4.2.2: Annoying Ways People Use Sources
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Annoying Ways People Use Sources
Read this chapter and take note of how sources can be used poorly. How can you avoid this in your own writing?
 When I am driving in the fast lane, maintaining the speed limit exactly, and I find myself behind someone who thinks the fast lane is for people who drive ten miles per hour below the speed limit, I get an annoyed feeling in my chest like hot water filling a heavy bucket. I wave my arms around and yell, "What . . . ? But, hey. . . oh come on! " There are at least two explanations for why some slow drivers fail to move out of the way:
- They do not know that the generally accepted practice of highway driving in the U.S. is to move to the right if an upcoming car wants to pass.
- They know the guidelines but do not care.
But here is the thing: writers can forget that their readers are sometimes just as annoyed at writing that fails to follow conventions as drivers are when stuck behind a car that fails to move over. In other words, there is something similar between these two people: the knowledgeable driver who thinks, "I thought all drivers knew that the left lane is for the fastest cars", and the reader who thinks, "I thought all writers knew that outside sources should be introduced, punctuated, and cited according to a set of standards".
One day, you may discover that something you have written has just been read by a reader who, unfortunately, was annoyed at some of the ways you integrated sources. She was reading along and then suddenly exclaimed, "What . . . ? But, hey . . . oh come on! " If you are lucky, this reader will try to imagine why you typed things the way you did, giving you the benefit of the doubt. But sometimes you will be slotted into positions that might not really be accurate. When this frustrated reader walks away from your work, trying to figure out, say, why you used so many quotations, or why you kept starting and ending paragraphs with them, she may come to the same conclusions I do about slow drivers:
- You do not know the generally accepted practices of using sources (especially in academic writing) in the United States. or,
- You know the guidelines but do not care.

This judgment, of course, will often be unfair. These readers might completely ignore the merits of your insightful, stylistically beautiful, or revolutionarily important language – just as my anger at another driver makes me fail to admire his custom paint job. But readers and writers do not always see eye to eye on the same text. In fact, some things I write about in this essay will only bother your pickiest readers (some teachers, some editors, some snobby friends), while many other readers might zoom past how you use sources without blinking.
But in my experience, I find that teachers do a disservice when we fail to alert students to the kind of things that some readers might be annoyed at – however illogical these things sometimes seem. People are often unreasonably picky, and writers have to deal with that – which they do by trying to anticipate and preemptively fix whatever might annoy a broad range of readers. Plus, the more effectively you anticipate that pickiness, the more likely it is that readers will interpret your quotations and paraphrases in the way you want them to – critically or acceptingly, depending on your writing context.
It helps me to remember that the conventions of writing have a fundamentally rhetorical nature. That is, I follow different conventions depending on the purpose and audience of my writing, because I know that I will come across differently to different people depending on how well I follow the conventions expected in any particular writing space. In a blog, I cite a source by hyperlinking; in an academic essay, I use a parenthetical citation that refers to a list of references at the end of the essay.
One of the fundamental ideas of rhetoric is that speakers/writers/composers shape what they say/write/create based on what they want it to do, where they are publishing it, and what they know about their audience/readers. And those decisions include nitty-gritty things like introducing quotations and citing paraphrases clearly: not everyone in the entire world approaches these things the same way, but when I strategically learn the expectations of my U.S. academic audience, what I really want to say comes across smoothly, without little annoying blips in my readers' experience.
Notice that I am not saying that there is a particular right or wrong way to use conventions in my writing – if the modern U.S. academic system had evolved from a primarily African or Asian or Latin American cultural consciousness instead of a European one, conventions for writing would probably be very different. That is why they are conventions and not rules.
 In the examples that follow, I wrote all of the annoying examples myself, but all the examples I use of good writing come from actual student papers in first year composition classes at my university; I have their permission to quote them.
Armadillo Roadkill: Dropping in a Quotation Without Introducing It First
Everyone in the car hears it: buh-BUMP. The driver insists to the passengers, "But that armadillo – I did not see it! It just came out of nowhere!"
Sadly, a poorly introduced quotation can lead readers to a similar exclamation: "It just came out of nowhere!" And though readers probably will not experience the same level of grief and regret when surprised by a quotation as opposed to an armadillo, I submit that there is a kinship between the experiences: both involve a normal, pleasant activity (driving; reading) stopped suddenly short by an unexpected barrier (a sudden armadillo; a sudden quotation).
Here is an example of what I am talking about:
We should all be prepared with a backup plan if a zombie invasion occurs. "Unlike its human counterparts, an army of zombies is completely independent of support" (Brooks 155). Preparations should be made in the following areas. . . .
Did you notice how the quotation is dropped in without any kind of warning? (Buh-BUMP.)
The Fix: The easiest way to effectively massage in quotations is by purposefully returning to each one in your draft to see if you set the stage for your readers – often, by signaling that a quote is about to come, stating who the quote came from, and showing how your readers should interpret it. In the above example, that could be done by introducing the quotation with something like this (new text bolded):
We should all be prepared with a backup plan if a zombie invasion occurs. Max Brooks suggests a number of ways to prepare for zombies' particular traits, though he underestimates the ability of humans to survive in harsh environments. For example, he writes, "Unlike its human counterparts, an army of zombies is completely independent of support" (155). His shortsightedness could have a number of consequences. . . .
In this version, I know a quotation is coming ("For example"), I know it is going to be written by Max Brooks, and I know I am being asked to read the quote rather skeptically ("he underestimates"). The sentence with the quotation itself also now begins with a "tag" that eases us into it ("he writes").
Here is an actual example from Alexsandra. Notice the way she builds up to the quotation and then explains it:
In the first two paragraphs, the author takes a defensive position when explaining the perception that the public has about scientists by saying that "there is anxiety that scientists lack both wisdom and social responsibility and are so motivated by ambition . . ." and "scientists are repeatedly referred to as 'playing God'" (Wolpert 345).
With this last sentence especially, his tone seems to demonstrate how he uses the ethos appeal to initially set a tone of someone that is tired of being misunderstood.
Alexsandra prepares us for the quotation, quotes, and then analyzes it. I love it. This is not a hard and fast rule – I have seen it broken by the best of writers, I admit – but it is a wise standard to hold yourself to unless you have a reason not to.
Dating Spider-Man: Starting or Ending a Paragraph with a Quotation
An annoyance that is closely connected to Armadillo Roadkill is the tendency writers sometimes have of starting or ending paragraphs with quotations. This is not technically wrong, and there are situations when the effect of surprise is what you are going for. But often, a paragraph-beginning or paragraph-closing quotation feels rushed, unexplained, disjointed.
It is like dating Spider-Man. You are walking along with him and he says something remarkably interesting – but then he tilts his head, hearing something far away, and suddenly shoots a web onto the nearest building and zooms away through the air. As if you had just read an interesting quotation dangling at the end of a paragraph, you wanted to hear more of his opinion, but it is too late – he is already moved on.
Later, he suddenly jumps off a balcony and is by your side again, and he starts talking about something you do not understand. You are confused because he just dropped in and expected you to understand the context of what was on his mind at that moment, much like when readers step into a paragraph that begins with a quotation. Here is an example:
[End of a preceding paragraph:] . . . Therefore, the evidence clearly suggests that we should be exceptionally careful about deciding when and where to rest.
 After all, consider the following scenario. . . .
There is a perfectly good reason why this feels odd – which should feel familiar after reading about the Armadillo Roadkill annoyance above. When you got to the quotation in the second paragraph, you did not know what you were supposed to think about it; there was no guidance.
The Fix is the same: in the majority of situations, readers appreciate being guided to and led away from a quotation by the writer doing the quoting. Readers get a sense of pleasure from the safe flow of hearing how to read an upcoming quotation, reading it, and then being told one way to interpret it. Prepare, quote, analyze.
I mentioned above that there can be situations where starting a paragraph with a quotation can have a strong effect. Personally, I usually enjoy this most at the beginning of essays or the beginning of sections – like in this example from the very beginning of Jennifer's essay:
"Nothing is ever simple: Racism and nobility can exist in the same man, hate and love in the same woman, fear and loyalty, compromise and idealism, all the yin-yang dichotomies that make the human species so utterly confounding, yet so utterly fascinating" (Hunter). The hypocrisy and complexity that Stephen Hunter from the Washington Post describes is the basis of the movie Crash (2004).
Instantly, her quotation hooks me. It does not feel thoughtless, like it would feel if I continued to be whisked to quotations without preparation throughout the essay. But please do not overdo it; any quotation that opens an essay or section ought to be integrally related to your topic (as is Jennifer's), not just a cheap gimmick.
Uncle Barry and his Encyclopedia of Useless Information: Using too Many Quotations in a Row
You probably know someone like this: a person (for me, my Uncle Barry) who constantly tries to impress me with how much he knows about just about everything. I might casually bring up something in the news ("Wow, these health care debates are getting really heated, aren't they?") and then find myself barraged by all of Uncle Barry's ideas on government- sponsored health care – which then drifts into a story about how his cousin Maxine died in an underfunded hospice center, which had a parking lot that he could have designed better, which reminds him of how good he is at fixing things, just like the garage door at my parents' house, which probably only needs a little. . . . You get the idea. I might even think to myself, "Wait, I want to know more about that topic, but you are zooming on before you contextualize your information at all".
This is something like reading an essay that relies too much on quotations. Readers get the feeling that they are moving from one quotation to the next without ever quite getting to hear the real point of what the author wants to say, never getting any time to form an opinion about the claims. In fact, this often makes it sound as if the author has almost no authority at all. You may have been annoyed by paragraphs like this before:
Addressing this issue, David M. Potter comments, "Whether Seward meant this literally or not, it was in fact a singularly accurate forecast for territorial Kansas" (199). Of course, Potter's view is contested, even though he claims, "Soon, the Missourians began to perceive the advantages of operating without publicity" (200). Interestingly, "The election was bound to be irregular in any case" (201).
Wait – huh? This author feels like Uncle Barry to me: grabbing right and left for topics (or quotes) in an effort to sound authoritative.
The Fix is to return to each quotation and decide why it is there and then massage it in accordingly. If you just want to use a quote to cite a fact, then consider paraphrasing or summarizing the source material (which I find is usually harder than it sounds but is usually worth it for the smoothness my paragraph gains). But if you quoted because you want to draw attention to the source's particular phrasing, or if you want to respond to something you agree with or disagree with in the source, then consider taking the time to surround each quotation with guidance to your readers about what you want them to think about that quote.
In the following passage, I think Jessica demonstrates a balance between source and analysis well. Notice that she only uses a single quotation, even though she surely could have chosen more. But instead, Jessica relies on her instincts and remains the primary voice of authority in the passage:
Robin Toner's article, "Feminist Pitch by a Democrat named Obama", was written a week after the video became public and is partially a response to it. She writes, "The Obama campaign is, in some ways, subtly marketing its candidate as a post-feminist man, a generation beyond the gender conflicts of the boomers". Subtly is the key word. Obama is a passive character throughout the video, never directly addressing the camera. Rather, he is shown indirectly through speeches, intimate conversations with supporters and candid interaction with family. This creates a sense of intimacy, which in turn creates a feeling of trust.
Toner's response to the Obama video is like a diving board that Jessica bounces off of before she gets to the really interesting stuff: the pool (her own observations). A bunch of diving boards lined up without a pool (tons of quotes with no analysis) would not please anyone – except maybe Uncle Barry.
Am I in the Right Movie? Failing to Integrate a Quotation into the Grammar of the Preceding Sentence
When reading drafts of my writing, this is a common experience: I start to read a sentence that seems interesting and normal, with everything going just the way I expect it to. But then the unexpected happens: a quotation blurts itself into the sentence in a way that does not fit with the grammar that built up to quotation. It feels like sitting in a movie theater, everything going as expected, when suddenly the opening credits start for a movie I did not plan to see. Here are two examples of what I am talking about. Read them out loud, and you will see how suddenly wrong they feel.
- Therefore, the author warns that a zombie's vision "are no different than those of a normal human" (Brooks 6).
- Sheila Anne Barry advises that "Have you ever wondered what it is like to walk on a tightrope – many feet up in the air?" (50)
In the first example, the quoter's build-up to the quotation uses a singular subject – a zombie's vision – which, when paired with the quotation, is annoyingly matched with the plural verb are. It would be much less jolting to write, "a zombie's vision is, " which makes the subject and verb agree. In the second example, the quoter builds up to the quotation with a third-person, declarative independent clause: Sheila Anne Barry advises. But then the quotation switches into second person – you – and unexpectedly asks a question – completely different from the expectation that was built up by the first part of the sentence.
The Fix is usually easy: you read your essay out loud to someone else, and if you stumble as you enter a quotation, there is probably something you can adjust in your lead-in sentence to make the two fit together well. Maybe you will need to choose a different subject to make it fit with the quote's verb ( reader instead of readers; each instead of all ), or maybe you will have to scrap what you first wrote and start over.
On occasion you will even feel the need to transparently modify the quotation by adding an [s] to one of its verbs, always being certain to use square brackets to show that you adjusted something in the quotation. Maybe you will even find a way to quote a shorter part of the quotation and squeeze it into the context of a sentence that is mostly your own, a trick that can have a positive effect on readers, who like smooth water slides more than they like bumpy slip-and-slides. Jennifer does this well in the following sentence, for example:
In Crash, no character was allowed to "escape his own hypocrisy" (Muller), and the film itself emphasized that the reason there is so much racial tension among strangers is because of the personal issues one cannot deal with alone.
She saw a phrase that she liked in Muller's article, so she found a way to work it in smoothly, without the need for a major break in her thought. Let's put ourselves in Jennifer's shoes for a moment: it is possible that she started drafting this sentence using the plural subject characters, writing "In Crash, no characters were allowed. . . ". But then, imagine she looked back at the quote from Muller and saw that it said "escape his own hypocrisy", which was a clue that she had to change the first part of her sentence to match the singular construction of the quote.
I Cannot Find the Stupid Link: No Connection Between the First Letter of a Parenthetical Citation and the First Letter of a Works Cited Entry
You have been in this situation: you are on a website that seems like it might be interesting and you want to learn more about it. But the home page does not tell you much, so you look for an "About Us" or "More Information" or "FAQ" link. But no matter where you search – Top of page? Bottom?  This is usually the fault of web designers, who do not always take the time to test their sites as much as they should with actual users. The communication failure here is simple: you are used to finding certain kinds of basic information in the places people usually put it. If it is not there, you are annoyed.
Similarly, a reader might see a citation and have a quick internal question about it: What journal was this published in? When was it published? Is this an article I could find online to skim myself? This author has a sexy last name – I wonder what his first name is? Just like when you look for a link to more information, this reader has a simple, quick question that he or she expects to answer easily.
And the most basic way for readers to answer those questions (when they are reading a work written in APA or MLA style) is (1) to look at the information in the citation, and (2) skim the references or works cited section alphabetically, looking for the first letter in the citation. There is an assumption that the first letter of a citation will be the letter to look for in the list of works cited.
In short, the following may annoy readers who want to quickly learn more about the citation:
[Essay Text:] A respected guide on the subject suggests, "If possible, always take the high ground and hold it" ( The Zombie Survival Guide 135).
[Works Cited Page:] Brooks, Max. The Zombie Survival Guide: Complete Protection from the Living Dead. New York: Three Rivers, 2003. Print.
The reader may wonder when The Zombie Survival Guide was published and flip back to the works cited page, but the parenthetical citation sends her straight to the Z 's in the works cited list (because initial A 's and The 's are ignored when alphabetizing). However, the complete works cited entry is actually with the B 's (where it belongs).
The Fix is to make sure that the first word of the works cited entry is the word you use in your in-text citation, every time. If the works cited entry starts with Brooks, use (Brooks) in the essay text.
Citations not including last names may seem to complicate this advice, but they all follow the same basic concept. For instance, you might have:
- A citation that only lists a title. For instance, your citation might read ("Gray Wolf General Information"). In this case, the assumption is that the citation can be found under the G section of the works cited page. Leah cites her paraphrase of a source with no author in the following way, indicating that I should head to the G 's if I want to learn more about her source:
Alaska is the only refuge that is left for the wolves in the United States, and once that is gone, they will more than likely become extinct in this country ("Gray Wolf General Information").
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A citation that only lists a page number. Maybe the citation simply says (25). That implies that somewhere in the surrounding text, the essay writer must have made it stupendously clear what name or title to look up in the works cited list. This happens a lot, since it is common to introduce a quotation by naming the person it came from, in which case it would be repetitive to name that author again in the citation.
- A quotation without a citation at all. This happens when you cite a work that is both A) from a web page that does not number the pages or paragraphs and B) is named in the text surrounding the quotation. Readers will assume that the author is named nearby. Stephanie wisely leaves off any citation in the example below, where it is already clear that I should head to the O 's on the works cited page to find information about this source, a web page written by Opotow:
To further this point, Opotow notes, "Do not imagine you will be unscathed by the methods you use. The end may justify the means. . . . But there is a price to pay, and the price does tend to be oneself".
I Swear I Did Some Research: Dropping in a Citation Without Making it Clear What Information Came from That Source
Let's look in-depth at this potentially annoying passage from a hypothetical student paper:
It is possible that a multidisciplinary approach to understanding the universe will open new doors of understanding. If theories from sociology, communication, and philosophy joined with physics, the possibilities would be boundless. This would inspire new research, much like in the 1970s when scientists changed their focus from grand-scale theories of the universe to the small concerns of quantum physics (Hawking 51).
In at least two ways, this is stellar material. First, the author is actually voicing a point of view; she sounds knowledgeable, strong. Second, and more to the point of this chapter, the author includes a citation, showing that she knows that ethical citation standards ask authors to cite paraphrases and summaries – not just quotations.
But on the other hand, which of these three sentences, exactly, came from Hawking's book? Did Hawking claim that physics experts should join up with folks in other academic disciplines, or is that the student writer? In other words, at which point does the author's point of view meld into material taken specifically from Hawking?
I recognize that there often are not clean answers to a question like that. What we read and what we know sometimes meld together so unnoticeably that we do not know which ideas and pieces of information are "ours" and which are not. Discussing "patchwriting", a term used to describe writing that blends words and phrases from sources with words and phrases we came up with ourselves, scholar Rebecca Moore Howard writes, "When I believe I am not patchwriting, I am simply doing it so expertly that the seams are no longer visible – or I am doing it so unwittingly that I cannot cite my sources" (91).
In other words, all the moves we make when writing came from somewhere else at some point, whether we realize it or not. Yikes. But remember our main purpose here: to not look annoying when using sources. And most of your instructors are not going to say, "I understand that I could n0t tell the difference between your ideas and your source's because we quite naturally patchwrite all the time. That's fine with me. Party on!" They are much more likely to imagine that you plopped in a few extra citations as a way of defensively saying, "I swear I did some research! See? Here is a citation right here! Doesn't that prove I worked really hard?"
The Fix: Write the sentences preceding the citation with specific words and phrases that will tell readers what information came from where. Like this (bolded words are new):
It is possible that a multidisciplinary approach to understanding the universe will open new doors of understanding. I believe that if theories from sociology, communication, and philosophy joined with physics, the possibilities would be boundless. This would inspire new research, much like the changes Stephen Hawking describes happening in the 1970s when scientists changed their focus from grand-scale theories of the universe to the small concerns of quantum physics (51).
Perhaps these additions could still use some stylistic editing for wordiness and flow, but the source-related job is done: readers know exactly which claims the essay writer is making and which ones Hawking made in his book. The last sentence and only the last sentence summarizes the ideas Hawking describes on page 51 of his book.
One warning: you will find that scholars in some disciplines (especially in the sciences and social sciences) use citations in the way I just warned you to avoid. You might see sentences like this one, from page 64 of Glenn Gordon Smith, Ana T. Torres-Ayala, and Allen J. Heindel's article in the Journal of Distance Education:
Some researchers have suggested "curriculum" as a key element in the design of web-based courses (Berge, 1998; Driscoll, 1998; Meyen, Tangen, & Lian, 1999; Wiens & Gunter, 1998).
Whoa – that is a lot of citations. Remember how the writer of my earlier example cited Stephen Hawking because she summarized his ideas? Well, a number of essays describing the results of experiments, like this one, use citations with a different purpose, citing previous studies whose general conclusions support the study described in this new paper, like building blocks.
It is like saying to your potentially skeptical readers, "Look, you might be wondering if I am a quack. But I can prove I am not! See, all these other people published in similar areas! Are you going to pick fights with all of them too?" You might have noticed as well that these citations are in APA format, reflecting the standards of the social sciences journal this passage was published in. Well, in this kind of context APA's requirement to cite the year of a study makes a lot of sense too – after all, the older a study, the less likely it is to still be relevant.
Conclusion: Use Your Turn Signals
You may have guessed the biggest weakness in an essay like this: what is annoying varies from person to person, with some readers happily skimming past awkward introductions to quotations without a blink, while others see a paragraph-opening quotation as something to complain about on Facebook. All I have given you here – all I can give you unless I actually get to know you and your various writing contexts – are the basics that will apply in a number of academic writing contexts.
Think of these as signals to your readers about your intentions, much as wise drivers rely on their turn signals to communicate their intentions to other drivers. In some cases when driving, signaling is an almost artistic decision, relying on the gut reaction of the driver to interpret what is best in times when the law does not mandate use one way or the other. I hope your writing is full of similar signals. Now if I could only convince the guy driving in front of me to use his blinker. . . .
Discussion
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Because so many of these guidelines depend on the writer's purpose, publication space, and audience, it can be difficult to know when to follow them strictly and when to bend them. What are some specific writing situations where a writer is justified to bend the standards of how to incorporate sources?
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Choose one of the annoyances. Then, look through a number of different pieces of writing from different genres and collect two examples of writers who followed your chosen guideline perfectly and two who did not. For each source you found, jot a sentence or two describing the context of that source and why you think its writer did or did not follow the guideline.
 Now, rank them from the point of view of a newspaper editor, a popular blogger, and another college student. What changes did you make in your rankings?
Works Cited
Barry, Sheila Anne. Tricks & Pranks to Fool Your Friends. New York: Sterling, 1984. Print.
Brooks, Max. The Zombie Survival Guide: Complete Protection from the Living Dead. New York: Three Rivers, 2003. Print.
"Gray Wolf General Information". Environmental Conservation Online System. U.S. Fish and Wildlife Service, 15 Oct. 2008. Web. 23 Oct. 2008.
Hawking, Stephen. A Brief History of Time: From the Big Bang to Black Holes. New York: Bantam, 1988. Print.
Howard, Rebecca Moore. "The New Abolitionism Comes to Plagiarism". Perspectives on Plagiarism and Intellectual Property in a Postmodern World. Ed. Lisa Buranen and Alice M. Roy. Albany: SUNY P, 1999. 87–95. Print.
Hunter, Stephen. "'Crash': The Collision Of Human Contradictions". The Washington Post. The Washington Post Company, 6 May 2005. Web. 21 Feb. 2008.
Muller, Bill. "Crash: LA Tale Confronts, Then Shatters, Stereotypes". The Arizona Republic. AZCentral.com, 6 May 2005. Web. 21 Feb. 2008.
Opotow, Susan. "Moral Exclusion and Torture: The Ticking Bomb Scenario and the Slippery Ethical Slope". Peace and Conflict Studies 13.4 (2007): 457–61. PsycINFO. Web. 27 Sept. 2008.
Piven, Joshua, and David Borgenicht. The Worst-Case Scenario Survival Handbook: Work. San Francisco: Chronicle, 2003. Print.
Potter, David M. The Impending Crisis: 1848–1861. Ed. Don E. Fehrenbacher. New York: Harper & Row, 1976. Print.
Smith, Glenn Gordon, Ana T. Torres-Ayala, and Allen J. Heindel. "Disciplinary Differences in E-learning Instructional Design: The Case of Mathematics". Journal of Distance Education 22.3 (2008): 63–88. Web. 10 Sept. 2009.
Toner, Robin. "Feminist Pitch by a Democrat Named Obama". The New York Times. The New York Times Company, 2 Dec. 2007. Web. 22 Oct. 2008.
Wolpert, Lewis. "Is Cell Science Dangerous?" Journal of Medical Ethics 33.6 (2007): 345–48. Academic Search Premier. Web. 28 Jan. 2009.


Obtaining a bachelor’s degree can be easily likened to taking a cross-country trip. Let’s say that a person wanted to travel from Los Angeles to New York City. This person could take a more northern route and experience mountains and cooler temperatures. Or, this person might elect to take a more southern route and experience deserts, beaches, and warmer temperatures. A really adventurous person might elect to traverse the globe and see oceans, China, and Europe along the way. Earning a degree should be seen as a journey—a long term commitment to self-improvement through education. Many decisions will be made along the way that can take you down discrete paths in your quest for a degree that may ultimately lead you down different career paths based on your strengths and interests. Academic advisors will ultimately help you navigate your academic journey.
Who Is Your Academic Advisor?
Typically in high school, students were automatically enrolled in a course schedule for a full academic year. These schedules were most likely created by a guidance counselor based on your student record and some administrative input. However, in college, students create their course schedule each semester with the help of an academic advisor, not a counselor. In your first academic year, you will be meeting with full-time professional academic advisors from the University Advising Center in the Division of Student Success who will help you navigate your first-year courses and provide additional support as you “learn the ropes.” Once you leave the Division of Student Success, you will be advised within your major academic department. In this case, your academic advisor could be a professional staff advisor, a faculty member who agrees to take on advising responsibilities, or even a graduate assistant who works under the supervision of an advising administrator. While being advised by the Division of Student Success, you are not discouraged from talking with the academic advisors of your major department. In fact, in some cases, the Division of Student Success advisors may request that you speak with a departmental advisor to answer specific questions. However, you should always check in with a Division of Student Success advisor before making any adjustments to your course schedule.
University Advising Center in the Division of Student Success
www.uta.edu/advising
Your academic advisors should have information available in the academic department or on the department Web site regarding their office hours and preferred methods for contacting them. It is your responsibility to seek out that information and have it available for when you need to make contact.
When Should I See an Advisor?
Quite simply, you should feel free to contact an academic advisor any time you have a question about your college experience or are experiencing a difficulty that is impeding your success in college. Then again, you do not have to simply see an advisor when you are having a problem; you can even contact your academic advisor when you want to celebrate successes no matter the size. Here are some more concrete instances in which you will want to speak with your advisor:
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For assistance in developing a 4–5-year academic plan for graduation.
Helpful Hint: Refer to Activity 2-3 in this chapter for assistance with this plan.
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Prior to the next registration period to discuss your schedule of classes for the following academic session.
Helpful Hint: Each semester, you will need to see an academic advisor to have an enrollment hold removed in order to register for classes. Beat the advising rush and schedule an appointment to meet with your advisor several weeks before your enrollment appointment specified in the Student Services Center in MyMav.
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When you are experiencing any type of difficulty (personal, academic, or social) that is impeding your ability to perform well in or attend class and need referrals to campus resources or study skills advice that may be able to help you perform better.
Helpful Hint: Dealing with issues as soon as they arise rather than waiting until it is impossible to recover will benefit you in the long run.
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When you are considering changing your major.
Helpful Hint: You may want to visit with many academic advisors on campus including your current advisor and also consider seeking advisement from a Majors Exploration advisor in the Division of Student Success.
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When you need assistance adding, dropping, or withdrawing from classes.
Helpful Hint: After Late Registration, you will be required to seek the assistance of an academic advisor to add or drop classes to complete the MyMav functions.
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Prior to the Last Drop Day during a given session if you are experiencing difficulty in a course.
Helpful Hint: It is in your best interest to check in with the faculty for your classes to determine your current grades and then talk with an advisor if you are considering dropping. Waiting until the last minute
is not recommended. Professors’ and advisors’ office hours may conflict with your schedule, and it may take several days to get all of the appropriate signatures.
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Any time you have a question about a UTA policy or procedure that is discussed in the Undergraduate Catalog.
Helpful Hint: It is the responsibility of every UTA student to have read their Undergraduate Catalog and understand the policies. “Nobody told me” is not an acceptable defense if you do not follow university policy.
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If your Academic Standing at the end of a session is anything other than “Good Standing.”
Helpful Hint: “Chapter 6: Avoiding Hazards Along the Way” discusses issues related to Academic Standing at UT Arlington more in depth.
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If you want some guidance on how to apply for graduate school or discuss career options for your major.
Helpful Hint: In addition to talking with your academic advisor, you can also talk with your professors, attend workshops given by the Graduate School, and do some research in the Lockheed Martin Career Development Center.
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When you reach 60 cumulative credit hours, so you can discuss having your intended major changed to a major if you have not done so already.
Helpful Hint: UT Arlington policy requires that all students are in a major by the time they have earned 75 credit hours. Students not able to make a major decision are required to meet with an advisor in the Division of Student Success.
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Before enrolling in any courses at another institution that you intend on transferring back to UT Arlington to meet degree requirements.
Helpful Hint: While courses may transfer into UTA, they may not meet the requirements of a major degree plan.
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The long semester before you plan to graduate from UT Arlington to double-check your degree progress and find out the procedures you need to follow to apply for graduation.
Student Responsibilities
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Understand and communicate personal values, abilities, and goals.
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Provide accurate and truthful information when being advised.
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Schedule and keep appointments or call ahead to reschedule if you are going to be late or need to cancel the appointment.
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Learn and understand UT Arlington policies, procedures, and requirements by reading the Undergraduate Catalog.
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Ask questions about UT Arlington policies, procedures, and requirements if you do not understand them.
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Come prepared to an advising appointment. For instance, if you are meeting with your advisor to discuss enrollment for the next term, bring your Academic Plan for Graduation and a list of courses you might consider taking.
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Be open to new possibilities that the academic advisor may sug- gest.
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Follow through on the plans- of-action agreed upon in the advising session.
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Understand and accept that students are ultimately responsible for their education and their own decisions. Take an active role in the advising process.
Academic Advisor Responsibilities
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Inform students of the nature of the advisor/advisee relationship and the expectations you have for your advisees.
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Develop purposeful and meaningful relationships with advisees.
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Provide and update contact information and posted office hours.
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Keep appointments or contact a student if it is necessary to change or cancel an appointment.
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Inform and refer students to campus resources and special services available to them.
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Assist students in defining and developing desired educational, career, and life plans.
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Listen and help students in developing a realistic academic plan for graduation that will help a student meet his or her goals.
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Monitor progress toward educational/career goals.
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Interpret and provide rationale for university policies, procedures, and requirements.
Why Is the Undergraduate Catalog Such an Important Document?
The policies and procedures set forth in the Undergraduate Catalog are in effect a contract or agreement with the students who enter UTA during that catalog period. If students complete the degree requirements set forth using the policies and procedures outlined from that catalog, UTA will confer a bachelor’s degree. Though policies and degree requirements may change from year to year, the policies and degree requirements set forth in the catalog that students entered under is what they have to follow.
Undergraduate Catalog
catalog.uta.edu
The official Undergraduate Catalog (and the archive of previous years’ catalogs) can only be accessed online. Students can access official information about the degree plans offered by the different academic departments as well as the current course catalog that describes each of the courses more in depth and explains the prerequisites for the courses. In addition to the basic degree requirements, the Undergraduate Catalog is considered the official source of information regarding the policies and procedures that students must follow as they are taking courses to obtain their degree.
It is highly recommended that students read and familiarize themselves with the policies and procedures of their entering catalog.
Components of UT Arlington Degree Plans
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Core Curriculum
—State-determined courses meant to provide all undergraduates with a well-rounded education.
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Major Coursework
—Courses specific to the discipline students have selected. Includes required courses, major electives, and possibly subplan specialties.
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Electives
—General courses that students can take to meet the total semester credit hours required to earn the degree.
Degree requirements can be found in the Undergraduate Catalog at catalog.uta.edu.
Students should always consult undergraduate academic advisors from their major department before enrolling in classes to ensure the proper selection of courses that will meet degree plan requirements.
How to Calculate a Grade Point Average (GPA)
Students will receive information regarding their cumulative and semester GPAs at the end of each term when grades are posted in MyMav. Any courses taken at another institution are not calculated into UTA’s semester or cumulative GPAs, so
do not take courses at another institution and assume that those grades will impact your UTA GPA and Academic Standing. They will not.
At times, it is helpful to be able to calculate your own GPA if:
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you want to determine what your major GPA is,
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you want to play a guessing game to determine what you might need to earn in order to change your GPA from one level to another, or
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you are interested in applying to graduate or professional school upon finishing your degree (you can calculate an overall GPA that would include those grades as the Graduate Admissions Offices would using this basic formula).
At UTA, students are graded on a 4-point scale:
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Grade
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Grade Value
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Meaning
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A
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4.0
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Excellent
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B
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3.0
|
Good
|
|
C
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2.0
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Fair
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|
D
|
1.0
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Poor
|
|
F
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0.0
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Failing
|
Note: Grades of I, P, Q, W, or Z are not included in the calculation of a GPA.
Let’s Do Some Math…
Example:
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List the courses attempted and the grades earned.
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Multiply the grade value by the number of credit hours of that course.
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Course
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Letter Grade Earned
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Grade Point Allocation
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Grade x Credit Hours of course
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= Grade Points Earned
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ENGL 1301
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A
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4.0
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4.0 x 3 hrs
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= 12
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MATH 1302
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D
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1.0
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1.0 x 3 hrs
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= 3
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UNIV 1301
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B
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3.0
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3.0 x 3 hrs
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= 9
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BIOL 1441
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C
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2.0
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2.0 x 4 hrs
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= 8
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13 credit hours attempted
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|
|
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= 32 Grade Points
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Add the total hours attempted and the total grade points earned.
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Divide the grade points earned by the credit hours attempted >> 32 grade points ÷ 13 hours
Grade point average of these courses: 2.46
Thought Question
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How are the academic advisors in college different from the guidance counselors in high school?
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Overall, how do you think college will be different than high school?
Activity 2-1:
Read all of the sections under “General Information” of the online Undergraduate Catalog. Also, select one department of interest to you in the “Academics” section and read about that department and the degree requirements. Look up at least one course listed in the degree requirements under the “Course Descriptions” section to see if there are any prerequisites for that course. As you are reading write down any questions you may have. Schedule an appointment with an academic advisor and ask the questions you have about the degree requirements and policies and procedures you read in the catalog.
Activity 2-2:
Using the Undergraduate Catalog, investigate the following policies:
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Schedule Changes (Adds and Drops)—How many courses can an undergraduate student drop during their academic career?
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Withdrawal for Non-Payment
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75 Hours to Undergraduate Major Policy
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Academic Probation and Dismissal Policy
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Freshman Potential for Academic Success Policy
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Grade Replacement and Grade Exclusion Policies
Write down any questions that you may have about these policies and talk about them with an academic advisor.
Activity 2-3:
Using the Undergraduate Catalog, complete this worksheet to the best of your ability.


10.2: Collisions
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In this section we go through a few examples of applying conservation of momentum to model collisions. Collisions can loosely be defined as events where the momenta of individual particles in a system are different before and after the event.
We distinguish between two types of collisions: elastic and inelastic collisions. Elastic collisions are those for which the total mechanical energy of the system is conserved during the collision (i.e. it is the same before and after the collision). Inelastic collisions are those for which the total mechanical energy of the system is not conserved. In either case, to model the system, one chooses to define the system such that there are no external forces on the system so that total momentum is conserved.
Inelastic collisions
In this section, we give a few examples of modelling inelastic collisions. Inelastic collisions are usually easier to handle mathematically, because one only needs to consider conservation of momentum and does not use conservation of energy (which usually involves equations that are quadratic in the speeds because of the kinetic energy term).
Example \(\PageIndex{1}\)
You (mass \(m_s\) ) and your friend (mass \(m_f\) ) face each other on ice skates on an ice surface that is slippery enough that friction can be considered negligible, as shown in Figure \(\PageIndex{1}\). You shove your friend away from you so that he moves with velocity \(\vec v_f\) away from you (the velocity is measured relative to the ice). Is the collision elastic? What is your speed relative to the ice after you shoved your friend?
Solution
We can consider the system as being comprised of you and your friend. There are no net external forces on the system (gravity and normal forces cancel each other), so the momentum of the system will be conserved.
The mechanical energy will not be conserved. You had to use chemical potential energy stored in your muscles to shove your friend. Thus, external energy (i.e. not mechanical energy from you or your friend) was injected into the system, and we should expect the total mechanical energy to be larger after the collision.
Before the collision, both you and your friend have zero speed, and thus zero kinetic energy and zero momentum. After the collision, your friend has a velocity \(\vec v_f\) . We can use conservation of total momentum, \(\vec P\) , to determine your velocity, \(\vec v_s\) , after the collision.
\[\begin{aligned} \vec P &=\vec P'\\[4pt] 0 &= m_s\vec v_s + m_f\vec v_f\\[4pt] \therefore \vec v_s &= -\frac{m_f}{m_s}\vec v_f\end{aligned}\]
where primes ( \('\) ) denote a quantity after the collision. We find that your velocity is in the opposite direction from that of your friend. Before the collision, the mechanical energy, \(E\) , of the system is zero (we can ignore gravitational potential energy, since everything is in the horizontal plane). After the collision, the mechanical energy, \(E'\) , is:
\[\begin{aligned} E' = \frac{1}{2}m_sv_s^2+\frac{1}{2}m_fv_f^2\end{aligned}\]
which is clearly bigger than the mechanical energy before the collision (i.e. 0), as we suspected it would be.
Discussion
We find that you recoil in the opposite direction, which makes sense. If you push your friend in one direction, Newton’s Third Law says that your friend pushes you in the opposite direction. Your speed furthermore depends on the ratio of your friend’s mass to yours. This also makes sense, because if you both feel the same force, the person with the smallest mass will have the highest speed; if your mass is higher than your friend’s, then your speed after the collision will be smaller than your friend’s.
We also saw that mechanical energy was not conserved. In terms of energy, we can explain this by saying that you burned up chemical potential energy stored in your muscles in order to shove your friend. Because we included both you and your friend in the system, the shove was an internal force and momentum is conserved. Of course, if we had considered only you as the system, then your momentum would not have been conserved during the collision.
The type of collision that we described here is also sometimes called an “explosion”. You can imagine all of the parts that make up a bomb as small particles. When the bomb explodes, chemical potential energy is converted into the kinetic energy of the bomb fragments. If you consider all of the particles/fragments of the bomb as a system, then the total momentum of all of the bomb fragments is conserved (and equal to zero if the bomb was initially at rest). Again, mechanical energy would not be conserved (and would increase) as the chemical potential energy is converted into mechanical energy.
Example \(\PageIndex{2}\)
A proton of mass \(m_p\) and initial velocity \(\vec v_p\) collides inelastically with a nucleus of mass \(m_N\) at rest, as shown in Figure \(\PageIndex{2}\). A coordinate system is set up as shown, such that the initial velocity of the proton is in the \(x\) direction. After the collision, the proton’s speed is measured to be \(v'_p\) and its velocity vector is found to make an angle \(\theta\) with the \(x\) axis as shown. What is the velocity vector of the nucleus after the collision? Assume that the collision takes place in vacuum.
Solution
As a system, we consider the proton and the nucleus together, so that the total momentum of the system is conserved during the collision, as no other external forces are exerted on the two particles (since they are in vacuum). Because momentum is a vector, each component of the total momentum, \(\vec P\) , is conserved during the collision:
\[\begin{aligned} \vec P &= \vec P'\\[4pt] \therefore P_x &= P'_x\\[4pt] \therefore P_y &= P'_y\end{aligned}\]
where, as usual, primes ( \('\) ) denote quantities after the collision. After the collision, both particles will have velocity vectors that have \(x\) and \(y\) components. Let the velocity vector of the nucleus after the collision be \(\vec v'_N\) and let \(\phi\) be the angle that it makes with the \(x\) axis, as shown in Figure \(\PageIndex{2}\).
We can start by considering the conservation of the \(x\) component of the total momentum. The initial and final momenta in the \(x\) direction are given by:
\[\begin{aligned} P_x &= m_p v_p\\[4pt] P'_x &= m_p v'_p\cos\theta + m_N v'_N\cos\phi\\[4pt] \therefore m_p v_p &= m_p v'_p\cos\theta + m_N v'_N\cos\phi\end{aligned}\]
which gives us a first equation to determine the final velocity of the nucleus.
The \(y\) component of the total momentum before the collision is zero since we chose the coordinate system such that the initial velocity of the proton is in the \(x\) direction. The initial and final momenta in the \(y\) direction are given by:
\[\begin{aligned} P_y &= 0\\[4pt] P'_y &= m_p v'_p\sin\theta - m_N v'_N\sin\phi\\[4pt] \therefore m_p v'_p\sin\theta &= m_N v'_N\sin\phi\end{aligned}\]
which gives us a second equation to solve for the velocity of the nucleus. With the two equations from momentum conservation, we can solve for the magnitude and direction of the velocity of the nucleus. From the \(y\) component of momentum conservation, we can find an expression for the speed of the nucleus:
\[\begin{aligned} m_p v'_p\sin\theta &= m_N v'_N\sin\phi\\[4pt] \therefore v'_N &= \frac{m_p}{m_N}v'_p\sin\theta \frac{1}{\sin\phi}\end{aligned}\]
which we can substitute into the \(x\) equation for momentum conservation to solve for the angle \(\phi\) :
\[\begin{aligned} m_p v_p &= m_p v'_p\cos\theta + m_N v'_N\cos\phi\\[4pt] m_p v_p &= m_p v'_p\cos\theta + m_N\frac{m_p}{m_N}v'_p\sin\theta \frac{\cos\phi}{\sin\phi} \\[4pt] v_p &= v'_p\cos\theta + v'_p\sin\theta \frac{1}{\tan\phi}\\[4pt] \therefore \tan\phi &= \frac{v'_p\sin\theta}{v_p-v'_p\cos\theta}\end{aligned}\]
If we were given numbers for the initial and final speed of the proton, as well as the angle \(\theta\) , we would be able to find a value for the angle \(\phi\) , which we could then use to determine the final speed of the nucleus:
\[\begin{aligned} v'_N &= \frac{m_p}{m_N}v'_p\sin\theta \frac{1}{\sin\phi}\end{aligned}\]
Discussion:
By using the conservation of momentum equation and writing out the \(x\) and \(y\) components, we were able to find two equations to determine the magnitude and direction of the nucleus’ velocity after the collision. In the limit where \(m_N >> m_p\) , the final speed of the nucleus would be very small (close to zero).
Elastic collisions
In this section, we give a few examples of modeling elastic collisions. Even though it is mechanical energy that is conserved in an elastic collision, one can almost always simplify this to only kinetic energy being conserved. If a collision takes place in a well localized position in space (i.e. before and after the collision are the same point in space), then the potential energies of the objects involved will not change, thus any change in their mechanical energy is due to a change in kinetic energy.
Example \(\PageIndex{3}\)
A block of mass \(M\) moves with velocity \(\vec v_M\) in the \(x\) direction, as shown in Figure \(\PageIndex{3}\). A block of mass \(m\) is moving with velocity \(\vec v_m\) also in the \(x\) direction and collides elastically with block \(M\) . Both blocks slide with no friction on the horizontal surface. What are the velocities of the two blocks after the collision?
Solution
Because this is an elastic collision, both the total momentum and total mechanical energy are conserved. Equating the total momentum before and after the collision, and considering only the \(x\) component gives the following equation:
\[\begin{aligned} \vec P &=\vec P'\\[4pt] Mv_M+mv_m&=Mv'_M+mv'_m\end{aligned}\]
where the primes ( \('\) ) correspond to the quantities after the collision. Note that, in principle, the \(x\) components of the velocities ( \(v_M\) , \(v'_M\) , \(v_m\) , \(v'_m\) ) could be negative numbers if the corresponding block is moving in the negative \(x\) direction.
For the mechanical energy of the two blocks, we only need to consider their kinetic energy since their gravitational potential energies are the same before and after the collision on the horizontal surface. The total mechanical energy of the system, before and after the collision is given by:
\[\begin{aligned} E &=E'\\[4pt] \frac{1}{2}Mv_M^2+\frac{1}{2}mv_m^2&=\frac{1}{2}Mv'^2_M+\frac{1}{2}mv'^2_m\\[4pt] \therefore Mv_M^2+mv_m^2&=Mv'^2_M+mv'^2_m\end{aligned}\]
where we canceled the factor of one half in the last line. This gives two equations (conservation of energy and momentum) and two unknowns (the two speeds after the collision). This is not a linear system of equations, because the equation from conservation of energy is quadratic in the speeds.
The following method allows many models for elastic collisions between two particles to be solved easily by converting the quadratic equation from energy conservation into an equation that is linear in the speeds. First, write both equations so that the quantities related to each particle are on opposite sides of the equation. For momentum, this gives:
\[Mv_{M}+mv_{m}=Mv_{M}^{'} +mv_{m}^{'}\]
\[\therefore M(v_{M}-v_{M}^{'})=m(v'm-v_{m})\]
For conservation of energy, this gives:
\[Mv_{M}^{2}+mv_{m}^{2}=Mv_{M}^{'2}+mv_{m}^{'2}\]
\[\therefore M(v_{M}^{2}-v_{M}^{'2}=M(v_{m}^{'2}-v_{m}^{2})\]
which we can re-write as:
\[\begin{aligned} M(v_M^2-v'^2_M)&= M(v'^2_m-v^2_m)\\[4pt] M(v_M-v'_M)(v_M+v'_M)&= M(v'_m-v_m)(v'_m+v_m)\end{aligned}\]
We can then divide Equation 10.2.3 and 10.2.4 by Equation 10.2.1 and 10.2.2 :
\[\begin{aligned} \frac{M(v_M-v'_M)(v_M+v'_M)}{M(v_M-v'_M)}&= \frac{M(v'_m-v_m)(v'_m+v_m)}{m(v'm-v_m)}\\[4pt] \therefore v_M+v'_M&=v'_m+v_m\end{aligned}\]
which gives us an equation that is much easier to work with, since it is linear in the speeds. If we re-arrange this last equation back so that quantities before and after the collision are on different sides of the equality:
\[v_{M}-v_{m}=-(v_{M}^{'}-v_{m}^{'})\]
we can see that the relative speed between \(M\) and \(m\) is the same before and after the collision. That is, if block \(M\) “saw” block \(m\) approaching with a speed of \(3\text{m/s}\) before the collision, it would “see” block \(m\) moving away with speed \(3\text{m/s}\) after the collision, regardless of the actual directions and velocities of the block, if the collision was elastic.
By using this equation with the original conservation of momentum equation, we now have two equations and two unknowns that are easy to solve:
\[\begin{aligned} v_M-v_m &= - (v'_M-v'_m)\\[4pt] Mv_M+mv_m&=Mv'_M+mv'_m\end{aligned}\]
Solving for \(v'_m\) in both equations gives:
\[\begin{aligned} v_M-v_m &= - (v'_M-v'_m)\\[4pt] \therefore v'_m &= v_M+v'_M-v_m\\[4pt] Mv_M+mv_m&=Mv'_M+mv'_m\\[4pt] \therefore v'_m&=\frac{1}{m}(Mv_M+mv_m-Mv'_M)\end{aligned}\]
Equating the two expressions for \(v'_m\) allows us to solve for \(v'_M\) :
\[\begin{aligned} \frac{1}{m}(Mv_M+mv_m-Mv'_M)&=v_M+v'_M-v_m\\[4pt] Mv_M+mv_m-Mv'_M&=mv_M+mv'_M-mv_m\\[4pt] (M-m)v_M+2mv_m&=(M+m)v'_M\\[4pt] \therefore v'_M&=\frac{M-m}{M+m}v_M+\frac{2m}{M+m}v_m\end{aligned}\]
One can easily solve for the other speed, \(v'_m\) :
\[\begin{aligned} \therefore v'_m &= \frac{m-M}{M+m}v_m+\frac{2M}{M+m}v_M\end{aligned}\]
And writing these together:
\[\begin{aligned} v'_M&=\frac{M-m}{M+m}v_M+\frac{2m}{M+m}v_m\\[4pt] v'_m &= \frac{m-M}{M+m}v_m+\frac{2M}{M+m}v_M\end{aligned}\]
Discussion
The formulas that we obtained above are valid for any one dimensional elastic collision.
Exercise \(\PageIndex{1}\)
Two trains of equal masses collide elastically on a track. If train \(A\) had a speed \(v\) and train \(B\) was at rest, what are the speeds of the trains after the collision?
- Both trains \(A\) and \(B\) travel away from each other with speeds \(\frac{1}{2}v\) .
- Train \(A\) will be at rest and train \(B\) will move away with a speed \(v\) .
- Both trains \(A\) and \(B\) will stick together and move at a speed of \(v\) .
- Train \(B\) will be at rest and train \(A\) will move away at a speed of \(v\) .
- Answer
Example \(\PageIndex{4}\)
A proton of mass \(m\) and initial velocity \(\vec v_1\) collides elastically with a second proton that is at rest. After the collision, the two protons have velocities \(\vec v'_1\) and \(\vec v'_2\) , as shown in Figure \(\PageIndex{4}\). Show that the velocity vectors of the two protons are perpendicular after the collision.
Solution
This example highlights a particular feature of elastic collisions when the two objects have the same mass and one of the objects is initially at rest. The conservation of momentum for the system comprised of the two protons can be written as:
\[\begin{aligned} m\vec v_1 &= m\vec v'_1 + m\vec v'_2\\[4pt] \vec v_1 &= \vec v'_1 + \vec v'_2\end{aligned}\]
where the left hand side corresponds to the initial total momentum and the right hand side to the total momentum after the collision. In the second line, we canceled out the mass, and obtained a vector relation between the velocity vectors. We can graphically illustrate the vector relation as in Figure \(\PageIndex{5}\) which shows the triangle that is formed by adding the two outgoing velocity vectors to obtain the initial velocity vector.
Conservation of kinetic energy for the collision can be written as:
\[\begin{aligned} \frac{1}{2}mv_1^2 &= \frac{1}{2}mv'^2_1+\frac{1}{2}mv'^2_2\\[4pt] v_1^2 &= v'^2_1+ v'^2_2\end{aligned}\]
where the left hand side corresponds to the initial kinetic energy and the right hand side to the final kinetic energy. We canceled the mass and factor of one half in the second line. This last equation gives a relation between the magnitudes of the velocity vectors. By comparing the equation above to Pythagoras’ theorem, and by inspecting the triangle in Figure \(\PageIndex{5}\), it is clear that the triangle must be a right angle triangle, and thus that \(\vec v'_1\) and \(\vec v'_2\) must be perpendicular.
Frames of reference
review topics
Before proceeding, you may wish to review Sections 3.4 and 4.1 on expressing velocities in different frames of reference.
Because the momentum of a particle is defined using the velocity of the particle, its value depends on the reference frame in which we chose to measure that velocity. In some cases, it is useful to apply momentum conservation in a frame of reference where the total momentum of the system is zero. For example, consider two particles of mass \(m_1\) and \(m_2\) , moving towards each other with velocities \(\vec v_1\) and \(\vec v_2\) , respectively, as measured in a frame of reference \(S\) , as illustrated in Figure 10.2.6.
In the frame of reference \(S\) , the total momentum, \(\vec P\) , of the two particles can be written:
\[\begin{aligned} \vec P = m_1\vec v_1 + m_2\vec v_2\end{aligned}\]
Consider a frame of reference, \(S'\) , that is moving with velocity, \(\vec v_{CM}\) , relative to the frame of reference \(S\) . In that frame of reference, the velocities of the two particles are different and given by:
\[\begin{aligned} \vec v'_1&=\vec v_1- \vec v_{CM}\\[4pt] \vec v'_2&=\vec v_2- \vec v_{CM}\end{aligned}\]
The total momentum, \(\vec P'\) , in the frame of reference \(S'\) is then given by 1 :
\[\begin{aligned} \vec P' &= m_1\vec v'_1 + m_2 \vec v'_2\\[4pt] &=m_1(\vec v_1- \vec v_{CM})+m_2(\vec v_2- \vec v_{CM})\\[4pt] &= m_1\vec v_1 + m_2\vec v_2 - (m_1+m_2) \vec v_{CM}\end{aligned}\]
We can choose the velocity of the frame \(S'\) , \(\vec v_{CM}\) , such that the total momentum in that frame of reference is zero:
\[\begin{aligned} \vec P' &= 0\\[4pt] m_1\vec v_1 + m_2\vec v_2 - (m_1+m_2) \vec v_{CM} &=0\\[4pt] \therefore \vec v_{CM} &= \frac{m_1\vec v_1 + m_2\vec v_2 }{m_1+m_2}\end{aligned}\]
This “special” frame of reference, in which the total momentum of the system is zero, is called the “center of mass frame of reference”. The velocity of center of mass frame of reference can easily be obtained if there are \(N\) particles involved instead of two:
\[\therefore \vec v_{CM}=\frac{m_{1}\vec v_{1}+m_{2}\vec v_{2}+m_{3}\vec v_{3}+...}{m_{1}+m_{2}+m_{3}+...}=\frac{\sum m_{i}\vec v_{i}}{\sum m_{i}}\]
Again, you should note that because the above equation is a vector equation, it represents one equation per component of the vectors. For example, the \(x\) component of the velocity of the center of mass frame of reference is given by:
\[\begin{aligned} \therefore v_{CMx} = \frac{m_1 v_{1x} + m_2v_{2x} + m_3 v_{3x} + \dots }{m_1+m_2+m_3+\dots}=\frac{\sum m_iv_{ix}}{\sum m_i}\end{aligned}\]
Example \(\PageIndex{5}\)
In the frame of reference of a lab, a block of mass \(m\) has a velocity \(\vec v_1\) directed along the positive \(x\) axis and is approaching a second block of mass \(m\) that is at rest ( \(\vec v_2=0\) ), as shown in Figure \(\PageIndex{7}\). What is the velocity of the center of mass frame? What is the velocity of each block in the center of mass frame? Verify that the total momentum is zero in the center of mass frame.
Solution
Since this is a one dimensional situation, we only need to evaluate the \(x\) component of the velocity of the center of mass:
\[\begin{aligned} \vec v_{CM} &= \frac{m_1\vec v_1 + m_2\vec v_2 }{m_1+m_2}\\[4pt] \therefore v_{CMx} &= \frac{m_1 v_{1x} + m_2 v_{2x}}{m_1+m_2}\\[4pt] &=\frac{mv_1 + m(0) }{m+m}\\[4pt] &=\frac{1}{2}v_1\end{aligned}\]
The center of mass frame of reference is thus also moving along the positive direction of the \(x\) axis, but with a speed that is half of that of the moving block. In the center of mass frame of reference, it appears that the block on the left is slower than in the lab frame and that the block on the right is moving in the negative \(x\) direction. The velocities of the two blocks in the center of mass frame of reference are given by:
\[\begin{aligned} v'_1&=v_1-v_{CMx}=\frac{1}{2}v_1\\[4pt] v'_2&=(0)-v_{CMx}=-\frac{1}{2}v_1\end{aligned}\]
Thus, in the reference frame of the center of mass, the two block are approaching each other with the same speed ( \(v_1/2\) ), which is only the case because the two blocks have the same mass. The blocks, as viewed in the center of mass frame of reference, are shown in Figure \(\PageIndex{8}\).
Clearly, the total momentum is zero in the center of mass frame of reference:
\[\begin{aligned} \vec P' = m\vec v'_1+ m\vec v'_2 = m \left(\frac{1}{2}\vec v_1 - \frac{1}{2}\vec v_1\right) = 0\end{aligned}\]
Discussion
As we have seen, in the center of mass frame of reference the total momentum is zero. If there are only two particles, and they have the same mass, then, in the center of mass frame of reference, they both have the same speed and move either towards or away from each other.
Footnotes
1. Note that we are using primes ( \('\) ) to denote quantities in a different reference frame, not after a collision.


6.4: Finding Orthogonal Bases
\(\newcommand{\threevec}[3]{\begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix} } \)
\(\newcommand{\twovec}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix} } \)
The last section demonstrated the value of working with orthogonal, and especially orthonormal, sets. If we have an orthogonal basis \(\mathbf w_1,\mathbf w_2,\ldots,\mathbf w_n\) for a subspace \(W\text{,}\) the Projection Formula 6.3.15 tells us that the orthogonal projection of a vector \(\mathbf b\) onto \(W\) is
An orthonormal basis \(\mathbf u_1,\mathbf u_2,\ldots,\mathbf u_n\) is even more convenient: after forming the matrix \(Q=\begin{bmatrix} \mathbf u_1 & \mathbf u_2 & \ldots & \mathbf u_n \end{bmatrix}\text{,}\) we have \(\widehat{\mathbf{b}} = QQ^T\mathbf b\text{.}\)
In the examples we've seen so far, however, orthogonal bases were given to us. What we need now is a way to form orthogonal bases. In this section, we'll explore an algorithm that begins with a basis for a subspace and creates an orthogonal basis. Once we have an orthogonal basis, we can scale each of the vectors appropriately to produce an orthonormal basis.
Suppose we have a basis for \(\mathbb R^2\) consisting of the vectors
as shown in Figure 6.4.1. Notice that this basis is not orthogonal.
- Find the vector \(\widehat{\mathbf{v}}_2\) that is the orthogonal projection of \(\mathbf v_2\) onto the line defined by \(\mathbf v_1\text{.}\)
- Explain why \(\mathbf v_2 - \widehat{\mathbf{v}}_2\) is orthogonal to \(\mathbf v_1\text{.}\)
- Define the new vectors \(\mathbf w_1=\mathbf v_1\) and \(\mathbf w_2=\mathbf v_2-\widehat{\mathbf{v}}_2\) and sketch them in Figure 6.4.2. Explain why \(\mathbf w_1\) and \(\mathbf w_2\) define an orthogonal basis for \(\mathbb R^2\text{.}\)
- Write the vector \(\mathbf b=\twovec8{-10}\) as a linear combination of \(\mathbf w_1\) and \(\mathbf w_2\text{.}\)
- Scale the vectors \(\mathbf w_1\) and \(\mathbf w_2\) to produce an orthonormal basis \(\mathbf u_1\) and \(\mathbf u_2\) for \(\mathbb R^2\text{.}\)
Subsection 6.4.1 Gram-Schmidt orthogonalization
The preview activity illustrates the main idea behind an algorithm, known as Gram-Schmidt orthogonalization , that begins with a basis for some subspace of \(\mathbb R^m\) and produces an orthogonal or orthonormal basis. The algorithm relies on our construction of the orthogonal projection. Remember that we formed the orthogonal projection \(\widehat{\mathbf{b}}\) of \(\mathbf b\) onto a subspace \(W\) by requiring that \(\mathbf b-\widehat{\mathbf{b}}\) is orthogonal to \(W\) as shown in Figure 6.4.3.
This observation guides our construction of an orthogonal basis for it allows us to create a vector that is orthogonal to a given subspace. Let's see how the Gram-Schmidt algorithm works.
Suppose that \(W\) is a three-dimensional subspace of \(\mathbb R^4\) with basis:
We can see that this basis is not orthogonal by noting that \(\mathbf v_1\cdot\mathbf v_2 = 8\text{.}\) Our goal is to create an orthogonal basis \(\mathbf w_1\text{,}\) \(\mathbf w_2\text{,}\) and \(\mathbf w_3\) for \(W\text{.}\)
To begin, we declare that \(\mathbf w_1=\mathbf v_1\text{,}\) and we call \(W_1\) the line defined by \(\mathbf w_1\text{.}\)
- Find the vector \(\widehat{\mathbf{v}}_2\) that is the orthogonal projection of \(\mathbf v_2\) onto \(W_1\text{,}\) the line defined by \(\mathbf w_1\text{.}\)
- Form the vector \(\mathbf w_2 = \mathbf v_2-\widehat{\mathbf{v}}_2\) and verify that it is orthogonal to \(\mathbf w_1\text{.}\)
-
Explain why \(Span \{\mathbf w_1,\mathbf w_2 \} = Span \{\mathbf v_1,\mathbf v_2 \}\) by showing that any linear combination of \(\mathbf v_1\) and \(\mathbf v_2\) can be written as a linear combination of \(\mathbf w_1\) and \(\mathbf w_2\) and vice versa.
- The vectors \(\mathbf w_1\) and \(\mathbf w_2\) are an orthogonal basis for a two-dimensional subspace \(W_2\) of \(\mathbb R^4\text{.}\) Find the vector \(\widehat{\mathbf{v}}_3\) that is the orthogonal projection of \(\mathbf v_3\) onto \(W_2\text{.}\)
- Verify that \(\mathbf w_3 = \mathbf v_3-\widehat{\mathbf{v}}_3\) is orthogonal to both \(\mathbf w_1\) and \(\mathbf w_2\text{.}\)
- Explain why \(\mathbf w_1\text{,}\) \(\mathbf w_2\text{,}\) and \(\mathbf w_3\) form an orthogonal basis for \(W\text{.}\)
- Now find an orthonormal basis for \(W\text{.}\)
As this activity illustrates, Gram-Schmidt orthogonalization begins with a basis \(\mathbf v_1\mathbf v_2,\ldots,\mathbf v_n\) for a subspace \(W\) of \(\mathbb R^m\) and creates an orthogonal basis for \(W\text{.}\) Let's work through a second example.
Let's start with the basis
which is a basis for \(\mathbb R^3\text{.}\)
To get started, we'll simply set \(\mathbf w_1=\mathbf v_1=\threevec{2}{-1}2\text{.}\) We construct \(\mathbf w_2\) from \(\mathbf v_2\) by subtracting its orthogonal projection onto \(W_1\text{,}\) the line defined by \(\mathbf w_1\text{.}\) This gives
Notice that we found \(\mathbf v_2 = -\mathbf w_1 + \mathbf w_2\text{.}\) Therefore, we can rewrite any linear combination of \(\mathbf v_1\) and \(\mathbf v_2\) as
a linear combination of \(\mathbf w_1\) and \(\mathbf w_2\text{.}\) This tells us that
In other words, \(\mathbf w_1\) and \(\mathbf w_2\) is a basis for the same 2-dimensional subsapce as \(\mathbf v_1\) and \(\mathbf v_2\text{.}\)
Finally, we form \(\mathbf w_3\) from \(\mathbf v_3\) by subtracting its orthogonal projection onto \(W_2\text{:}\)
We can now check that
is an orthogonal set. Furthermore, we find that, as before, \(Span \{\mathbf w_1,\mathbf w_2,\mathbf w_3 \} = Span \{\mathbf v_1,\mathbf v_2,\mathbf v_3 \}\) so that we have found a new orthogonal basis for \(\mathbb R^3\text{.}\)
To create an orthonormal basis, we form unit vectors parallel to each of the vectors in the orthogonal basis:
More generally, if we have a basis \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) for a subspace \(W\) of \(\mathbb R^m\text{,}\) the Gram-Schmidt algorithm creates an orthogonal basis for \(W\) in the following way:
From here, we may form an orthonormal basis by constructing a unit vector parallel to each vector in the orthogonal basis: \(\mathbf u_j = 1/\len{\mathbf w_j}~\mathbf w_j\text{.}\)
Sage can automate these computations for us. Before we begin, however, it will be helpful to understand how we can combine things using a
list
in Python. For instance, if the vectors
v1
,
v2
, and
v3
form a basis for a subspace, we can bundle them together using square brackets:
[v1, v2, v3]
. Furthermore, we could assign this to a variable, such as
basis = [v1, v2, v3]
.
Evaluating the following cell will load in some special commands.
-
There is a command to apply the projection formula:
projection(b, basis)
returns the orthogonal projection ofb
onto the subspace spanned bybasis
, which is a list of vectors. -
The command
unit(w)
returns a unit vector parallel tow
. -
Given a collection of vectors, say,
v1
andv2
, we can form the matrix whose columns arev1
andv2
usingmatrix([v1, v2]).T
. When given alist
of vectors, Sage constructs a matrix whose rows are the given vectors. For this reason, we need to apply the tranpose.
Let's now consider \(W\text{,}\) the subspace of \(\mathbb R^5\) having basis
-
Apply the Gram-Schmidt algorithm to find an orthogonal basis \(\mathbf w_1\text{,}\) \(\mathbf w_2\text{,}\) and \(\mathbf w_3\) for \(W\text{.}\)
-
Find \(\widehat{\mathbf{b}}\text{,}\) the orthogonal projection of \(\mathbf b = \fivevec{-5}{11}0{-1}5\) onto \(W\text{.}\)
-
Explain why we know that \(\widehat{\mathbf{b}}\) is a linear combination of the original vectors \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\) and then find weights so that
\begin{equation*} \widehat{\mathbf{b}} = c_1\mathbf v_1 + c_2\mathbf v_2 + c_3\mathbf v_3. \end{equation*}
-
Find an orthonormal basis \(\mathbf u_1\text{,}\) \(\mathbf u_2\text{,}\) for \(\mathbf u_3\) for \(W\) and form the matrix \(Q\) whose columns are these vectors.
- Find the product \(Q^TQ\) and explain the result.
- Find the matrix \(P\) that projects vectors orthogonally onto \(W\) and verify that \(P\mathbf b\) gives \(\widehat{\mathbf{b}}\text{,}\) the orthogonal projection that you found earlier.
Subsection 6.4.2 \(QR\) factorizations
Now that we've seen how the Gram-Schmidt algorithm forms an orthonormal basis for a given subspace, we will explore how the algorithm leads to an important matrix factorization known as the \(QR\) factorization.
Suppose that \(A\) is the \(4\times3\) matrix whose columns are
These vectors form a basis for \(W\text{,}\) the subspace of \(\mathbb R^4\) that we encountered in Activity 6.4.2. Since these vectors are the columns of \(A\text{,}\) we have \(Col(A) = W\text{.}\)
-
When we implemented Gram-Schmidt, we first found an orthogonal basis \(\mathbf w_1\text{,}\) \(\mathbf w_2\text{,}\) and \(\mathbf w_3\) using
\begin{equation*} \begin{aligned} \mathbf w_1 & = \mathbf v_1 \\ \mathbf w_2 & = \mathbf v_2 - \frac{\mathbf v_2\cdot\mathbf w_1}{\mathbf w_1\cdot\mathbf w_1}\mathbf w_1 \\ \mathbf w_3 & = \mathbf v_3 - \frac{\mathbf v_3\cdot\mathbf w_1}{\mathbf w_1\cdot\mathbf w_2}\mathbf w_1 - \frac{\mathbf v_3\cdot\mathbf w_2}{\mathbf w_2\cdot\mathbf w_2}\mathbf w_2\text{.} \\ \end{aligned} \end{equation*}
Use these expressions to write \(\mathbf v_1\text{,}\) \(\mathbf v_1\text{,}\) and \(\mathbf v_3\) as linear combinations of \(\mathbf w_1\text{,}\) \(\mathbf w_2\text{,}\) and \(\mathbf w_3\text{.}\)
-
We next normalized the orthogonal basis \(\mathbf w_1\text{,}\) \(\mathbf w_2\text{,}\) and \(\mathbf w_3\) to obtain an orthonormal basis \(\mathbf u_1\text{,}\) \(\mathbf u_2\text{,}\) and \(\mathbf u_3\text{.}\)
Write the vectors \(\mathbf w_i\) as scalar multiples of \(\mathbf u_i\text{.}\) Then use these expressions to write \(\mathbf v_1\text{,}\) \(\mathbf v_1\text{,}\) and \(\mathbf v_3\) as linear combinations of \(\mathbf u_1\text{,}\) \(\mathbf u_2\text{,}\) and \(\mathbf u_3\text{.}\)
-
Suppose that \(Q = \left[ \begin{array}{ccc} \mathbf u_1 & \mathbf u_2 & \mathbf u_3 \end{array} \right]\text{.}\) Use the result of the previous part to find a vector \(\rvec_1\) so that \(Q\rvec_1 = \mathbf v_1\text{.}\)
-
Then find vectors \(\rvec_2\) and \(\rvec_3\) such that \(Q\rvec_2 = \mathbf v_2\) and \(Q\rvec_3 = \mathbf v_3\text{.}\)
-
Construct the matrix \(R = \left[ \begin{array}{ccc} \rvec_1 & \rvec_2 & \rvec_3 \end{array} \right]\text{.}\) Remembering that \(A = \left[ \begin{array}{ccc} \mathbf v_1 & \mathbf v_2 & \mathbf v_3 \end{array} \right]\text{,}\) explain why \(A = QR\text{.}\)
- What is special about the shape of \(R\text{?}\)
- Suppose that \(A\) is a \(10\times 6\) matrix whose columns are linearly independent. This means that the columns of \(A\) form a basis for \(W=Col(A)\text{,}\) a 6-dimensional subspace of \(\mathbb R^{10}\text{.}\) Suppose that we apply Gram-Schmidt orthogonalization to create an orthonormal basis whose vectors form the columns of \(Q\) and that we write \(A=QR\text{.}\) What are the dimensions of \(Q\) and what are the dimensions of \(R\text{?}\)
When the columns of a matrix \(A\) are linearly independent, they form a basis for \(Col(A)\) so that we can perform the Gram-Schmidt algorithm. The previous activity shows how this leads to a factorization of \(A\) as the product of a matrix \(Q\) whose columns are an orthonormal basis for \(Col(A)\) and an upper triangular matrix \(R\text{.}\)
We'll consider the matrix \(A=\begin{bmatrix} 2 & -3 & -2 \\ -1 & 3 & 7 \\ 2 & 0 & 1 \\ \end{bmatrix}\) whose columns, which we'll denote \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{,}\) are the basis of \(\mathbb R^3\) that we considered in Example 6.4.4. There we found an orthogonal basis \(\mathbf w_1\text{,}\) \(\mathbf w_2\text{,}\) and \(\mathbf w_3\) that satisfied
In terms of the resulting orthonormal basis \(\mathbf u_1\text{,}\) \(\mathbf u_2\text{,}\) and \(\mathbf u_3\text{,}\) we had
so that
Therefore, if \(Q=\begin{bmatrix} \mathbf u_1 & \mathbf u_2 & \mathbf u_3 \end{bmatrix}\text{,}\) we have the \(QR\) factorization
As before, we would like to use Sage to automate the process of finding and using the \(QR\) factorization of a matrix \(A\text{.}\) Evaluating the following cell provides a command
QR(A)
that returns the factorization, which may be stored using, for example,
Q, R = QR(A)
.
Suppose that \(A\) is the following matrix whose columns are linearly independent.
- If \(A=QR\text{,}\) what are the dimensions of \(Q\) and \(R\text{?}\) What is special about the form of \(R\text{?}\)
-
Find the \(QR\) factorization using
Q, R = QR(A)
and verify that \(R\) has the predicted shape and that \(A=QR\text{.}\) - Find the matrix \(P\) that orthogonally projects vectors onto \(Col(A)\text{.}\)
- Find \(\widehat{\mathbf{b}}\text{,}\) the orthogonal projection of \(\mathbf b=\fourvec4{-17}{-14}{22}\) onto \(Col(A)\text{.}\)
- Explain why the equation \(A\mathbf x=\widehat{\mathbf{b}}\) must be consistent and then find \(\mathbf x\text{.}\)
In fact, Sage provides its own version of the \(QR\) factorization that is a bit different than the way we've developed the factorization here. For this reason, we have provided our own version of the factorization.
Subsection 6.4.3 Summary
This section explored the Gram-Schmidt orthogonalization algorithm and how it leads to the matrix factorization \(A=QR\) when the columns of \(A\) are linearly independent.
-
Beginning with a basis \(\mathbf v_1, \mathbf v_2,\ldots,\mathbf v_n\) for a subspace \(W\) of \(\mathbb R^m\text{,}\) the vectors
\begin{align*} \mathbf w_1 & = \mathbf v_1\\ \mathbf w_2 & = \mathbf v_2 - \frac{\mathbf v_2\cdot\mathbf w_1}{\mathbf w_1\cdot\mathbf w_1}\mathbf w_1\\ \mathbf w_3 & = \mathbf v_3 - \frac{\mathbf v_3\cdot\mathbf w_1}{\mathbf w_1\cdot\mathbf w_1}\mathbf w_1 - \frac{\mathbf v_3\cdot\mathbf w_2}{\mathbf w_2\cdot\mathbf w_2}\mathbf w_2\\ & \vdots\\ \mathbf w_n & = \mathbf v_n - \frac{\mathbf v_n\cdot\mathbf w_1}{\mathbf w_1\cdot\mathbf w_1}\mathbf w_1 - \frac{\mathbf v_n\cdot\mathbf w_2}{\mathbf w_2\cdot\mathbf w_2}\mathbf w_2 - \ldots - \frac{\mathbf v_n\cdot\mathbf w_{n-1}} {\mathbf w_{n-1}\cdot\mathbf w_{n-1}}\mathbf w_{n-1} \end{align*}form an orthogonal basis for \(W\text{.}\)
- We may scale each vector \(\mathbf w_i\) appropriately to obtain an orthonormal basis \(\mathbf u_1,\mathbf u_2,\ldots,\mathbf u_n\text{.}\)
- Expressing the Gram-Schmidt algorithm in matrix form shows that, if the columns of \(A\) are linearly independent, then we can write \(A=QR\text{,}\) where the columns of \(Q\) form an orthonormal basis for \(Col(A)\) and \(R\) is upper triangular.
Exercises 6.4.4 Exercises
Suppose that a subspace \(W\) of \(\mathbb R^3\) has a basis formed by
- Find an orthogonal basis for \(W\text{.}\)
- Find an orthonormal basis for \(W\text{.}\)
- Find the matrix \(P\) that projects vectors orthogonally onto \(W\text{.}\)
- Find the orthogonal projection of \(\threevec34{-2}\) onto \(W\text{.}\)
Find the \(QR\) factorization of \(A=\begin{bmatrix} 4 & 7 \\ -2 & 4 \\ 4 & 4 \end{bmatrix} \text{.}\)
Consider the basis of \(\mathbb R^3\) given by the vectors
- Apply the Gram-Schmit orthogonalization algorithm to find an orthonormal basis \(\mathbf u_1\text{,}\) \(\mathbf u_2\text{,}\) \(\mathbf u_3\) for \(\mathbb R^3\text{.}\)
- If \(A\) is the \(3\times3\) whose columns are \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{,}\) find the \(QR\) factorization of \(A\text{.}\)
-
Suppose that we want to solve the equation \(A\mathbf x=\mathbf b = \threevec{-9}17\text{,}\) which we can rewrite as \(QR\mathbf x = \mathbf b\text{.}\)
- If we set \(\yvec=R\mathbf x\text{,}\) explain why the equation \(Q\yvec=\mathbf b\) is computationally easy to solve.
- Explain why the equation \(R\mathbf x=\yvec\) is computationally easy to solve.
- Find the solution \(\mathbf x\text{.}\)
Consider the vectors
and the subspace \(W\) of \(\mathbb R^5\) that they span.
- Find an orthonormal basis for \(W\text{.}\)
- Find the \(5\times5\) matrix that projects vectors orthogonally onto \(W\text{.}\)
- Find \(\widehat{\mathbf{b}}\text{,}\) the orthogonal projection of \(\mathbf b=\fivevec{-8}3{-12}8{-4}\) onto \(W\text{.}\)
- Express \(\widehat{\mathbf{b}}\) as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{.}\)
Consider the set of vectors
- What happens when we apply the Gram-Schmit orthogonalization algorithm?
- Why does the algorithm fail to produce an orthogonal basis for \(\mathbb R^3\text{?}\)
Suppose that \(A\) is a matrix with linearly independent columns and having the factorization \(A=QR\text{.}\) Determine whether the following statements are true or false and explain your thinking.
- It follows that \(R=Q^TA\text{.}\)
- The matrix \(R\) is invertible.
- The product \(Q^TQ\) projects vectors orthogonally onto \(Col(A)\text{.}\)
- The columns of \(Q\) are an orthogonal basis for \(Col(A)\text{.}\)
-
The orthogonal complement \(Col(A)^\perp = Nul(Q^T)\text{.}\)
Suppose we have the \(QR\) factorization \(A=QR\text{,}\) where \(A\) is a \(7\times 4\) matrix.
- What are the dimensions of the product \(QQ^T\text{?}\) Explain the significance of this product.
- What are the dimensions of the product \(Q^TQ\text{?}\) Explain the significance of this product.
- What are the dimensions of the matrix \(R\text{?}\)
- If \(R\) is a diagonal matrix, what can you say about the columns of \(A\text{?}\)
Suppose we have the \(QR\) factorization \(A=QR\) where the columns of \(A\) are \(\avec_1,\avec_2,\ldots,\avec_n\) and the columns of \(R\) are \(\rvec_1,\rvec_2,\ldots,\rvec_n\text{.}\)
- How can the matrix product \(A^TA\) be expressed in terms of dot products?
- How can the matrix product \(R^TR\) be expressed in terms of dot products?
- Explain why \(A^TA=R^TR\text{.}\)
-
Explain why the dot product \(\avec_i\cdot\avec_j = \rvec_i\cdot\rvec_j\text{.}\)


11.6: The Probabilistic Method
At the outset of this chapter, we presented Erdős' original proof for the lower bound for the Ramsey number \(R(n,n)\) using counting. Later, we recast the proof in a probabilistic setting. History has shown that this second perspective is the right one. To illustrate the power of this approach, we present a classic theorem, which is also due to Erdős, showing that there are graphs with large girth and large chromatic number.
The girth \(g\) of a graph \(G\) is the smallest integer for which \(G\) contains a cycle on \(g\) vertices. The girth of a forest is taken to be infinite, while the girth of a graph is three if and only if it has a triangle. You can check the families of triangle-free, large chromatic number, graphs constructed in Chapter 5 and see that each has girth four.
For every pair \(g,t\) of integers with \(g \geq 3\) , there exists a graph \(G\) with \(\chi (G)>t\) and the girth of \(G\) greater than \(g\) .
- Proof
-
Before proceeding with the details of the argument, let's pause to get the general idea behind the proof. We choose integers \(n\) and \(s\) with \(n>s\), and it will eventually be clear how large they need to be in terms of \(g\) and \(t\). We will then consider a random graph on vertex set \(\{1,2,…,n\}\), and just as before, for each \(i\) and \(j\) with \(1≤i<j≤n\), the probability that the pair \(ij\) is an edge is \(p\), but now \(p\) will depend on \(n\). Of course, the probability that any given pair is an edge is completely independent of all other pairs.
Our first goal is to choose the values of \(n, s\) and \(p\) so that with high probability, a random graph does not have an independent set of size \(s\). You might think as a second goal, we would try to get a random graph without small cycles. But this goal is too restrictive. Instead, we just try to get a graph in which there are relatively few small cycles. In fact, we want the number of small cycles to be less than \(n/2\). Then we will remove one vertex from each small cycles, resulting in a graph with at least \(n/2\) vertices, having no small cycles and no independent set of size \(s\). The chromatic number of this graph is at least \(n/2s\), so we will want to have the inequality \(n>2st\).
Now for some details. Let \(X_1\) be the random variable that counts the number of \(s\)-element independent sets. Then
\(E(X_1) = \dbinom{n}{s} (1-p)^{C(s,2)}\)
Now we want \(E(X_1)<1/4\). Since \(C(n,s)≤n^s=e^{s \ln n}\) and \((1−p)^{C(s,2)}≤e^{−ps^2/2}\), it suffices to set \(s=2 \ln n/p\). By Markov's Inequality , the probability that \(X_1\) exceeds \(1/2≥2E(X_1)\) is less than 1/2.
Now let \(X_2\) count the number of cycles in \(G\) of size at most \(g\). Then
\(E(X_2) \leq \displaystyle \sum_{i=3}^g n(n-1)(n-2) ... (n-i+1)p^i \leq g(pn)^g\).
Now, we want \(E(X_2)≤n/4\), and an easy calculation shows that \(g(np)^g≤n/4\) when \(p=n^1/g^{−1}/10\). Again by Markov's Inequality , the probability that \(X_2\) exceeds \(n/2≥2E(X_2)\) is less than 1/2.
We conclude that there is a graph \(G\) for which \(X_1=0\) and \(X_2≤n/2\). Remove a vertex from each of the small cycles in \(G\) and let \(H\) be the graph that remains. Clearly, \(H\) has at least \(n/2\) vertices, no cycle of size at most \(g\) and no independent set of size \(s\). Finally, the inequality \(n>2st\) requires \(n^{1/g}/(40 \ln n) > t\).
11.6.1 Gaining Intuition with the Probabilistic Method
Experienced researchers are able to simplify the calculations in an argument of this type, as they know what can safely be discarded and what can not. Here's a quick tour of the essential steps. We want \(E(X_1)\) to be small, so we set \(n^se^{−ps^2}=1\) and get \(s= \ln n/p\). We want the number of small cycles to be about \(n\) so we set \((gp)^g=n\) and get \(p=n^1/g^{−1}\). Finally, we want \(n=st\) which requires \(n^{1/g}=t\). The rest is just paying attention to details.


4.1: Introduction to Solvents
In a very general sense, solvents are a class of chemical compounds that allow chemistry to occur. The concept of a solvent has significant ramifications because they serve as the matrix, medium, or carrier for solutes. They are necessary in a number of processes, reactions, and systems. We tend to think of a substance like “water” as a universal solvent because it is so useful in so many disciplines. Water cleans up everything, allows biochemical reactions to occur, is used in paints, coatings, and films, allows cooking to occur (or else everything would catch fire), and provides lubrication and ease of movement for a great many devices.
Interestingly, in the business world, the word “solvent” has another meaning: A state of financial soundness characterized by the ability of an entity to meet its monetary obligations when they fall due. This latter definition does indeed apply in a certain sense to the solvent concept we are espousing in this chapter. A solvent ($olvent) carries materials for a reaction or specific function in the same sense that solvent from a financial sense carries funding/money to meet its obligations. Thus, using an analogy:
Solvent : Solute :: $olvent : Money
Chemically, a solvent will dissolve or “solvate” a solute. What does that mean? It means that solvent molecules will surround the solute in such a way that a solution is formed; in other words, a homogeneous system is generated in which the solute is part of/indistinguishable from the solvent network. A solute/solution conceptcan be visualized in the Figure below:
Solubility
The ability of one a compound to dissolve in some other (likely different) compound is termed “solubility”. Miscibility is another term that characterizes the facility of compound A to dissolve in compound B. When the two compounds can completely dissolve or combine to form a homogeneous solution, the two liquids are said to be miscible. Two that can never blend well enough to form a solution are called immiscible.
All solutions have a positive entropy of mixing whereas the interactions between different compounds may or may not be energetically favored. If interactions are unfavorable, then the free energy decreases with increasing solute concentration. The energy loss may at some point in time outweighs the entropy gain, and no more solute can be dissolved – this is a condition solution in which the solution is said to be saturated. This condition can change with different environmental factors, such temperature, pressure, and purity of the system. A supersaturated solution can be prepared by raising the solubility (e.g., increasing temperature) to dissolve more solute, and then lowering it by cooling. However, most gases and several compounds exhibit solubilities that tend to drop with raised temperatures. The solubility of liquids in liquids, however, is less temperaturesensitive than solids or gases.


1.13: Energy and Momentum Conservation
The lack of energy conservation in an expanding universe is quite surprising to people with any training in physics and therefore merits some discussion, which we present here in this chapter. The student could skip this chapter and proceed to 15 without serious harm. If, subsequently, the lack of energy conservation becomes too troubling, know that this chapter is here for you.
We begin by reminding the reader of the deep connection between symmetries of the action and conserved quantities. For example, if the action (time-integral of the Lagrangian) is invariant under time translations (and hence a symmetry of the action) then energy is conserved. Likewise, if spatial translations do not change the action, then momentum is conserved.
To review, let us consider a single particle moving in one dimension in a possibly space and time-dependent external potential with Lagrangian:
\[L(x, \dot x, t) = \dfrac{1}{2}m{\dot x}^2 - V(x,t)\]
The action is given by integrating along a trajectory between two points fixed in space and time, points 1 and 2:
\[S = \int_{1}^{2}L(x,\dot x, t) dt\]
Although it is not obvious from the notation, it depends on an assumed \(x(t)\). Invariance of the action under time translation means that if we send \(t\) to \(t+\delta t\), the resulting change in \(S\), \(\delta S\), will be zero. It should be clear that this will be the case as long as \(L\) has no explicit time dependence; i.e., if \(\partial L/\partial t = 0\). Likewise, invariance of the action under spatial translation means that if we send \(x\) to \(x + \delta x\), the resulting change in \(S\), \(\delta S\), will be zero. This will be the case if \(\partial L/\partial x = 0\).
From the Euler-Lagrange equations one can now see directly that space-translation invariance leads to momentum conservation. By definition, the momentum conjugate to x is \(p = \partial L/\partial \dot x\). For our particular Lagrangian, that gives \(p=m\dot x\) as expected. The Euler Lagrange equations are:
\[\frac{d}{dt}\left(\dfrac{\partial L}{\partial \dot x}\right) = \dfrac{\partial L}{\partial x}\]
which for our Lagrangian becomes:
\[\frac{d}{dt}p = -\dfrac{\partial V(x,t)}{\partial x}\]
So we see that \(p\) does not change with time if the potential has no dependence on \(x\). Of course! If the potential has no dependence on \(x\) then it is not applying any force to the particle.
Seeing the consequences of time-translation invariance takes a little bit more work. However, from the Euler-Lagrange equations one can derive:
\[\dfrac{d}{dt}\left(\dot x \partial L/\partial \dot x -L\right) = -\partial L/\partial t\]
so we see that the quantity after \(d/dt\) on the left-hand side is conserved if \(L\) has no explicit time dependence. As long as the kinetic energy is quadratic in the velocity, and the potential does not depend on the velocity, then that quantity is equal to the total energy. Thus, given these conditions, energy conservation follows from time translation invariance.
For our Lagrangian the above equation becomes:
\[\dfrac{d}{dt}\left(\frac{1}{2}m{\dot x}^2+V(x,t)\right) = \partial V(x,t)/\partial t\]
Thus we see total energy is conserved if the potential energy function does not have a dependence on \(t\). Note that this is not a statement that the potential energy of the particle cannot change with time (which would be \(dV/dt = 0\)) but is instead a statement about the functional form of \(V(x,t)\), that it cannot have an explicit dependence on time. For example, \(V(x) = 1/2 kx^2\), has no explicit time dependence (\(\partial V/\partial t = 0\)) even though \(x\) will change with time (so \(dV/dt = kx \dot x \ne 0\)), so for this potential, energy is conserved. However, if the spring constant were to change with time so \(V(x) = 1/2 k(t)x^2\) then energy would not be conserved. Imagine the behavior of a mass attached to a spring with time-dependent spring constant. Can you see how energy would not be conserved?
Let's now consider how this works in an expanding universe. Consider a free non-relativistic particle of mass \(m\). If we adopt the point of view that the comoving grid everywhere specifies local rest, then we will naturally define kinetic energy based on departures from local rest.This leads us to write the velocity of the particle in terms of \(\dot x\) so that the velocity is \(a \dot x\) and the Lagrangian is \(L = 1/2 m a^2 {\dot x}^2\).
One might be tempted to use the physical coordinates \(y=ax\) and write \(v=\dot y\) and \(L=1/2m{\dot y}^2\) and therefore \(m\dot y\) is conserved. However, this Lagrangian would give kinetic energy to a particle that is stationary with respect to the local rest frame, as long as it is not at the origin. Further, i n a universe with \(\ ddot a \ne 0\) we run into the same problem as discussed in chapter 12 that the coordinate system is locally inertial near the origin but not globally inertial. The comoving coordinate system, in contrast, is globally inertial.
Let's therefore return to \(L = 1/2 m a^2 {\dot x}^2\) as our free-particle Lagrangian. Note that for this Lagrangian there is no explicit dependence on \(x\) so the momentum conjugate to \(x\) will be conserved. That momentum is \(p = \partial L/\partial \dot x = m a^2 \dot x\) = \(amv\) which is \(a\) times the usual momentum. Since \(amv\) is conserved, we find \(mv \propto 1/a\).
Thus, in an expanding homogeneous universe, spatial homogeneity; i.e., translational invariance, leads to a conserved quantity which is the usual momentum divided by the scale factor. Thus we see once again that free particles in an expanding universe have a momentum that decreases as \(1/a\).
What about energy, the conserved quantity associated with time translation invariance? We do not have time translation invariance because of the expansion! Consequently, energy is not conserved as one can see explicitly for our free particle from \(dE/dt = -\partial L/\partial t = -ma\dot a {\dot x}^2 \ne 0\). Of course this result is consistent with our finding that the momentum of the free particle decreases as the universe expands.
I wrote at the beginning of this chapter that the lack of energy conservation in the expanding universe comes as a surprise to many. I ran up against this when I wrote an article for Physics Today about cosmic inflation. In an initial draft I wrote, about the patch that eventually becomes our observable universe, "The total mass-energy in this patch was about \(10^4\) Joules, the caloric content of two diet Cokes." This line, which I loved, was then removed by the editor! I was very disappointed! The editor explained he was skeptical it was true and was afraid the readers would be as well. I tried to address the skepticism with this suggested replacement:
"The total mass-energy in this patch was about \(10^4\) Joules, the caloric content of two diet Cokes. (Surprising to most physicists is that there is no global conservation of energy in general relativity, a fact that allows for the existence today of regular Coke, too---as well as the other \(\sim 10^{70}\) Joules of mass-energy in the currently observable universe.)"
which had the added benefit of further play on the Coke theme! Fortunately, the editor was open minded and we settled on this for the final published version:
"The total mass-energy in that patch was about \(10^4\) Joules, the caloric content of two diet Cokes. Today's universe includes regular Cokes and about \(10^{70}\) joules of mass energy from other sources, a result compatible with the perhaps surprising fact that energy is not conserved in general relativity."
I also ran up against it a previous time I taught this class, when students pointed out to me that the textbook we were using for the course explicitly claimed energy was conserved. I pointed it out to the author, who quickly agreed with me it was an error. He was in fact quite unhappy he was finding this out just a little too late to keep the error from propagating into the next edition of his textbook.