7.2: Diagonalization
- Determine when it is possible to diagonalize a matrix.
- When possible, diagonalize a matrix.
Similarity and Diagonalization
We begin this section by recalling the definition of similar matrices. Recall that if \(A,B\) are two \(n\times n\) matrices, then they are similar if and only if there exists an invertible matrix \(P\) such that \[A=P^{-1}BP\nonumber \]
In this case we write \(A \sim B\). The concept of similarity is an example of an equivalence relation .
Similarity is an equivalence relation, i.e. for \(n \times n\) matrices \(A,B,\) and \(C\),
- \(A \sim A\) (reflexive)
- If \(A \sim B\), then \(B \sim A\) (symmetric)
- If \(A \sim B\) and \(B \sim C\), then \(A \sim C\) (transitive)
- Proof
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It is clear that \(A\sim A\), taking \(P=I\).
Now, if \(A\sim B,\) then for some \(P\) invertible, \[A=P^{-1}BP\nonumber \] and so \[PAP^{-1}=B\nonumber \] But then \[\left( P^{-1}\right) ^{-1}AP^{-1}=B\nonumber \] which shows that \(B\sim A\).
Now suppose \(A\sim B\) and \(B\sim C\). Then there exist invertible matrices \(P,Q\) such that \[A=P^{-1}BP,\ B=Q^{-1}CQ\nonumber \] Then, \[A=P^{-1} \left( Q^{-1}CQ \right)P=\left( QP\right) ^{-1}C\left( QP\right)\nonumber \] showing that \(A\) is similar to \(C\).
Another important concept necessary to this section is the trace of a matrix. Consider the definition.
If \(A=[a_{ij}]\) is an \(n\times n\) matrix, then the trace of \(A\) is \[\mathrm{trace}(A) = \sum_{i=1}^n a_{ii}.\nonumber \]
In words, the trace of a matrix is the sum of the entries on the main diagonal.
For \(n\times n\) matrices \(A\) and \(B\), and any \(k\in\mathbb{R}\),
- \(\mathrm{trace}(A+B)=\mathrm{trace}(A) + \mathrm{trace}(B)\)
- \(\mathrm{trace}(kA)=k\cdot\mathrm{trace}(A)\)
- \(\mathrm{trace}(AB)=\mathrm{trace}(BA)\)
The following theorem includes a reference to the characteristic polynomial of a matrix. Recall that for any \(n \times n\) matrix \(A\), the characteristic polynomial of \(A\) is \(c_A(x)=\det(xI-A)\).
If \(A\) and \(B\) are \(n\times n\) matrices and \(A\sim B\), then
- \(\det(A) = \det(B)\)
- \(\mathrm{rank}(A) = \mathrm{rank}(B)\)
- \(\mathrm{trace}(A)= \mathrm{trace}(B)\)
- \(c_A(x)=c_B(x)\)
- \(A\) and \(B\) have the same eigenvalues
We now proceed to the main concept of this section. When a matrix is similar to a diagonal matrix, the matrix is said to be diagonalizable. We define a diagonal matrix \(D\) as a matrix containing a zero in every entry except those on the main diagonal. More precisely, if \(d_{ij}\) is the \(ij^{th}\) entry of a diagonal matrix \(D\), then \(d_{ij}=0\) unless \(i=j\). Such matrices look like the following. \[D = \left[ \begin{array}{ccc} \ast & & 0 \\ & \ddots & \\ 0 & & \ast \end{array} \right]\nonumber \] where \(\ast\) is a number which might not be zero.
The following is the formal definition of a diagonalizable matrix.
Let \(A\) be an \(n\times n\) matrix. Then \(A\) is said to be diagonalizable if there exists an invertible matrix \(P\) such that \[P^{-1}AP=D\nonumber \] where \(D\) is a diagonal matrix.
Notice that the above equation can be rearranged as \(A=PDP^{-1}\). Suppose we wanted to compute \(A^{100}\). By diagonalizing \(A\) first it suffices to then compute \(\left(PDP^{-1}\right)^{100}\), which reduces to \(PD^{100}P^{-1}\). This last computation is much simpler than \(A^{100}\). While this process is described in detail later, it provides motivation for diagonalization.
Diagonalizing a Matrix
The most important theorem about diagonalizability is the following major result.
An \(n\times n\) matrix \(A\) is diagonalizable if and only if there is an invertible matrix \(P\) given by \[P=\left[\begin{array}{cccc} X_{1} & X_{2} & \cdots & X_{n} \end{array} \right]\nonumber\] where the \(X_{k}\) are eigenvectors of \(A\).
Moreover if \(A\) is diagonalizable, the corresponding eigenvalues of \(A\) are the diagonal entries of the diagonal matrix \(D\).
- Proof
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Suppose \(P\) is given as above as an invertible matrix whose columns are eigenvectors of \(A\). Then \(P^{-1}\) is of the form \[P^{-1}=\left[\begin{array}{c} W_{1}^{T} \\ W_{2}^{T} \\ \vdots \\ W_{n}^{T} \end{array} \right]\nonumber \] where \(W_{k}^{T}X_{j}=\delta _{kj},\) which is the Kronecker’s symbol defined by \[\delta _{ij}=\left\{ \begin{array}{c} 1 \text{ if }i=j \\ 0\text{ if }i\neq j \end{array} \right.\nonumber \]
Then \[\begin{aligned} P^{-1}AP & = \left[\begin{array}{c} W_{1}^{T} \\ W_{2}^{T} \\ \vdots \\ W_{n}^{T} \end{array} \right] \left[\begin{array}{cccc} AX_{1} & AX_{2} & \cdots & AX_{n} \end{array} \right] \\ & = \left[\begin{array}{c} W_{1}^{T} \\ W_{2}^{T} \\ \vdots \\ W_{n}^{T} \end{array} \right] \left[\begin{array}{cccc} \lambda _{1}X_{1} & \lambda _{2}X_{2} & \cdots & \lambda _{n}X_{n} \end{array} \right] \\ &= \left[\begin{array}{ccc} \lambda _{1} & & 0 \\ & \ddots & \\ 0 & & \lambda _{n} \end{array} \right] \end{aligned}\]
Conversely, suppose \(A\) is diagonalizable so that \(P^{-1}AP=D.\) Let \[P=\left[\begin{array}{cccc} X_{1} & X_{2} & \cdots & X_{n} \end{array} \right]\nonumber \] where the columns are the \(X_{k}\) and \[D=\left[\begin{array}{ccc} \lambda _{1} & & 0 \\ & \ddots & \\ 0 & & \lambda _{n} \end{array} \right]\nonumber \] Then \[AP=PD=\left[\begin{array}{cccc} X_{1} & X_{2} & \cdots & X_{n} \end{array} \right] \left[\begin{array}{ccc} \lambda _{1} & & 0 \\ & \ddots & \\ 0 & & \lambda _{n} \end{array} \right]\nonumber\] and so \[\left[\begin{array}{cccc} AX_{1} & AX_{2} & \cdots & AX_{n} \end{array} \right] =\left[\begin{array}{cccc} \lambda _{1}X_{1} & \lambda _{2}X_{2} & \cdots & \lambda _{n}X_{n} \end{array} \right]\nonumber\] showing the \(X_{k}\) are eigenvectors of \(A\) and the \(\lambda _{k}\) are eigenvectors.
Below is a video on the diagonalization of matrices
Notice that because the matrix \(P\) defined above is invertible it follows that the set of eigenvectors of \(A\), \(\left\{ X_1, X_2, \cdots, X_n \right\}\), form a basis of \(\mathbb{R}^n\).
We demonstrate the concept given in the above theorem in the next example. Note that not only are the columns of the matrix \(P\) formed by eigenvectors, but \(P\) must be invertible so must consist of a wide variety of eigenvectors. We achieve this by using basic eigenvectors for the columns of \(P\).
Let \[A=\left[\begin{array}{rrr} 2 & 0 & 0 \\ 1 & 4 & -1 \\ -2 & -4 & 4 \end{array} \right]\nonumber\] Find an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(P^{-1}AP=D\).
Solution
By Theorem \(\PageIndex{2}\) we use the eigenvectors of \(A\) as the columns of \(P\), and the corresponding eigenvalues of \(A\) as the diagonal entries of \(D\).
First, we will find the eigenvalues of \(A\). To do so, we solve \(\det \left( \lambda I -A \right) =0\) as follows. \[\det \left( \lambda \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] - \left[\begin{array}{rrr} 2 & 0 & 0 \\ 1 & 4 & -1 \\ -2 & -4 & 4 \end{array} \right] \right) = 0\nonumber \]
This computation is left as an exercise, and you should verify that the eigenvalues are \(\lambda_1 =2, \lambda_2 = 2\), and \(\lambda_3 = 6\).
Next, we need to find the eigenvectors. We first find the eigenvectors for \(\lambda_1, \lambda_2 = 2\). Solving \(\left(2I - A \right)X = 0\) to find the eigenvectors, we find that the eigenvectors are \[t\left[\begin{array}{r} -2 \\ 1 \\ 0 \end{array} \right] +s\left[\begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right]\nonumber \] where \(t,s\) are scalars. Hence there are two basic eigenvectors which are given by \[X_1 = \left[\begin{array}{r} -2 \\ 1 \\ 0 \end{array} \right], X_2 = \left[\begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right]\nonumber \]
You can verify that the basic eigenvector for \(\lambda_3 =6\) is \(X_3 = \left[\begin{array}{r} 0 \\ 1 \\ -2 \end{array} \right]\)
Then, we construct the matrix \(P\) as follows. \[P= \left[\begin{array}{rrr} X_1 & X_2 & X_3 \end{array} \right] = \left[\begin{array}{rrr} -2 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -2 \end{array} \right]\nonumber \] That is, the columns of \(P\) are the basic eigenvectors of \(A\). Then, you can verify that \[P^{-1}=\left[\begin{array}{rrr} - \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{2} & - \frac{1}{4} \end{array} \right]\nonumber \] Thus, \[\begin{aligned} P^{-1}AP &=\left[\begin{array}{rrr} - \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{2} & - \frac{1}{4} \end{array} \right] \left[\begin{array}{rrr} 2 & 0 & 0 \\ 1 & 4 & -1 \\ -2 & -4 & 4 \end{array} \right] \left[\begin{array}{rrr} -2 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -2 \end{array} \right] \\ &=\left[\begin{array}{rrr} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{array} \right] \end{aligned}\]
You can see that the result here is a diagonal matrix where the entries on the main diagonal are the eigenvalues of \(A\). We expected this based on Theorem \(\PageIndex{2}\) . Notice that eigenvalues on the main diagonal must be in the same order as the corresponding eigenvectors in \(P\).
Below is a video on the diagonalization of matrices given eigenvalues and eigenvectors of a 3x3 matrix
Below is a video on the diagonalization of a 2x2 matrix.
Below is a video on the diagonalization of a 3x3 matrix.
Consider the next important theorem.
Let \(A\) be an \(n\times n\) matrix, and suppose that \(A\) has distinct eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_m\). For each \(i\), let \(X_i\) be a \(\lambda_i\)-eigenvector of \(A\). Then \(\{ X_1, X_2, \ldots, X_m\}\) is linearly independent.
The corollary that follows from this theorem gives a useful tool in determining if \(A\) is diagonalizable.
Let \(A\) be an \(n \times n\) matrix and suppose it has \(n\) distinct eigenvalues. Then it follows that \(A\) is diagonalizable.
It is possible that a matrix \(A\) cannot be diagonalized. In other words, we cannot find an invertible matrix \(P\) so that \(P^{-1}AP=D\).
Consider the following example.
Let \[A = \left[\begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array} \right]\nonumber\] If possible, find an invertible matrix \(P\) and diagonal matrix \(D\) so that \(P^{-1}AP=D\).
Solution
Through the usual procedure, we find that the eigenvalues of \(A\) are \(\lambda_1 =1, \lambda_2=1.\) To find the eigenvectors, we solve the equation \(\left(\lambda I - A \right) X = 0\). The matrix \(\left(\lambda I -A \right)\) is given by \[\left[\begin{array}{cc} \lambda - 1 & -1 \\ 0 & \lambda - 1 \end{array} \right]\nonumber\]
Substituting in \(\lambda = 1\), we have the matrix \[\left[\begin{array}{cc} 1 - 1 & -1 \\ 0 & 1 - 1 \end{array} \right] = \left[\begin{array}{rr} 0 & -1 \\ 0 & 0 \end{array} \right]\nonumber\]
Then, solving the equation \(\left(\lambda I - A\right) X = 0\) involves carrying the following augmented matrix to its reduced row-echelon form. \[\left[\begin{array}{rr|r} 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[\begin{array}{rr|r} 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right]\nonumber\]
Then the eigenvectors are of the form \[t\left[\begin{array}{r} 1 \\ 0 \end{array} \right]\nonumber\] and the basic eigenvector is \[X_1 = \left[\begin{array}{r} 1 \\ 0 \end{array} \right]\nonumber\]
In this case, the matrix \(A\) has one eigenvalue of multiplicity two, but only one basic eigenvector. In order to diagonalize \(A\), we need to construct an invertible \(2\times 2\) matrix \(P\). However, because \(A\) only has one basic eigenvector, we cannot construct this \(P\). Notice that if we were to use \(X_1\) as both columns of \(P\), \(P\) would not be invertible. For this reason, we cannot repeat eigenvectors in \(P\).
Hence this matrix cannot be diagonalized.
The idea that a matrix may not be diagonalizable suggests that conditions exist to determine when it is possible to diagonalize a matrix. We saw earlier in Corollary \(\PageIndex{1}\) that an \(n \times n\) matrix with \(n\) distinct eigenvalues is diagonalizable. It turns out that there are other useful diagonalizability tests.
First we need the following definition.
Let \(A\) be an \(n\times n\) matrix and \(\lambda\in\mathbb{R}\). The eigenspace of \(A\) corresponding to \(\lambda\), written \(E_{\lambda}(A)\) is the set of all eigenvectors corresponding to \(\lambda\).
In other words, the eigenspace \(E_{\lambda}(A)\) is all \(X\) such that \(AX = \lambda X\). Notice that this set can be written \(E_{\lambda}(A) = \mathrm{null}(\lambda I - A)\), showing that \(E_{\lambda}(A)\) is a subspace of \(\mathbb{R}^n\).
Recall that the multiplicity of an eigenvalue \(\lambda\) is the number of times that it occurs as a root of the characteristic polynomial.
Consider now the following lemma.
If \(A\) is an \(n\times n\) matrix, then \[\dim(E_{\lambda}(A))\leq m\nonumber\] where \(\lambda\) is an eigenvalue of \(A\) of multiplicity \(m\).
This result tells us that if \(\lambda\) is an eigenvalue of \(A\), then the number of linearly independent \(\lambda\)-eigenvectors is never more than the multiplicity of \(\lambda\). We now use this fact to provide a useful diagonalizability condition.
Let \(A\) be an \(n \times n\) matrix \(A\). Then \(A\) is diagonalizable if and only if for each eigenvalue \(\lambda\) of \(A\), \(\dim(E_{\lambda}(A))\) is equal to the multiplicity of \(\lambda\).
Complex Eigenvalues
In some applications, a matrix may have eigenvalues which are complex numbers. For example, this often occurs in differential equations. These questions are approached in the same way as above.
Consider the following example.
Let \[A=\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ]\nonumber \] Find the eigenvalues and eigenvectors of \(A\).
Solution
We will first find the eigenvalues as usual by solving the following equation.
\[\det \left( \lambda \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right ] - \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ] \right) =0\nonumber \] This reduces to \(\left( \lambda -1\right) \left( \lambda^{2}-4 \lambda +5\right) =0.\) The solutions are \(\lambda_1 =1,\lambda_2 = 2+i\) and \(\lambda_3 =2-i.\)
There is nothing new about finding the eigenvectors for \(\lambda_1 =1\) so this is left as an exercise.
Consider now the eigenvalue \(\lambda_2 =2+i.\) As usual, we solve the equation \(\left(\lambda I -A \right) X = 0\) as given by \[\left( \left( 2+i\right) \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right ] - \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ] \right) X =\left [ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right ]\nonumber \] In other words, we need to solve the system represented by the augmented matrix \[\left [ \begin{array}{crr|r} 1+i & 0 & 0 & 0 \\ 0 & i & 1 & 0 \\ 0 & -1 & i & 0 \end{array} \right ]\nonumber \]
We now use our row operations to solve the system. Divide the first row by \(\left( 1+i\right)\) and then take \(-i\) times the second row and add to the third row. This yields \[\left [ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & i & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ]\nonumber\] Now multiply the second row by \(-i\) to obtain the reduced row-echelon form, given by \[\left [ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & 1 & -i & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ]\nonumber \] Therefore, the eigenvectors are of the form \[t\left [ \begin{array}{r} 0 \\ i \\ 1 \end{array} \right ]\nonumber\] and the basic eigenvector is given by \[X_2 = \left [ \begin{array}{r} 0 \\ i \\ 1 \end{array} \right ]\nonumber\]
As an exercise, verify that the eigenvectors for \(\lambda_3 =2-i\) are of the form \[t\left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ]\nonumber\] Hence, the basic eigenvector is given by \[X_3 = \left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ]\nonumber\]
As usual, be sure to check your answers! To verify, we check that \(AX_3 = \left(2 - i \right) X_3\) as follows. \[\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 0 & 1 & 2 \end{array} \right ] \left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ] = \left [ \begin{array}{c} 0 \\ -1-2i \\ 2-i \end{array} \right ] =\left( 2-i\right) \left [ \begin{array}{r} 0 \\ -i \\ 1 \end{array} \right ]\nonumber \]
Therefore, we know that this eigenvector and eigenvalue are correct.
Notice that in Example \(\PageIndex{3}\) , two of the eigenvalues were given by \(\lambda_2 = 2 + i\) and \(\lambda_3 = 2-i\). You may recall that these two complex numbers are conjugates . It turns out that whenever a matrix containing real entries has a complex eigenvalue \(\lambda\), it also has an eigenvalue equal to \(\overline{\lambda}\), the conjugate of \(\lambda\).
Below is a video on determining Ax using a diagonalized matrix.


7.2: Linear Programming - Maximization Applications
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Learning Objectives
In this section, you will learn to:
- Recognize the typical form of a linear programming problem.
- Formulate maximization linear programming problems.
- Graph feasible regions for maximization linear programming problems.
- Determine optimal solutions for maximization linear programming problems.
Prerequisite Skills
Before you get started, take this prerequisite quiz.
1. Graph this system of inequalities:
\(\left\{\begin{array} {l} 2x+4y\leq 10\\3x−5y<15\end{array}\right.\)
- Click here to check your answer
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If you missed this problem, review Section 2.3 . (Note that this will open in a new window.)
2. Graph this system of inequalities:
\(\left\{\begin{array} {l} y\leq 2x+1\\y\leq-3x+6\\x\geq0\\y\geq0\end{array}\right.\)
- Click here to check your answer
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(The solution is in the center region.)
If you missed this problem, review Section 2.3 . (Note that this will open in a new window.)
Application problems in business, economics, and social and life sciences often ask us to make decisions on the basis of certain conditions. The conditions or constraints often take the form of inequalities. In this section, we will begin to formulate, analyze, and solve such problems, at a simple level, to understand the many components of such a problem.
A typical linear programming problem consists of finding an extreme value of a linear equation subject to certain constraints. We are either trying to maximize or minimize the value of this linear equation, such as to maximize profit or revenue, or to minimize cost. That is why these linear programming problems are classified as maximization or minimization problems , or just optimization problems . The value we are trying to optimize is called an objective function , and the conditions that must be satisfied are called constraints .
When we graph all constraints, the area of the graph that satisfies all constraints is called the feasible region . The Fundamental Theorem of Linear Programming states that the maximum (or minimum) value of the objective function always takes place at the vertices of the feasible region. We call these vertices critical points. These are found using any methods from Section 1.4 as we are looking for the points where any two of the boundary lines intersect.
A typical example is to maximize profit from producing several products, subject to limitations on materials or resources needed for producing these items; the problem requires us to determine the amount of each item produced. Another type of problem involves scheduling; we need to determine how much time to devote to each of several activities in order to maximize income from (or minimize cost of) these activities, subject to limitations on time and other resources available for each activity.
In this chapter, we will work with problems that involve only two variables, and therefore, can be solved by graphing. Here are the steps we'll follow:
The Maximization Linear Programming Problems
- Define the unknowns.
- Write the objective function that needs to be maximized.
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Write the constraints.
- For the standard maximization linear programming problems, constraints are of the form: \(ax + by ≤ c\)
- When the variables represent quantities that cannot be negative, we include the constraints: \(x ≥ 0\) and \(y ≥ 0\).
- Graph the constraints.
- Determine the feasible region.
- Find the corner points.
- Find the value of the objective function at each corner point to determine the corner point that gives the maximum value.
- State the answer to the original question.
Example \(\PageIndex{1}\)
Niki holds two part-time jobs, Job I and Job II. She never wants to be at work more than a total of 12 hours a week. She has determined that for every hour she works at Job I, she needs 2 hours of preparation time, and for every hour she works at Job II, she needs one hour of preparation time, and she cannot spend more than 16 hours for preparation.
If Niki makes $40 an hour at Job I, and $30 an hour at Job II, how many hours should she work per week at each job to maximize her income?
Solution
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We start by defining our unknowns.
- Let the number of hours per week Niki will work at Job I = \(x\).
- Let the number of hours per week Niki will work at Job II = \(y\).
- Now we write the objective function. We're trying to maximize her income, so the equation will represent the calculation for finding her income. Since Niki gets paid $40 an hour at Job I, and $30 an hour at Job II, her total income I is given by the equation \[I = 40x + 30y \nonumber\]
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Our next task is to find the constraints. Each limitation in the problem will give a constraint.
- The second sentence in the problem states, "She never wants to work more than a total of 12 hours a week." This translates into the following constraint: \[x + y \leq 12 \nonumber \]
- The third sentence states, "For every hour she works at Job I, she needs 2 hours of preparation time, and for every hour she works at Job II, she needs one hour of preparation time, and she cannot spend more than 16 hours for preparation." Therefore the next constraint is \[2x + y \leq 16 \nonumber\]
- The fact that \(x\) and \(y\) can never be negative is represented by the following two constraints: \[x \geq 0 \text{, and } y \geq 0 \nonumber.\]
- In summary, our entire problem can be written as:
\[\begin{array}{ll}
\textbf { Maximize } & \mathrm{I}=40 \mathrm{x}+30 \mathrm{y} \\
\textbf { Subject to: } & \mathrm{x}+\mathrm{y} \leq 12 \\
& 2 \mathrm{x}+\mathrm{y} \leq 16 \\
& \mathrm{x} \geq 0 ; \mathrm{y} \geq 0
\end{array}\nonumber\]
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In order to solve the problem, we graph the constraints and shade the region that satisfies
all
the inequality constraints. Any appropriate method can be used to graph the lines for the constraints.
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Graph each inequality.
- When doing this by hand, refer to the strategies in Section 2.2 .
- This may also be done in the Desmos app or on https://www.desmos.com/calculator . On a computer, the \( \leq \) symbol can be written by typing the \(<\) symbol followed by the \(=\) symbol.
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Determine which direction to shade for each line. There are three main strategies to decide how much shading is put on the graph:
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Strategy A: Graph all inequalities exactly as written. The negative of this method is that it is often difficult to see exactly which region is covered by all colors. In the graphic below, the feasible region is the darkest gray four-sided figure in the lower left portion of the graph.
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Strategy B: Graph ALMOST all inequalities, omitting \(x ≥ 0\) and \(y ≥ 0\). The feasible region will be slightly easier to identify, but the negative of this method is that students must remember that the feasible region actually stops at the axes.
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Strategy C: Graph the OPPOSITE of each inequality. The feasible region will be clearly visible as the void or the white space between the graphs. The negative of this strategy is that it can be easy to make a mistake when typing in the inequalities, as they will be different than what is in the problem. Often students switch some inequalities but accidentally don't switch ALL inequalities.
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Strategy A: Graph all inequalities exactly as written. The negative of this method is that it is often difficult to see exactly which region is covered by all colors. In the graphic below, the feasible region is the darkest gray four-sided figure in the lower left portion of the graph.
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Graph each inequality.
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Determine the feasible region, which is the region that makes all inequalities true. No matter which strategy is used, the inequalities above should yield the feasible region shown below.
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Determine the corner points of the feasible region. These are known as the critical points or critical values.
- When graphing by hand, this can be done using any of the methods discussed in Section 1.3.
- When graphing in Desmos, this can be accomplished by clicking on the point of intersection.
- The Fundamental Theorem of Linear Programming states that the maximum (or minimum) value of the objective function always takes place at the vertices of the feasible region. Therefore, we will substitute these points in the objective function to see which point gives us the maximum value. In this case, we are looking for the highest income per week. We list the results below.
| Critical Points | Income = \(40x+30y\) |
|---|---|
| (0, 0) | I = 40(0) + 30(0) = $0 |
| (0, 12) | I = 40(0) + 30(12) = $360 |
| (4, 8) | I = 40(4) + 30(8) = $400 |
| (8, 0) | I = 40(8) + 30(0) = $320 |
8. The point (4, 8) gives the most profit: $400. Therefore, we conclude that Niki should work 4 hours at Job I, and 8 hours at Job II.
Example \(\PageIndex{2}\)
A factory manufactures two types of gadgets, regular and premium. Each gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing, while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other restrictions, the company can make at most 7 gadgets a day. If a profit of $20 is realized for each regular gadget and $30 for a premium gadget, how many of each should be manufactured to maximize profit?
Solution
1. We start by defining our unknowns:
- Let the number of regular gadgets manufactured each day = \(x\).
- and the number of premium gadgets manufactured each day = \(y\).
2. Now we write the objective function. We are trying to maximize profit, which will be $20 for however many regular gadgets we manufacture (\(x\) and $30 for however many premium gadgets we manufacture (\(y\). Therefore profit is is given by the equation
\[P = 20x + 30y \nonumber\]
3. We now write the constraints.
- The fourth sentence states that the company can make at most 7 total gadgets a day. This translates as
\[x + y \leq 7 \nonumber\]
- One limitation given is the hours for the assembly operation. Since the regular gadget requires one hour of assembly and the premium gadget requires two hours of assembly, and there are at most 12 hours available for this operation, we get
\[x + 2y \leq 12 \nonumber\]
- Similarly, we limited by the finishing operation. The regular gadget requires two hours of finishing and the premium gadget one hour. Again, there are at most 12 hours available for finishing. This gives us the following constraint.
\[2x + y \leq 12 \nonumber \]
- We are also limited by reality. The fact that \(x\) and \(y\) can never be negative is represented by the following two constraints:
\[x \geq 0 \text{, and } y \geq 0 \nonumber.\]
- In summary, our entire problem can be written as follows:
\[\begin{array}{ll}
\textbf { Maximize } & \mathrm{P}=20 \mathrm{x}+30 \mathrm{y} \\
\textbf { Subject to: } & \mathrm{x}+\mathrm{y} \leq 7 \\
& \mathrm{x}+2\mathrm{y} \leq 12 \\
& 2\mathrm{x} +\mathrm{y} \leq 12 \\
& \mathrm{x} \geq 0 ; \mathrm{y} \geq 0
\end{array} \nonumber\]
4. In order to solve the problem, we next graph each of the constraints.
By using Strategy A shown in the example above, we get a graph like this:
5. Determine the feasible region, which is the region that makes all inequalities true.
6. Determine the corner points of the feasible region.
- When graphing by hand, this can be done using any of the methods discussed in Section 1.3.
- When graphing in Desmos, this can be accomplished by clicking on the point of intersection.
7. We substitute these points in the objective function to see which point gives us the maximum profit each day. The results are listed below.
| Critical Point | Profit = \(20x + 30y\) |
|---|---|
| (0, 0) | P = 20(0) + 30(0) = $0 |
| (0, 6) | P = 20(0) + 30(6) = $180 |
| (2, 5) | P = 20(2) + 30(5) = $190 |
| (5, 2) | P = 20(5) + 30(2) = $160 |
| (6,0) | P = 20(6) + 30(0) = $120 |
8. The point (2, 5) gives the most profit, and that profit is $190. Therefore, we conclude that we should manufacture 2 regular gadgets and 5 premium gadgets daily to obtain the maximum profit of $190.
So far we have focused on “ standard maximization problems ” in which
- The objective function is to be maximized
- All constraints are of the form \(ax + by \leq c\)
- All variables are constrained to by non-negative (\(x ≥ 0\), \(y ≥ 0\))
We will next consider an example where that is not the case. Our next problem is said to have “ mixed constraints ”, since some of the inequality constraints are of the form \(ax + by ≤ c\) and some are of the form \(ax + by ≥ c\). The non-negativity constraints are still an important requirement in any linear program.
Example \(\PageIndex{3}\)
Solve the following maximization problem graphically.
\[\begin{array}{ll}
\textbf { Maximize } & \mathrm{Z}=10 \mathrm{x}+15 \mathrm{y} \\
\textbf { Subject to: } & \mathrm{x}+\mathrm{y} \geq 1 \\
& \mathrm{x}+2\mathrm{y} \leq 6 \\
& 2\mathrm{x} +\mathrm{y} \leq 6 \\
& \mathrm{x} \geq 0 ; \mathrm{y} \geq 0
\end{array} \nonumber\]
Solution
The graph is shown below.
The five critical points are listed in the above figure.
| Critical point | Z = \(10x + 15y\) |
|---|---|
| (1, 0) | Z = 10(1) + 15(0) = 10 |
| (3, 0) | Z = 10(3) + 15(0) = 30 |
| (2, 2) | Z = 10(2) + 15(2) = 50 |
| (0, 3) | Z = 10(0) + 15(3) = 45 |
| (0,1) | Z = 10(0) + 15(1) = 15 |
The point (2, 2) maximizes the objective function to a maximum z-value of 50.
We summarize:
The Maximization Linear Programming Problems
- Define the unknowns.
- Write the objective function that needs to be maximized.
-
Write the constraints.
- For the standard maximization linear programming problems, constraints are of the form: \(ax + by ≤ c\)
- Since the variables are non-negative, we include the constraints: \(x ≥ 0\); \(y ≥ 0\).
- Graph the constraints.
- Determine the feasible region.
- Find the corner points.
- Find the value of the objective function at each corner point to determine the corner point that gives the maximum value.
- State the answer to the original question.


7.3: Vaginal Infections and Other Conditions
By the end of this section, you will be able to:
- Describe the common vaginal infections of bacterial vaginitis and candidiasis
- Educate patients regarding the diagnosis, treatment, and prevention of bacterial vaginitis and candidiasis
Bacterial vaginitis and vulvovaginal candidiasis are common infectious causes of vaginitis. Anaerobic bacteria cause bacterial vaginitis, while vulvovaginal candidiasis is caused by a fungal infection (Paladine & Desai, 2018). Group B strep infection is a bacterial infection that can colonize the vagina and infect a newborn during delivery (CDC, 2022b).
Bacterial Vaginitis
The condition bacterial vaginitis (BV) occurs when there is an imbalance between the good bacteria and the harmful bacteria in the vagina, resulting in a larger amount of anaerobic bacteria. Symptoms may be minimal but include a homogeneous, thin vaginal discharge with a fishy odor (CDC, 2021b).
Incidence
Bacterial vaginitis is the most common cause of vaginal symptoms in persons assigned female at birth and affects over 21 million of them in the United States each year (CDC, 2021b). It affects 23 percent to 29 percent of persons AFAB across the world.
Screening and Diagnosis
Routine screening for BV is not indicated, but persons experiencing symptoms should report them to their provider. A person assigned female at birth whose partner is BV positive should get tested (CDC, 2021b).
Diagnosis can be made using clinical criteria or a gram stain. The most common diagnostic test is the Amsel diagnostic criteria , which requires that three out of four of the following be present:
- thin, homogeneous discharge
- positive whiff test
- clue cells present on microscopy (Figure 7.8)
- vaginal pH > 4.5
A whiff test is done by mixing a few drops of potassium hydroxide (KOH) with a vaginal sample. The KOH kills bacteria and leaves only yeast behind, revealing if there is a yeast infection. In addition, there is a fishy odor before or after the addition of 10 percent KOH to the sample (Paladine & Desai, 2018). Newer lab tests use DNA probes to detect BV (CDC, 2021b).
Vaginal Culture and Pap Smear
Disclaimer: Always follow the agency’s policy for vaginal culture and Pap smear.
Definition: Reduce the risk of harm to patients through effective, efficient, and competent performance.
Knowledge: The nurse will analyze basic safety principles, understand evidence-based practice standards, and reflect on unsafe nursing practices to ensure that patients are screened properly.
Skill: Demonstrate effective strategies to reduce the risk of harm. The nurse will do the following:
- Perform hand hygiene before patient contact. Don appropriate personal protective equipment (PPE) based on the patient’s need for isolation precautions or the risk of exposure to bodily fluids.
- Introduce yourself to the patient.
- Verify the correct patient using two identifiers.
- Determine whether the patient has had any previous pelvic examinations, procedures, or surgeries. Also determine if there are any questions the patient needs answered, as well as the patient’s ability to cooperate for the examination.
- Have the patient empty the bladder.
- Ensure that the light source works.
- Set up supplies for the examination and specimen collection.
- Assist the patient into the lithotomy position with the buttocks at the edge of the table and place a pillow under the head. Drape the patient’s abdomen and lower extremities so that only the perineal area is exposed. Do not place the patient into this position until just before the physical examination begins. Be prepared to provide assistance if a weak or dizzy patient is not able to maintain this position.
-
Ensure that a chaperone is present for all breast, genital, and rectal examinations performed by the practitioner.
- Papanicolaou (Pap) smear: Provider to obtain this specimen first.
- Obtain a specimen to diagnose Candida (yeast) organisms and submit it on a slide or on a swab per the organization’s practice. This may be done by a nurse or provider, depending on the organization.
- KOH dissolves other types of cells and protein, which makes it easier to see the Candida cells. On exam, will see clue cells—vaginal epithelial cells studded with adherent bacteria.
- Assist the patient with removing the feet from the footrests.
- Offer the patient a damp washcloth and towel to use to cleanse self after the examination.
- In the presence of the patient, label the specimen per the organization’s practice.
- Place the labeled specimen in a biohazard bag and transport it to the laboratory immediately per the organization’s practice.
- Assess, treat, and reassess pain.
- Discard supplies, remove PPE, and perform hand hygiene.
Attitude: The nurse will respect their role in assisting with vaginal cultures and Pap smears by adhering to safe, evidence-based practice standards.
Management and Treatment
Bacterial vaginitis can be treated with metronidazole (Flagyl) 500 mg orally 2 times a day for 7 days or metronidazole gel 0.75 percent one full applicator (5 g) intravaginally once a day for 5 days or clindamycin cream (Cleocin Vaginal) 2 percent one full applicator (5 g) intravaginally at bedtime for 7 days (CDC, 2021b). Douching can increase the risk of BV and should be avoided. Regular condom use can help prevent BV. The exact cause of the infection is not well understood (CDC, 2021b).
Complications
Bacterial vaginitis can cause discomfort, such as vaginal itching, copious discharge from the vagina that may be foul smelling, and burning during urination (WHO, 2023). It can also increase the risk of HIV, gonorrhea , chlamydia , and herpes infections (Paladine & Desai, 2018). Bacterial vaginitis during pregnancy can cause preterm delivery (CDC, 2021b).
Patient Education
Partners AMAB do not require treatment, but partners AFAB can spread the infection to each other. Bacterial vaginitis may go away without treatment, but it can increase the risk of complications (CDC, 2021b). The patient should be educated that douching can change the pH of the vagina and lead to BV and should always be avoided (CDC, 2021b).
Vaginitis in Those over 65
Vaginitis in patients over the age of 65 requires normal evaluation in addition to evaluation for vulvovaginal atrophy and genital neoplasia. Vulvovaginal atrophy is common in menopausal patients who may experience watery, white, or yellow discharge; vaginal burning or irritation; itching; and other urinary symptoms. Microscopy findings are nonspecific, and treatment includes topical estrogen therapy (Sobel, 2023).
Vulvovaginal Candidiasis
The yeast Candida causes the fungal infection vulvovaginal candidiasis (VVC) . This yeast normally lives on the skin and within the body without causing problems. It can cause a problem when there is overgrowth caused by hormones, medications, or changes in the immune system. Obesity and pregnancy can increase endogenous estrogen, which increases the risk. Diabetes mellitus, immunosuppressant medications, and broad-spectrum antibiotics can also raise the risk of acquiring VVC (Jeanmonod et al., 2023). VVC can cause vaginal itching and soreness, dyspareunia, external dysuria, and abnormal vaginal discharge (CDC, 2022b).
Incidence
About 75 percent of all persons assigned female at birth will have at least one occurrence of VVC in their lifetime, and 40 percent to 45 percent will have two or more episodes (CDC, 2022b). VVC is classified as noncomplicated or complicated. VVC is considered noncomplicated when it is infrequent and mild to moderate in a person who is not immunocompromised, and it is likely caused by Candida albicans . VVC is complicated when it is recurrent or severe, a non- albicans candidiasis, or in a person with diabetes or other immunocompromising conditions, such as HIV , or who is on immunosuppressive therapy (CDC, 2022b).
Screening and Diagnosis
There is no routine screening for VVC. Patients may complain of dysuria, pruritis, pain, swelling, redness, dyspareunia, and postcoital bleeding. They may also have a thick discharge with curds. The diagnosis can be made by the health-care provider noting the thick discharge on the vaginal walls or with a wet preparation : A slide of vaginal discharge under a microscope will show yeast (Figure 7.9). A 10 percent KOH preparation added to the slide will improve visualization of yeast. A vaginal culture can also be used to make the diagnosis (CDC, 2022b).
Management and Treatment
Fluconazole (Diflucan) is the only prescription oral treatment, which consists of one 150 mg dose. There are numerous over-the-counter vaginal creams, ointments, or suppositories used to treat VVC. These include clotrimazole , miconazole , and tioconazole . Prescription vaginal creams, ointments, or suppositories that are available include butoconazole (Gynazole-1) and terconazole (Terazol 3) (CDC, 2022b). During pregnancy, any of the topical azole medications is appropriate for treatment, but oral fluconazole should not be used, as it has been shown to increase the risk of spontaneous abortion and congenital anomalies (CDC, 2022b).
Fluconazole
- Generic Name: fluconazole
- Trade Name: Diflucan
- Class/Action: antifungal agent
- Route/Dosage: oral, 150 mg one time; severe infection, 150 mg every 72 hours for 2 to 3 doses
- High Alert/Black Box Warning: may occasionally cause dizziness or seizures, caution with driving
- Indications: used to treat vulvovaginal candidiasis
- Mechanism of Action: interferes with fungal activity, decreasing synthesis and cell formation
- Contraindications: hypersensitivity to fluconazole; coadministration with CYP3A4 substrates
- Adverse Reactions/Side Effects: headache, rash, abdominal pain, diarrhea, nausea, vomiting, dizziness, prolonged cardiac QT interval
- Parent/Family Education: Tell your doctor if you are pregnant or may be pregnant. Fluconazole can cause fetal harm.
Complications
Patients with complicated VVC should have a vaginal culture or PCR test to confirm the diagnosis and determine if the cause is non- albicans Candida . Patients with complicated VVC usually need a longer course of treatment. A longer course of a non-fluconazole treatment is recommended for non- albicans VVC (CDC, 2022b).
Patient Education
Poorly controlled diabetes can increase the risk of VVC. HIV-positive patients and patients taking immunosuppressant medications are at an increased risk of getting VVC, as are patients taking antibiotics. These patients may need longer treatment of VVC (CDC, 2022b). Creams and suppositories used to treat VVC are oil based and can weaken latex condoms and diaphragms. Sex partners do not usually require treatment (CDC, 2022b). The nurse should instruct the AFAB patient to follow these guidelines to prevent VVC:
- Do not douche because it alters normal bacteria.
- Do not use scented feminine products.
- Change tampons, pads, and panty liners often.
- Do not wear tight underwear, pantyhose, or clothes, which can increase the temperature.
- Wear underwear with a cotton crotch to stay dry.
- Change out of wet swimsuits and workout clothes as soon as possible.
- Always wipe from front to back when using the bathroom.
- Avoid hot tubs and taking hot baths (U.S. Department of Health and Human Services, 2021).
Group B Streptococcus
The bacterium group B streptococcus (GBS) can live in a person’s gastrointestinal and genital tracts without causing problems; it is not an STI pathogen. Sometimes, the bacteria can invade the body and cause infection. They also can be passed to a fetus during a vaginal delivery. Newborns infected with GBS can have severe complications, including death (CDC, 2022a).
Incidence
More than 28,000 cases of invasive group B streptococcus (GBS) are diagnosed in the United States each year (CDC, 2022a). About 1 in 4 pregnant persons have GBS in their body (CDC, 2022a). While the incidence in the United States has decreased due to maternal treatment, there are still 320,000 cases of newborns affected with GBS each year in the world (Berardi et al., 2021).
Screening and Diagnosis
Cultures should be taken to determine if an infection is present. These cultures can consist of blood, urine, or spinal fluid (CDC, 2022a). All pregnant persons should have a lower vaginal and rectal swab to test for GBS at 36 to 37 weeks’ gestation. Universal screening has yielded an 80 percent reduction in early-onset GBS in newborns. The swab should test for sensitivity if the patient has a severe penicillin allergy (Baker, 2023).
Management and Treatment
Treatment depends on the type of infection that the patient has. In pregnant persons with a positive GBS culture, intrapartum antibiotics should be given. Patients who have a history of giving birth to an infant with early-onset disease or who had GBS bacteriuria during the current pregnancy should also receive intrapartum antibiotics. Pregnant persons who have an unknown culture status (culture not performed or result not available) should receive antibiotics if they have
- intrapartum fever (≥100.4° F [≥38° C]) or
- preterm labor (<37+0 weeks of gestation) or
- preterm prelabor rupture of membranes or
- prolonged rupture of membranes (≥18 hours) or
- intrapartum nucleic acid amplification test (NAAT) positive for GBS.
Patients who have a scheduled cesarean section do not need intrapartum antibiotics (Baker, 2023).
Penicillin is the treatment of choice for pregnant persons with group B strep. The usual dose is 5 million units administered intravenously (IV) for the initial dose, followed by 2.5 to 3 million units IV every 4 hours until delivery. Ampicillin 2 g IV initial dose, followed by 1 g every 4 hours until delivery, can also be used. If there is a high risk for anaphylaxis to penicillin, susceptibility should be performed to see if the bacteria are susceptible to erythromycin or clindamycin (Baker, 2023).
Complications
Adults with a group B strep infection can develop sepsis. Infants born to persons with untreated GBS can experience serious consequences. Newborns can exhibit fever, difficulty in feeding, irritability or lethargy, difficulty in breathing, or blue color of the skin. These infants can also develop meningitis, which can cause death. Other long-term complications include deafness and developmental disabilities (CDC, 2022a).
Patient Education
Patients should be educated on the importance of testing for GBS during pregnancy. Patients should also be aware of their GBS results during pregnancy. Results should be at the place of birth where the patient delivers the baby, but in case they are not, the patient should be able to tell the nurse and provider their GBS status. Table 7.6 summarizes information about vaginal infections and other conditions.
| Disease | Organism | Signs and Symptoms | Lab Diagnosis | Treatment per CDC Guidelines |
|---|---|---|---|---|
| Bacterial vaginitis | Gardnerella. vaginalis, Prevotella species, Mobiluncus species, Atopobium vaginae , and other BV-associated anaerobic bacteria | Thin white or gray vaginal discharge; pain, itching, or burning in vagina; strong fish-like odor, especially after sex; burning with urination; itching around outside of vagina | Microscopy, vaginal pH, and Whiff test |
Metronidazole 500 mg orally 2 times/day for 7 days
OR Metronidazole gel 0.75% one full applicator (5 g) intravaginally, once a day for 5 days OR Clindamycin cream 2% one full applicator (5 g) intravaginally at bedtime for 7 days |
| Vulvovaginal candidiasis | Candida albicans | Vaginal itching or soreness, pain during sexual intercourse, pain or discomfort when urinating, or abnormal vaginal discharge | Wet preparation (saline, 10% KOH) of vaginal discharge demonstrates budding yeasts, hyphae, or pseudohyphae; and vaginal culture |
Clotrimazole 2% cream 5 g intravaginally daily for 3 days
OR Miconazole cream intravaginally daily for 7 days OR Miconazole vaginal suppository OR Tioconazole 6.5% ointment intravaginally in a single application OR other prescription antifungals |
| Group B streptococcus | Group B Streptococcus | Usually no symptoms; with cystitis: may have urinary frequency, urgency, or dysuria; with pyelonephritis: may have fever, urinary symptoms, nausea and vomiting, flank pain; infected newborn can have fever, difficulty in feeding, difficulty in breathing, lethargy, irritability, or bluish skin | Urine culture or vaginal culture |
Penicillin G (during labor) 5 million units IV initial dose, then 2.5 to 3 million units every 4 hours until delivery
OR Ampicillin 2 g IV initial dose, then 1 g IV every 4 hours until delivery |


4.15: Calculating Return on Investment
Learning Outcomes
Calculate return on investment
Liquidity ratios measure the financing that is supplied by the owners of the business versus the financing that is provided by creditors such as suppliers and banks. Leverage ratios also are indicators of the financial risk a business is exposed to. Low leverage ratios indicate that a downturn in the economy will have minimal impact on the financial position of a business. For example, if a business is funding its operations primarily through bank loans, a downturn in the economy will decrease revenues, but the business will still be responsible for payments to its creditors and lenders.
- Current Ratio. The current ratio measures a company’s ability to pay its current liabilities from its current assets.
Current Ratio= (Current Assets)/(Current Liabilities)
Generally, the higher the current ratio, the stronger the financial position of the business and typically financial analysts consider a current ratio of 2:1 to be ensure that a business has sufficient working capital to fund ongoing operations.
- Quick Ratio or Return on Investment (ROI). The quick ratio or return on investment (sometimes referred to as the acid test ratio) is more conservative than the current ratio because it does not include inventory in current assets. The quick ratio measures a company’s ability to pay its current debts if all revenue ceased.
Quick Ratio= (Current Assets — Inventory)/(Current Liabilities)
A quick ratio of 1:1 is considered satisfactory and a ratio higher than this is an indicator of greater financial security.
practice questions


2.1.6: Addressing the Drug Problem and Reducing Drug Use
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Learning Objectives
- Explain the problems associated with arresting hundreds of thousands of people for drug possession.
- List any three specific measures that may help deal with the drug problem.
As you may have already noticed and will notice again, the other chapters in this book usually present a fairly optimistic assessment when they discuss prospects for addressing the social problem discussed in each chapter. They point to the experience of other nations that do a good job of addressing the social problem, they cite social science evidence that points to solutions for addressing the problem, and they generally say that the United States could address the problem if it had the wisdom to approach it appropriately and to spend sufficient sums of money.
This chapter will not end with an optimistic assessment for addressing the drug problem. The reason for this lack of optimism is that what’s past is prologue: People have enthusiastically used drugs since prehistoric times and show no signs of reducing their drug use. Many and perhaps most scholars think the legal war on drugs has had little, if any, impact on drug use (Walker, 2011), and many scholars recognize that this war brought with it the many disadvantages cited in the previous section. As Kleiman et al. (2011, p. xvi) observe, “Our current drug policies allow avoidable harm by their ineffectiveness and create needless suffering by their excesses.”
A growing number of people in the political world agree. In 2011, the Global Commission on Drug Policy issued a major report on the world’s antidrug efforts. The commission comprised nineteen members, including a former United Nations secretary general, a former US secretary of state, a former chair of the US Federal Reserve, and former presidents or prime ministers of Brazil, Colombia, Greece, Mexico, and Switzerland. The commission’s report called for a drastic rethinking of current drug policy: “The global war on drugs has failed, with devastating consequences for individuals and societies around the world…Fundamental reforms in national and global drug control policies are urgently needed” (Global Commission on Drug Policy, 2011, p. 3). Decriminalization and even legalization of illegal drugs should be seriously considered, the report concluded.
Given this backdrop, many drug experts question whether our current drug policies make sense. They add that the best approach our society could take would be to expand the prevention, treatment, and harm reduction approaches discussed earlier; because drugs will always be with us, our society should do what it can to minimize the many harms that drugs cause. Thus drug education prevention and drug treatment programs should be expanded, sterile needles should be made available for drug addicts who inject their drugs, and drug courts should be used for a greater number of drug offenders.
Beyond these approaches, some experts say marijuana use should be decriminalized and that decriminalization of other drugs should be seriously considered. If marijuana were not only decriminalized but also legalized and taxed, it is estimated that this new tax revenue would amount to $8.7 billion annually and that about $8.7 billion annually would also be saved in reduced law enforcement costs, for a total of more than $17 billion in new funds that could be used for drug prevention, drug treatment, and other needs (Kristof, 2010). Many Americans agree with these experts: In a 2011 Gallup poll, 50 percent of the public favored legalizing marijuana, while 46 percent opposed legalizing it (Graves, 2011).
More generally, these experts say, it makes little sense to arrest more than 1.3 million people each year for drug possession and to put many of them in jail or prison. We do not arrest and imprison alcoholics and cigarette smokers; instead we try to offer them various kinds of help, and we should do the same for people who are addicted to other kinds of drugs. If arrest and imprisonment must continue, these measures should be reserved for sellers of large quantities of illegal drugs, not for the people who use the drugs or for those who sell only small quantities. When low-level drug dealers are imprisoned, they are simply replaced on the street by new dealers. Providing low-level dealers with alternative sentencing would reduce the number of imprisoned dealers over time by several hundred thousand annually without making illegal drugs more available.
In addition to all these measures, several other steps might well reduce certain kinds of drug use or at least reduce the harm that both drugs and our current drug policies cause (Kleiman et al., 2011). These steps include the following:
- Providing legally prescribed heroin and/or substitute opiates, including methadone, for heroin addicts. This provision has proven effective in several other nations.
- Encouraging primary care physicians and other health-care providers to screen more carefully for substance abuse.
- Basing drug sentencing less on the quantity of illegal drugs sold and more on the level of violence in which some drug sellers engage.
- Abandoning DARE. According to Kleiman et al. (2011, p. 201), “the continued dominance in school-based drug education of DARE—a program that has never been shown to actually reduce drug use—is a scandal.” They instead recommend school-based programs that help children develop self-control and prosocial behavior, as these programs have also been shown to reduce children’s subsequent drug use.
- Following the psychological principle of operant conditioning by providing drug addicts small cash payments for clean drug tests, as these rewards have been shown to be effective.
- Fully reintegrating former drug dealers and recovering drug addicts into society. They should have full access to public housing, educational loans, and other benefits, and they should be allowed to vote in states that now do not let them vote.
- Raising alcohol taxes. According to Kleiman et al. (2011), tripling the alcohol tax would especially reduce drinking by heavy drinkers and by minors, and it would reduce the number of homicides by 1,000 annually and the number of motor vehicle accidents by 2,000 annually. The new tax money could also help fund alcohol treatment and prevention programs. “In the entire field of drug-abuse control,” Kleiman et al. (2011, p. 204) write, “there is no bargain as attractive as a higher alcohol tax.”
- Prohibiting alcohol sales to anyone who has engaged in drunk driving or who has committed violence under the influence of alcohol. For this ban to work, everyone who wants to buy alcohol would have to show an ID, and those prohibited from buying alcohol would have that indicated on their ID. This ban would certainly be unpopular among the many drinkers who drink responsibly, but it would reduce the great harm that alcohol causes.
- Allowing marijuana users to grow their own cannabis or to buy it from small growers. This would reduce the sales of cannabis, and thus its profits, from the organized crime groups and the Mexican cartels that now provide much of the marijuana used in the United States.
- Raising the cigarette tax. Some states already have high cigarette taxes, but several states have low cigarette taxes. Raising the taxes in the low-tax states would reduce cigarette smoking in these states. The new tax revenue could be used to fund treatment programs that help reduce smoking.
Key Takeaways
- Critics of the war on drugs say that people who use illegal drugs should be treated, not arrested, just as people who use alcohol and tobacco are treated, if they seek treatment, rather than arrested.
- Specific measures that could help address the drug problem include providing legally prescribed heroin or substitute opiates for heroin addicts and raising the alcohol tax.
For Your Review
- Do you think that alcohol taxes should be raised? Why or why not?
- Do you favor decriminalization of marijuana? Explain your answer.
References
Global Commission on Drug Policy. (2011). War on drugs: Report of the Global Commission on Drug Policy . Rio de Janeiro, Brazil: Author.
Graves, L. (2011, October 18). Marijuana legalization receives 50 percent support in new poll. Huffington Post . Retrieved from http://www.huffingtonpost.com/2011/2...tml?utm_source =DailyBrief&utm_campaign=1101811&utm_medium=email&utm_content=NewsEntry &utm_term=Daily%1016420Brief.
Kleiman, M. A. R., Caulkins, J. P., & Hawken, A. (2011). Drugs and drug policy: What everyone needs to know . New York, NY: Oxford University Press.
Kristof, N. D. (2010, October 28). End the war on pot. New York Times , p. A33.
Walker, S. (2011).  Belmont, CA: Wadsworth.


18.3: Measuring Stellar Masses
By the end of this section, you will be able to:
- Distinguish the different types of binary star systems
- Understand how we can apply Newton’s version of Kepler’s third law to derive the sum of star mass es in a binary star system
- Apply the relationship between stellar mass and stellar luminosity to determine the physical characteristics of a star
The mass of a star—how much material it contains—is one of its most important characteristics. If we know a star’s mass, as we shall see, we can estimate how long it will shine and what its ultimate fate will be. Yet the mass of a star is very difficult to measure directly. Somehow, we need to put a star on the cosmic equivalent of a scale.
Luckily, not all stars live like the Sun, in isolation from other stars. About half the stars are binary stars —two stars that orbit each other, bound together by gravity. Masses of binary stars can be calculated from measurements of their orbits, just as the mass of the Sun can be derived by measuring the orbits of the planets around it (see Orbits and Gravity).
Binary Stars
Before we discuss in more detail how mass can be measured, we will take a closer look at stars that come in pairs. The first binary star was discovered in 1650, less than half a century after Galileo began to observe the sky with a telescope. John Baptiste Riccioli (1598–1671), an Italian astronomer, noted that the star Mizar, in the middle of the Big Dipper’s handle, appeared through his telescope as two stars. Since that discovery, thousands of binary stars have been cataloged. (Astronomers call any pair of stars that appear to be close to each other in the sky double stars , but not all of these form a true binary, that is, not all of them are physically associated. Some are just chance alignments of stars that are actually at different distances from us.) Although stars most commonly come in pairs, there are also triple and quadruple systems.
One well-known binary star is Castor, located in the constellation of Gemini. By 1804, astronomer William Herschel , who also discovered the planet Uranus, had noted that the fainter component of Castor had slightly changed its position relative to the brighter component. (We use the term “component” to mean a member of a star system.) Here was evidence that one star was moving around another. It was actually the first evidence that gravitational influences exist outside the solar system. The orbital motion of a binary star is shown in Figure 18.4. A binary star system in which both of the stars can be seen with a telescope is called a visual binary .
Edward C. Pickering (1846–1919), at Harvard, discovered a second class of binary stars in 1889—a class in which only one of the stars is actually seen directly. He was examining the spectrum of Mizar and found that the dark absorption lines in the brighter star’s spectrum were usually double. Not only were there two lines where astronomers normally saw only one, but the spacing of the lines was constantly changing. At times, the lines even became single. Pickering correctly deduced that the brighter component of Mizar, called Mizar A, is itself really two stars that revolve about each other in a period of 104 days. A star like Mizar A, which appears as a single star when photographed or observed visually through the telescope, but which spectroscopy shows really to be a double star, is called a spectroscopic binary .
Mizar, by the way, is a good example of just how complex such star systems can be. Mizar has been known for centuries to have a faint companion called Alcor, which can be seen without a telescope. Mizar and Alcor form an optical double —a pair of stars that appear close together in the sky but do not orbit each other. Through a telescope, as Riccioli discovered in 1650, Mizar can be seen to have another, closer companion that does orbit it; Mizar is thus a visual binary. The two components that make up this visual binary, known as Mizar A and Mizar B, are both spectroscopic binaries. So, Mizar is really a quadruple system of stars.
Strictly speaking, it is not correct to describe the motion of a binary star system by saying that one star orbits the other. Gravity is a mutual attraction. Each star exerts a gravitational force on the other, with the result that both stars orbit a point between them called the center of mass . Imagine that the two stars are seated at either end of a seesaw. The point at which the fulcrum would have to be located in order for the seesaw to balance is the center of mass, and it is always closer to the more massive star (Figure 18.5).
Figure 18.6 shows two stars (A and B) moving around their center of mass, along with one line in the spectrum of each star that we observe from the system at different times. When one star is approaching us relative to the center of mass, the other star is receding from us. In the top left illustration, star A is moving toward us, so the line in its spectrum is Doppler-shifted toward the blue end of the spectrum. Star B is moving away from us, so its line shows a redshift. When we observe the composite spectrum of the two stars, the line appears double. When the two stars are both moving across our line of sight (neither away from nor toward us), they both have the same radial velocity (that of the pair’s center of mass); hence, the spectral lines of the two stars come together. This is shown in the two bottom illustrations in Figure 18.6.
A plot showing how the velocities of the stars change with time is called a radial velocity curve ; the curve for the binary system in Figure 18.6 is shown in Figure 18.7.
This animation lets you follow the orbits of a binary star system in various combinations of the masses of the two stars.
Masses from the Orbits of Binary Stars
We can estimate the masses of binary star systems using Newton’s reformulation of Kepler’s third law (discussed in Newton’s Universal Law of Gravitation). Kepler found that the time a planet takes to go around the Sun is related by a specific mathematical formula to its distance from the Sun. In our binary star situation, if two objects are in mutual revolution, then the period ( P ) with which they go around each other is related to the semimajor axis ( D ) of the orbit of one with respect to the other, according to this equation
where D is in astronomical units, P is measured in years, and M 1 + M 2 is the sum of the masses of the two stars in units of the Sun’s mass. This is a very useful formula for astronomers; it says that if we can observe the size of the orbit and the period of mutual revolution of the stars in a binary system, we can calculate the sum of their masses.
Most spectroscopic binaries have periods ranging from a few days to a few months, with separations of usually less than 1 AU between their member stars. Recall that an AU is the distance from Earth to the Sun, so this is a small separation and very hard to see at the distances of stars. This is why many of these systems are known to be double only through careful study of their spectra.
We can analyze a radial velocity curve (such as the one in Figure 18.7) to determine the masses of the stars in a spectroscopic binary. This is complex in practice but not hard in principle. We measure the speeds of the stars from the Doppler effect . We then determine the period—how long the stars take to go through an orbital cycle—from the velocity curve. Knowing how fast the stars are moving and how long they take to go around tells us the circumference of the orbit and, hence, the separation of the stars in kilometers or astronomical units. From Kepler’s law, the period and the separation allow us to calculate the sum of the stars’ masses.
Of course, knowing the sum of the masses is not as useful as knowing the mass of each star separately. But the relative orbital speeds of the two stars can tell us how much of the total mass each star has. As we saw in our seesaw analogy, the more massive star is closer to the center of mass and therefore has a smaller orbit. Therefore, it moves more slowly to get around in the same time compared to the more distant, lower-mass star. If we sort out the speeds relative to each other, we can sort out the masses relative to each other. In practice, we also need to know how the binary system is oriented in the sky to our line of sight, but if we do, and the just-described steps are carried out carefully, the result is a calculation of the masses of each of the two stars in the system.
To summarize, a good measurement of the motion of two stars around a common center of mass, combined with the laws of gravity, allows us to determine the masses of stars in such systems. These mass measurements are absolutely crucial to developing a theory of how stars evolve. One of the best things about this method is that it is independent of the location of the binary system. It works as well for stars 100 light-years away from us as for those in our immediate neighborhood.
To take a specific example, Sirius is one of the few binary stars in Appendix J for which we have enough information to apply Kepler’s third law:
In this case, the two stars, the one we usually call Sirius and its very faint companion, are separated by about 20 AU and have an orbital period of about 50 years. If we place these values in the formula we would have
This can be solved for the sum of the masses:
Therefore, the sum of masses of the two stars in the Sirius binary system is 3.2 times the Sun’s mass. In order to determine the individual mass of each star, we would need the velocities of the two stars and the orientation of the orbit relative to our line of sight.
The Range of Stellar Masses
How large can the mass of a star be? Stars more massive than the Sun are rare. None of the stars within 30 light-years of the Sun has a mass greater than four times that of the Sun. Searches at large distances from the Sun have led to the discovery of a few stars with masses up to about 100 times that of the Sun, and a handful of stars (a few out of several billion) may have masses as large as 250 solar masses. However, most stars have less mass than the Sun.
According to theoretical calculations, the smallest mass that a true star can have is about 1/12 that of the Sun. By a “true” star, astronomers mean one that becomes hot enough to fuse protons to form helium (as discussed in The Sun: A Nuclear Powerhouse). Objects with masses between roughly 1/100 and 1/12 that of the Sun may produce energy for a brief time by means of nuclear reactions involving deuterium, but they do not become hot enough to fuse protons. Such objects are intermediate in mass between stars and planets and have been given the name brown dwarfs (Figure 18.8). Brown dwarfs are similar to Jupiter in radius but have masses from approximately 13 to 80 times larger than the mass of Jupiter. 2
Still-smaller objects with masses less than about 1/100 the mass of the Sun (or 10 Jupiter masses) are called planets. They may radiate energy produced by the radioactive elements that they contain, and they may also radiate heat generated by slowly compressing under their own weight (a process called gravitational contraction). However, their interiors will never reach temperatures high enough for any nuclear reactions, to take place. Jupiter, whose mass is about 1/1000 the mass of the Sun, is unquestionably a planet, for example. Until the 1990s, we could only detect planets in our own solar system, but now we have thousands of them elsewhere as well. (We will discuss these exciting observations in The Birth of Stars and the Discovery of Planets outside the Solar System.)
The Mass-Luminosity Relation
Now that we have measurements of the characteristics of many different types of stars, we can search for relationships among the characteristics. For example, we can ask whether the mass and luminosity of a star are related. It turns out that for most stars, they are: The more massive stars are generally also the more luminous. This relationship, known as the mass-luminosity relation , is shown graphically in Figure 18.9. Each point represents a star whose mass and luminosity are both known. The horizontal position on the graph shows the star’s mass, given in units of the Sun’s mass, and the vertical position shows its luminosity in units of the Sun’s luminosity.
We can also say this in mathematical terms.
It’s a reasonably good approximation to say that luminosity (expressed in units of the Sun’s luminosity) varies as the fourth power of the mass (in units of the Sun’s mass). (The symbol ~ means the two quantities are proportional.) If two stars differ in mass by a factor of 2, then the more massive one will be 2 4 , or about 16 times brighter; if one star is 1/3 the mass of another, it will be approximately 81 times less luminous.
The mass-luminosity formula can be rewritten so that a value of mass can be determined if the luminosity is known.
Solution
First, we must get our units right by expressing both the mass and the luminosity of a star in units of the Sun’s mass and luminosity:
Now we can take the 4th root of both sides, which is equivalent to taking both sides to the 1/4 = 0.25 power. The formula in this case would be:
In the previous section, we determined the sum of the masses of the two stars in the Sirius binary system (Sirius and its faint companion) using Kepler’s third law to be 3.2 solar masses. Using the mass-luminosity relationship, calculate the mass of each individual star.
- Answer
-
In Appendix J, Sirius is listed with a luminosity 23 times that of the Sun. This value can be inserted into the mass-luminosity relationship to get the mass of Sirius:
The mass of the companion star to Sirius is then solar mass.
Notice how good this mass-luminosity relationship is. Most stars (see Figure 18.9) fall along a line running from the lower-left (low mass, low luminosity) corner of the diagram to the upper-right (high mass, high luminosity) corner. About 90% of all stars obey the mass-luminosity relation. Later, we will explore why such a relationship exists and what we can learn from the roughly 10% of stars that “disobey” it.
This visualization illustrates how all the properties of a star are determined by the mass that the star happened to form with and how these properties change when you change the mass.
Footnotes
- 2 Exactly where to put the dividing line between planets and brown dwarfs is a subject of some debate among astronomers as we write this book (as is, in fact, the exact definition of each of these objects). Even those who accept deuterium fusion (see The Birth of Stars and the Discovery of Planets outside the Solar System) as the crucial issue for brown dwarfs concede that, depending on the composition of the star and other factors, the lowest mass for such a dwarf could be anywhere from 11 to 16 Jupiter masses.


3.23: Linear Relationships (4 of 4)
A link to an interactive elements can be found at the bottom of this page.
To see how an outlier affects the correlation, do the following:
- Fill the scatterplot with a hypothetical positive linear relationship between X and Y (by clicking on the graph about a dozen times starting at the lower left and going up diagonally to the top right). Pay attention to the correlation coefficient calculated at the top left of the simulation.
- Once you are satisfied with your hypothetical data, create an outlier by clicking on one of the data points in the upper right of the graph and dragging it down along the right side of the graph. Again, pay attention to what happens to the value of the correlation.
What did this activity illustrate? This activity illustrates that the correlation decreases when the outlier deviates from the pattern of the relationship. By dragging a data point from the upper right to the lower right, you created an outlier that does not fit the positive association in the rest of the data. This decreases the strength of the linear relationship and causes a decrease in .
In the next activity, you will see how the correlation increases when the outlier is consistent with the direction of the linear relationship.
Let’s Summarize
- A special case of the relationship between two quantitative variables is the linear relationship in which a straight line simply and adequately summarizes the relationship.
- When the scatterplot displays a linear relationship, we supplement it with the correlation coefficient ( r ), which measures the strength and direction of a linear relationship between two quantitative variables. The correlation ranges between −1 and 1. Values near −1 indicate a strong negative linear relationship, values near 0 indicate a weak linear relationship, and values near 1 indicate a strong positive linear relationship.
- The correlation is an appropriate numerical measure only for linear relationships and is sensitive to outliers. Therefore, the correlation should be used only as a supplement to a scatterplot (after we look at the data).
Contributors and Attributions
CC licensed content, Shared previously
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution


29.1: Chordates
Skills to Develop
- Describe the distinguishing characteristics of chordates
- Identify the derived character of craniates that sets them apart from other chordates
- Describe the developmental fate of the notochord in vertebrates
Vertebrates are members of the kingdom Animalia and the phylum Chordata (Figure \(\PageIndex{1}\)). Recall that animals that possess bilateral symmetry can be divided into two groups—protostomes and deuterostomes—based on their patterns of embryonic development. The deuterostomes, whose name translates as “second mouth,” consist of two phyla: Chordata and Echinodermata. Echinoderms are invertebrate marine animals that have pentaradial symmetry and a spiny body covering, a group that includes sea stars, sea urchins, and sea cucumbers. The most conspicuous and familiar members of Chordata are vertebrates, but this phylum also includes two groups of invertebrate chordates.
Characteristics of Chordata
Animals in the phylum Chordata share four key features that appear at some stage during their development: a notochord, a dorsal hollow nerve cord, pharyngeal slits, and a post-anal tail (Figure \(\PageIndex{2}\)). In some groups, some of these are present only during embryonic development.
The chordates are named for the notochord , which is a flexible, rod-shaped structure that is found in the embryonic stage of all chordates and in the adult stage of some chordate species. It is located between the digestive tube and the nerve cord, and provides skeletal support through the length of the body. In some chordates, the notochord acts as the primary axial support of the body throughout the animal’s lifetime. In vertebrates, the notochord is present during embryonic development, at which time it induces the development of the neural tube and serves as a support for the developing embryonic body. The notochord, however, is not found in the postnatal stage of vertebrates; at this point, it has been replaced by the vertebral column (that is, the spine).
Art Connection
Which of the following statements about common features of chordates is true?
- The dorsal hollow nerve cord is part of the chordate central nervous system.
- In vertebrate fishes, the pharyngeal slits become the gills.
- Humans are not chordates because humans do not have a tail.
- Vertebrates do not have a notochord at any point in their development; instead, they have a vertebral column.
The dorsal hollow nerve cord derives from ectoderm that rolls into a hollow tube during development. In chordates, it is located dorsal to the notochord. In contrast, other animal phyla are characterized by solid nerve cords that are located either ventrally or laterally. The nerve cord found in most chordate embryos develops into the brain and spinal cord, which compose the central nervous system.
Pharyngeal slits are openings in the pharynx (the region just posterior to the mouth) that extend to the outside environment. In organisms that live in aquatic environments, pharyngeal slits allow for the exit of water that enters the mouth during feeding. Some invertebrate chordates use the pharyngeal slits to filter food out of the water that enters the mouth. In vertebrate fishes, the pharyngeal slits are modified into gill supports, and in jawed fishes, into jaw supports. In tetrapods, the slits are modified into components of the ear and tonsils. Tetrapod literally means “four-footed,” which refers to the phylogenetic history of various groups that evolved accordingly, even though some now possess fewer than two pairs of walking appendages. Tetrapods include amphibians, reptiles, birds, and mammals.
The post-anal tail is a posterior elongation of the body, extending beyond the anus. The tail contains skeletal elements and muscles, which provide a source of locomotion in aquatic species, such as fishes. In some terrestrial vertebrates, the tail also helps with balance, courting, and signaling when danger is near. In humans, the post-anal tail is vestigial, that is, reduced in size and nonfunctional.
Link to Learning
Click for a video discussing the evolution of chordates and five characteristics that they share.
Chordates and the Evolution of Vertebrates
Chordata also contains two clades of invertebrates: Urochordata and Cephalochordata. Members of these groups also possess the four distinctive features of chordates at some point during their development.
Urochordata
Members of Urochordata are also known as tunicates (Figure \(\PageIndex{3}\)). The name tunicate derives from the cellulose-like carbohydrate material, called the tunic, which covers the outer body of tunicates. Although adult tunicates are classified as chordates, they do not have a notochord, a dorsal hollow nerve cord, or a post-anal tail, although they do have pharyngeal slits. The larval form, however, possesses all four structures. Most tunicates are hermaphrodites. Tunicate larvae hatch from eggs inside the adult tunicate’s body. After hatching, a tunicate larva swims for a few days until it finds a suitable surface on which it can attach, usually in a dark or shaded location. It then attaches via the head to the surface and undergoes metamorphosis into the adult form, at which point the notochord, nerve cord, and tail disappear.
Most tunicates live a sessile existence on the ocean floor and are suspension feeders. The primary foods of tunicates are plankton and detritus. Seawater enters the tunicate’s body through its incurrent siphon. Suspended material is filtered out of this water by a mucous net (pharyngeal slits) and is passed into the intestine via the action of cilia. The anus empties into the excurrent siphon, which expels wastes and water. Tunicates are found in shallow ocean waters around the world.
Cephalochordata
Members of Cephalochordata possess a notochord, dorsal hollow nerve cord, pharyngeal slits, and a post-anal tail in the adult stage (Figure \(\PageIndex{4}\)). The notochord extends into the head, which gives the subphylum its name. Extinct members of this subphylum include Pikaia , which is the oldest known cephalochordate. Pikaia fossils were recovered from the Burgess shales of Canada and dated to the middle of the Cambrian age, making them more than 500 million years old.
Extant members of Cephalochordata are the lancelets , named for their blade-like shape. Lancelets are only a few centimeters long and are usually found buried in sand at the bottom of warm temperate and tropical seas. Like tunicates, they are suspension feeders.
Craniata and Vertebrata
A cranium is a bony, cartilaginous, or fibrous structure surrounding the brain, jaw, and facial bones (Figure \(\PageIndex{5}\)). Most bilaterally symmetrical animals have a head; of these, those that have a cranium compose the clade Craniata . Craniata includes the hagfishes (Myxini), which have a cranium but lack a backbone, and all of the organisms called “vertebrates.”
Vertebrates are members of the clade Vertebrata . Vertebrates display the four characteristic features of the chordates; however, members of this group also share derived characteristics that distinguish them from invertebrate chordates. Vertebrata is named for the vertebral column , composed of vertebrae, a series of separate bones joined together as a backbone (Figure \(\PageIndex{6}\)). In adult vertebrates, the vertebral column replaces the notochord, which is only seen in the embryonic stage.
Based on molecular analysis, vertebrates appear to be more closely related to lancelets (cephalochordates) than to tunicates (urochordates) among the invertebrate chordates. This evidence suggests that the cephalochordates diverged from Urochordata and the vertebrates subsequently diverged from the cephalochordates. This hypothesis is further supported by the discovery of a fossil in China from the genus Haikouella . This organism seems to be an intermediate form between cephalochordates and vertebrates. The Haikouella fossils are about 530 million years old and appear similar to modern lancelets. These organisms had a brain and eyes, as do vertebrates, but lack the skull found in craniates. 1 This evidence suggests that vertebrates arose during the Cambrian explosion. Recall that the “Cambrian explosion” is the name given to a relatively brief span of time during the Cambrian period during which many animal groups appeared and rapidly diversified. Most modern animal phyla originated during the Cambrian explosion.
Vertebrates are the largest group of chordates, with more than 62,000 living species. Vertebrates are grouped based on anatomical and physiological traits. More than one classification and naming scheme is used for these animals. Here we will consider the traditional groups Agnatha, Chondrichthyes, Osteichthyes, Amphibia, Reptilia, Aves, and Mammalia, which constitute classes in the subphylum Vertebrata. Many modern authors classify birds within Reptilia, which correctly reflects their evolutionary heritage. We consider them separately only for convenience. Further, we will consider hagfishes and lampreys together as jawless fishes, the agnathans, although emerging classification schemes separate them into chordate jawless fishes (the hagfishes) and vertebrate jawless fishes (the lampreys).
Animals that possess jaws are known as gnathostomes, which means “jawed mouth.” Gnathostomes include fishes and tetrapods—amphibians, reptiles, birds, and mammals. Tetrapods can be further divided into two groups: amphibians and amniotes. Amniotes are animals whose eggs are adapted for terrestrial living, and this group includes mammals, reptiles, and birds. Amniotic embryos, developing in either an externally shed egg or an egg carried by the female, are provided with a water-retaining environment and are protected by amniotic membranes.
 Chordata contains two clades of invertebrates: Urochordata (tunicates) and Cephalochordata (lancelets), together with the vertebrates in Vertebrata. Most tunicates live on the ocean floor and are suspension feeders. Lancelets are suspension feeders that feed on phytoplankton and other microorganisms. Vertebrata is named for the vertebral column, which is a feature of almost all members of this clade.
Art Connections
Figure \(\PageIndex{2}\): Which of the following statements about common features of chordates is true?
- The dorsal hollow nerve cord is part of the chordate central nervous system.
- In vertebrate fishes, the pharyngeal slits become the gills.
- Humans are not chordates because humans do not have a tail.
- Vertebrates do not have a notochord at any point in their development; instead, they have a vertebral column.
- Answer
-
A
Footnotes
- 1 Chen, J. Y., Huang, D. Y., and Li, C. W., “An early Cambrian craniate-like chordate,” Nature 402 (1999): 518–522, doi:10.1038/990080.
Glossary
- Cephalochordata
- chordate clade whose members possess a notochord, dorsal hollow nerve cord, pharyngeal slits, and a post-anal tail in the adult stage
- Chordata
- phylum of animals distinguished by their possession of a notochord, a dorsal hollow nerve cord, pharyngeal slits, and a post-anal tail at some point during their development
- Craniata
- clade composed of chordates that possess a cranium; includes Vertebrata together with hagfishes
- cranium
- bony, cartilaginous, or fibrous structure surrounding the brain, jaw, and facial bones
- dorsal hollow nerve cord
- hollow, tubular structure derived from ectoderm, which is located dorsal to the notochord in chordates
- lancelet
- member of Cephalochordata; named for its blade-like shape
- notochord
- flexible, rod-shaped support structure that is found in the embryonic stage of all chordates and in the adult stage of some chordates
- pharyngeal slit
- opening in the pharynx
- post-anal tail
- muscular, posterior elongation of the body extending beyond the anus in chordates
- tetrapod
- phylogenetic reference to an organism with a four-footed evolutionary history; includes amphibians, reptiles, birds, and mammals
- tunicate
- sessile chordate that is a member of Urochordata
- Urochordata
- clade composed of tunicates
- vertebral column
- series of separate bones joined together as a backbone
- Vertebrata
- members of the phylum Chordata that possess a backbone


1.4: The Lorentz Transformation
Learning Objectives
- Study of the Lorentz transformation
Philosophically, coordinates are unnecessary, but they are convenient. They are arbitrary, so we can change from one set to another. For example, we can change the units used to measure time and position, as in the first and second panels of figure \(\PageIndex{1}\). Nothing changes about the underlying events; only the labels are different. The third panel shows a convenient convention we will use to depict such changes visually. The gray rectangle represents the original grid from the first panel, while the grid of black lines represents the new version from the second panel. Omitting the grid from the gray rectangle makes the diagram easier to decode visually.
In special relativity it is of interest to convert between the Minkowski coordinates of observers who are in motion relative to one another. The result, shown in figure \(\PageIndex{1}\), is a kind of stretching and smooshing of the diagonals. Since the area is invariant, one diagonal grows by the same factor by which the other shrinks. This change of coordinates is called the Lorentz transformation .
Figure \(\PageIndex{3}\) shows how time dilation and length contraction come about in this picture. It should be emphasized here that the Lorentz transformation includes more effects than just length contraction and time dilation. Many beginners at relativity get confused and come to erroneous conclusions by trying to reduce everything to a matter of inserting factors of \(\gamma\) in various equations. If the Lorentz transformation amounted to nothing more than length contraction and time dilation, it would be merely a change of units like the one shown in figure \(\PageIndex{1}\).
The Lorentz transformation can be notated algebraically:
\[t' = \gamma t - \nu\gamma x \\ x' = -\nu\gamma t + \gamma x \label{\(\PageIndex{1}\)}\]
The fact that this is the correct relativistic transformation can be verified by noting that
- the speed-of-light lines \(x = \pm t\) are preserved, and
- the determinant equals \(1\).
so that areas are preserved. Alternatively, it is sufficient to check the invariance of the spacetime interval under this transformation.
Equations \(\PageIndex{1}\) treat space and time in a perfectly symmetric way, but this should not be taken as implying that special relativity perfectly embodies such a symmetry. For example, I can easily revisit a place that I’ve been to before, but I can’t go back in time. And of course we have three dimensions of space; our use of \(1+1\) dimensions rather than \(3+1\) is just a matter of convenience for the moment. Note also that there is no exact analogy between figure \(\PageIndex{3}\) (1), where the clock is a pointlike object tracing a line through spacetime, and figure \(\PageIndex{3}\) (2), where the ruler is an extended body that sweeps out a parallel-sided ribbon.
Example \(\PageIndex{1}\): Observers agree on their relative speeds
Observer \(A\) says observer \(B\) is moving away from her at velocity \(v\); is it true, as in Galilean relativity, that \(B\) measures the same speed for \(A\)?
Yes, it is true, but not completely obvious. One way to verify this fact is to check that Lorentz transformations with velocities \(v\) and \(-v\) are inverses. A more physically transparent justification is shown in figure \(\PageIndex{4}\). In figure \(\PageIndex{4}\) (1), \(A\) determines \(B’s\) velocity relative to her by sending out two round-trip signals at the speed of light, and measuring the difference between the two round-trip times. Because space is the same in all directions, the experimental data are exactly the same when \(B\) carries out the measurement, figure \(\PageIndex{4}\) (2), and therefore \(B\) infers the same speed.
Example \(\PageIndex{2}\): Motion in the opposite direction
Figure \(\PageIndex{5}\) shows the case where the observer whose frame is represented by the grid is moving to the left relative to the one whose frame is represented by the gray square.
Example \(\PageIndex{3}\): Other quadrants
So far I’ve been arbitrarily choosing to draw only the first quad- rant of each coordinate system. Figure \(\PageIndex{6}\) shows a region that includes all four quadrants.
Example \(\PageIndex{4}\): A numerical example of invariance
Figure \(\PageIndex{7}\) shows two frames of reference in motion relative to one another at \(v = \frac{3}{5}\). (For this velocity, the stretching and squishing of the main diagonals are both by a factor of \(2\)). Events are marked at coordinates that in the frame represented by the square are
\[(t,x) = (0,0) \\ (t,x) = (13,11)\]
The interval between these events is \(13^2 - 11^2 = 48\). In the frame represented by the parallelogram, the same two events lie at coordinates
\[(t',x') = (0,0) \\ (t',x') = (8,4)\]
Calculating the interval using these values, the result is \(8^2 - 4^2 = 48\), which comes out the same as in the other frame.
Example \(\PageIndex{5}\): The garage paradox
One of the most famous of all the so-called relativity paradoxes has to do with our incorrect feeling that simultaneity is well defined. The idea is that one could take a schoolbus and drive it at relativistic speeds into a garage of ordinary size, in which it normally would not fit. Because of the length contraction, the bus would supposedly fit in the garage. The driver, however, will perceive the garage as being contracted and thus even less able to contain the bus.
The paradox is resolved when we recognize that the concept of fitting the bus in the garage “all at once” contains a hidden assumption, the assumption that it makes sense to ask whether the front and back of the bus can simultaneously be in the garage. Observers in different frames of reference moving at high relative speeds do not necessarily agree on whether things happen simultaneously. As shown in figure aw. the person in the garage’s frame can shut the door at an instant B he perceives to be simultaneous with the front bumper’s arrival A at the back wall of the garage, but the driver would not agree about the simultaneity of these two events, and would perceive the door as having shut long after she plowed through the back wall.
Example \(\PageIndex{6}\): Shifting clocks
The top row of clocks in the figure \(\PageIndex{9}\) are located in three different places. They have been synchronized in the frame of reference of the earth, represented by the paper. This synchronization is carried out by exchanging light signals (Einstein synchronization). For example, if the front and back clocks both send out flashes of light when they think it’s \(2\) o’clock, the one in the middle will receive them both at the same time. Event \(A\) is the one at which the back clock \(A\) reads \(2\) o’clock, etc.
The bottom row of clocks are aboard the train, and have been synchronized in a similar way. For the reasons discussed in example 1.1.4 in section 1.1, their synchronization differs from that of the earth-based clocks. By referring to the diagram of the Lorentz transformation shown on the right, we see that in the frame of the train, \(2\), \(C\) happens first, then \(B\), then \(A\).
This is an example of the interpretation of the term \(t' = \dots - \nu\gamma x\) in the Lorentz transformation (eq. \(\PageIndex{1}\)). Because the events occur at different \(x’s\), each is shifted in time relative to the next, according to clocks synchronized in frame 2 (\(t'\), the train).
Example \(\PageIndex{7}\): Deja vu?
The grids we’ve been drawing are mere conventions, as elaborate and arbitrary as dress in the court of Louis XIV. They come from a surveying process, which may need to be planned in advance and whose results we might not be able to see until later. The dark line in figure \(\PageIndex{10}\) is the world-line of an observer \(O\), who moves inertially for a while, accelerates to the left, and then moves inertially again. On the right are pasted two coordinate grids adapted to the two inertial segments. At event \(E\), Bush steals the 2000 election, and this is depicted as being simultaneous with both event \(A\) and event \(B\). Does poor O see it happen twice? No, even if the bad news is transmitted by a signal moving at the speed of light (dashed line), \(O\) receives it only once, at event \(C\).
The only problem here is a poor choice of labels, which causes \(E\) to have more than one label. Something similar happens in a constant-acceleration frame, section 7.1,
Many mistakes by beginners at relativity revolve around a set of unexamined preconceptions about what it means to observe things. One imagines that effects such as length contraction and time dilation are what an observer actually sees, and perhaps that this process of seeing is instantaneous. Or one thinks of Minkowski coordinates as if they were the result of a simple and automatic process of perception by an observer. This is the kind of thinking that will lead one to believe that example \(\PageIndex{7}\) is crazy or paradoxical.
As another example, it should not be imagined that the length contraction of a stick by \(\frac{1}{\gamma}\) is what an observer actually sees when looking at the stick. Optical observations are influenced, for example, by the unequal times taken for light to propagate from the ends of the stick to the eye. A simulation of this type of effect is drawn in example 1.3.3.
Length contraction, time dilation, the observer-dependence of simultaneity, and Minkowski coordinates are all sophisticated results of a laborious process of collecting and analyzing data obtained by techniques such as Einstein synchronization, which require actions such as consulting atomic clocks or exchanging signals between different points at the speed of light. Figure \(\PageIndex{11}\) outlines such a process in a cartoonish way. A fleet of rocket ships, carrying surveyors, is sent out from Earth and dispersed throughout a vast region of space. Surveyors look through their theodolites at images, which are formed by light rays (dashed lines) that have arrived after traveling at the finite speed c. Such light rays carry old, stale information about various events. A nuclear war has broken out. Rock and roll music has arrived on Saturn. The resulting data are then transmitted by various means (passenger pigeon, Morse-coded radio, paper mail) and consolidated at the surveying office. At the office, workers at a long row of desks crunch the numbers and produce a chart of Minkowski coordinates with the events marked in.