force, only capable of acting over distances about the size of the atomic nucleus. A quick thought experiment shows how important this force is. Take a look at your finger and consider the atoms composing it. Among them is carbon, one of the basic elements of life. Focus your imagination on the nucleus of one of your carbon atoms. It contains six protons, which have a positive charge, and six neutrons, which are neutral. Thus, the nucleus has a net charge of six positives. If only the electrical force were acting, the protons in this and every carbon atom would find each other very repulsive and fly apart.

The strong nuclear force is an attractive force, stronger than the electrical force, and it keeps the particles of the nucleus tightly bound together. We saw earlier that if under the force of gravity a star “shrinks”—bringing its atoms closer together—gravitational energy is released. In the same way, if particles come together under the strong nuclear force and unite to form an atomic nucleus, some of the nuclear energy is released. The energy given up in such a process is called the binding energy of the nucleus.

When such binding energy is released, the resulting nucleus has slightly less mass than the sum of the masses of the particles that came together to form it. In other words, the energy comes from the loss of mass. This slight deficit in mass is only a small fraction of the mass of one proton. But because each bit of lost mass can provide a lot of energy (remember, E = mc2), this nuclear energy release can be quite substantial.

Measurements show that the binding energy is greatest for atoms with a mass near that of the iron nucleus (with a combined number of protons and neutrons equal to 56) and less for both the lighter and the heavier nuclei. Iron, therefore, is the most stable element: since it gives up the most energy when it forms, it would require the most energy to break it back down into its component particles.

What this means is that, in general, when light atomic nuclei come together to form a heavier one (up to iron), mass is lost and energy is released. This joining together of atomic nuclei is called nuclear fusion.

Energy can also be produced by breaking up heavy atomic nuclei into lighter ones (down to iron); this process is called nuclear fission. Nuclear fission was the process we learned to use first—in atomic bombs and in nuclear reactors used to generate electrical power—and it may therefore be more familiar to you. Fission also sometimes occurs spontaneously in some unstable nuclei through the process of natural radioactivity. But fission requires big, complex nuclei, whereas we know that the stars are made up predominantly of small, simple nuclei. So we must look to fusion first to explain the energy of the Sun and the stars (Figure 16.5).

Figure 16.5 Fusion and Fission. (a) In fusion, light atomic nuclei join together to form a heavier nuclei, releasing energy in the process. (b) In fission, energy is produced by the breaking up of heavy, complex nuclei into lighter ones.

### Nuclear Attraction versus Electrical Repulsion

So far, we seem to have a very attractive prescription for producing the energy emitted by the Sun: “roll” some nuclei together and join them via nuclear fusion. This will cause them to lose some of their mass, which then turns into energy. However, every nucleus beyond simple hydrogen has two or more protons—and protons all have positive charges. Since like charges repel via the electrical force, the closer we get two nuclei to each other, the more they repel. It’s true that if we can get them within “striking distance” of the nuclear force, they will then come together with a much stronger attraction. But that striking distance is very tiny, about the size of a nucleus. How can we get nuclei close enough to participate in fusion?

The answer turns out to be heat—tremendous heat—which speeds the protons up enough to overcome the electrical forces that try to keep protons apart. Inside the Sun, as we saw, the most common element is hydrogen, whose nucleus contains only a single proton. Two protons can fuse only in regions where the temperature is greater than about 12 million K, and the speed of the protons average around 1000 kilometers per second or more. (In old-fashioned units, that’s over 2 million miles per hour!)

In our Sun, such extreme temperatures are reached only in the regions near its center, which has a temperature of 15 million K. Calculations show that nearly all of the Sun’s energy is generated within about 150,000 kilometers of its core, or within less than 10% of its total volume.

Even at these high temperatures, it is exceedingly difficult to force two protons to combine. On average, a proton will rebound from other protons in the Sun’s crowded core for about 14 billion years, at the rate of 100 million collisions per second, before it fuses with a second proton. This is, however, only the average waiting time. Some of the enormous numbers of protons in the Sun’s inner region are “lucky” and take only a few collisions to achieve a fusion reaction: they are the protons responsible for producing the energy radiated by the Sun. Since the Sun is about 4.5 billion years old, most of its protons have not yet been involved in fusion reactions.

### Nuclear Reactions in the Sun’s Interior

The Sun, then, taps the energy contained in the nuclei of atoms through nuclear fusion. Let’s look at what happens in more detail. Deep inside the Sun, a three-step process takes four hydrogen nuclei and fuses them together to form a single helium nucleus. The helium nucleus is slightly less massive than the four hydrogen nuclei that combine to form it, and that mass is converted into energy.

The initial step required to form one helium nucleus from four hydrogen nuclei is shown in Figure 16.6. At the high temperatures inside the Sun’s core, two protons combine to make a deuterium nucleus, which is an isotope (or version) of hydrogen that contains one proton and one neutron. In effect, one of the original protons has been converted into a neutron in the fusion reaction. Electric charge has to be conserved in nuclear reactions, and it is conserved in this one. A positron (antimatter electron) emerges from the reaction and carries away the positive charge originally associated with one of the protons.

Figure 16.6 Proton-Proton Chain, Step 1. This is the first step in the process of fusing hydrogen into helium in the Sun. High temperatures are required because this reaction starts with two hydrogen nuclei, which are protons (shown in blue at left) that must overcome electrical repulsion to combine, forming a hydrogen nucleus with a proton and a neutron (shown in red). Note that hydrogen containing one proton and one neutron is given its own name: deuterium. Also produced in this reaction are a positron, which is an antielectron, and an elusive particle named the neutrino.

Since it is antimatter, this positron will instantly collide with a nearby electron, and both will be annihilated, producing electromagnetic energy in the form of gamma-ray photons. This gamma ray, which has been created in the center of the Sun, finds itself in a world crammed full of fast-moving nuclei and electrons. The gamma ray collides with particles of matter and transfers its energy to one of them. The particle later emits another gamma-ray photon, but often the emitted photon has a bit less energy than the one that was absorbed.

Such interactions happen to gamma rays again and again and again as they make their way slowly toward the outer layers of the Sun, until their energy becomes so reduced that they are no longer gamma rays but X-rays (recall what you learned in The Electromagnetic Spectrum). Later, as the photons lose still more energy through collisions in the crowded center of the Sun, they become ultraviolet photons.

By the time they reach the Sun’s surface, most of the photons have given up enough energy to be ordinary light—and they are the sunlight we see coming from our star. (To be precise, each gamma-ray photon is ultimately converted into many separate lower-energy photons of sunlight.) So, the sunlight given off by the Sun today had its origin as a gamma ray produced by nuclear reactions deep in the Sun’s core. The length of time that photons require to reach the surface depends on how far a photon on average travels between collisions, and the travel time depends on what model of the complicated solar interior we accept. Estimates are somewhat uncertain but indicate that the emission of energy from the surface of the Sun can lag its production in the interior by 100,000 years to as much as 1,000,000 years.

In addition to the positron, the fusion of two hydrogen atoms to form deuterium results in the emission of a neutrino. Because neutrinos interact so little with ordinary matter, those produced by fusion reactions near the center of the Sun travel directly to the Sun’s surface and then out into space, in all directions. Neutrinos move at nearly the speed of light, and they escape the Sun about two seconds after they are created.

Figure 16.7 Proton-Proton Chain, Step 2. This is the second step of the proton-proton chain, the fusion reaction that converts hydrogen into helium in the Sun. This step combines one hydrogen nucleus, which is a proton (shown in blue), with the deuterium nucleus from the previous step (shown as a red and blue particle). The product of this is an isotope of helium with two protons (blue) and one neutron (red) and energy in the form of gamma-ray radiation.

The second step in forming helium from hydrogen is to add another proton to the deuterium nucleus to create a helium nucleus that contains two protons and one neutron (Figure 16.7). In the process, some mass is again lost and more gamma radiation is emitted. Such a nucleus is helium because an element is defined by its number of protons; any nucleus with two protons is called helium. But this form of helium, which we call helium-3 (and write in shorthand as 3He) is not the isotope we see in the Sun’s atmosphere or on Earth. That helium has two neutrons and two protons and hence is called helium-4 (4He).

To get to helium-4 in the Sun, helium-3 must combine with another helium-3 in the third step of fusion (illustrated in Figure 16.8). Note that two energetic protons are left over from this step; each of them comes out of the reaction ready to collide with other protons and to start step 1 in the chain of reactions all over again.

Figure 16.8 Proton-Proton Chain, Step 3. This is the third step in the fusion of hydrogen into helium in the Sun. Note that the two helium-3 nuclei from the second step (see Figure 16.7) must combine before the third step becomes possible. The two protons that come out of this step have the energy to collide with other protons in the Sun and start step one again.

### The Proton-Proton Chain

The nuclear reactions in the Sun that we have been discussing can be described succinctly through the following nuclear formulas:

$1H+1H⟶2H+e++v2H+1H⟶3He+γ3He+3He⟶4He+1H+1H1H+1H⟶2H+e++v2H+1H⟶3He+γ3He+3He⟶4He+1H+1H$

Here, the superscripts indicate the total number of neutrons plus protons in the nucleus, e+ is the positron, v is the neutrino, and γ indicates that gamma rays are emitted. Note that the third step requires two helium-3 nuclei to start; the first two steps must happen twice before the third step can occur.

Although, as we discussed, the first step in this chain of reactions is very difficult and generally takes a long time, the other steps happen more quickly. After the deuterium nucleus is formed, it survives an average of only about 6 seconds before being converted into 3He. About a million years after that (on average), the 3He nucleus will combine with another to form 4He.

We can compute the amount of energy these reactions generate by calculating the difference in the initial and final masses. The masses of hydrogen and helium atoms in the units normally used by scientists are 1.007825 u and 4.00268 u, respectively. (The unit of mass, u, is defined to be 1/12 the mass of an atom of carbon, or approximately the mass of a proton.) Here, we include the mass of the entire atom, not just the nucleus, because electrons are involved as well. When hydrogen is converted into helium, two positrons are created (remember, the first step happens twice), and these are annihilated with two free electrons, adding to the energy produced.

$4×1.007825u=4.03130u(mass of initial hydrogen atoms)−4.00268u(mass of final helium atoms)=0.02862u(mass lost in the transformation)4×1.007825u=4.03130u(mass of initial hydrogen atoms)−4.00268u(mass of final helium atoms)=0.02862u(mass lost in the transformation)$

The mass lost, 0.02862 u, is 0.71% of the mass of the initial hydrogen. Thus, if 1 kilogram of hydrogen is converted into helium, then the mass of the helium is only 0.9929 kilogram, and 0.0071 kilogram of material is converted into energy. The speed of light (c) is 3 × 108 meters per second, so the energy released by the conversion of just 1 kilogram of hydrogen into helium is:

$E=mc2E=0.0071kg×(3 ×108m/s)2=6.4×1014JE=mc2E=0.0071kg×(3 ×108m/s)2=6.4×1014J$

This amount, the energy released when a single kilogram (2.2 pounds) of hydrogen undergoes fusion, would supply all the electricity used by the typical U.S. household for roughly 17,000 years.

To produce the Sun’s luminosity of 4 × 1026 watts, some 600 million tons of hydrogen must be converted into helium each second, of which about 4 million tons are converted from matter into energy. As large as these numbers are, the store of hydrogen (and thus of nuclear energy) in the Sun is still more enormous, and can last a long time—billions of years, in fact.

At the temperatures inside the stars with masses smaller than about 1.2 times the mass of our Sun (a category that includes the Sun itself), most of the energy is produced by the reactions we have just described, and this set of reactions is called the proton-proton chain (or sometimes, the p-p chain). In the proton-proton chain, protons collide directly with other protons to form helium nuclei.

In hotter stars, another set of reactions, called the carbon-nitrogen-oxygen (CNO) cycle, accomplishes the same net result. In the CNO cycle, carbon and hydrogen nuclei collide to initiate a series of reactions that form nitrogen, oxygen, and ultimately, helium. The nitrogen and oxygen nuclei do not survive but interact to form carbon again. Therefore, the outcome is the same as in the proton-proton chain: four hydrogen atoms disappear, and in their place, a single helium atom is created. The CNO cycle plays only a minor role in the Sun but is the main source of energy for stars with masses greater than about the mass of the Sun.

So you can see that we have solved the puzzle that so worried scientists at the end of the nineteenth century. The Sun can maintain its high temperature and energy output for billions of years through the fusion of the simplest element in the universe, hydrogen. Because most of the Sun (and the other stars) is made of hydrogen, it is an ideal “fuel” for powering a star. As will be discussed in the following chapters, we can define a star as a ball of gas capable of getting its core hot enough to initiate the fusion of hydrogen. There are balls of gas that lack the mass required to do this (Jupiter is a local example); like so many hopefuls in Hollywood, they will never be stars.

### Making Connections

#### Fusion on Earth

Wouldn’t it be wonderful if we could duplicate the Sun’s energy mechanism in a controlled way on Earth? (We have already duplicated it in an uncontrolled way in hydrogen bombs, but we hope our storehouses of these will never be used.) Fusion energy would have many advantages: it would use hydrogen (or deuterium, which is heavy hydrogen) as fuel, and there is abundant hydrogen in Earth’s lakes and oceans. Water is much more evenly distributed around the world than oil or uranium, meaning that a few countries would no longer hold an energy advantage over the others. And unlike fission, which leaves dangerous byproducts, the nuclei that result from fusion are perfectly safe.

The problem is that, as we saw, it takes extremely high temperatures for nuclei to overcome their electrical repulsion and undergo fusion. When the first hydrogen bombs were exploded in tests in the 1950s, the “fuses” to get them hot enough were fission bombs. Interactions at such temperatures are difficult to sustain and control. To make fusion power on Earth, after all, we have to do what the Sun does: produce temperatures and pressures high enough to get hydrogen nuclei on intimate terms with one another.

The European Union, the United States, South Korea, Japan, China, Russia, Switzerland, and India are collaborating on the International Thermonuclear Experimental Reactor (ITER), a project to demonstrate the feasibility of controlled fusion (Figure 16.9). The facility is being built in France. Construction will require over 10,000,000 components and 2000 workers for assembly. The date for the start of operations is yet to be determined.

ITER is based on the Tokamak design, in which a large doughnut-shaped container is surrounded by superconducting magnets to confine and control the hydrogen nuclei in a strong magnetic field. Previous fusion experiments have produced about 15 million watts of energy, but only for a second or two, and they have required 100 million watts to produce the conditions necessary to achieve fusion. The goal of ITER is to build the first fusion device capable of producing 500 million watts of fusion energy for up to 1000 seconds. The challenge is keeping the deuterium and tritium—which will participate in fusion reactions—hot enough and dense enough, for a long enough time to produce energy.

Figure 16.9 ITER Design. The bright yellow areas in this model show where the superconducting magnets will circle the chamber within which fusion will take place. A huge magnet will keep the charged nuclei of heavy hydrogen confined. The goal is to produce 500 megawatts of energy. (credit: modification of work by Stephan Mosel)
Do you know how you learn best?


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# Jet nozzles effect in seawater

1. Apr 14, 2009

### TorMcOst

Is it possible to calculate a high-pressure nozzles effect/carrying length in seawater if the medium used in the high-pressure nozzle is seawater?

2. Apr 14, 2009

### Staff: Mentor

It is not clear what one is asking. Is the high-pressure nozzle operating underwater (as opposed to shooting into the air)?

If one uses the appropriate equation, then one simply uses the density and viscosity of sea-water.

A water jet underwater will dissipate quickly in the surrounding water, but it depends on the velocity of the jet. If the nozzle is used to propel a craft then the crafts forward speed would need to be considered.

3. Apr 14, 2009

### TorMcOst

Thats right Astronuc, the high pressure nozzle operates under water and is not moving.
The diameter of the nozzle is Ø1,2mm and the media flushed through it is seawater (the same as the media it is submerged in). The flow is 5-10L/min.
What equation would you recommend to figure:
- The length of the spray?
- The effect of the spray(on a surface) if it has a distance to a surface of i.e. 10mm?

4. Apr 14, 2009

### Staff: Mentor

Sounds like a jet cleaning system to clean underwater surfaces.

I see if I can find an example. I would imagine one wants some equation that is a function of distance and angle to surface, both of which will effect the momentum of the fluid at the surface.

For seawater, I'd recommend a 6 Moly SS, like ALX-6N or 254 SMO or the more recent Avesta 654 SMO.

http://www.alleghenyludlum.com/Ludlum/documents/AL_6XN_SourceBook.pdf

Stepping outside the standards for higher corrosion resistance

Last edited by a moderator: May 4, 2017
5. Apr 14, 2009

### minger

Oh man, you are lucky I'm in the middle of a numerical run right now.

From Viscous Fluid Flow(3rd ed.), Frank White, Chapter 4, Section 10.6.

If a round jet emerges from a circular hole with sufficient momentum, it remains narrow and grows slowly, the radial changes $$\partial /\partial r$$ being much larger than axial changes $$\partial / \partial x$$
Continuity:
$$\frac{\partial u}{\partial x} + \frac{1}{r}\frac{\partial}{\partial r}(rv) = 0$$
x momentum
$$u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = \frac{v}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$$

Schlichting reasoned that the jet thickness grew linearly, so the similarity variable is $$r/x$$. he defined the stream function
$$\psi(r,x) = \nu x F(\eta)\,\, \eta = \frac{r}{x}$$
From which the axisymmetrical velocity components are:
$$u = \frac{1}{r}\frac{\partial \psi}{\partial r} = \frac{\nu F'}{r}$$
$$v = -\frac{1}{r}\frac{\partial \psi}{\partial x} = \frac{\nu}{r}(\eta F' - F)$$
Substitution into the x-momentum equation gives the following third-order non-linear differential equation.
$$\frac{d}{d\eta}\left( F'' - \frac{F'}{\eta}\right) = \frac{1}{\eta^2} (FF'' - \eta F'^2 - \eta FF'')$$
The boundary conditions are F(0) = F'(0) = F'(infinity) = 0. The exact solution is:
$$F = \frac{(C\eta)^2}{1 + (C\eta /2)^2}$$
Where C is a constant determined from the momentum of the jet
$$J = \rho \int^\infty_0 u^2 2\pi r\,dr = \frac{16\pi}{3}\rho C^2 \nu^2$$
$$C = \left( \frac{3J}{16\pi \rho \nu^2}\right)^{1/2}$$
The axial jet velocity is then:
$$u = \frac{3J}{8\pi \mu x}\left( 1 + \frac{C^2 \eta^2}{4}\right)^{-2}$$
The term in parenthesis is the shape of the jet profile. The jet centerline velocity drops off as $$x^{-1}$$. The mass flow rate across any axial section of the jet is:
$$\dot{m} = \rho \int^{\infty}_0 u2\pi r\,dr = 8\pi \mu x$$

IF the jet is laminar, and we assume a simple plane laminar jet, we find that the velocity distribution is:
$$u(x,y) = u_{max} \sech^2 a\eta$$ Or:

$$u(x,y) = u_{max}\sech^2 \left[ 0.2752 \left(\frac{J\rho}{\mu^2 x^2}\right)^{1/3} y \right]$$

Where J is the momentum flux. At this point, if we define width of the jet as twice the distance y where $$u = 0.001u_{max}$$ then:
$$\mbox{Width} = 21.8 \left(\frac{x^2 \mu^2}{J \rho}\right)^{1/3}$$

That may be slightly more helpful on the width of the jet question. On the effect of the surface....no idea.

Last edited: Apr 14, 2009
6. Apr 15, 2009

### TorMcOst

Thanks a lot Minger and Astronuc! This were of great help!


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# Interpolating process that follows geometric Brownian motion

So I have a set of time series data that follows (at least assumed) GBM but there are missing data. To interpolate, I'm thinking about simulation and create some sample paths. I have two options in mind and would like to know which one makes more sense.

(1) Estimate the parameters of the GBM using available data then interpolate with those parameters (GBM model).

(2) Modify the Brownian bridge model so that the bridge follows a GBM but then the parameters (mu and sigma) will be different from those in (1). E.g. the mu is log(BT/B0)/T+sigma^2/2 with B0 and BT the start and end of the bridge and T the length of increments.