#physics
Transport theorem, or Transport Equation, or Bour's formula is a vector formula that relates the time-derivatives of a vector in rotating and non-rotating reference frames.

The equation is given as

$$\left(\frac{d\mathbf{A}}{dt}\right){I} = \left( \frac{d\mathbf{A}}{dt} \right) + \vec{\omega} \times \mathbf{A}$$

where $\overrightarrow{\omega }$ is the angular velocity of the body in the inertial frame.
The subscripts $I$ and $B$ denote inertial frame and body frame respectively.

Body frame is generally aligned with the principal axes of the body to make calculations easy.

Proof

Consider a vector $$\mathbf{A} = a_{1}\mathbf{e}{1} + a\mathbf{e}{2} + a \mathbf{e}_{3}$$where ${\mathbf{e}}_i$'s are the basis vector of the coordinate system we are considering.

Differentiating, we consider we are looking from an inertial frame, $$\left( \frac{d\mathbf{A}}{dt} \right){I} = \sum\left( \frac{da{i}}{dt}\mathbf{e}{i} + a \frac{d\mathbf{e}_{i}}{dt} \right)$$
The first part of this sum is the rate of change of $\mathbf{A}$ with respect to the initial axes, the body frame.
The second part of this sum is what accounts for the rotation of the frame itself.

If the basis vectors are rotating with a angular velocity $\overrightarrow{\omega }$, then $$\frac{d\mathbf{e}{i}}{dt} = \vec{\omega} \times \mathbf{e}$$
So combining these two, we obtain the final closed form of the transport theorem

This is often written as a operator itself, showing for any vector $\mathbf{A}$ $$\left( \frac{d}{dt} \right){I} = \left( \frac{d}{dt} \right) + \omega \times $$
We apply a vector to both sides of the equation to obtain the transport theorem for that vector.