What does the notation of the subscript behind the brackets in the differential mean?

From "Theoretical Mechanics of Particles and Continua" by A. Fetter and J. Walecka.

As emphasized in the preceding section, the general expression $$(7.11)$$ can be applied to the coordinate vector $$\mathbf r$$ of a moving particle, relating the velocity seen by observers fixed in the inertial frame and in the rotating frame: $$\left(\frac{\text d \mathbf{r}}{\text d t}\right)_{inertial}=\left(\frac{\text d\mathbf{r}}{\text d t}\right)_{body}+\mathbf{\omega}\times\mathbf{r}$$

What does the notation of the subscripts in the above equation mean?

• Can you link to a text or article that uses this notation? – rob Feb 27 at 14:15
• The text above the equation explains it. They don't say "body", but you can use process of elimination, right? – Aaron Stevens Feb 27 at 14:34

For example (using cylindrical basis vectors $$\hat{z}, \hat{r},\hat{\varphi}$$):
Let's say the earth is rotating counter-clockwise with $${\boldsymbol \omega} = \omega\hat{z}$$. And there's a motionless object at $${\boldsymbol r}= r \hat{r}$$. The object is motionless so for a inertial observer: $$\left(\frac{\text d \mathbf{r}}{\text d t}\right)_{\rm inertial}=0$$
However, an observer on earth should see the object moving clockwise at velocity $$\omega r$$, and indeed: $$\left(\frac{\text d \mathbf{r}}{\text d t}\right)_{\rm body} =\left(\frac{\text d \mathbf{r}}{\text d t}\right)_{\rm inertial}-{\boldsymbol \omega}\times{\boldsymbol r}= -\omega r \hat{\varphi}$$