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Homework Help: Norm of a vector

  1. Mar 24, 2009 #1
    Ok, so I have no idea how to take the norm of a vector composed of two vectors. I have
    [tex]\vec{q}[/tex]=[tex]\vec{pi}[/tex] - [tex]\vec{pf}[/tex]

    we are given:

    so i know that
    |[tex]\vec{q}[/tex]| [tex]\neq[/tex] 0, that would be too easy, and it doesn't make sense.

    now, is the following right? it just doesn't seem to be

  2. jcsd
  3. Mar 24, 2009 #2


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    You can try working it out explicitly through the inner product.
    Note that
    [tex]|\vec q|^2 = \vec q \cdot \vec q[/tex]
    [tex](\vec p_i - \vec p_f) \cdot (\vec p_i - \vec p_f) = \ldots? [/tex]
  4. Mar 24, 2009 #3
    ok so that gets me |[tex]\vec{q}[/tex]|=[tex]\sqrt{|\vec{p_{i}^2}|+|\vec{p_{f}^2}|}[/tex],
    would that then be [tex]\sqrt{2}[/tex]p?
  5. Mar 24, 2009 #4

    Tom Mattson

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    No, it doesn't get you that. Not unless [itex]p_i[/itex] is orthogonal to [itex]p_f[/itex], and the problem statement (as you gave it) doesn't say that.
  6. Mar 24, 2009 #5


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    Please stop guessing and work it out properly.

    (\vec p_i - \vec p_f) \cdot (\vec p_i - \vec p_f) =
    (\vec p_i - \vec p_f) \cdot \vec p_i - (\vec p_i - \vec p_f) \cdot \vec p_f =
    \vec p_i \cdot \vec p_i + \cdots
  7. Mar 24, 2009 #6
    not guessing, just not typing it out entirely. i wrongly assumed pi and pf were orthogonal, but i was trying to make use of the fact that
    |p|=|pi|= |pf|.

    i know |q|=[tex]\sqrt{\vec{p_{i}^2} + \vec{p_{f}^2 }-2*\vec{p_i}*\vec{p_f}}[/tex]

    but i know |p|=|pi|= |pf|, so |p|[tex]^2[/tex] must equal |p|2=|pi|2 = |pf|2, hence, possibly with faulty reasoning,
  8. Mar 24, 2009 #7


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    You are getting closer, it's just not entirely clear to me what
    [itex]2 * \vec p_i * \vec p_f [/itex]
    means. In particular, what do the asterisks mean?

    After clearing that up, can you write that expression in terms of p as well?
  9. Mar 24, 2009 #8
    apologies, I began to get sloppy, as I'm still trying to get used to Latex, it should be
    |q|=[tex]\sqrt{2 p^2-2 \vec{p_i}\cdot\vec{p_f}}[/tex]. I would not know how to write it out in terms of p, and it might be useless in this case, as we are supposed to plug in a known value for p. [tex]\vec{q}[/tex] was mainly momentum transfer, and we were given that for this case [tex]\vec{q} = \vec{p_i} - \vec{p_f}[/tex]. i already turned in the assignment, just wanted to reach the answer for the sake of learning.
  10. Mar 25, 2009 #9


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    Yes, at first you will work more slowly when you have to do the math and learn how to write it out in [itex]\LaTeX[/itex] - compliments for doing it anyway :)

    What I was hinting at was to define [itex]\theta[/itex] as the angle between the ingoing and outgoing momentum, and writing
    [tex]\vec p_i \cdot \vec p_f = |\vec p_i| |\vec p_f| \cos\theta = 2 |\vec p| \cos\theta[/tex].

    Then you can take [itex]\sqrt{2}p[/itex] outside the square root, if you want to write it more beautifully.
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