Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help understanding rotation relations in different frame

  1. Dec 17, 2008 #1

    KFC

    User Avatar

    In every textbook about analytic mechanics, it will give the relation of time derivative of some variable between the space coordinate and body coordinate

    [tex]\left(\dfrac{d\vec{v}}{dt}\right)_{space} = \left(\dfrac{d\vec{v}}{dt}\right)_{body} + \vec{\omega}\times\vec{v}[/tex]

    I don't really understand that relation, especially the [tex]\vec{\omega}[/tex]. Here is my understanding, please tell me if it is correct or not.

    Let that the earth as example, we take the sun as the origin of space coordinate system, and setup a 'body' coordinate system to the earth, if there is a vector changing with time in body system, and we want to find out what does that vector look like to an observer sitting in space coordinate, so we can apply the relation above to find it out. Is that right? My doubt is: what is the [tex]\vec{\omega}[/tex] here? Should it be the angular velocity of the body which is rotating around some axis in space coordinate system or the its own spinning angular velocity?
     
    Last edited: Dec 17, 2008
  2. jcsd
  3. Dec 17, 2008 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    The equation is correct. One way to look at it as as the difference between the acceleration of some object as measured by an observer fixed and not rotating with respect to space and the acceleration of that object as measured by an observer fixed and not rotating with respect to the body frame.

    The cited equation (called the "transport theorem" in some texts) is very general:

    [tex]\left(\frac{d\boldsymbol q}{dt}\right)_A
    = \left(\frac{d\boldsymbol q}{dt}\right)_B +
    \boldsymbol{\omega}_{A \to B}\times \boldsymbol q[/tex]

    where q is any vector quantity, the A/B subscripts on the derivative indicate the derivative of the vector quantity q as observed by fixed and not rotating with respect to frames A and B, and [itex]\boldsymbol{\omega}_{A \to B}[/itex] is the angular velocity of frame B with respect to frame A.

    How to derive it? The representations of the vector quantity q in frames A and B are related by the transformation matrix from frame B and A:

    [tex]\boldsymbol q_A = \mathbf T_{B\to A} \boldsymbol q_B[/tex]

    Differentiating these representations with respect to time,

    [tex]\frac{d \boldsymbol q_A}{dt} =
    \mathbf T_{B\to A} \frac{d \boldsymbol q_B}{dt} +
    \frac {d \mathbf T_{B\to A}}{dt}\boldsymbol q_B[/tex]

    The time derivative of the transformation matrix (I am not going to derive it here; too messy) is

    [tex]\frac {d \mathbf T_{B\to A}}{dt} = \mathbf T_{B\to A} \mathbf X(\boldsymbol{\omega}_{A \to B:B})[/tex]

    where X(a) is the skew-symmetric cross product matrix generated by the vector a and [itex]\boldsymbol{\omega}_{A \to B:B}[/itex] is the angular velocity of frame B with respect to frame A expressed in the coordinates of frame B.

    Since matrix multiplication is associative and distributes over addition,

    [tex]\frac{d \boldsymbol q_A}{dt} =
    \mathbf T_{B\to A} \left(\frac{d \boldsymbol q_B}{dt} +
    \mathbf X(\boldsymbol{\omega}_{A \to B}) \boldsymbol q_B\right)
    = \mathbf T_{B\to A} \left(\frac{d \boldsymbol q_B}{dt} +
    \boldsymbol{\omega}_{A \to B:B} \times \boldsymbol q_B\right)[/tex]

    Without the transformation matrix,

    [tex]\left(\frac{d\boldsymbol q}{dt}\right)_A
    = \left(\frac{d\boldsymbol q}{dt}\right)_B +
    \boldsymbol{\omega}_{A \to B}\times \boldsymbol q[/tex]
     
  4. Dec 17, 2008 #3

    KFC

    User Avatar

    Thank you so much for your detailed reply. Though I am not understanding everything, it does help me to understand bit more.
     
  5. Dec 17, 2008 #4

    KFC

    User Avatar

    sorry to bother again. I just reconsider the formula and apply it to an actual case. However, I stuck with the following

    Suppose an object is rotating along some axis with constant angular frequency [tex]omega[/tex]. We setup a body coordinate system attaching to the body with the z axis along the rotating axis. We then setup another axis call space coordinate (i.e. laboratory coordinate system). I want to find the torque in the space coordinate, I apply

    [tex]
    \left(\frac{d\vec{L}}{dt}\right)_{space} = \left(\frac{d\vec{L}}{dt}\right)_{body} + \vec{\omega}\times\vec{L}
    [/tex]

    Now, I totally lose in above formula. As the previous thread's reply, I know [tex]\vec{\omega}[/tex] is a angular velocity with respect to the space coordinate system. But there are three [tex]\vec{L}[/tex] occurs in above formula! Which frame should these [tex]\vec{L}[/tex] be with respect to? The first term in the LHS is what I am solving for, so just left it along. The second term is zero because I choose the body coordinate rotating with the body. What about the last term? What [tex]\vec{L}[/tex] I should use there?

    By the way, if I choose a body coordinate with z axis not align with the rotating axis so the second term is no more ZERO, then what [tex]\vec{L}[/tex] I should use there?

    Thanks
     
  6. Dec 17, 2008 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    In freshman-level physics texts one often sees Newton's second law for rotation written in the form

    [tex]\boldsymbol{\tau} = \mathbf I \boldsymbol{\alpha}[/tex]

    There are a slew of problems with this representation. First, it implicitly assumes the torque, inertia tensor, and angular acceleration are expressed in inertial coordinates. All three are typically expressed in body coordinates. Even worse, it is not valid in general; it implicitly assumes the inertia tensor (expressed in inertial coordinates no less) is constant. Freshman-level physics rotational motion problems have to be carefully constructed to whitewash over these problems.

    The quoted equation enables one to overcome these problems. With it, one can compute the rotational response of a solid object to some external torque without whitewashing over the issues raised above.

    Newton's second law of motion is often written as F=ma. A more general form is F= dp/dt. The rotational equivalent: [itex]\tau = dL/dt[/tex]. Newton's second law of motion is strictly valid only in an inertial frame. The same is true for the rotational form of Newton's second law. Note that the dL/dt term on the left-hand side of the quoted equation is from the perspective of an inertial observer. (The dL/dt term on the right-hand side is not.) The left-hand side is just the total external torque:

    [tex]\left(\frac {d\boldsymbol L}{dt}\right)_{\text{space}} =
    \boldsymbol{\tau}_{\text{ext}}[/tex]

    from which

    [tex]\left(\frac {d\boldsymbol L}{dt}\right)_{\text{body}} =
    \boldsymbol{\tau}_{\text{ext}} -
    \boldsymbol{\omega}\times \boldsymbol L[/tex]

    Another way to express the angular momentum is as the product of the inertia tensor and the angular velocity. Using this,

    [tex]\left(\frac {d\mathbf I \boldsymbol \omega}{dt}\right)_{\text{body}} =
    \boldsymbol{\tau}_{\text{ext}} -
    \boldsymbol{\omega}\times (\mathbf I \boldsymbol \omega)[/tex]

    The inertia tensor for a constant mass solid body is constant in the body frame. In such case, the inertia tensor can be taken out of the derivative:

    [tex]\mathbf I\left(\frac {d \boldsymbol \omega}{dt}\right)_{\text{body}} =
    \boldsymbol{\tau}_{\text{ext}} -
    \boldsymbol{\omega}\times (\mathbf I \boldsymbol \omega)[/tex]

    Typically the inertia tensor, angular velocity, and external torques are expressed in the body frame. In this case, the body subscript on the above derivative can be removed (as being superfluous):

    [tex]\mathbf I \frac {d \boldsymbol \omega}{dt}\right =
    \boldsymbol{\tau}_{\text{ext}} -
    \boldsymbol{\omega}\times (\mathbf I \boldsymbol \omega)[/tex]
     
  7. Dec 17, 2008 #6

    KFC

    User Avatar

    Thank you so much for your reply. Well, I am afraid I don't fully understand this, but I will try my best to read it again. According to your explanation, everything in the left hand side of the quoted equation should be defined in space frame and everything on the right hand side are defined in the body frame, is that what you mean?

    Well, may I give an example to illustrate that? Assume an object is rotating around some axis with constant angular velocity [tex]\vec{\omega}[/tex]. We setup a body frame with the z axis aligned with the rotating axis. Since the body frame is rotating along the body, [tex]\left(\frac{d\vec{L}}{dt}\right)_{body} = 0[/tex], what I have to do is to find out the angular momentum [tex]\vec{L}_{body}[/tex]defined in body frame, so the total external torque with respect with space frame would be

    [tex]\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}_{space} = \vec{\omega}\times\vec{L}_{body}[/tex]

    Is this correct?

    By the way, if I choose the so called principal axes as the coordinate and use Euler equation to find the torque, that equation can only tell me the torque with respect to the principal axes, right? How can I change that torque back to that defined in space frame?

     
  8. Dec 18, 2008 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Whoa! You have a significant misunderstanding here. Time to go back to the start. In the general equation cited earlier, the "transport equation",

    [tex]\left(\frac{d\boldsymbol q}{dt}\right)_A
    = \left(\frac{d\boldsymbol q}{dt}\right)_B +
    \boldsymbol{\omega}_{A \to B}\times \boldsymbol q[/tex]

    The vector q in the derivatives (and the cross product) is (*must be*) the same vector on the left and right hand sides. It is the derivatives that are substantially different vectors.

    What I mean by "the same vector": Suppose an external force F acts on some body and suppose there are two inertial frames of interest, A and A', one rotated (not rotating) with respect to the other (i.e., there is a constant, non-identity transformation matrix from frame A to frame A'). While the representation of the force vector will be different in the two frames, both representations are describing the same vector. Similarly, consider a displacement vector from point 1 to point 2. That vector is the same vector in all frames (in classical mechanics); it just has different representations.

    When you said "Since the body frame is rotating along the body, [itex]\left(\frac{d\vec{L}}{dt}\right)_{body} = 0[/itex]," you fell into the trap of applying the transport equation to two different vectors. Doing so will lead to erroneous results. In particular, when applying the transport equation to the angular momentum, you must use the angular momentum of the body with respect to inertial space.
    No. The angular momentum of a body with respect to inertial space expressed in body coordinates is

    [tex]\boldsymbol L = \mathbf I \boldsymbol{\omega}[/tex]

    where [itex]\boldsymbol{\omega}[/itex] is the angular velocity of the body with respect to inertial space. Typically, the inertia tensor and the angular velocity are expressed in body frame coordinates. That the inertia tensor is expressed in body frame coordinates makes sense. The inertia tensor of a rigid, constant-mass object is constant in the body frame. If the object is rotating, the inertia tensor expressed in inertial coordinates is in general time-varying. That the angular velocity is expressed in body frame coordinates is a bit counterintuitive. One reason for doing so is because the inertia tensor is constant in the body frame. Another reason is that rotation sensors (e.g., rate gyros) measure angular velocity with respect to inertial expressed in body frame coordinates. (More precisely, they measure angular velocity with respect to inertial expressed in the sensor housing frame (case frame). Since the sensor housing is fixed with respect to the body, transforming those "case frame" coordinates to body frame is a trivial matter.)

    Don't do that! The Euler equations are a special case of the equation I derived in the my post. From post #5,

    [tex]\mathbf I \frac {d \boldsymbol \omega}{dt}\right =
    \boldsymbol{\tau}_{\text{ext}} -
    \boldsymbol{\omega}\times (\mathbf I \boldsymbol \omega)[/tex]

    Rearranging,

    [tex]\mathbf I \frac {d \boldsymbol \omega}{dt}\right +
    \boldsymbol{\omega}\times (\mathbf I \boldsymbol \omega) =
    \boldsymbol{\tau}_{\text{ext}}[/tex]

    This is Euler's equations, written in a more general form. Suppose the inertia tensor is diagonal (i.e., principal axes):

    [tex]I=\bmatrix I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3\endbmatrix[/tex]

    Denoting the angular velocity as [itex]\boldsymbol{\omega} = [\omega_1 \,\, \omega_2 \,\, \omega_3]^T[/itex], the ith component of the angular momentum will be [itex]I_i \omega_i[/itex] and the cross product [itex]\boldsymbol{\omega}\times (\mathbf I \boldsymbol \omega)[/itex] will be

    [tex]
    \boldsymbol{\omega}\times (\mathbf I \boldsymbol \omega) =
    \bmatrix \omega_1 \\ \omega_2 \\ \omega_3 \endbmatrix
    \times
    \bmatrix I_1\omega_1 \\ I_2\omega_2 \\ I_3\omega_3 \endbmatrix
    =
    \bmatrix
    (I_3-I_2)\omega_2\omega_3 \\
    (I_1-I_3)\omega_3\omega_1 \\
    (I_2-I_1)\omega_1\omega_2
    \endbmatrix
    [/tex]

    Applying this to the equation in question,

    [tex]\aligned
    I_1 \frac{d\boldsymbol{\omega}_1}{dt}\right + (I_3-I_2)\omega_2\omega_3 &= \tau_1 \\
    I_2 \frac{d\boldsymbol{\omega}_2}{dt}\right + (I_1-I_3)\omega_3\omega_1 &= \tau_2 \\
    I_3 \frac{d\boldsymbol{\omega}_3}{dt}\right + (I_2-I_1)\omega_1\omega_2 &= \tau_3
    \endaligned[/tex]

    Ta da! Euler's equations!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?