Abosolute derivative of a vector represented in a rotating frame

29 Jul 2019

\(XYZ\) is an inertial frame. \(xyz\) is a frame that is rotating with an angular velocity \(\mathbf{\Omega}\) relative to the inertial frame. \(\mathbf{\hat{I}}\), \(\mathbf{\hat{J}}\), and \(\mathbf{\hat{K}}\) are unit vectors along \(X\), \(Y\), and \(Z\) directions respectively. \(\mathbf{\hat{i}}\), \(\mathbf{\hat{j}}\), and \(\mathbf{\hat{k}}\) are unit vectors along \(x\), \(y\), and \(z\) axes respectively. The orientation of the two frames are shown in the figure below.

Our goal is to find the time derivative of a time-dependent vector relative to the inertial frame. It’s a bit tricky because we know the components of the vector in the non-inertial frame \(xyz\). The derivative relative to an inertial frame is called the absolute derivative.

xyz frame is rotating with a constant angular velocity relative to XYZ frame

Let \(\mathbf{Q}\) be an arbitrary time-dependent vector. In the inertial frame, its components are,

\[\begin{equation} \mathbf{Q} = Q_{x}\mathbf{\hat{I}} + Q_{y}\mathbf{\hat{J}} + Q_{z}\mathbf{\hat{K}} \end{equation}\]

where \(Q_{x}\), \(Q_{y}\), and \(Q_{z}\) are functions of time. Since \(XYZ\) is an inertial frame, \(\mathbf{\hat{I}}\) \(\mathbf{\hat{J}}\), and \(\mathbf{\hat{K}}\) are fixed.

Therefore,

\[\begin{equation} \frac{d\mathbf{Q}}{dt} = \frac{dQ_{X}}{dt}\mathbf{\hat{I}} + \frac{dQ_{Y}}{dt}\mathbf{\hat{J}} + \frac{dQ_{Z}}{dt}\mathbf{\hat{K}}. \end{equation}\]

\(\mathbf{Q}\) can also be resolved into components in the \(xyz\) frame. Let

\[\begin{equation} \mathbf{Q} = Q_{x}\mathbf{\hat{i}} + Q_{y}\mathbf{\hat{j}} + Q_{z}\mathbf{\hat{k}}. \end{equation} \label{eq:q-xyz}\]

Since \(xyz\) frame is rotating relative to the inertial frame, its unit vectors are not fixed and are constantly changing their direction. However, since they are unit vectors, their magnitudes remain unchanged.

Therefore, the absolute time derivative of equation \eqref{eq:q-xyz} is given by,

\[\begin{equation} \frac{d\mathbf{Q}}{dt} = \frac{dQ_{X}}{dt}\mathbf{\hat{i}} + \frac{dQ_{Y}}{dt}\mathbf{\hat{j}} + \frac{dQ_{Z}}{dt}\mathbf{\hat{k}} + Q_{x}\frac{d\mathbf{\hat{i}}}{dt} + Q_{y}\frac{d\mathbf{\hat{j}}}{dt} + Q_{z}\frac{d\mathbf{\hat{k}}}{dt} \end{equation} \label{eq:dQ}\]

The absolute time derivative of a vector with constant magnitude and varying direction is given by,

\[\begin{equation} \frac{d\mathbf{\hat{i}}}{dt} = \mathbf{\Omega} \times \mathbf{\hat{i}} \qquad \frac{d\mathbf{\hat{j}}}{dt} = \mathbf{\Omega} \times \mathbf{\hat{j}} \qquad \frac{d\mathbf{\hat{k}}}{dt} = \mathbf{\Omega} \times \mathbf{\hat{k}}, \qquad \end{equation} \label{eq:di}\]

where \(\mathbf{\Omega}\) is the angular velocity of the vector relative to an inertial frame.

Substituting equation \eqref{eq:di} into equation \eqref{eq:dQ} yields,

\[\begin{aligned} \frac{d\mathbf{Q}}{dt} &= \frac{dQ_{x}}{dt}\mathbf{\hat{i}} + \frac{dQ_{y}}{dt}\mathbf{\hat{j}} + \frac{dQ_{z}}{dt}\mathbf{\hat{k}} + Q_{x}\mathbf{\Omega} \times \mathbf{\hat{i}} + Q_{y}\mathbf{\Omega} \times \mathbf{\hat{j}} + Q_{z}\mathbf{\Omega} \times \mathbf{\hat{k}} \\ &= \frac{dQ_{x}}{dt}\mathbf{\hat{i}} + \frac{dQ_{y}}{dt}\mathbf{\hat{j}} + \frac{dQ_{z}}{dt}\mathbf{\hat{k}} + (\mathbf{\Omega} \times Q_{x}\mathbf{\hat{i}}) + (\mathbf{\Omega} \times Q_{y}\mathbf{\hat{j}}) + (\mathbf{\Omega} \times Q_{z}\mathbf{\hat{k}}) \\ &= \frac{dQ_{x}}{dt}\mathbf{\hat{i}} + \frac{dQ_{y}}{dt}\mathbf{\hat{j}} + \frac{dQ_{z}}{dt}\mathbf{\hat{k}} + \mathbf{\Omega} \times (Q_{x}\mathbf{\hat{i}} + Q_{y}\mathbf{\hat{j}} + Q_{z}\mathbf{\hat{k}}) \end{aligned}\] \[\begin{equation} \boxed{\frac{d\mathbf{Q}}{dt} = \frac{d\mathbf{Q}}{dt}\Bigg)_{\substack{\text{Moving}\\ \text{frame}}} + \mathbf{\Omega} \times \mathbf{Q}} \end{equation}\]

where,

\[\begin{equation} \frac{d\mathbf{Q}}{dt}\Bigg)_{\substack{\text{Moving}\\ \text{frame}}}= \frac{dQ_{x}}{dt}\mathbf{\hat{i}} + \frac{dQ_{y}}{dt}\mathbf{\hat{j}} + \frac{dQ_{z}}{dt}\mathbf{\hat{k}} \end{equation}\]

is the time derivative of \(\mathbf{Q}\) relative to \(xyz\) frame. Note that this is not the absolute time derivative.

The absolute time derivative and the time derivative in \(xyz\) frame are equal when \(\mathbf{\Omega} = 0\). This situation corresponds to pure translation with no acceleration and the invarience of time derivative is in accordance with the special principle of relativity.

Reference

Howard Curtis. Orbital Mechanics for Engineering Students. Elsevier Butterworth - Heinemann