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Time derivative of a vector

  1. Feb 18, 2015 #1
    In chapter 1 of the book "Introduction to Mechanics" by Kleppner and Kolenkow, the derivative of a generic vector ##\vec{A}## is discussed in terms of decomposing an increment in ##\vec{A}##, ##Δ\vec{A}##, into two perpendicular vector vectors; one parallel to ##\vec{A}## and the other perpendicular to ##\vec{A}##
    Two equations are then derived:
    $$|\frac{d\vec{A}_{perp}}{dt}| = A\frac{dθ}{dt}$$
    $$|\frac{d\vec{A}_{par}}{dt}| = \frac{dA}{dt}$$
    Do these equations fail if ##θ## or ##A## decrease with time (since the magnitude of a vector is always positive or zero)?
     
  2. jcsd
  3. Feb 18, 2015 #2

    Svein

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    No. A is allowed to decrease with time. When you use polar coordinates, you introduce a sort of singularity when A = 0. When A tries to go through 0, θ changes value abruptly (by π). A will keep on being positive.
     
  4. Feb 18, 2015 #3
    If A decreases with time, then ##\frac{dA}{dt}## is negative, right? But it also happens to be the magnitude of the vector I described earlier.
     
  5. Feb 18, 2015 #4
    Just because dA/dt is negative doesn't mean that A is negative.

    Chet
     
  6. Feb 18, 2015 #5
    Of course. But the vector I was referring to wasn't ##\vec{A}##, I was talking about the derivative of the vector component of ##\vec{A}## parallel to ##\vec{A}## (left hand side).
     
  7. Feb 18, 2015 #6
    The component of the derivative of ##\vec{A}## parallel to ##\vec{A}## also has a direction. That direction is either ##+\frac{\vec{A}}{|\vec{A}|}## or ##-\frac{\vec{A}}{|\vec{A}|}##, depending on the sign of dA/dt. The vector component of the derivative of the vector ##\vec{A}## with respect to time in the direction parallel to ##\vec{A}## is:

    $$\frac{dA}{dt}\frac{\vec{A}}{|\vec{A}|}$$

    Even if dA/dt is negative, the magnitude of this vector component is |dA/dt| (positive). The unit vector describing the direction of this component is equal to this vector divided by |dA/dt|. So the negative sign is associated with the unit vector and not the magnitude.

    Chet
     
  8. Feb 18, 2015 #7
    So if we want to be more technical, we should put absolute value bars on the right hand side of the equation, right?
    What about ##A\frac{d\theta}{dt}##? Does the same argument hold?
     
  9. Feb 18, 2015 #8
    Yes.
     
  10. Feb 18, 2015 #9
    Thanks! I am very curious as to where (and why) the author left out the absolute value bars in the derivation. I have attached two screenshots to this post.
     

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  11. Feb 18, 2015 #10
    He probably didn't think of it in the way that you have thought about it.

    Chet
     
  12. Feb 18, 2015 #11

    vanhees71

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    Ok, you forgot to tell us the all-clarifying point given in the book. The author considers a function ##\vec{A}:\mathbb{R} \rightarrow V## with ##|\vec{A}|=\text{const}##. This implies that
    $$|\vec{A}|^2=\vec{A} \cdot \vec{A}=\text{const} \; \Rightarrow \; \dot{\vec{A}} \cdot \vec{A}=0,$$
    i.e., that ##\vec{A} \perp \dot{\vec{A}}##.

    In more physical terms: If you take ##\vec{A}## as the position vector of a particle, ##|\vec{A}|=\text{const}## means that the particle is moving along a circle (or spherical shell). Then it's velocity is always perpendicular to the position, i.e., the radius vector of the circle/shell. Or more geometrically: The tangent vector to a circle or spherical shell is always perpendicular to the radius vector of this circle/shell. This is also well known from elementary Euclidean geometry!
     
  13. Feb 18, 2015 #12
    I agree. But how does this change anything?
     
  14. Feb 18, 2015 #13

    vanhees71

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    Change what? It's just the short and elegant analytical-geometry proof of the fact that the radius vector of a circle or a spherical shell is always perpendicular to all tangent vectors on this curve/surface.
     
  15. Feb 18, 2015 #14
    My bad, I misunderstood your post; I thought you were hinting at some relationship between what you were saying and what I was discussing earlier regarding absolute values.
     
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