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Derivative of a unit vector

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data
    the first derivative of a vector of constant magnitude is the cross product of the angular velocity
    of the vector(i.e , the angular velocity of the moving coordinate system ) and the vector itself.)

    2. Relevant equations

    di/dt= w x i , here w is the angular velocity and x is the cross product symbol

    3. The attempt at a solution
    dİ/dt=di/dΘ * dΘ/dt because i(unit vector in the direction i) is changing only in direction i.e it is
    rotating, only it's direction will change.its direction changes with Θ(the angle) and Θ changes with
    with the time. so there must be a chain rule. but I cant show that di/dt=wxi
  2. jcsd
  3. Jan 27, 2014 #2
    What is the definition of "the angular velocity of the vector"?
  4. Jan 27, 2014 #3


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    would be more explicit to write that unit vector using trig functions of the angle theta ... then take derivatives.
  5. Jan 28, 2014 #4
    unit vector is rotating so angular velocity must be dΘ/dt
  6. Jan 28, 2014 #5


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    It's much easier to prove that the time derivative of a unit vector is always perpendicular to itself. Then it's clear that the derivative must be of the form [itex]\dot{\vec{i}}=\vec{\omega} \times \vec{i}[/itex], which defines the angular velocity.
  7. Jan 28, 2014 #6
    Even assuming that ##\theta## is defined (which it is not), that does not define a vector.

    I'd suggest that you follow the advice of vanhees71.
  8. Jan 28, 2014 #7
    yes , it is much easier to show that, but it can be another orthogonal forms such ixw or ixj or
    jxi these three are also perpendicular to it.
  9. Jan 28, 2014 #8
    but there are two coordinates in the plane which unit vector i lies , Θ and L (polar coordinates),
    since it is a unit vector its magnitude is constant and thus L isnt a coordinate. this shows that
    there must be an angle which describes the rotation of unit vector i so we must assume that Θ is defined. Mean while I have found a geometrical approach but I am much interested in analytical or algebraic methods.
    Last edited: Jan 28, 2014
  10. Jan 28, 2014 #9
    That is not true. If you have two orthogonal vectors, there is only one (in 3D) vector, up to a scaling factor, that is orthogonal to both of them. Because the scaling factor can be negative, there are only two unit vectors with that property; they have opposite directions. For angular velocity, a particular direction is chosen, that is part of its definition.
  11. Feb 5, 2014 #10

    What is a scaling factor for vectors or of vectors? I cant find any basic info about it on the web.
  12. Feb 5, 2014 #11
    This is a true question related with my question. How does w appear in the equation as i'=i x w
    and does it appear from other relations between w, direction angle Θ, and the time?

    Last edited: Feb 5, 2014
  13. Feb 5, 2014 #12
    If two vectors ##\vec a## and ##\vec b## are such that ##\vec a = c \vec b ##, where ##c## is a number, then we say that the vectors are equal up to a scaling factor, and the scaling factor is ##c## Geometrically this means the vectors are collinear.

    If both vectors are unit vectors, then ##c = \pm 1 ##.
  14. Feb 5, 2014 #13
    You need to be able to answer this clearly and unambiguously. Otherwise nothing will ever make sense.
  15. Feb 5, 2014 #14

    D H

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    voko is spot on here.

    mech-eng, you have been a bit sloppy in this thread. What do you mean by ##\vec{\omega}##? By ##\hat \imath##? When you ask about ##\frac {d \hat \imath}{dt}##, with respect to what is your ##\hat \imath## expressed?

    We can't guess at the underlying nature of these questions, but they're just guesses. It would be much better if you filled in those blanks for us. If you can't, is there a textbook whose notation is confusing to you? If that's the case, a reference to the text and the pages within it would be helpful.
  16. Feb 5, 2014 #15
    I get the general idea mech-eng wants to grasp.

    Suppose we have a vector dependent on time, with constant (unit) length but with variable angle. The vector is wandering happily over the surface of the unit sphere.

    The question: what will be the time derivative of the variable vector? What mathematical object will it be?

    The problem will be easier if we rewrite it slightly. Suppose we have the variable vector:
    Now, let's take the value of the vector in the time 0.
    [itex]v_0 = v(0)[/itex]
    Now let's consider rotation matrices. There is always a rotation matrix that transforms the original vector into the current vector
    [itex]v(t) = R(t) v_0[/itex]

    Now let's consider the time derivative of this expression. Since [itex]v_0[/itex] is constant, we have:
    [itex]\frac{∂v(t)}{∂t} = \frac{∂R(t)}{∂t} v_0[/itex]

    To calculate the derivative of the rotation matrix, let's consider the following:
    First, rotation matrices form a group. Multiplication of two rotation matrices yields another rotation matrix.
    [itex]R(t + Δt) = (1 + S(t) Δt) R(t) [/itex] (where 1 is the unit matrix and [itex]S(t)[/itex] is some other matrix)
    [itex]R(t + Δt) = R(t) + S(t) R(t) Δt[/itex]
    [itex]R(t + Δt) - R(t) = S(t) R(t) Δt[/itex]
    [itex]\frac{R(t + Δt) - R(t)}{Δt} = S(t) R(t)[/itex]
    By taking the limit [itex]Δt → 0[/itex], we get the derivative:
    [itex]\frac{∂R(t)}{∂t} = S(t) R(t)[/itex]

    Now what is the S(t) matrix?
    It must satisfy the first equation:
    [itex]R(t + Δt) = (1 + S(t) Δt) R(t) [/itex]

    R(t + Δt) and R(t) are rotation matrices. If this equation has to hold, then S(t) must be a skew-symmetric matrix.

    Let's go back to the original problem:
    [itex]v(t) = R(t) v_0[/itex]
    We can write now (in the limit Δt → 0):
    [itex]v(t + Δt) = R(t + Δt) v_0[/itex]
    [itex]v(t + Δt) = (1 + S(t) Δt) R(t) v_0[/itex]
    [itex]v(t + Δt) = (1 + S(t) Δt) v(t)[/itex]
    [itex]v(t + Δt) = v(t) + S(t) v(t) Δt[/itex]
    [itex]v(t + Δt) - v(t) = S(t) v(t) Δt[/itex]
    [itex]\frac{v(t + Δt) - v(t)}{Δt} = S(t) v(t)[/itex]
    [itex]\frac{∂v(t)}{∂t} = S(t) v(t)[/itex]

    So, the derivative of the vector wandering over the surface of the sphere will be equal to the vector multiplied by the skew-symmetric matrix that can be computed from the relevant rotation matrix.

    Now, in 3 dimensions we have a special situation. The skew-symmetric matrix has exactly 3 independent components that can be rearranged into something that resembles a vector:
    [itex]S(t) = \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix}[/itex]
    [itex]s(t) = [a, b, c][/itex]
    This is called a pseudovector or an axial vector and means what is known under the name "angular velocity".

    The operation of the pseudovector on the original vector is given by the cross product:
    [itex]S(t) v(t) = s(t) \times v(t)[/itex]

    So, in the Δt → 0 limit:
    [itex]v(t + Δt) = v(t) + (s(t) \times v(t)) Δt[/itex]

    This is what you mean by saying "vector v(t) has the angular velocity s(t)".
  17. Feb 5, 2014 #16
    it should, of course, be changing of direction with respect to time and using trig function(which lightgrav said) we can define a unit vector. Let it be in xy plane and in positive x direction. we should specify this by defining angle θ whose derivative should be size of angular velocity with respect to time. For that unit vector being in position x direction we should choose θ as 0 and then unit vector m=cos θi+sinθj, in terms of fixed coordinate vectors, then we can take its derivative which is,dm/dt=-sinΘi*dΘ/dt+cosΘj*dΘ/dt.Then by factoring out dm/dt=dΘ/dt(-sinΘi+cosΘj) and inserting Θ=0 and dΘ/dt=w it becomes dm/dt=wj which is equal of w x i. Is it a true approach?
    Last edited: Feb 5, 2014
  18. Feb 5, 2014 #17
    I cannot follow that. I do not know what you have as a given and what you are supposed to derive. I do not even know what definition of "angular velocity" you have, or even if you have one to begin with. It's been a week but we have not made much progress.

    Please give us the exact definition of "angular velocity of a vector" that you are supposed to use.
  19. Feb 5, 2014 #18
    There is no need to jump back and forth between vectors and matrices. It is a good exercise after you have learned both, but it only confuses matters while you are under way.

    It is sufficient that for a unit vector ## \vec u \cdot \vec u = 1 ##. Then you have immediately ## \vec u \cdot \dot {\vec u} = 0 ## which means ## \dot {\vec u} ## is orthogonal to ## \vec u ##. This in turn means that we can define another unit vector orthogonal to these two vectors (using either the right-hand or the left-hand rule). This is the direction of the angular velocity. The magnitude of the angular velocity is simply ## \dot u / u ##. The rules of cross product then imply that ## \dot {\vec u} = \vec \omega \times \vec u ##.
  20. Feb 5, 2014 #19
    Yes, but where do the rules of cross product come from?

    In my opinion, one must first learn vectors and translations, then matrices, then the rotation group, then skew-symmetric matrices and then the cross-product approach to angular motion.
    This was the way I understood this. Your path may be different.

    I think it's wrong to teach about cross products without any justification where they come from.
  21. Feb 5, 2014 #20

    D H

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    Picky detail: ##\vec u \cdot \dot {\vec u}## means that ##\dot {\vec u}## is either zero or orthogonal to ##\vec u##. If ##\dot{\vec u}## is zero (i.e., ##\vec u## is constant, period, not just constant magnitude), you can't use ##\vec u##, ##\dot {\vec u}##, and ##\vec u \times \dot {\vec u}## to span three dimensional space.
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