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Concerning Vectors in Scalar Form

  1. Jun 21, 2015 #1
    Hey, all. I have a question concerning the treatment and use of vectors when solving problems (or in general, really).

    I know that vectors have both magnitude and direction, while scalars only have magnitude. However, in solving problems and looking at how others have solved them, I've noticed something: vectors seem to be often written in "scalar form," if you will. For example, consider circular motion

    [tex] \vec r = r(\cos \omega t \hat{\imath}+ \sin \omega t \hat{\jmath}) [/tex]
    [tex] | \vec r| = r = constant [/tex]
    [tex] \vec v = \frac {d \vec r}{dt} = r \omega (-\sin \omega t \hat{\imath} + \cos \omega t \hat{\jmath}) [/tex]

    This all makes total sense to me. Where I find trouble, however, is that when I look around, I see people solve problems by simply saying ##v = r \omega##. No vector notation, no mention of direction, etc. How do they know to do this, and how can I know when to do this as well? This is causing me an unbelievable amount of frustration. Thanks!
     
  2. jcsd
  3. Jun 21, 2015 #2

    ShayanJ

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    Gold Member

    They're working with the magnitude of the vectors i.e. ## \vec S=S_x \hat i+S_y \hat j \Rightarrow S\equiv |\vec S|=\sqrt{\vec S\cdot \vec S}=\sqrt{S_x^2+S_y^2} ## which is a scalar. Sometimes you don't need to care about the direction of the vector and considering its magnitude is enough. I suggest you post one problem where you've seen this in the homework section and ask people why you don't need to consider the direction of the vector. You'll understand things better.
     
  4. Jun 21, 2015 #3
    Right, I could tell they were dealing with the magnitude. I'll find a problem and post it later to figure out why this is the case. Thanks.
     
  5. Jun 21, 2015 #4
    In addition to what Shyan had to say, your initial equation can be expressed in polar coordinates using unit vectors in the r and theta directions, ##\hat{r}## and ##\hat{\theta}##.

    Using ##r=\sqrt{x^2+y^2}## and ##\tan{\theta}=x / y##, a generic vector in polar coordinates is expressed as ##\vec{r}=r \hat{r} + r \theta \hat{\theta}##.

    In your problem, ##\vec{r}=r \hat{r} + r \omega t \hat{\theta}##, where ##\omega = \frac{d\theta}{dt}##.

    If both ##r## and ##\omega## remain constant in time, ##\vec{v} = 0\hat{r} + r\omega \hat{\theta}## or ##\vec{v} = r\omega \hat{\theta}##.

    ##\vec{v}## and ##\hat{\theta}## point in the same direction so that ##v \hat{\theta}= r\omega \hat{\theta}##.

    We can divide through by the unit vector ##\hat{\theta}##, so that ##v=r\omega## but in the process lose directional information.

    The rather subtle point I am trying to relate, is that ##v=r\omega## is more than just an equation in vector magnitudes, it is also an equation between the elements of two vectors.
     
    Last edited: Jun 21, 2015
  6. Jun 23, 2015 #5
    Could you elaborate on what you mean by the "elements" of the vectors? Perhaps that's just terminology I'm not familiar with.
     
  7. Jun 23, 2015 #6
    I'm sorry. Element is very nonstandard usage. I find only two textual references: one to the components of a tensor and another to the components of a symbol. The former is probably in error, ;p.

    I should have said "components of a vector". For ##V = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}##, I refer to ##V_x##, ##V_y##, and ##V_z##.
     
  8. Jun 23, 2015 #7
    No worries! I figured that was what you meant, based on the equations. I just wanted to be sure.

    Out of curiosity, why do we do this if we do lose the directional information?
     
  9. Jun 23, 2015 #8
    No, problem. Thank's for asking. I'll wash my mouth out with proper-physics-terminology soap.

    Given just ##v=r\omega##, all we can infer is that v is a speed, not really a velocity in the ##\phi## direction. It could mean ##v\hat{r}=r\omega \hat{r}##, for all we know. We only know these are components of a vector pointing in the ##\phi## direction because of the problem set-up.

    Also, it could be for clockwise rather than counter clockwise rotation: ##\hat{v (-\phi)} = r\omega \hat{(-\phi)}##.
     
    Last edited: Jun 23, 2015
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