Lagrangian of a free particle in a rotating coordinate system

In this report, we derive the Lagrangian for a free particle in 3 dimensions, in a rotating coordinate system. We then use the Euler-Lagrange equation in 3 dimensions to find formulae for the fictitious centrifugal and Coriolis forces, which have applications for particles in Earth's rotating atmosphere.

The derivation

We start by considering a system that has an axis through the origin \( O = \left(0,0,0\right) \) parallel to a vector \( \mathbf{\Omega} \). We can then define a rotation matrix \( R(\theta) \) which rotates points about the aforementioned axis by an angle \( \theta \) in the direction assigned by the "right-hand screw rule" for \( \mathbf{\Omega} \). Given this, if \( \mathbf{r} \) is the position of a particle relative to \( O \) in the rotated frame, then the actual position vector of the particle \( \mathbf{r}' \) relative to the \( O \) is\[ \mathbf{r}' = R(\theta)\,\mathbf{r}. \] Differentiating with respect to time means that for instantaneous velocity vectors, a particle with a velocity \( \mathbf{v} \) in the rotated frame will have an actual velocity \( \mathbf{v}' \) where \[ \mathbf{v}' = R(\theta)\,\mathbf{v}. \] If the frame is actively rotating with angular velocity \( \Omega = | \mathbf{\Omega}| \), so \( \theta = wt \) and \( R = R(wt) \), then we must also include the contribution of the rotation, \( \mathbf{\Omega} \times \mathbf{r}' \), in the final velocity of the particle. Superposing these gives the resultant velocity \( \mathbf{V} \) given by \begin{align*} \mathbf{V} &= R\,\mathbf{v} +\mathbf{\Omega} \times \mathbf{r}' \\ &= R\,\mathbf{v} +\mathbf{\Omega} \times R\,\mathbf{r}. \end{align*} Now we can use this velocity in the Lagrangian \( \mathcal{L} \) of the free particle, given it has a mass \( m \), which is \begin{align*} \mathcal{L} &= \frac{1}{2}m|\mathbf{V}|^2 \\ & = \frac{1}{2}m|R\,\mathbf{v} +\mathbf{\Omega} \times R\,\mathbf{r}|^2 \\ & = \frac{1}{2}m|R\,\mathbf{v} +R\,\mathbf{\Omega} \times R\,\mathbf{r}|^2 \quad \{ R\,\mathbf{\Omega} = \mathbf{\Omega} \}\\ & = \frac{1}{2}m|R\,\mathbf{v} +R\left(\mathbf{\Omega} \times \mathbf{r}\right)|^2 \quad \{ R\,\mathbf{u} \times R\,\mathbf{v} = R \left( \mathbf{u} \times \mathbf{v} \right) \} \\ & = \frac{1}{2}m|R\,\left(\mathbf{v} +\mathbf{\Omega} \times \mathbf{r}\right)|^2 \\ & = \frac{1}{2}m|\mathbf{v} +\mathbf{\Omega} \times \mathbf{r}|^2 \quad \{ |R\,\mathbf{u}|=|\mathbf{u}| \}.\end{align*} We continue by expanding the modulus. \begin{align*} \mathcal{L} & = \frac{1}{2}m|\mathbf{v} +\mathbf{\Omega} \times \mathbf{r}|^2 \\ &= \frac{1}{2}m\left(|\mathbf{v}|^2+|\mathbf{\Omega} \times \mathbf{r}|^2+2\,\mathbf{v}\cdot\left(\mathbf{\Omega} \times \mathbf{r}\right)\right) \\ &= \frac{1}{2}m\left(|\mathbf{v}|^2+|\mathbf{\Omega}|^2|\mathbf{r}|^2-\left(\mathbf{\Omega}\cdot\mathbf{r}\right)^2+2\, \mathbf{v}\cdot\left(\mathbf{\Omega} \times \mathbf{r}\right)\right). \end{align*} We now seek to apply the 3-dimensional Euler-Lagrange equation, being \[ \frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial \mathcal{L} }{\partial \mathbf{v}} \right)= \frac{\partial \mathcal{L}}{\partial \mathbf{r}} \quad \text{where} \quad \displaystyle \left(\frac{\partial \mathcal{L}}{\partial \mathbf{r} }\right)_i = \frac{\partial \mathcal{L}}{\partial r_i} \quad \text{for} \quad i \in \{ x,y,z \} . \] By analysing components, we find: \begin{align*} \left(\frac{\partial \mathcal{L} }{\partial \mathbf{v}}\right)_{i} &= \frac{1}{2}m\left(2v_{i}+2\left(\mathbf{\Omega} \times \mathbf{r}\right)_{i}\right), \\ \therefore\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial \mathcal{L} }{\partial \mathbf{v}} \right) &= m\left(\mathbf{a}+\mathbf{\Omega} \times \mathbf{v}\right) \quad \text{and} \\ & \\ \left(\frac{\partial \mathcal{L} }{\partial \mathbf{r}}\right)_{i} &= \frac{1}{2}m\left(2|\mathbf{\Omega}|^2r_{i}-2\left(\mathbf{\Omega}\cdot\mathbf{r}\right)w_{i}+2\frac{\partial}{\partial r_{i}} \left( \mathbf{r}\cdot\left(\mathbf{v} \times \mathbf{\Omega}\right) \right)\right) \\ &=m\left(| \mathbf{\Omega}|^{2}r_{i} - \left( \mathbf{\Omega} \cdot \mathbf{r} \right) w_{i} + \left(\mathbf{v} \times \mathbf{\Omega} \right)_{i} \right),\\ \therefore\frac{\partial \mathcal{L} }{\partial \mathbf{r}} &= m\left(\left( \mathbf{\Omega} \cdot \mathbf{\Omega} \right) \mathbf{r} - \left( \mathbf{\Omega} \cdot \mathbf{r} \right) \mathbf{\Omega} + \mathbf{v} \times \mathbf{\Omega} \right) \\ &=m\left( -\mathbf{\Omega} \times \left( \mathbf{\Omega} \times \mathbf{r} \right)- \mathbf{\Omega} \times \mathbf{v} \right). \end{align*} Finally, substituting into the Euler-Lagrange equation yields: \[ \mathbf{F}_{fictitious} = m\,\mathbf{a} = -m\,\mathbf{\Omega} \times \left( \mathbf{\Omega} \times \mathbf{r} \right)-2m\, \mathbf{\Omega} \times \mathbf{v}.\] The first of the two fictitious force terms is the centrifugal force, always pointing away from the axis of rotation and perpendicular to it. The second is the Coriolis force, which causes Earthly phenomena such as the different rotation direction of storms in the Northern and Southern hemispheres, as well as the Eötvös effect.