I'm studying the book "Classical Mechanics" by Goldstein together with a coursebook my professor provided.

I'm having trouble grasping how to intuitively determine what the rate of change of a vector is affected by.

Earlier on in the book, I saw a body with 3 frames of reference:
$Oxyz$ which is just an arbitrary coordinate system.
$O'x'y'z'$ which has a fixed origin in a point of the body and fixed axes that rotate corresponding to the body.(Body fixed?)
$O'xyz$ which has a fixed origin in a point of the body and fixed axes that are parallel with the original coordinate system.(Space fixed?)

Now, in my professors book it says the following:

It is clear that there are 2 sources of time-dependence in the Carthesian components of a general vector in the (body-fixed) coordinate system:
On one hand: The intrinsic time-dependence of the vector.
On the other hand: The time dependence of the basisvectors $\vec n'_i$ of the moving coordinate system, with respect to which the Carthesian components (orthogonal projection of the vector onto $\vec n'_i$) will be determined.

I'm confused as to what exactly is meant by this paragraph.
What exactly is the intrinsic time-dependence of the vector? To my knowledge, in the body-fixed coordinate system every vector stays constant, so there is no rate of change.
I understand that the $\vec n'_i$ vectors move through space, but during a translation they wouldn't change since they are their orientation stays the same. During a rotation however I understand they would change.

What am I getting wrong here? I have a feeling I'm mixing these frames of reference up since it's not so clear in my book. I think that I'm misunderstanding how these frames of reference behave as the body moves.


1 Answer 1

  1. Consider a changing vector $\vec{A}(t)$ on a fixed coordinate system. The rate of change of the vector components are $$\frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \frac{\partial \vec{A}(t)}{\partial t}= \dot{\vec{A}}(t) $$
  2. Consider a fixed vector $\vec{A}$ on a rotating coordinate system, with angular velocity $\vec\Omega(t)$. The rate of change of the vector components with respect to the attached frame are zero, but with respect to an inertial frame the rate of change is $$\frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \vec{\Omega}(t) \times \vec{A} $$
  3. Combined the changing $\vec{A}(t)$ vector on a rotating frame $\vec\Omega(t)$ is $$\boxed{ \frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \frac{\partial \vec{A}(t)}{\partial t} + \vec{\Omega}(t) \times \vec{A}(t)}$$

The first part is the intrinsic change along the rotating frame, and the second part the change due to the rotation of the coordinates.


Two bodies are connected with a slider joint, defined by an axis $\vec{e}$ fixed to the first body which is rotating by $\Omega(t)$. If the joint distance is $\chi(t)$ then the position of the 2nd body is defined relative to the first body as $$\vec{r}_2(t) = \vec{r}_1(t) + \vec{e}\, \chi(t)$$

The position vectors are differentiated to derive the velocity kinematics as:

$$ \vec{v}_1 = \frac{{\rm d}\vec{r}_1(t)}{{\rm d}t} \\ \vec{v}_2 = \frac{{\rm d}\vec{r}_2(t)}{{\rm d}t} $$

$$\begin{align} \vec{v}_2 & = \vec{v}_1 + \frac{{\rm d}(\vec{e}\,\chi(t))}{{\rm d}t} \\ & = \vec{v}_1 + \frac{\partial \vec{e} \chi(t)}{\partial t} + \vec{\Omega}(t) \times \vec{e} \,\chi(t) \\ & = \vec{v}_1 + \vec{e} \dot\chi (t) + \vec{\Omega}(t) \times \left( \vec{r}_2(t) - \vec{r}_1(t) \right) \end{align}$$

See related answer: Derivation of Euler's equations for rigid body rotation

  • $\begingroup$ It still hasn't completely clicked, but first things first: $\endgroup$
    – Joshua
    Commented Jun 10, 2014 at 12:33
  • $\begingroup$ Body fixed means that the origin is a certain point of the body and that the axes rotate along with the body. Space fixed has the same origin as body fixed has on t=0, but different axes. It is seen as completely fixed (immovable) as time proceeds. Are these assumptions correct? Now secondly, I understand part 1 and 2 of your answer, but part 3 confuses me: In part 2 of your answer you said the change in the rotating system is 0, in part 3 you say that the change in the rotating system is now the sum of the previous two. $\endgroup$
    – Joshua
    Commented Jun 10, 2014 at 12:42
  • $\begingroup$ Your assumptions are correct. You also have to distinguish between changing vector, and changing the components (representation) of a vector. A vector fixed in space will have changing components when viewed from a rotating frame. Conversely a vector fixed on a rotating body will have changing components when viewed from a fixed frame (the $\Omega \times $ part) and non-changing components when viewed from a co-moving frame. In part 2, I talk about non changing components of a rotating vector. $\endgroup$ Commented Jun 10, 2014 at 13:50
  • $\begingroup$ So If I understand correctly, Part 3 is the total change when viewed from a space fixed (inertial) frame of reference? $\endgroup$
    – Joshua
    Commented Jun 10, 2014 at 14:02
  • $\begingroup$ Correct. Since forces and momenta are typically defined in inertial frames, the kinematics must also be viewed on inertial frames for the equations to work out. The last difficulty in this is expressing the mass moment of inertia tensor components in inertial coordinates, when it is defined on body fixed coordinates. See here on how to deal with this. $\endgroup$ Commented Jun 10, 2014 at 14:26

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