rotating coordinate systems: equations

velocity

There are three key equations for rotational mechanics. Let me refer to them as “the equations”. Goldstein writes a general equation for “some arbitrary vector G”:

\left(\frac{\text{dG}}{\text{dt}}\right)_{\text{space}}=\left(\frac{\text{dG}}{\text{dt}}\right)_{\text{rotating}} + \omega \times G

a specific equation for velocity:

v_s=v_r + \omega \times r

(that’s the first equation with G = position vector r) and an equation for acceleration:

a_s=a_r + 2\ \omega \times v_r +\omega \times (\omega \times r)

Let’s look at all that. I will derive these 3, and a general second-derivative equation, but I will have to return and derive some auxiliary facts.

I’m going to take a very special case. Some things should pop off the page.

Let’s suppose we have two coordinate systems, with a common origin, and one of them (new) is rotating CCW wrt the other (old). Their relationship can be described by a transition matrix T which is a rotation. I will also assume that the axis of rotation is fixed, and that the speed of rotation is constant.

  • a common origin
  • new is CCW rotation of axes wrt old
  • fixed axis of rotation
  • constant speed of rotation

It may be that the equations are valid under less restrictive assumptions; conversely, it is certainly true that we can write more general equations for non-constant axis of rotation and/or non-constant speed of rotation.

Let me not beat around the bush. The equations are used for gravity problems near the surface of the earth, with two coordinate systems: an “inertial frame” = “old” at the center of the earth with z axis along the axis of rotation; and a rotating frame on the surface of the earth, with its z-axis pointing “up”. I do not yet see why the equations should be applicable. What bothers me is that the displacement vector between the two origins is a function of time.

Later. For now, the simple – clear – case: the one that hammers home what’s going on, for me anyway.

If we have a vector R, let r and \rho be its old and new components. Then I can write a matrix-vector equation for the components:

r = T\ \rho

Differentiate wrt time.

\dot{r}=\rho\ \dot{T}+T \dot{\rho }

What the heck is \dot{T} ? For now, please trust me: just as the attitude matrix A of a rotation can be written

A = 1 + \sin (\theta )\ N + (1-\cos (\theta ))\ N^2

the transition matrix T can be written

T =  1 - \sin (\theta )\ N + (1-\cos (\theta ))\ N^2

where we change one sign and N is a skew matrix constructed from a unit vector along the axis of rotation.

Now let \theta = \omega\ t\ , compute the derivative of T, and we get (really!)

\dot{T} = - \omega\ N\ T\ .

I will prove all that, starting with that equation for A which we discussed here. In addition, I will prove that - \omega\ N\ T\ corresponds to a vector cross product.

Granting my formula for \dot{T}\ , we have

old & new components:

r = T\ \rho

differentiate:

\dot{r} = T\ \dot{\rho} + \dot{T}\ \rho

substitute for \dot{T}\ :

\dot{r} = T\ \dot{\rho} - \omega\ N\ T\ \rho

define v = \dot{r} and \nu = \dot{\rho}\ :

v = T\ \nu  - \omega\ N\ T\ \rho

and finally (the “large equation” because I don’t know how to color it) T\ \rho = r\ gives me a form I find useful:

v = T\ \nu  - \omega\ N\ r

and I claim that turns out to be (note the sign change)

v = T\ \nu + \omega \times r

(I have changed notation: I started with \omega a scalar; for the cross product form, I need \omega to be the angular velocity vector; in addition, I am using r for what I called the position vector R, because r is almost universally used.)

Let’s look at what we got. First, I do get the usual cross product formula, even though I sometimes, but not always, prefer the N matrix to the cross product. The crucial starting point was that r and \rho are components of a common vector. We compute the time derivatives of those components, calling them v and \nu\ .

The key implication of the large equation is that, by contrast, v and \nu\ are not the components of a common vector. (If they were, we would have v = T\ \nu\ !)

In fact, I can’t even be sure that either set of components v or \nu\ is the derivative of the position vector R.

In practice we may assume that the vector V whose old components are v is the derivative of the vector R. That is, we usually assume that the old basis vectors are constant.

We usually call the old frame “inertial”, but the derivation is perfectly valid for two coordinate systems both rotating wrt a third, with the new one rotating faster than the old. In that case, the equation

v = T\ \nu - \omega\ N\ r

is still valid, even though both coordinate systems are rotating; it still describes the relationship between the derivatives of components, but we didn’t really get the derivative of the vector R: our basis vectors for the two rotating systems are functions of time.

I don’t want to get too far afield here. It is crucial that v and \nu\ are the components of distinct vectors, called the velocities wrt the old and new coordinate systems; in fact, usually called the velocities wrt the space and body, or wrt fixed and rotating, coordinate systems.

Finally, the T\ \nu term in

v = T\ \nu - \omega\ N\ r

merely says that to write equations for the components of vectors, we’d better be using components wrt one coordinate system: \nu\ are new components, and must be transformed to old, T\ \nu\ . In fact, if we go back to

v = T\ \nu - \omega\ N\ T\ \rho

the T\ \rho term tells us the same thing: \rho are new components and must be transformed to old.

I find it enlightening to have components r and \rho, v and \nu\ , where

v = \dot{r}

and

\nu = \dot{\rho}\

I know exactly what “the velocity wrt the body axes” is: it is a vector whose old components are T\ \nu\ and whose new components are \nu = \dot{\rho}\ . Similarly, “the velocity wrt the space axes” is a vector whose old components are v = \dot{r} and whose new components are T^{-1}\ v = T^T\ v = A\ v\ (where A is the attitude matrix corresponding to the transition matrix T.)

If I have to construct vectors from components, well, I know how to do that.

I think I should warn you that sooner or later, in practice, you’re going to substitute v for T\ \nu\ . Check for that, if ever something’s gone wrong in your calculations. I do.

acceleration

Let’s take another derivative.

start with:

v = T\ \nu - \omega\ N\ r

differentiate:

\dot{v} = T\ \dot{\nu} + \dot{T}\ \nu - \omega\ N\ \dot{r}

let \dot{r}\rightarrow v\ :

\dot{v} = T\ \dot{\nu} + \dot{T}\ \nu - \omega\ N\ v

let \dot{T}\rightarrow - \omega\ N\ T\ :

\dot{v} = T\ \dot{\nu} - \omega\ N\ T\ \nu - \omega\ N\ v

let v \rightarrow T\ \nu - \omega\ N\ r \ :

\dot{v} = T\ \dot{\nu} - \omega\ N\ T\ \nu - \omega\ N\ \left(  T\ \nu - \omega\ N\ r \right)

collect terms:

\dot{v} = T\ \dot{\nu} - 2\ \omega\ N\ T\ \nu + \omega^2\ N^2\ r

define \alpha = \dot{\nu}\ \text{and } a = \dot{v}\ :

a = T\ \alpha - 2\ \omega\ N\ T\ \nu + \omega^2\ N^2\ r

and that, I claim, is equivalent to (note the sign change, one not two)

a = T\ \alpha + 2\ \omega \times \left(T\ \nu\right) + \omega \times \left(\omega \times r\right)

We have the same kind of result. a is the derivative of the old components v, hence a is the second derivative of the old components r of the vector R.

\alpha is the second derivative of \rho\ , hence it is new components, and we convert to the old coordinate system: T\  \alpha\ . \nu is still new components, so we convert to old: T\  \nu\ .

general, quickly

There was nothing special about our initial vector being the position vector. What was special about it was that we had two sets of components for one vector. If G is any vector, and g and \gamma are its components wrt old and new coordinate systems with T the transition matrix, then we proceed exactly as before. (This time I left out all the commentary in order to emphasize just how short the derivations are.)

g = T\ \gamma

\dot{g} = T\ \dot{\gamma} +\dot{T}\ \gamma

\dot{g} = T\ \dot{\gamma} - \omega\ N\ T\ \gamma

\dot{g} = T\ \dot{\gamma} - \omega\ N\ g

\ddot{g} = T\ \ddot{\gamma} +\dot{T}\ \dot{\gamma}- \omega\ N\ \dot{g}

\ddot{g} = T\ \ddot{\gamma} - \omega\ N\ T\ \dot{\gamma}- \omega\ N\ \left(T\ \dot{\gamma} - \omega\ N\ T\ \gamma\right)

\ddot{g} = T\ \ddot{\gamma} - 2\ \omega\ N\ T\ \dot{\gamma} + \omega^2\ N^2\ T\ \gamma

So, we have derived all three of goldstein’s equations (for v, a, and \dot{g}\ ), and one more, for \ddot{g}\ . The equations for v and a, of course, are the special case G = R, the position vector. And the general equations could be written

\dot{g} = T\ \dot{\gamma} + \omega \times g

\ddot{g} = T\ \ddot{\gamma} + 2\ \omega \times \left(T\ \dot{\gamma}\right) + \omega \times \left(\omega \times g\right)

Next, I owe you some supporting detail.

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