11.1: The one-sample z-test
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In this section I’ll describe one of the most useless tests in all of statistics: the z-test . Seriously – this test is almost never used in real life. Its only real purpose is that, when teaching statistics, it’s a very convenient stepping stone along the way towards the t-test, which is probably the most (over)used tool in all statistics.
inference problem that the test addresses
To introduce the idea behind the z-test, let’s use a simple example. A friend of mine, Dr Zeppo, grades his introductory statistics class on a curve. Let’s suppose that the average grade in his class is 67.5, and the standard deviation is 9.5. Of his many hundreds of students, it turns out that 20 of them also take psychology classes. Out of curiosity, I find myself wondering: do the psychology students tend to get the same grades as everyone else (i.e., mean 67.5) or do they tend to score higher or lower? He emails me the
zeppo.Rdata
file, which I use to pull up the
grades
of those students,
load( "./rbook-master/data/zeppo.Rdata" )
print( grades )
## [1] 50 60 60 64 66 66 67 69 70 74 76 76 77 79 79 79 81 82 82 89
and calculate the mean:
mean( grades )
## [1] 72.3
Hm. It might be that the psychology students are scoring a bit higher than normal: that sample mean of \(\bar{X}\) = 72.3 is a fair bit higher than the hypothesised population mean of μ=67.5, but on the other hand, a sample size of N=20 isn’t all that big. Maybe it’s pure chance.
To answer the question, it helps to be able to write down what it is that I think I know. Firstly, I know that the sample mean is \(\bar{X}\) =72.3. If I’m willing to assume that the psychology students have the same standard deviation as the rest of the class then I can say that the population standard deviation is σ=9.5. I’ll also assume that since Dr Zeppo is grading to a curve, the psychology student grades are normally distributed.
Next, it helps to be clear about what I want to learn from the data. In this case, my research hypothesis relates to the population mean μ for the psychology student grades, which is unknown. Specifically, I want to know if μ=67.5 or not. Given that this is what I know, can we devise a hypothesis test to solve our problem? The data, along with the hypothesised distribution from which they are thought to arise, are shown in Figure 13.1. Not entirely obvious what the right answer is, is it? For this, we are going to need some statistics.
Constructing the hypothesis test
The first step in constructing a hypothesis test is to be clear about what the null and alternative hypotheses are. This isn’t too hard to do. Our null hypothesis, H 0 , is that the true population mean μ for psychology student grades is 67.5%; and our alternative hypothesis is that the population mean isn’t 67.5%. If we write this in mathematical notation, these hypotheses become,
H 0 :μ=67.5
H 1 :μ≠67.5
though to be honest this notation doesn’t add much to our understanding of the problem, it’s just a compact way of writing down what we’re trying to learn from the data. The null hypotheses H 0 and the alternative hypothesis H 1 for our test are both illustrated in Figure 13.2. In addition to providing us with these hypotheses, the scenario outlined above provides us with a fair amount of background knowledge that might be useful. Specifically, there are two special pieces of information that we can add:
1 The psychology grades are normally distributed. 1 The true standard deviation of these scores σ is known to be 9.5.
For the moment, we’ll act as if these are absolutely trustworthy facts. In real life, this kind of absolutely trustworthy background knowledge doesn’t exist, and so if we want to rely on these facts we’ll just have make the assumption that these things are true. However, since these assumptions may or may not be warranted, we might need to check them. For now though, we’ll keep things simple.
The next step is to figure out what we would be a good choice for a diagnostic test statistic; something that would help us discriminate between H 0 and H 1 . Given that the hypotheses all refer to the population mean μ, you’d feel pretty confident that the sample mean \(\bar{X}\) would be a pretty useful place to start. What we could do, is look at the difference between the sample mean \(\bar{X}\) and the value that the null hypothesis predicts for the population mean. In our example, that would mean we calculate \(\bar{X}\) - 67.5. More generally, if we let μ 0 refer to the value that the null hypothesis claims is our population mean, then we’d want to calculate
\(\bar{X}-\mu_{0}\)
If this quantity equals or is very close to 0, things are looking good for the null hypothesis. If this quantity is a long way away from 0, then it’s looking less likely that the null hypothesis is worth retaining. But how far away from zero should it be for us to reject H 0 ?
To figure that out, we need to be a bit more sneaky, and we’ll need to rely on those two pieces of background knowledge that I wrote down previously, namely that the raw data are normally distributed, and we know the value of the population standard deviation σ. If the null hypothesis is actually true, and the true mean is μ 0 , then these facts together mean that we know the complete population distribution of the data: a normal distribution with mean μ 0 and standard deviation σ. Adopting the notation from Section 9.5, a statistician might write this as:
X∼Normal(μ 0 ,σ 2 )
Okay, if that’s true, then what can we say about the distribution of \(\bar{X}\)? Well, as we discussed earlier (see Section 10.3.3), the sampling distribution of the mean \(\bar{X}\) is also normal, and has mean μ. But the standard deviation of this sampling distribution SE (\(\bar{X}\)), which is called the standard error of the mean , is
\(\operatorname{SE}(\bar{X})=\dfrac{\sigma}{\sqrt{N}}\)
In other words, if the null hypothesis is true then the sampling distribution of the mean can be written as follows:
\(\bar{X}\)∼Normal(μ 0 ,SE(\(\bar{X}\)))
Now comes the trick. What we can do is convert the sample mean \(\bar{X}\) into a standard score (Section 5.6). This is conventionally written as z, but for now I’m going to refer to it as \(z_{\bar{X}}\). (The reason for using this expanded notation is to help you remember that we’re calculating standardised version of a sample mean, not a standardised version of a single observation, which is what a z-score usually refers to). When we do so, the z-score for our sample mean is
\(\ z_{\bar{X}} = {{\bar{X}-\mu_{0}} \over SE(\bar{X})}\)
or, equivalently
\(\ z_{\bar{X}} = {{\bar{X}-\mu_{0}} \over \sigma/ \sqrt{N} }\)
This z-score is our test statistic. The nice thing about using this as our test statistic is that like all z-scores, it has a standard normal distribution:
\(\ z_{\bar{X}}\)∼Normal(0,1)
(again, see Section 5.6 if you’ve forgotten why this is true). In other words, regardless of what scale the original data are on, the z-statistic iteself always has the same interpretation: it’s equal to the number of standard errors that separate the observed sample mean\(\bar{X}\) from the population mean μ 0 predicted by the null hypothesis. Better yet, regardless of what the population parameters for the raw scores actually are, the 5% critical regions for z-test are always the same, as illustrated in Figures 13.4 and 13.3. And what this meant, way back in the days where people did all their statistics by hand, is that someone could publish a table like this:
| desired α level | two-sided test | one-sided test |
|---|---|---|
| .1 | 1.644854 | 1.281552 |
| .05 | 1.959964 | 1.644854 |
| .01 | 2.575829 | 2.326348 |
| .001 | 3.290527 | 3.090232 |
which in turn meant that researchers could calculate their z-statistic by hand, and then look up the critical value in a text book. That was an incredibly handy thing to be able to do back then, but it’s kind of unnecessary these days, since it’s trivially easy to do it with software like R.
worked example using R
Now, as I mentioned earlier, the z-test is almost never used in practice. It’s so rarely used in real life that the basic installation of R doesn’t have a built in function for it. However, the test is so incredibly simple that it’s really easy to do one manually. Let’s go back to the data from Dr Zeppo’s class. Having loaded the
grades
data, the first thing I need to do is calculate the sample mean:
sample.mean <- mean( grades )
print( sample.mean )
## [1] 72.3
Then, I create variables corresponding to known population standard deviation (σ=9.5), and the value of the population mean that the null hypothesis specifies (μ 0 =67.5):
mu.null <- 67.5
sd.true <- 9.5
Let’s also create a variable for the sample size. We could count up the number of observations ourselves, and type
N <- 20
at the command prompt, but counting is tedious and repetitive. Let’s get R to do the tedious repetitive bit by using the
length()
function, which tells us how many elements there are in a vector:
N <- length( grades )
print( N )
## [1] 20
Next, let’s calculate the (true) standard error of the mean:
sem.true <- sd.true / sqrt(N)
print(sem.true)
## [1] 2.124265
And finally, we calculate our z-score:
z.score <- (sample.mean - mu.null) / sem.true
print( z.score )
## [1] 2.259606
At this point, we would traditionally look up the value 2.26 in our table of critical values. Our original hypothesis was two-sided (we didn’t really have any theory about whether psych students would be better or worse at statistics than other students) so our hypothesis test is two-sided (or two-tailed) also. Looking at the little table that I showed earlier, we can see that 2.26 is bigger than the critical value of 1.96 that would be required to be significant at α=.05, but smaller than the value of 2.58 that would be required to be significant at a level of α=.01. Therefore, we can conclude that we have a significant effect, which we might write up by saying something like this:
With a mean grade of 73.2 in the sample of psychology students, and assuming a true population standard deviation of 9.5, we can conclude that the psychology students have significantly different statistics scores to the class average (z=2.26, N=20, p<.05).
However, what if want an exact p-value? Well, back in the day, the tables of critical values were huge, and so you could look up your actual z-value, and find the smallest value of α for which your data would be significant (which, as discussed earlier, is the very definition of a p-value). However, looking things up in books is tedious, and typing things into computers is awesome. So let’s do it using R instead. Now, notice that the α level of a z-test (or any other test, for that matter) defines the total area “under the curve” for the critical region, right? That is, if we set α=.05 for a two-sided test, then the critical region is set up such that the area under the curve for the critical region is .05. And, for the z-test, the critical value of 1.96 is chosen that way because the area in the lower tail (i.e., below −1.96) is exactly .025 and the area under the upper tail (i.e., above 1.96) is exactly .025. So, since our observed z-statistic is 2.26, why not calculate the area under the curve below −2.26 or above 2.26? In R we can calculate this using the
pnorm()
function. For the upper tail:
upper.area <- pnorm( q = z.score, lower.tail = FALSE )
print( upper.area )
## [1] 0.01192287
The
lower.tail = FALSE
is me telling R to calculate the area under the curve from 2.26
and upwards
. If I’d told it that
lower.tail = TRUE
, then R would calculate the area from 2.26
and below
, and it would give me an answer 0.9880771. Alternatively, to calculate the area from −2.26 and below, we get
lower.area <- pnorm( q = -z.score, lower.tail = TRUE )
print( lower.area )
## [1] 0.01192287
Thus we get our p-value:
p.value <- lower.area + upper.area
print( p.value )
## [1] 0.02384574
Assumptions of the z-test
As I’ve said before, all statistical tests make assumptions. Some tests make reasonable assumptions, while other tests do not. The test I’ve just described – the one sample z-test – makes three basic assumptions. These are:
- Normality . As usually described, the z-test assumes that the true population distribution is normal. 186 is often pretty reasonable, and not only that, it’s an assumption that we can check if we feel worried about it (see Section 13.9).
- Independence . The second assumption of the test is that the observations in your data set are not correlated with each other, or related to each other in some funny way. This isn’t as easy to check statistically: it relies a bit on good experimetal design. An obvious (and stupid) example of something that violates this assumption is a data set where you “copy” the same observation over and over again in your data file: so you end up with a massive “sample size”, consisting of only one genuine observation. More realistically, you have to ask yourself if it’s really plausible to imagine that each observation is a completely random sample from the population that you’re interested in. In practice, this assumption is never met; but we try our best to design studies that minimise the problems of correlated data.
- Known standard deviation . The third assumption of the z-test is that the true standard deviation of the population is known to the researcher.  In no real world data analysis problem do you know the standard deviation σ of some population, but are completely ignorant about the mean μ. In other words, this assumption is always wrong.
In view of the stupidity of assuming that σ is known, let’s see if we can live without it. This takes us out of the dreary domain of the z-test, and into the magical kingdom of the t-test, with unicorns and fairies and leprechauns, and um…


8.6: Transform Plate Boundaries
Transform boundaries exist where one plate slides past another without production or destruction of crustal material. Most transform faults connect segments of mid-ocean ridges and are thus ocean-ocean plate boundaries. Some transform faults connect continental parts of plates. An example is the San Andreas Fault, which connects the southern end of the Juan de Fuca Ridge with the northern end of the East Pacific Rise (ridge) in the Gulf of California (Figure \(\PageIndex{1}\)). The part of California west of the San Andreas Fault and all of Baja California are on the Pacific Plate. Transform faults do not just connect divergent boundaries. For example, the Queen Charlotte Fault connects the north end of the Juan de Fuca Ridge, starting at the north end of Vancouver Island, to the Aleutian subduction zone.
As we will see in the next section, earthquakes are common along transform faults, as the two plates slide past each other.
*”Physical Geology” by Steven Earle used under a CC-BY 4.0 international license. Download this book for free at http://open.bccampus.ca


16.2B: Physical Properties of Soil
- Describe the physical properties or profile of soil
Soils are named and classified based on their horizons. The soil profile has four distinct layers:
- The O horizon has freshly-decomposing organic matter, humus, at its surface, with decomposed vegetation at its base. Humus enriches the soil with nutrients, enhancing soil moisture retention. Topsoil, the top layer of soil, is usually two to three inches deep, but this depth can vary considerably. For instance, river deltas, such as the Mississippi River delta, have deep layers of topsoil. Topsoil is rich in organic material. Microbial processes occur there; it is responsible for plant production.
- The A horizon consists of a mixture of organic material with inorganic products of weathering; it is the beginning of true mineral soil. This horizon is typically darkly colored because of the presence of organic matter. In this area, rainwater percolates through the soil and carries materials from the surface.
- The B horizon, or subsoil, is an accumulation of mostly fine material that has moved downward, resulting in a dense layer in the soil. In some soils, the B horizon contains nodules or a layer of calcium carbonate.
- The C horizon, or soil base, includes the parent material, plus the organic and inorganic material that is broken down to form soil. The parent material may be either created in its natural place or transported from elsewhere to its present location. Beneath the C horizon lies bedrock.
Some soils may have additional layers, or lack one of these layers. The thickness of the layers is also variable, depending on the factors that influence soil formation. In general, immature soils may have O, A, and C horizons, whereas mature soils may display all of these, plus additional layers.
Key Points
- The O horizon, or topsoil, is made of decaying organisms and plant life; it is responsible for plant production.
- The A horizon is of a mixture of organic material and inorganic products of weathering; it is the beginning of true mineral soil.
- The B horizon, or subsoil, is a dense layer of mostly fine material that has been pushed down from the topsoil.
- The C horizon, or soil base, is located just above bedrock and is made of parent, organic, and inorganic material.
Key Terms
- topsoil : top layer of soil containing humus at its surface and decomposing vegetation at its base; the most fertile soil
- subsoil : dense layer of soil containing fine material that has moved downward; the layer of earth that is below the topsoil


5.3: The Stress-energy Tensor
In general, the curvature of spacetime will contain contributions from both tidal forces and local sources, superimposed on one another. To develop the right formulation for the Einstein field equations, we need to eliminate the tidal part. Roughly speaking, we will do this by averaging the sectional curvature over all three of the planes \(t−x\), \(t−y\), and \(t−z\), giving a measure of curvature called the Ricci curvature . The “roughly speaking” is because such a prescription would treat the time and space coordinates in an extremely asymmetric manner, which would violate local Lorentz invariance.
To get an idea of how this would work, let’s compare with the Newtonian case, where there really is an asymmetry between the treatment of time and space. In the Cartan curved-spacetime theory of Newtonian gravity ( Chapter 2 ), the field equation has a kind of scalar Ricci curvature on one side, and on the other side is the density of mass, which is also a scalar. In relativity, however, the source term in the equation clearly cannot be the scalar mass density. We know that mass and energy are equivalent in relativity, so for example the curvature of spacetime around the earth depends not just on the mass of its atoms but also on all the other forms of energy it contains, such as thermal energy and electromagnetic and nuclear binding energy. Can the source term in the Einstein field equations therefore be the mass-energy E? No, because E is merely the timelike component of a particle’s momentum four-vector. To single it out would violate Lorentz invariance just as much as an asymmetric treatment of time and space in constructing a Ricci measure of curvature. To get a properly Lorentz invariant theory, we need to find a way to formulate everything in terms of tensor equations that make no explicit reference to coordinates. The proper generalization of the Newtonian mass density in relativity is the stress-energy tensor T ij , whose 16 elements measure the local density of mass-energy and momentum, and also the rate of transport of these quantities in various directions. If we happen to be able to find a frame of reference in which the local matter is all at rest, then \(T^{tt}\) represents the mass density. The reason for the word “stress” in the name is that, for example, the flux of x-momentum in the x direction is a measure of pressure.
For the purposes of the present discussion, it’s not necessary to introduce the explicit definition of T; the point is merely that we should expect the Einstein field equations to be tensor equations, which tells us that the definition of curvature we’re seeking clearly has to be a rank-2 tensor, not a scalar. The implications in four-dimensional spacetime are fairly complex. We’ll end up with a rank-4 tensor that measures the sectional curvature, and a rank-2 Ricci tensor derived from it that averages away the tidal effects. The Einstein field equations then relate the Ricci tensor to the energy-momentum tensor in a certain way. The stress-energy tensor is discussed further in Section 8.1 .


8.1: Drug Use and Drug Addiction
Drug addiction is a complex illness, many people don’t understand why or how other people become addicted to drugs. They may mistakenly think that those who use drugs lack moral principles or the willpower to quit without recognizing that drugs change the brain in ways that make quitting hard, even for those who want to. In reality, drug addiction is a complex disease, and quitting usually takes more than good intentions or a strong will. A person may initially choose to take the drug, however over time they are incapable of choosing not to take the drug.
Fortunately, researchers know more than ever about how drugs affect the brain and have found treatments that can help people recover from drug addiction and lead productive lives.
Drug Use and Overdose
The National Survey on Drug Use and Health (NSDUH) [1] is published each year to share statistics on drug use in the U.S. Key findings from the 2020 annual report show that drug use is fairly common.
Among people aged 12 or older in year 2020:
- 58.7 percent (or 162.5 million people) used tobacco, alcohol, or an illicit drug in the past month (current user).
- 21.4 percent (or 59.3 million people) used illicit drugs in the past year.
- Marijuana was the most commonly used illicit drug, with 17.9 percent (or 49.6 million people) using it in the past year.
- 3.7 percent (or 10.3 million people) misused central nervous system (CNS) stimulants in the past year.
- 3.4 percent (or 9.5 million people) misused opioids (heroin or prescription pain relievers) in the past year.
- 3.3 percent (or 9.3 million people) misused prescription pain relievers in the past year.
- 2.6 percent (or 7.1 million people) used hallucinogens in the past year.
- 14.5 percent (or 40.3 million people) had a Substance Use Disorder in the past year
Unfortunately both first time users and those who have an addiction are both at risk of possible drug overdose. Between 1999-2000 more than 932,000 people have died from a drug overdose. Most recently, the majority of overdoses are caused by synthetic opioids. For example in 2020, about 92,000 drug overdose deaths occurred in the United States and 75% of the overdoses involved opioid use. Data from just 24 states from January–June 2019 showed that Illicitly manufactured fentanyl, heroin, cocaine, or methamphetamine (alone or in combination) were involved in nearly 85% of drug overdose deaths.
Drug Addiction
Addiction is a chronic disease characterized by drug seeking and use that is compulsive, or difficult to control, despite harmful consequences. The initial decision to take drugs is voluntary for most people, but repeated drug use can lead to brain changes that challenge an addicted person’s self-control and interfere with their ability to resist intense urges to take drugs. These brain changes can be persistent, which is why drug addiction is considered a “relapsing” disease—people in recovery from drug use disorders are at increased risk for returning to drug use even after years of not taking the drug.
It’s common for a person to relapse, but relapse doesn’t mean that treatment doesn’t work. As with other chronic health conditions, treatment should be ongoing and should be adjusted based on how the patient responds. Treatment plans need to be reviewed often and modified to fit the patient’s changing needs.
What happens to the brain when a person takes drugs?
Most drugs affect the brain’s “reward circuit” by flooding it with the chemical messenger dopamine. This reward system controls the body’s ability to feel pleasure and motivates a person to repeat behaviors needed to thrive, such as eating and spending time with loved ones. This overstimulation of the reward circuit causes the intensely pleasurable “high” that can lead people to take a drug again and again.
As a person continues to use drugs, the brain adjusts to the excess dopamine by making less of it and/or reducing the ability of cells in the reward circuit to respond to it. This reduces the high that the person feels compared to the high they felt when first taking the drug, an effect known as tolerance , which involves taking more of the drug to try to achieve the same dopamine high. It can also cause them to get less pleasure from other things they once enjoyed, like food or social activities.
Long-term use can cause changes in other brain chemical systems and circuits that can negatively impact a persons learning, judgment, decision-making, stress, memory, and behavior. Long-term use can also lead to drug dependence, where your body is used to functioning with the drug in your system. If you are dependent on a drug and the drug is removed, it can cause withdrawals symptoms. Withdrawal symptoms vary depending on the type of drug dependence and may include tremors (shaking), cravings, agitation, nausea, sweating, or changes in mood.
Despite being aware of these harmful outcomes, many people who use drugs continue to take them, which is the nature of addiction.
Why do some people become addicted to drugs while others don’t?
No one factor can predict if a person will become addicted to drugs. A combination of factors influences risk for addiction. The more risk factors a person has, the greater the chance that taking drugs can lead to addiction. For example:
-
Biology
- The genes that people are born with account for about half of a person’s risk for addiction. Gender, ethnicity, and the presence of other mental disorders may also influence risk for drug use and addiction.
-
Environment
- A person’s environment includes many different influences, from family and friends to economic status and general quality of life. Factors such as peer pressure, physical and sexual abuse, early exposure to drugs, stress, and parental guidance can greatly affect a person’s likelihood of drug use and addiction.
-
Developmental stages
- Genetic and environmental factors interact with critical developmental stages in a person’s life to affect addiction risk. Although taking drugs at any age can lead to addiction, the earlier that drug use begins, the more likely it will progress to addiction. This is particularly problematic for teens. Because areas in their brains that control decision-making, judgment, and self-control are still developing, teens may be especially prone to risky behaviors, including trying drugs.
What are some consequences of Drug Abuse?
Drug abuse is a serious public health problem that affects many communities and families in some way. Each year drug abuse causes millions of serious illnesses or injuries among Americans. Drug abuse also plays a role in many major social problems, such as drugged driving, violence, stress, and child abuse. Drug abuse can lead to homelessness, crime, and missed work or problems with keeping a job. It harms unborn babies and destroys families.
Some people who are addicted don’t believe that they are sick and out of control, so they don’t look for treatment. They don’t see the problems they are causing themselves and those around them. Other people who are addicted are aware of the problem, but may be so upset and confused that they do not know how to ask for or get help.
Drugs don’t just hurt the person taking them. Everyone connected to the person can get hurt. When you or a loved one abuse drugs, everyday life can feel out of control.
Drug abuse affects the family in many ways, including:
- When a person has a drug problem, they have a disease that can hurt the family.
- Drug abuse puts a lot of stress on parents, brothers and sisters, children, grandparents—anyone who is part of the home.
- Family members might fight a lot because of the problems the drug abuse is causing.
- The drug user might do and say things that upset neighbors and friends, and make the family ashamed.
When family members take drugs, you may experience situations such as:
- You generally can’t count on them to do what they say they will do.
- They may forget or get distracted because their focus is on getting and taking drugs.
- They might lie or steal money to buy drugs.
- They might get fired from their jobs.
- They might not come home at night.
- They may do bad things they would never do if they weren’t abusing drugs.
Drug abuse can cause many problems including: Fighting and violence in and outside the home; money problems; trouble at school; trouble at work, losing a job; trouble in relationships; child abuse, neglect; driving accidents; and arrests and jail.


14.2E: Parasympathetic (Craniosacral) Division
Parasympathetic ganglia are the autonomic ganglia of the parasympathetic nervous system that lie near or within the organs they innervate.
- Describe features of the parasympathetic division of the autonomic nervous system
Key Points
- Each PSNS ganglion has three roots: a motor root, a sympathetic root, and a sensory root, as well as a number of exiting branches.
- Most are small terminal ganglia or intramural ganglia, so named because they lie near or within the organs they innervate.
- The parasympathetic system is referred to as having craniosacral outflow because of the location of PSNS fiber origins.
Key Terms
- lacrimal gland : One of a pair almond-shaped glands, one for each eye, that secrete the aqueous layer of the tear film.
- sympathetic root : This carries postsynaptic sympathetic fibers that traverse the ganglion without crossing a synapse.
- motor root : This carries presynaptic parasympathetic nerve fibers that terminate in the ganglion and create synapses for the postsynaptic fibers to travel to their target organs.
- sensory root : The proximal end of a dorsal afferent nerve that is attached to the spinal cord.
- parasympathetic : Relating to the part of the autonomic nervous system that inhibits or opposes the effects of the sympathetic nervous system.
EXAMPLES
Nerves that supply parasympathetic fibers to the parasympathetic ganglia of the head include the oculomotor nerve (ciliary ganglion); the facial nerve (pterygopalatine ganglion, submandibular ganglion); the glossopharyngeal nerve (otic ganglion); the vagus nerve (no named ganglion); and the pelvic splanchnic nerves (no named ganglion).
Parasympathetic ganglia are the autonomic ganglia of the parasympathetic nervous system, blue fibers). Most are small terminal ganglia or intramural ganglia, so named because they lie near or within (respectively) the organs they innervate.
The exceptions are the four paired parasympathetic ganglia of the head and neck. These paired ganglia supply all parasympathetic innervation to the head and neck: ciliary ganglion (spincter pupillae, ciliary muscle), pterygopalatine ganglion (lacrimal gland, glands of nasal cavity), submandibular ganglion (submandibular and sublingual glands), and otic ganglion (parotid gland).
Nerve innervation of the autonomic nervous system : The parasympathetic nervous system, shown in blue, is a division of the autonomic nervous system.
Each has three roots entering the ganglion (motor, sympathetic, and sensory roots) and a variable number of exiting branches.
- The motor root carries presynaptic parasympathetic nerve fibers (general visceral efferent fibers) that terminate in the ganglion by creating a synapse for the postsynaptic fibers traveling to target organs.
- The sympathetic root carries postsynaptic sympathetic fibers (general visceral efferent fibers) that traverse the ganglion without creating a synapse.
- The sensory root carries general sensory fibers (general somatic afferent fibers) that also do not create a synapse in the ganglion.
Some ganglia also carry special sensory fibers (special visceral afferent) for taste sensation.
The nerves that supply parasympathetic fibers to the parasympathetic ganglia of the head include the oculomotor nerve (ciliary ganglion), the facial nerve (pterygopalatine ganglion, submandibular ganglion), the glossopharyngeal nerve (otic ganglion), the vagus nerve, and the pelvic splanchnic nerves.
Because of its location, the parasympathetic system is commonly referred to as having craniosacral outflow, in contrast to the sympathetic nervous system, which is said to have thoracolumbar outflow.
Parasympathetic ganglia of the head : The parasympathetic division has craniosacral outflow, meaning that the neurons begin at the cranial nerves (CN3, CN7, CN9, CN10) and the sacral spinal cord (S2–S4). Pre- and post-ganglionic fibers and targets are depicted.


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3.2: Basic search techniques
Earlier, we discussed developing a topic into searchable keywords, and introduced key finding tools for library research. This chapter combines those topics to explore how you can use specific techniques to build an effective search strategy for your projects. Keep in mind that no single strategy, set of keywords, or finding tool will work perfectly every time. You’ll need to use different strategies for different situations.
Simple searching techniques
Starting simple is often a useful strategy, especially when you have a broad topic or don’t have a specific source in mind. The easiest way to get started in most finding tools is to type a few keywords into the search box. In the following examples we will use Quick Search, but these strategies will work in almost any finding tool.
Let’s say you are doing a project about solar energy and power storage. Just putting these words in the simple search box uses the default settings without restricting the results that come up. Sometimes that’s what you want, like when you’re doing an initial exploration of what’s available. However, doing this may retrieve far too many results. If that’s the case, you may have to try different search terms or strategies to narrow your results down to the most relevant sources.
Here are some simple ways you can narrow your search results:
Phrase searching is when you take a specific phrase and enter it into the search box using quotations around it. It’s helpful when you want to find a specific term or phrase, for example “solar energy.” By putting quotations around the phrase, that ensures the database will search for those words together, in the order you specify, rather than sprinkled anywhere in the resource.
Truncation is when you use an asterisk (*) to search for multiple endings for a word’s base, or “trunk.” Using a truncation symbol tells the search tool to find any word that starts with the letters leading up to your asterisk, regardless of what letters come after that point. This is a simplified way to search for different forms of the same word with a single search. For example, by searching for electric* you can find results containing all or any of these terms: electric, electrical, and electricity.
If there is a specific set of words you want to find, you can pinpoint where to truncate by spelling out your keywords and highlighting the last place where the letters of the words overlap. You can see an example of this in the image above.
Watch out! Truncating your term too early will increase the likelihood that you’ll retrieve irrelevant results (i.e., before it is specific enough, like elect* may retrieve elections and not electricity). Truncating too late defeats the whole purpose of trying to find variations on your keyword because it won’t find much else (i.e., after useful variations might occur, like electricit* will only retrieve electricity and not electrical).
Check your understanding
Filtering your results
Filtering is the process of narrowing your results by applying modifiers to your existing search. Think of it like online shopping: after you search for something you want, you can limit your results to only items within your price range, or those with free shipping. Filtering your search results when doing research is a similar process. Finding tools like Quick Search include various filters that let you do things like restrict your results to a specific date range (e.g. published in the past 5 years) or format (e.g. books instead of articles).
Where the filters are located will depend on which tool you’re using. In Quick Search, they are located in a column on the right side of the page, but many databases and indexes place them on the left. Once you have selected one or more filters and applied them to your search, they will typically appear in a list near the top of your search results page. One thing to be aware of is how your search tool handles filters when you change your search or start a new one. Filters from your previous searches may still be active, or not. If a filter was removed, you might want to add it back in; alternatively, if the filters are still active, you may want to remove one or more of them to broaden your results.
When you’re considering your search terms, take a few moments to reflect on whether modifying the phrases and keywords you’ve used will be useful for your search. Phrase searching, truncation, and filtering are not the only techniques you should know, but they are useful to know. In the next section, we’ll explore a technique that gives you more control over your results from the beginning.


1.8: Intervals and Interval Notation
Media
| Title | Level of Approach | Type | Length |
|---|---|---|---|
| Interval Notation | Elementary Algebra | Lecture | 15:05 |
| A Review of Interval Notation | Intermediate Algebra and Beyond | Review | 13:30 |
Using Interval Notation
Indicating the solution to an inequality such as \(x \geq 4\) can be achieved in several ways.
- We can use a number line as shown in Figure \(\PageIndex{1}\). The blue ray begins at \(x = 4\) and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to \(4\).
- We can use set-builder notation : \(\{x \mid x \geq 4\}\), which translates to "all real numbers \(x\) such that \(x\) is greater than or equal to \(4\)." Notice that braces are used to indicate a set.
- The third method is interval notation , in which solution sets are indicated with parentheses or brackets. The solutions to \(x \geq 4\) are represented as \([4,\infty)\).
Figure \(\PageIndex{1}\)
The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be "equaled." A few examples of an interval, or a set of numbers in which a solution falls, are \([−2,6)\), or all numbers between \(−2\) and \(6\), including \(−2\), but not including \(6\); \((−1,0)\), all real numbers between, but not including \(−1\) and \(0\); and \((−\infty,1]\), all real numbers less than and including \(1\). Table \(\PageIndex{1}\) outlines the possibilities.
| Set Indicated | Set-Builder Notation | Interval Notation |
|---|---|---|
| All real numbers between \(a\) and \(b\), but not including \(a\) or \(b\) | \(\{x \mid a<x<b\}\) | \((a,b)\) |
| All real numbers greater than \(a\), but not including \(a\) | \(\{x \mid x>a\}\) | \((a,\infty)\) |
| All real numbers less than \(b\), but not including \(b\) | \(\{x \mid x<b\}\) | \((−\infty,b)\) |
| All real numbers greater than \(a\), including \(a\) | \(\{x \mid x \geq a\}\) | \([a,\infty)\) |
| All real numbers less than \(b\), including \(b\) | \(\{x \mid x \leq b\}\) | \((−\infty,b]\) |
| All real numbers between \(a\) and \(b\), including \(a\) | \(\{x \mid a \leq x<b\}\) | \([a,b)\) |
| All real numbers between \(a\) and \(b\), including \(b\) | \(\{x \mid a<x \leq b\}\) | \((a,b]\) |
| All real numbers between \(a\) and \(b\), including \(a\) and \(b\) | \(\{x \mid a \leq x \leq b\}\) | \([a,b]\) |
| All real numbers less than \(a\) or greater than \(b\) | \(\{x \mid x<a\space and\space x>b\}\) | \((−\infty,a)\cup(b,\infty)\) |
| All real numbers | \(\{x \mid x\space is\space all\space real\space numbers\}\) | \((−\infty,\infty)\) |
Use interval notation to indicate all real numbers greater than or equal to \(−2\).
- Solution
-
Use a bracket on the left of \(−2\) and parentheses after infinity: \([−2,\infty)\). The bracket indicates that \(−2\) is included in the set with all real numbers greater than \(−2\) to infinity.
Use interval notation to indicate all real numbers between and including \(−3\) and \(5\).
- Answer
-
\([−3,5]\)
Write the interval expressing all real numbers less than or equal to \(−1\) or greater than or equal to \(1\).
- Solution
-
We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to \(1\). So, this interval begins at \(−\infty\) and ends at \(−1\), which is written as \((−\infty,−1]\).
The second interval must show all real numbers greater than or equal to \(1\), which is written as \([1,\infty)\). However, we want to combine these two sets. We accomplish this by inserting the union symbol, \cup , between the two intervals.
\[(−\infty,−1]\cup[1,\infty) \nonumber\]
Express all real numbers less than \(−2\) or greater than or equal to \(3\) in interval notation.
- Answer
-
\((−\infty,−2)\cup[3,\infty)\)


6.5: E- Coherent States
Coherent states are special states of bosonic systems (including the quantum harmonic oscillator, whose excitation quanta can be regarded as bosonic particles) whose dynamics are highly similar to classical oscillator trajectories. They provide an important link between quantum and classical harmonic oscillators.
E.1 Definition
The Hamiltonian of a simple harmonic oscillator (with \(\hbar = m = \omega_0 = 1\) for simplicity) is
\[\hat{H} = \frac{\hat{p}^2}{2} + \frac{\hat{x}^2}{2}, \label{hsho}\]
where \(\hat{x}\) and \(\hat{p}\) are the position and momentum operators. The ladder operators are
\[\begin{align} \hat{a} &= \frac{1}{\sqrt{2}} \left(\hat{x} + i\hat{p}\right) \label{a} \\ \hat{a}^\dagger &= \frac{1}{\sqrt{2}} \left(\hat{x} - i\hat{p}\right). \label{adagger}\end{align}\]
These obey the commutation relation
\[\left[\hat{a}, \hat{a}^\dagger\right] = 1. \label{commutator}\]
As a result, we can also regard these as the creation and annihilation operators for a bosonic particle that has only one single-particle state.
The Hamiltonian for the harmonic oscillator, Equation \(\eqref{hsho}\), can be written as
\[\hat{H} = \hat{a}^\dagger\hat{a} + 1/2.\]
The annihilation operator \(\hat{a}\) kills off the ground state \(|\varnothing\rangle\) :
\[\hat{a} |\varnothing\rangle = 0. \label{annihilation}\]
Thus, \(|\varnothing\rangle\) is analogous to the “vacuum state” for a bosonic particle.
Returning to the Hamiltonian \(\eqref{hsho}\), suppose we add a term proportional to \(\hat{x}\) :
\[\hat{H}' = \frac{\hat{p}^2}{2} + \frac{\hat{x}^2}{2} - \sqrt{2}\alpha_1\hat{x}. \label{hshift}\]
The coefficient of \(-\sqrt{2}\alpha_1\) , where \(\alpha_1 \in \mathbb{R}\) , is for later convenience. By completing the square, we see that this additional term corresponds to a shift in the center of the potential, plus an energy shift:
\[\hat{H}' = \frac{\hat{p}^2}{2} + \frac{1}{2}\left(\hat{x} - \sqrt{2}\alpha_1\right)^2 - \alpha_1^2.\]
Let \(|\alpha_1\rangle\) denote the ground state for the shifted harmonic oscillator.
By analogy with how we solved the original harmonic oscillator problem, let us define a new annihilation operator with displaced \(x\) :
\[\begin{align} \hat{a}' = \frac{1}{\sqrt{2}}\left(\hat{x} - \sqrt{2}\alpha_1 + i \hat{p}\right).\end{align}\]
This is related to the original annihilation operator by
\[\hat{a}' = \hat{a} - \alpha_1. \label{aprime}\]
We can easily show that \([\hat{a}',\hat{a}'^\dagger] = 1\) , and that \(\hat{H}' = \hat{a}'^\dagger \hat{a}' + 1/2 - \alpha_1^2\) . Hence,
\[\hat{a}' \, |\alpha_1 \rangle = 0.\]
But Equation \(\eqref{aprime}\) implies that in terms of the original annihilation operator,
\[\hat{a}\, |\alpha_1\rangle = \alpha_1 \,|\alpha_1\rangle. \label{aeigenv}\]
In other words, \(|\alpha_1\rangle\) is an eigenstate of the original harmonic oscillator’s annihilation operator, with the displacement parameter \(\alpha_1\) as the corresponding eigenvalue! For reasons that will become clear later, we call \(|\alpha_1\rangle\) a coherent state of the original harmonic oscillator \(\hat{H}\) .
E.2 Explicit Expression for the Coherent State
Let us derive an explicit expression for the coherent state in terms of \(\hat{a}\) and \(\hat{a}^\dagger\) , the creation and annihilation operators of the original harmonic oscillator. Consider the translation operator
\[\hat{T}(\Delta x) = \exp(-i\hat{p}\Delta x).\]
Since \(|\alpha_1\rangle\) is the ground state of a displaced harmonic oscillator, it can be generated by performing a displacement of the original oscillator’s ground state \(|\varnothing\rangle\) . The displacement is \(\Delta x = \sqrt{2}\alpha_1\) :
\[\begin{align} |\alpha_1\rangle &= \hat{T}\big(\sqrt{2}\alpha_1\big)\; |\varnothing\rangle \\ &= \exp\left[\alpha_1\left(\hat{a}^\dagger - \hat{a}\right)\right] |\varnothing\rangle.\end{align}\]
In deriving the second line, we have used Equations \(\eqref{a}\)–\(\eqref{adagger}\) to express \(\hat{p}\) in terms of \(\hat{a}\) and \(\hat{a}^\dagger\) . We can further simplify the result by using the Baker-Campbell-Hausdorff formula for operator exponentials:
\[\mathrm{If} \;\; [[\hat{A},\hat{B}],\hat{A}] = [[\hat{A},\hat{B}],\hat{B}] = 0 \;\;\Rightarrow \;\; e^{\hat{A}+\hat{B}} = e^{-[\hat{A},\hat{B}]/2}\, e^{\hat{A}} e^{\hat{B}}.\]
The result is
\[|\alpha_1\rangle = e^{-\alpha_1^2/2} \, e^{\alpha_1 \hat{a}^\dagger} |\varnothing\rangle.\]
If we write the exponential in its series form,
\[|\alpha_1\rangle = e^{-\alpha_1^2/2} \, \left(1 + \alpha_1 \hat{a}^\dagger + \frac{\alpha_1^2}{2} \left(\hat{a}^\dagger\right)^2 + \cdots\right) |\varnothing\rangle,\]
then we see that from the point of view of the bosonic excitations of the original Hamiltonian \(\hat{H}\) , the state \(|\alpha_1\,\rangle\) has an indeterminate number of bosons . It is a superposition of the zero-boson (vacuum) state, a one-boson state, a two-boson state, etc.
We can generalize the coherent state by performing a shift not just in space, but also in momentum. Instead of Equation \(\eqref{hshift}\), let us define
\[\begin{align} \hat{H}' &= \frac{1}{2}\left(\hat{p} - \sqrt{2}\alpha_2\right)^2 + \frac{1}{2}\left(\hat{x} - \sqrt{2}\alpha_1\right)^2 \\ &= \hat{H} - \left(\alpha \hat{a}^\dagger + \alpha^*\hat{a}\right) + \textrm{constant},\end{align}\]
where
\[\alpha \equiv \alpha_1 + i \alpha_2 \;\in \;\mathbb{C}.\]
It can then be shown that the ground state of \(\hat{H}'\) , which we denote by \(|\alpha\rangle\) , satisfies
\[\hat{a} \, |\alpha\rangle = \alpha \,|\alpha\rangle. \label{aaction}\]
(Note that \(\hat{a}\) is not Hermitian, so its eigenvalue \(\alpha\) need not be real.) In explicit terms,
\[|\alpha\rangle \,=\, \exp\left[\alpha\hat{a}^\dagger - \alpha^*\hat{a}\right] |\varnothing\rangle \,=\, e^{-|\alpha|^2/2} e^{\alpha\hat{a}^\dagger} |\varnothing\rangle. \label{alphaexp}\]
E.3 Basic Properties
There is one coherent state \(|\alpha\rangle\) for each complex number \(\alpha \in \mathbb{C}\) . They have the following properties:
-
They are normalized:
\[\langle\alpha|\alpha\rangle = 1. \label{normalization}\]
This follows from the fact that they are ground states of displaced harmonic oscillators.
-
They form a complete set, meaning that the identity operator can be resolved as
\[\hat{I} = C \int d^2\alpha |\alpha\rangle\langle\alpha|,\]
where \(C\) is some numerical constant and \(\int d^2\alpha\) denotes an integral over the complex plane. However, the coherent states do not form an orthonormal set, as they are over-complete: \(\langle\alpha|\alpha'\rangle \ne 0\) for \(\alpha \ne \alpha'\) .
-
The expected number of particles in a coherent state is
\[\langle\alpha| \hat{a}^\dagger\hat{a} | \alpha\rangle = |\alpha|^2.\]
-
The probability distribution of the number of particles follows a
Poisson distribution
:
\[P(n) = |\langle n | \alpha\rangle|^2 = e^{-|\alpha|^2} \frac{|\alpha|^{2n}}{n!}.\]
The mean and variance of this distribution are both \(|\alpha|^2\) .
-
The mean position and momentum are
\[\begin{align} \langle \alpha | \hat{x}|\alpha\rangle &= \sqrt{2} \, \mathrm{Re}(\alpha) \label{x} \\ \langle \alpha | \hat{p}|\alpha\rangle &= \sqrt{2} \, \mathrm{Im}(\alpha). \label{p} \end{align}\]
E.4 Dynamical Properties
Take the harmonic oscillator Hamiltonian with zero-point energy omitted for convenience:
\[\hat{H} = \hat{a}^\dagger \hat{a}.\]
Suppose we initialize the system in a coherent state \(|\alpha_0\rangle\) for some \(\alpha_0 \in \mathbb{C}\) . This is not an energy eigenstate of \(\hat{H}\) , so how will it subsequently evolve?
It turns out that the dynamical state has the form
\[|\psi(t)\rangle = |\alpha(t)\rangle, \;\;\;\mathrm{where}\;\;\alpha(0) = \alpha_0.\]
In other words, the system is always in a coherent state, but the complex parameter \(\alpha(t)\) varies with time. To find \(\alpha(t)\) , plug the ansatz into the time-dependent Schrödinger equation:
\[\begin{align} i \frac{d}{dt} |\alpha(t)\rangle &= \hat{a}^\dagger \hat{a} |\alpha(t)\rangle\\ i \,\Big\langle\alpha(t)\Big| \frac{d}{dt} \Big|\alpha(t)\Big\rangle &= |\alpha(t)|^2.\end{align}\]
We can calculate the left-hand side using Equations \(\eqref{aaction}\), \(\eqref{alphaexp}\), and \(\eqref{normalization}\):
\[\begin{align} i \, \Big\langle\alpha(t)\Big| \frac{d}{dt} \Big|\alpha(t)\Big\rangle &= i\, \langle \alpha(t) | \; \frac{d}{dt} \exp\left[ e^{-|\alpha|^2/2} e^{\alpha\hat{a}^\dagger} \right] |\varnothing\rangle \\ &= i\, \langle \alpha(t) | \left(-\frac{1}{2} \frac{d}{dt}(\alpha\alpha^*) + \frac{d\alpha}{dt} \hat{a}^\dagger \right) |\alpha(t)\rangle \\ &= i \left(-\frac{1}{2} \dot{\alpha}\alpha^* - \frac{1}{2}\alpha\dot{\alpha}^* + \dot{\alpha}\alpha^*\right).\end{align}\]
Hence,
\[\frac{i}{2}\left(\dot{\alpha}\alpha^* - \alpha\dot{\alpha}^*\right) = |\alpha|^2. \label{eom}\]
This looks more complicated than it actually is. Dividing both sides by \(\alpha \alpha^*\) gives
\[\frac{1}{2}\, \left(\frac{i\dot{\alpha}}{\alpha} + \left[\frac{i\dot{\alpha}}{\alpha}\right]^*\right) = \mathrm{Re}\left[\frac{i\dot{\alpha}}{\alpha}\right] = 1.\]
This reduces to
\[\dot{\alpha} = \big[-i + \gamma(t)\big] \alpha(t), \;\;\; \mathrm{\gamma}\in\mathbb{R}.\]
This is the equation for a complex harmonic oscillator with an arbitrary damping or amplification factor \(\gamma\) . For \(\gamma = 0\) , the oscillator is energy-conserving and the solutions are
\[\alpha(t) = \alpha_0 \; e^{-it}.\]
Referring back to Equations \(\eqref{x}\)–\(\eqref{p}\), this implies that the mean position and momentum have the following time-dependence:
\[\begin{align} \langle x\rangle &=\, \;\;\sqrt{2} |\alpha_0| \cos\left[t - \mathrm{arg}(\alpha_0)\right] \\ \langle p\rangle &= -\sqrt{2} |\alpha_0| \sin\left[t - \mathrm{arg}(\alpha_0)\right].\end{align}\]
The dynamics of a coherent state therefore reproduces the motion of a classical harmonic oscillator with \(m = \omega_0 = 1\) .


4.4: Surface Integrals and the Divergence Theorem
In Section 4.1, we learned how to integrate along a curve. We will now learn how to perform integration over a surface in \(\mathbb{R}^3\), such as a sphere or a paraboloid. Recall from Section 1.8 how we identified points \((x, y, z)\) on a curve \(C\) in \(\mathbb{R}^3\), parametrized by \(x = x(t), y = y(t), z = z(t), a ≤ t ≤ b\), with the terminal points of the position vector
\[\nonumber \textbf{r}(t)= x(t)\textbf{i}+ y(t)\textbf{j}+ z(t)\textbf{k} \text{ for }t \text{ in }[a,b].\]
The idea behind a parametrization of a curve is that it “transforms” a subset of \(\mathbb{R}^ 1\) (normally an interval \([a,b]\)) into a curve in \(\mathbb{R}^ 2\) or \(\mathbb{R}^ 3\) (Figure \(\PageIndex{1}\)).
Similar to how we used a parametrization of a curve to define the line integral along the curve, we will use a parametrization of a surface to define a surface integral . We will use two variables, \(u \text{ and }v\), to parametrize a surface \(Σ\) in \(\mathbb{R}^ 3\) : \(x = x(u,v), y = y(u,v), z = z(u,v), \text{ for }(u,v)\) in some region \(R\) in \(\mathbb{R}^ 2\) (Figure \(\PageIndex{1}\) ).
In this case, the position vector of a point on the surface \(Σ\) is given by the vector-valued function
\[\nonumber \textbf{r}(u,v) = x(u,v)\textbf{i} + y(u,v)\textbf{j} + z(u,v)\textbf{k} \text{ for }(u,v) \text{ in }R.\]
Since \(\textbf{r}(u,v)\) is a function of two variables, define the partial derivatives \(\dfrac{∂r}{ ∂u} \text{ and }\dfrac{∂r }{∂v}\) for \((u,v) \text{ in }R\) by
\[\nonumber \begin{align}&\dfrac{∂\textbf{r}}{ ∂u }(u,v)=\dfrac{∂x }{∂u} (u,v)\textbf{i}+\dfrac{∂y}{ ∂u} (u,v)\textbf{j}+\dfrac{∂z }{∂u} (u,v)\textbf{k},\text{ and} \\[4pt] \nonumber & \dfrac{∂\textbf{r}}{ ∂v }(u,v) = \dfrac{∂x }{∂v} (u,v)\textbf{i}+\dfrac{∂y}{ ∂v} (u,v)\textbf{j}+\dfrac{∂z }{∂v} (u,v)\textbf{k} \\[4pt] \end{align}\]
The parametrization of \(Σ\) can be thought of as “transforming” a region in \(\mathbb{R}^2\) (in the \(uv\)-plane) into a 2-dimensional surface in \(\mathbb{R}^3\). This parametrization of the surface is sometimes called a patch , based on the idea of “patching” the region \(R\) onto \(Σ\) in the grid-like manner shown in Figure \(\PageIndex{2}\).
In fact, those gridlines in \(R\) lead us to how we will define a surface integral over \(Σ\). Along the vertical gridlines in \(R\), the variable \(u\) is constant. So those lines get mapped to curves on \(Σ\), and the variable \(u\) is constant along the position vector \(\textbf{r}(u,v)\). Thus, the tangent vector to those curves at a point \((u,v)\) is \(\dfrac{∂r}{ ∂v}\). Similarly, the horizontal gridlines in \(R\) get mapped to curves on \(Σ\) whose tangent vectors are \(\dfrac{∂r}{ ∂u}\).
Now take a point \((u,v)\) in \(R\) as, say, the lower left corner of one of the rectangular grid sections in \(R\), as shown in Figure \(\PageIndex{2}\). Suppose that this rectangle has a small width and height of \(∆u \text{ and }∆v\), respectively. The corner points of that rectangle are \((u,v), (u + ∆u,v), (u +∆u,v +∆v)\) and \((u,v +∆v)\). So the area of that rectangle is \(A = ∆u∆v\). Then that rectangle gets mapped by the parametrization onto some section of the surface \(Σ\) which, for \(∆u \text{ and }∆v\) small enough, will have a surface area (call it \(dσ\)) that is very close to the area of the parallelogram which has adjacent sides \(\textbf{r}(u+∆u,v)−\textbf{r}(u,v)\) (corresponding to the line segment from \((u,v) \text{ to }(u + ∆u,v) \text{ in }R\)) and \(\textbf{r}(u,v + ∆v) − \textbf{r}(u,v)\) (corresponding to the line segment from \((u,v) \text{ to }(u,v + ∆v) \text{ in }R\)). But by combining our usual notion of a partial derivative (Definition 2.3 in Section 2.2) with that of the derivative of a vector-valued function (Definition 1.12 in Section 1.8) applied to a function of two variables, we have
\[\nonumber \begin{align} &\dfrac{∂\textbf{r}}{ ∂u} \approx \dfrac{\textbf{r}(u +∆u,v)−\textbf{r}(u,v)}{∆u},\text{ and } \\[4pt] \nonumber &\dfrac{∂\textbf{r}}{ ∂v} \approx \dfrac{\textbf{r}(u,v+∆v)−\textbf{r}(u,v)}{∆v}, \\[4pt] \end{align}\]
and so the surface area element \(dσ\) is approximately
\[\nonumber \lVert (\textbf{r}(u +∆u,v)−\textbf{r}(u,v))\times (\textbf{r}(u,v+∆v)−\textbf{r}(u,v)) \rVert \approx \Big \lVert (∆u \dfrac{∂\textbf{r}}{ ∂u} )\times (∆v \dfrac{∂\textbf{r}}{ ∂v} ) \Big \rVert = \Big \lVert \dfrac{∂\textbf{r}}{ ∂u}\times \dfrac{∂\textbf{r}} {∂v} \Big \rVert \,∆u\,∆v\]
by Theorem 1.13 in Section 1.4. Thus, the total surface area \(S \text{ of }Σ\) is approximately the sum of all the quantities \(\lVert \dfrac{ ∂\textbf{r}}{ ∂u} \times \dfrac{∂\textbf{r}}{ ∂v} \rVert \,∆u\,∆v\), summed over the rectangles in \(R\). Taking the limit of that sum as the diagonal of the largest rectangle goes to 0 gives
\[S = \iint\limits_R \Big \lVert \dfrac{∂\textbf{r}} {∂u} \times \dfrac{∂\textbf{r}} {∂v} \Big \rVert \,du \,dv.\label{Eq4.26}\]
We will write the double integral on the right using the special notation
\[\iint\limits_Σ dσ = \iint\limits_R \Big \lVert \dfrac{∂\textbf{r}}{ ∂u} \times \dfrac{∂\textbf{r}} {∂v} \Big \rVert \,du \,dv .\label{Eq4.27}\]
This is a special case of a surface integral over the surface \(Σ\), where the surface area element \(dσ\) can be thought of as \(1dσ\). Replacing 1 by a general real-valued function \(f (x, y, z)\) defined in \(\mathbb{R}^ 3\), we have the following:
Definition: Surface Integral
Let \(Σ\) be a surface in \(\mathbb{R}^ 3\) parametrized by \(x = x(u,v), y = y(u,v), z = z(u,v), \text{ for }(u,v)\) in some region \(R\) in \(\mathbb{R}^ 2\). Let \(\textbf{r}(u,v) = x(u,v)\textbf{i} + y(u,v)\textbf{j} + z(u,v)\textbf{k}\) be the position vector for any point on \(Σ\), and let \(f (x, y, z)\) be a real-valued function defined on some subset of \(\mathbb{R}^ 3\) that contains \(Σ\). The surface integral of \(f (x, y, z) \text{ over }Σ\) is
\[\iint\limits_Σ f (x, y, z)\,dσ = \iint\limits_R f (x(u,v), y(u,v), z(u,v)) \Big \lVert \dfrac{∂\textbf{r}}{∂u}\times \dfrac{∂\textbf{r}}{ ∂v} \Big \rVert \,du \,dv.\label{Eq4.28}\]
In particular, the surface area \(S\) of \(Σ\) is
\[S = \iint\limits_Σ 1\,dσ .\label{Eq4.29}\]
Example 4.9
A torus \(T\) is a surface obtained by revolving a circle of radius \(a\) in the \(yz\)-plane around the \(z\)-axis, where the circle’s center is at a distance \(b\) from the \(z\)-axis \((0 < a < b)\), as in Figure \(\PageIndex{3}\). Find the surface area of \(T\).
Solution
For any point on the circle, the line segment from the center of the circle to that point makes an angle \(u\) with the \(y\)-axis in the positive \(y\) direction (Figure \(\PageIndex{3}\)(a)). And as the circle revolves around the \(z\)-axis, the line segment from the origin to the center of that circle sweeps out an angle \(v\) with the positive \(x\)-axis (Figure \(\PageIndex{3}\)(b)). Thus, the torus can be parametrized as:
\[\nonumber x = (b + a\cos u)\cos v ,\quad y = (b + a\cos u)\sin v ,\quad z = a\sin u,\quad 0 ≤ u ≤ 2π ,\quad 0 ≤ v ≤ 2π\]
So for the position vector
\[\nonumber \begin{align} \textbf{r}(u,v) &= x(u,v)\textbf{i} + y(u,v)\textbf{j} + z(u,v)\textbf{k} \\[4pt] \nonumber &= (b + a\cos u)\cos v\textbf{i} + (b + a\cos u)\sin v\textbf{j} + a\sin u\textbf{k} \\[4pt] \end{align}\]
we see that
\[\nonumber \begin{align}\dfrac{∂\textbf{r}}{ ∂u} &=−a\sin u \cos v\textbf{i} − a\sin u \sin v\textbf{j} + a\cos u\textbf{k} \\[4pt] \nonumber \dfrac{∂\textbf{r}}{ ∂v} &= −(b + a\cos u)\sin v\textbf{i} + (b + a\cos u)\cos v\textbf{j} + 0\textbf{k} , \\[4pt] \end{align}\]
and so computing the cross product gives
\[\nonumber \dfrac{∂\textbf{r}}{ ∂u}\times \dfrac{∂\textbf{r}}{ ∂v} = −a(b + a\cos u)\cos v \cos u\textbf{i} − a(b + a\cos u)\sin v \cos u\textbf{j} − a(b + a\cos u)\sin u\textbf{k} ,\]
which has magnitude
\[\nonumber \Big \lVert \dfrac{∂\textbf{r}}{ ∂u}\times \dfrac{∂\textbf{r}}{ ∂v} \Big \rVert = a(b + a\cos u) .\]
Thus, the surface area of \(T\) is
\[\nonumber \begin{align} S&= \iint\limits_Σ 1\,dσ \\[4pt] \nonumber &=\int_0^{2\pi} \int_0^{2\pi} \Big \lVert \dfrac{∂\textbf{r}}{ ∂u}\times \dfrac{∂\textbf{r}}{ ∂v} \Big \rVert \,du\, dv \\[4pt] \nonumber &=\int_0^{2\pi}\int_0^{2\pi} a(b + a\cos u)\,du \,dv \\[4pt] \nonumber &=\int_0^{2\pi} \left (abu + a^2 \sin u \Big |_{u=0}^{u=2\pi} \right ) \,dv \\[4pt] \nonumber &=\int_0^{2\pi} 2πab\, dv \\[4pt] \nonumber &=4π ^2 ab \\[4pt] \end{align}\]
Since \(\dfrac{∂\textbf{r}}{ ∂u}\text{ and }\dfrac{∂\textbf{r}}{ ∂v}\) are tangent to the surface \(Σ\) (i.e. lie in the tangent plane to \(Σ\) at each point on \(Σ\)), then their cross product \(\dfrac{∂\textbf{r}}{ ∂u}\times \dfrac{∂\textbf{r}}{ ∂v}\) is perpendicular to the tangent plane to the surface at each point of \(Σ\). Thus,
\[\nonumber \iint\limits_Σ f (x, y, z)\,dσ = \iint\limits_R f (x(u,v), y(u,v), z(u,v))\lVert \textbf{n} \rVert \,dσ ,\]
where \(\textbf{n} = \dfrac{∂\textbf{r}}{ ∂u} \times \dfrac{∂\textbf{r}}{ ∂v}\). We say that n is a normal vector to \(Σ\).
Recall that normal vectors to a plane can point in two opposite directions. By an outward unit normal vector to a surface \(Σ\), we will mean the unit vector that is normal to \(Σ\) and points away from the “top” (or “outer” part) of the surface. This is a hazy definition, but the picture in Figure \(\PageIndex{4}\) gives a better idea of what outward normal vectors look like, in the case of a sphere. With this idea in mind, we make the following definition of a surface integral of a 3-dimensional vector field over a surface:
Definition: surface integral of f over \(Σ\)
Let \(Σ\) be a surface in \(\mathbb{R}^ 3\) and let \(\textbf{f}(x, y, z) = f_1(x, y, z)\textbf{i}+ f_2(x, y, z)\textbf{j}+ f_3(x, y, z)\textbf{k}\) be a vector field defined on some subset of \(\mathbb{R}^ 3\) that contains \(Σ\). The surface integral of f over \(Σ\) is
\[\iint\limits_Σ \textbf{f}\cdot d\textbf{σ} = \iint\limits_Σ \textbf{f}\cdot \textbf{n}\,dσ,\label{Eq4.30}\]
where, at any point on \(Σ\), n is the outward unit normal vector to \(Σ\).
Note in the above definition that the dot product inside the integral on the right is a real-valued function, and hence we can use Definition 4.3 to evaluate the integral.
Example \(\PageIndex{1}\)
Evaluate the surface integral \(\iint\limits_Σ \textbf{f}\cdot d\textbf{σ}\), where \(\textbf{f}(x, y, z) = yz\textbf{i}+xz\textbf{j}+x y\textbf{k}\text{ and }Σ\) is the part of the plane \(x+ y+z = 1 \text{ with }x ≥ 0, y ≥ 0, \text{ and }z ≥ 0\), with the outward unit normal n pointing in the positive \(z\) direction (Figure \(\PageIndex{5}\)).
Solution:
Since the vector v \(= (1,1,1)\) is normal to the plane \(x+ y+ z = 1\) (why?), then dividing v by its length yields the outward unit normal vector n \(= \left ( \dfrac{ 1 }{\sqrt{ 3}} , \dfrac{1}{\sqrt{ 3}} , \dfrac{1}{ \sqrt{ 3}}\right ) \). We now need to parametrize \(Σ\). As we can see from Figure \(\PageIndex{5}\), projecting \(Σ\) onto the \(x y\)-plane yields a triangular region \(R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x}\). Thus, using \((u,v)\) instead of \((x, y)\), we see that
\[\nonumber x = u, y = v, z = 1−(u + v),\text{ for }0 ≤ u ≤ 1,0 ≤ v ≤ 1− u\]
is a parametrization of \(Σ\) over \(R\) (since \(z = 1−(x+ y)\) on \(Σ\)). So on \(Σ\),
\[\nonumber \begin{align} \textbf{f}\cdot \textbf{n} &=(yz, xz, x y)\cdot \left ( \dfrac{ 1 }{\sqrt{ 3}} , \dfrac{1}{\sqrt{ 3}} , \dfrac{1}{ \sqrt{ 3}} \right ) = \dfrac{1}{\sqrt{3}}(yz + xz + x y) \\[4pt] \nonumber &=\dfrac{1}{\sqrt{3}}((x+ y)z + x y) = \dfrac{1}{\sqrt{3}}((u + v)(1−(u + v))+ uv) \\[4pt] \nonumber &=\dfrac{1}{\sqrt{3}}((u + v)−(u + v)^2 + uv) \\[4pt] \end{align}\]
for \((u,v)\) in \(R\), and for \(\textbf{r}(u,v) = x(u,v)\textbf{i}+ y(u,v)\textbf{j}+ z(u,v)\textbf{k} = u\textbf{i}+ v\textbf{j}+(1−(u + v))\textbf{k}\) we have
\[\nonumber \dfrac{∂\textbf{r}}{ ∂u} \times \dfrac{∂\textbf{r}}{ ∂v}=(1,0,−1)\times (0,1,−1) = (1,1,1)\, \Rightarrow \, \Big \lVert \dfrac{∂\textbf{r}}{ ∂u} \times \dfrac{∂\textbf{r}}{ ∂v} \Big \rVert = \sqrt{3}.\]
Thus, integrating over \(R\) using vertical slices (e.g. as indicated by the dashed line in Figure \(\PageIndex{5}\)) gives
\[\nonumber \begin{align} \iint\limits_Σ \textbf{f}\cdot d\textbf{σ} &= \iint\limits_Σ \textbf{f}\cdot \textbf{n}\,dσ \\[4pt] \nonumber &=\iint\limits_R (\textbf{f}(x(u,v), y(u,v), z(u,v))\cdot \textbf{n}) \Big \lVert \dfrac{∂\textbf{r}}{ ∂u} \times \dfrac{∂\textbf{r}}{ ∂v} \Big \rVert \,dv\, du \\[4pt] \nonumber &=\int_0^1 \int_0^{1-u} \dfrac{1}{\sqrt{3}}((u + v)−(u + v)^2 + uv) \sqrt{ 3}\,dv\, du \\[4pt] \nonumber &=\int_0^1 \left ( \dfrac{(u + v)^2}{ 2}-\dfrac{(u + v)^3}{ 3}+\dfrac{uv^2}{ 2} \Big |_{v=0}^{v=1-u} \right )\,du \\[4pt] \nonumber &=\int_0^1 \left ( \dfrac{1}{6}+\dfrac{u}{2} -\dfrac{3u^2}{2}+\dfrac{5u^3}{6} \right )\,du \\[4pt] \nonumber &=\dfrac{u}{6} +\dfrac{u^2}{4}-\dfrac{u^3}{2}+\dfrac{5u^4}{24} \Big |_0^1 =\dfrac{1}{8} \\[4pt] \end{align}\]
Computing surface integrals can often be tedious, especially when the formula for the outward unit normal vector at each point of \(Σ\) changes. The following theorem provides an easier way in the case when \(Σ\) is a closed surface , that is, when \(Σ\) encloses a bounded solid in \(\mathbb{R}^ 3\). For example, spheres, cubes, and ellipsoids are closed surfaces, but planes and paraboloids are not.
Divergence Theorem
Let \(Σ\) be a closed surface in \(\mathbb{R}^ 3\) which bounds a solid \(S\), and let \(\textbf{f}(x, y, z) = f_1(x, y, z)\textbf{i}+ f_2(x, y, z)\textbf{j}+ f_3(x, y, z)\textbf{k}\) be a vector field defined on some subset of \(\mathbb{R}^ 3\) that contains \(Σ\). Then
\[\iint\limits_Σ \textbf{f}\cdot d\textbf{σ} = \iiint\limits_S \text{div}\,\textbf{f}\,dV,\label{Eq4.31}\]
where
\[\text{div}\,\textbf{f} = \dfrac{∂f_1}{ ∂x}+\dfrac{∂f_2}{ ∂y}+\dfrac{∂f_3}{ ∂z}\label{Eq4.32}\]
is called the divergence of f .
The proof of the Divergence Theorem is very similar to the proof of Green’s Theorem, i.e. it is first proved for the simple case when the solid \(S\) is bounded above by one surface, bounded below by another surface, and bounded laterally by one or more surfaces. The proof can then be extended to more general solids.
Example \(\PageIndex{2}\)
Evaluate \(\iint\limits_Σ \textbf{f}\cdot d\textbf{σ}\), where \(\textbf{f}(x, y, z) = x\textbf{i} + y\textbf{j} + z\textbf{k}\text{ and }Σ\) is the unit sphere \(x^2 + y^2 + z^2 = 1\).
Solution:
We see that div f = 1+1+1 = 3, so
\[\nonumber \begin{align} \iint\limits_Σ \textbf{f}\cdot d\textbf{σ}&=\iiint\limits_S \text{ div }\textbf{f}\,dV = \iiint\limits_S 3\,dV \\[4pt] \nonumber &=3\iiint\limits_S 1\,dV = 3\text{ vol}(S) = 3\cdot \dfrac{4π(1)^3}{3} = 4\pi. \\[4pt] \end{align}\]
In physical applications, the surface integral \(\iint\limits_Σ \textbf{f}\cdot d\textbf{σ}\) is often referred to as the flux of f through the surface \(Σ\). For example, if f represents the velocity field of a fluid, then the flux is the net quantity of fluid to flow through the surface \(Σ\) per unit time. A positive flux means there is a net flow out of the surface (i.e. in the direction of the outward unit normal vector n ), while a negative flux indicates a net flow inward (in the direction of − n ).
The term divergence comes from interpreting div f as a measure of how much a vector field “diverges” from a point. This is best seen by using another definition of div f which is equivalent to the definition given by Equation \ref{Eq4.32}. Namely, for a point \((x, y, z)\) in \(\mathbb{R}^ 3\),
\[\text{div }\textbf{f}(x, y, z) =\lim\limits_{V \to 0} \dfrac{1}{V} \iint\limits_Σ \textbf{f}\cdot d\textbf{σ},\label{Eq4.33}\]
where \(V\) is the volume enclosed by a closed surface \(Σ\) around the point \((x, y, z)\). In the limit, \(V → 0\) means that we take smaller and smaller closed surfaces around \((x, y, z)\), which means that the volumes they enclose are going to zero. It can be shown that this limit is independent of the shapes of those surfaces. Notice that the limit being taken is of the ratio of the flux through a surface to the volume enclosed by that surface, which gives a rough measure of the flow “leaving” a point, as we mentioned. Vector fields which have zero divergence are often called solenoidal fields.
The following theorem is a simple consequence of Equation \ref{Eq4.33}.
Theorem
If the flux of a vector field f is zero through every closed surface containing a given point, then div f = 0 at that point.
Proof: By Equation \ref{Eq4.33}, at the given point \((x, y, z)\) we have
\[\nonumber \begin{align}\text{div }\textbf{f}(x, y, z) &= \lim\limits_{V \to 0} \dfrac{1}{V}\iint\limits_Σ \textbf{f}\cdot d\textbf{σ} \text{ for closed surfaces }Σ\text{ containing }(x, y, z),\text{ so} \\[4pt] \nonumber &=\lim\limits_{V \to 0}\dfrac{1}{V}(0) \text{ by our assumption that the flux through each }Σ\text{ is zero, so} \\[4pt] \nonumber &=\lim\limits_{V \to 0} 0 \\[4pt] \nonumber &=0 \\[4pt] \end{align}\]
\(\tag{\(\textbf{QED}\)}\)
Lastly, we note that sometimes the notation
is used to denote surface integrals of scalar and vector fields, respectively, over closed surfaces. Especially in physics texts, it is more common to see \(\oint\limits_Σ\) instead.


3: Understanding Plate Tectonics
Scientists now have a fairly good understanding of how the plates move and how such movements relate to earthquake activity. Most movement occurs along narrow zones between plates where the results of plate-tectonic forces are most evident.
There are four types of plate boundaries:
- Divergent boundaries -- where new crust is generated as the plates pull away from each other.
- Convergent boundaries -- where crust is destroyed as one plate dives under another.
- Transform boundaries -- where crust is neither produced nor destroyed as the plates slide horizontally past each other.
- Plate boundary zones -- broad belts in which boundaries are not well defined and the effects of plate interaction are unclear.
Artist's cross section illustrating the main types of plate boundaries (see text); East African Rift Zone is a good example of a continental rift zone. (Cross section by José F. Vigil from This Dynamic Planet -- a wall map produced jointly by the U.S. Geological Survey, the Smithsonian Institution, and the U.S. Naval Research Laboratory.)
Divergent boundaries
Divergent boundaries occur along spreading centers where plates are moving apart and new crust is created by magma pushing up from the mantle. Picture two giant conveyor belts, facing each other but slowly moving in opposite directions as they transport newly formed oceanic crust away from the ridge crest.
Perhaps the best known of the divergent boundaries is the Mid-Atlantic Ridge. This submerged mountain range, which extends from the Arctic Ocean to beyond the southern tip of Africa, is but one segment of the global mid-ocean ridge system that encircles the Earth. The rate of spreading along the Mid-Atlantic Ridge averages about 2.5 centimeters per year (cm/yr), or 25 km in a million years. This rate may seem slow by human standards, but because this process has been going on for millions of years, it has resulted in plate movement of thousands of kilometers. Seafloor spreading over the past 100 to 200 million years has caused the Atlantic Ocean to grow from a tiny inlet of water between the continents of Europe, Africa, and the Americas into the vast ocean that exists today.
The Mid-Atlantic Ridge, which splits nearly the entire Atlantic Ocean north to south, is probably the best-known and most-studied example of a divergent-plate boundary. (Illustration adapted from the map
This Dynamic Planet.
)
The volcanic country of Iceland, which straddles the Mid-Atlantic Ridge, offers scientists a natural laboratory for studying on land the processes also occurring along the submerged parts of a spreading ridge. Iceland is splitting along the spreading center between the North American and Eurasian Plates, as North America moves westward relative to Eurasia.
Map showing the Mid-Atlantic Ridge splitting Iceland and separating the North American and Eurasian Plates. The map also shows Reykjavik, the capital of Iceland, the Thingvellir area, and the locations of some of Iceland's active volcanoes (red triangles), including Krafla.
The consequences of plate movement are easy to see around Krafla Volcano, in the northeastern part of Iceland. Here, existing ground cracks have widened and new ones appear every few months. From 1975 to 1984, numerous episodes of rifting (surface cracking) took place along the Krafla fissure zone. Some of these rifting events were accompanied by volcanic activity; the ground would gradually rise 1-2 m before abruptly dropping, signalling an impending eruption. Between 1975 and 1984, the displacements caused by rifting totalled about 7 m.
Lava fountains (5p;10 m high) spouting from eruptive fissures during the October 1980 eruption of Krafla Volcano. (Photograph by Gudmundur E. Sigvaldason, Nordic Volcanological Institute, Reykjavik, Iceland.)
Aerial view of the area around Thingvellir, Iceland, showing a fissure zone (in shadow) that is an on-land exposure of the Mid-Atlantic Ridge. Right of the fissure, the North American Plate is pulling westward away from the Eurasian Plate (left of fissure). This photograph encompasses the historical tourist area of Thingvellir, the site of Iceland's first parliament, called the Althing, founded around the year A.D. 930. Large building (upper center) is a hotel for visitors. (Photograph by Oddur Sigurdsson, National Energy Authority, Iceland.)
In East Africa, spreading processes have already torn Saudi Arabia away from the rest of the African continent, forming the Red Sea. The actively splitting African Plate and the Arabian Plate meet in what geologists call a triple junction, where the Red Sea meets the Gulf of Aden. A new spreading center may be developing under Africa along the East African Rift Zone. When the continental crust stretches beyond its limits, tension cracks begin to appear on the Earth's surface. Magma rises and squeezes through the widening cracks, sometimes to erupt and form volcanoes. The rising magma, whether or not it erupts, puts more pressure on the crust to produce additional fractures and, ultimately, the rift zone.
Map of East Africa showing some of the historically active volcanoes(red triangles) and the Afar Triangle (shaded, center) -- a so-called
triple junction
(or
triple point
), where three plates are pulling away from one another: the Arabian Plate, and the two parts of the African Plate (the Nubian and the Somalian) splitting along the East African Rift Zone.
East Africa may be the site of the Earth's next major ocean. Plate interactions in the region provide scientists an opportunity to study first hand how the Atlantic may have begun to form about 200 million years ago. Geologists believe that, if spreading continues, the three plates that meet at the edge of the present-day African continent will separate completely, allowing the Indian Ocean to flood the area and making the easternmost corner of Africa (the Horn of Africa) a large island.
Helicopter view (in February 1994) of the active lava lake within the summit crater of 'Erta 'Ale (Ethiopia), one of the active volcanoes in the East African Rift Zone. Two helmeted, red-suited volcanologists -- observing the activity from the crater rim -- provide scale. Red color within the crater shows where molten lava is breaking through the lava lake's solidified, black crust. (Photograph by Jacques Durieux, Groupe Volcans Actifs.)
Oldoinyo Lengai, another active volcano in the East African Rift Zone, erupts explosively in 1966. (Photograph by Gordon Davies, courtesy of Celia Nyamweru, St. Lawrence University, Canton, New York.)
Convergent boundaries
The size of the Earth has not changed significantly during the past 600 million years, and very likely not since shortly after its formation 4.6 billion years ago. The Earth's unchanging size implies that the crust must be destroyed at about the same rate as it is being created, as Harry Hess surmised. Such destruction (recycling) of crust takes place along convergent boundaries where plates are moving toward each other, and sometimes one plate sinks (is subducted) under another. The location where sinking of a plate occurs is called a subduction zone.
The type of convergence -- called by some a very slow "collision" -- that takes place between plates depends on the kind of lithosphere involved. Convergence can occur between an oceanic and a largely continental plate, or between two largely oceanic plates, or between two largely continental plates.
Oceanic-continental convergence
If by magic we could pull a plug and drain the Pacific Ocean, we would see a most amazing sight -- a number of long narrow, curving trenches thousands of kilometers long and 8 to 10 km deep cutting into the ocean floor. Trenches are the deepest parts of the ocean floor and are created by subduction.
Off the coast of South America along the Peru-Chile trench, the oceanic Nazca Plate is pushing into and being subducted under the continental part of the South American Plate. In turn, the overriding South American Plate is being lifted up, creating the towering Andes mountains, the backbone of the continent. Strong, destructive earthquakes and the rapid uplift of mountain ranges are common in this region. Even though the Nazca Plate as a whole is sinking smoothly and continuously into the trench, the deepest part of the subducting plate breaks into smaller pieces that become locked in place for long periods of time before suddenly moving to generate large earthquakes. Such earthquakes are often accompanied by uplift of the land by as much as a few meters.
The convergence of the Nazca and South American Plates has deformed and pushed up limestone strata to form towering peaks of the Andes, as seen here in the Pachapaqui mining area in Peru. (Photograph by George Ericksen, USGS.)
On 9 June 1994, a magnitude-8.3 earthquake struck about 320 km northeast of La Paz, Bolivia, at a depth of 636 km. This earthquake, within the subduction zone between the Nazca Plate and the South American Plate, was one of deepest and largest subduction earthquakes recorded in South America. Fortunately, even though this powerful earthquake was felt as far away as Minnesota and Toronto, Canada, it caused no major damage because of its great depth.
Volcanic arcs and oceanic trenches partly encircling the Pacific Basin form the so-called
Ring of Fire,
a zone of frequent earthquakes and volcanic eruptions. The trenches are shown in blue-green. The volcanic island arcs, although not labelled, are parallel to, and always landward of, the trenches. For example, the island arc associated with the Aleutian Trench is represented by the long chain of volcanoes that make up the Aleutian Islands.
Oceanic-continental convergence also sustains many of the Earth's active volcanoes, such as those in the Andes and the Cascade Range in the Pacific Northwest. The eruptive activity is clearly associated with subduction, but scientists vigorously debate the possible sources of magma: Is magma generated by the partial melting of the subducted oceanic slab, or the overlying continental lithosphere, or both?
Oceanic-oceanic convergence
As with oceanic-continental convergence, when two oceanic plates converge, one is usually subducted under the other, and in the process a trench is formed. The Marianas Trench (paralleling the Mariana Islands), for example, marks where the fast-moving Pacific Plate converges against the slower moving Philippine Plate. The Challenger Deep, at the southern end of the Marianas Trench, plunges deeper into the Earth's interior (nearly 11,000 m) than Mount Everest, the world's tallest mountain, rises above sea level (about 8,854 m).
Subduction processes in oceanic-oceanic plate convergence also result in the formation of volcanoes. Over millions of years, the erupted lava and volcanic debris pile up on the ocean floor until a submarine volcano rises above sea level to form an island volcano. Such volcanoes are typically strung out in chains called island arcs. As the name implies, volcanic island arcs, which closely parallel the trenches, are generally curved. The trenches are the key to understanding how island arcs such as the Marianas and the Aleutian Islands have formed and why they experience numerous strong earthquakes. Magmas that form island arcs are produced by the partial melting of the descending plate and/or the overlying oceanic lithosphere. The descending plate also provides a source of stress as the two plates interact, leading to frequent moderate to strong earthquakes.
Continental-continental convergence
The Himalayan mountain range dramatically demonstrates one of the most visible and spectacular consequences of plate tectonics. When two continents meet head-on, neither is subducted because the continental rocks are relatively light and, like two colliding icebergs, resist downward motion. Instead, the crust tends to buckle and be pushed upward or sideways. The collision of India into Asia 50 million years ago caused the Eurasian Plate to crumple up and override the Indian Plate. After the collision, the slow continuous convergence of the two plates over millions of years pushed up the Himalayas and the Tibetan Plateau to their present heights. Most of this growth occurred during the past 10 million years. The Himalayas, towering as high as 8,854 m above sea level, form the highest continental mountains in the world. Moreover, the neighboring Tibetan Plateau, at an average elevation of about 4,600 m, is higher than all the peaks in the Alps except for Mont Blanc and Monte Rosa, and is well above the summits of most mountains in the United States.
Above: The collision between the Indian and Eurasian plates has pushed up the Himalayas and the Tibetan Plateau. Below: Cartoon cross sections showing the meeting of these two plates before and after their collision. The reference points (small squares) show the amount of uplift of an imaginary point in the Earth's crust during this mountain-building process.
Transform boundaries
The zone between two plates sliding horizontally past one another is called a transform-fault boundary, or simply a transform boundary. The concept of transform faults originated with Canadian geophysicist J. Tuzo Wilson, who proposed that these large faults or fracture zones connect two spreading centers (divergent plate boundaries) or, less commonly, trenches (convergent plate boundaries). Most transform faults are found on the ocean floor. They commonly offset the active spreading ridges, producing zig-zag plate margins, and are generally defined by shallow earthquakes. However, a few occur on land, for example the San Andreas fault zone in California. This transform fault connects the East Pacific Rise, a divergent boundary to the south, with the South Gorda -- Juan de Fuca -- Explorer Ridge, another divergent boundary to the north.
The Blanco, Mendocino, Murray, and Molokai fracture zones are some of the many fracture zones (transform faults) that scar the ocean floor and offset ridges (see text). The San Andreas is one of the few transform faults exposed on land.
The San Andreas fault zone, which is about 1,300 km long and in places tens of kilometers wide, slices through two thirds of the length of California. Along it, the Pacific Plate has been grinding horizontally past the North American Plate for 10 million years, at an average rate of about 5 cm/yr. Land on the west side of the fault zone (on the Pacific Plate) is moving in a northwesterly direction relative to the land on the east side of the fault zone (on the North American Plate).
Aerial view of the San Andreas fault slicing through the Carrizo Plain in the Temblor Range east of the city of San Luis Obispo. (Photograph by Robert E. Wallace, USGS.)
Oceanic fracture zones are ocean-floor valleys that horizontally offset spreading ridges; some of these zones are hundreds to thousands of kilometers long and as much as 8 km deep. Examples of these large scars include the Clarion, Molokai, and Pioneer fracture zones in the Northeast Pacific off the coast of California and Mexico. These zones are presently inactive, but the offsets of the patterns of magnetic striping provide evidence of their previous transform-fault activity.
Plate-boundary zones
Not all plate boundaries are as simple as the main types discussed above. In some regions, the boundaries are not well defined because the plate-movement deformation occurring there extends over a broad belt (called a plate-boundary zone). One of these zones marks the Mediterranean-Alpine region between the Eurasian and African Plates, within which several smaller fragments of plates (microplates) have been recognized. Because plate-boundary zones involve at least two large plates and one or more microplates caught up between them, they tend to have complicated geological structures and earthquake patterns.
Rates of motion
We can measure how fast tectonic plates are moving today, but how do scientists know what the rates of plate movement have been over geologic time? The oceans hold one of the key pieces to the puzzle. Because the ocean-floor magnetic striping records the flip-flops in the Earth's magnetic field, scientists, knowing the approximate duration of the reversal, can calculate the average rate of plate movement during a given time span. These average rates of plate separations can range widely. The Arctic Ridge has the slowest rate (less than 2.5 cm/yr), and the East Pacific Rise near Easter Island, in the South Pacific about 3,400 km west of Chile, has the fastest rate (more than 15 cm/yr).
One of the mysterious, imposing stone monoliths -- some standing 5 m tall and weighing 14 tons -- on Easter Island (Chile), carved by ancient Polynesians out of volcanic rock. Easter Island, which lies on the Nazca Plate close to the East Pacific Rise, is moving eastward toward South America by seafloor spreading at the fastest rate known in the world (see text). (Photograph by Carlos Capurro, U.S. Embassy, Santiago, Chile.)
Evidence of past rates of plate movement also can be obtained from geologic mapping studies. If a rock formation of known age -- with distinctive composition, structure, or fossils -- mapped on one side of a plate boundary can be matched with the same formation on the other side of the boundary, then measuring the distance that the formation has been offset can give an estimate of the average rate of plate motion. This simple but effective technique has been used to determine the rates of plate motion at divergent boundaries, for example the Mid-Atlantic Ridge, and transform boundaries, such as the San Andreas Fault.
Above:
Artist's conception of a Global Positioning System (GPS) satellite in orbit. (Illustration courtesy of NASA.)
Below:
A GPS ground receiver -- here set up on the flank of Augustine Volcano (Cook Inlet, Alaska) -- recording the signals sent by four or more of the orbiting GPS satellites. (Photograph by Jerry Svarc, USGS.)
Current plate movement can be tracked directly by means of ground-based or space-based geodetic measurements; geodesy is the science of the size and shape of the Earth. Ground-based measurements are taken with conventional but very precise ground-surveying techniques, using laser-electronic instruments. However, because plate motions are global in scale, they are best measured by satellite-based methods. The late 1970s witnessed the rapid growth of space geodesy, a term applied to space-based techniques for taking precise, repeated measurements of carefully chosen points on the Earth's surface separated by hundreds to thousands of kilometers. The three most commonly used space-geodetic techniques -- very long baseline interferometry (VLBI), satellite laser ranging (SLR), and the Global Positioning System (GPS) -- are based on technologies developed for military and aerospace research, notably radio astronomy and satellite tracking.
Among the three techniques, to date the GPS has been the most useful for studying the Earth's crustal movements. Twenty-one satellites are currently in orbit 20,000 km above the Earth as part of the NavStar system of the U.S. Department of Defense. These satellites continuously transmit radio signals back to Earth. To determine its precise position on Earth (longitude, latitude, elevation), each GPS ground site must simultaneously receive signals from at least four satellites, recording the exact time and location of each satellite when its signal was received. By repeatedly measuring distances between specific points, geologists can determine if there has been active movement along faults or between plates. The separations between GPS sites are already being measured regularly around the Pacific basin. By monitoring the interaction between the Pacific Plate and the surrounding, largely continental plates, scientists hope to learn more about the events building up to earthquakes and volcanic eruptions in the circum-Pacific Ring of Fire. Space-geodetic data have already confirmed that the rates and direction of plate movement, averaged over several years, compare well with rates and direction of plate movement averaged over millions of years.
Contributors and Attributions
W. Jacquelyne Kious and Robert Tilling (" The Dynamic Earth " via the U.S. Geological Survey )


11.1.3: Rights Granted by Copyright
Author's Bundle of Rights
Activity: Author Rights
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So, now that you know what kinds of works are covered by copyright, what exactly are the rights granted to a copyright holder?
Six exclusive rights are granted to the creator of a copyrighted work. We call these the Author’s Bundle of Rights. This means the copyright holder is the only person who has the right to do these things and has the authority to grant permission for others to do these things, with some important exceptions that we will discuss later in this chapter.
If you are not the copyright holder and want to do any of the examples, you may need to get permission to do so from the holder of the copyright.
Author's Bundle of Rights
To Reproduce
To Prepare Derivative Works
To Distribute
To Perform Publicly
To Display Publicly
To Perform Publicly a sound recording by means of a digital audio transmission
Activity: Author Rights
Under current U.S. law, copyright applies as soon as an original work is fixed in a tangible medium of expression. This means that when you save a file, take a photograph, record a song, or paint a picture your work has copyright protection.
As the creator, provided that the work is not a work made for hire, you are the owner of the copyright on your work. You do not have to register the work with the U.S. Copyright Office, publish it, or put a copyright notice on it.
If you wish to give away, sell or license any or all of the copyright on your work, you have the right to do so.
If you give away or sell your exclusive copyright to someone else, you no longer have the rights mentioned above and need to treat the work the same as any other copyrighted work created by someone else.
See Public Domain and Term of Copyright later in this section for details about the duration of copyright.