# Potential for a general fictitious force?

In a general non inertial system four fictitious forces arise:

1. Coriolis Force,

2. centrifugal Force,

3. azimuthal Force,

4. "translational" Force (due to linear acceleration of the origin of the system)

I know that I can apply Newtonian laws to the system if I take these forces into account. I also know that I can apply the work- energy principle, if I take the work of these forces into account.

What I am not sure about is: can I associate a potential energy to all of these forces? (I.e. consider a fictitious potential field). This is not trivial since the fictitious forces do not obey to Newton third law (--> no reaction). I also don't understand if I should consider there forces as internal or external to the system.

I ask this question because I read somewhere about centrifugal force potential.

PS only thing I am sure: the Coriolis force does no work --> no potential associated with It.

• There is a potential associated to the centrifugal force, which you can write as $\mathbf{F}_{\text{centrifugal}}/m = \mathbf{\omega}\times\mathbf{\omega}\times\mathbf{r} = (\mathbf{\omega}\cdot\mathbf{r})\mathbf{\omega} - |\mathbf{\omega}|^2\mathbf{r} = (\mathbf{\omega}\cdot\mathbf{r})\nabla(\mathbf{\omega}\cdot\mathbf{r}) - \frac{1}{2}|\mathbf{\omega}|^2\nabla|\mathbf{r}|^2 = -\frac{1}{2}\nabla\left[(\mathbf{\omega}\cdot\mathbf{r})^2 - |\mathbf{\omega}|^2|\mathbf{r}|^2\right] = -\frac{1}{2}\nabla |\mathbf{\omega}\times\mathbf{r}|^2$. – Tob Ernack Aug 22 at 14:26
• So the potential associated to the centrifugal force is $\frac{1}{2}m|\mathbf{\omega}\times\mathbf{r}|^2$. – Tob Ernack Aug 22 at 14:26
• Thank you. I guess this is only valid if angular velocity is constant... what if it is not the case? And what about the other fictitious forces? – Federico Toso Aug 22 at 15:01
• The formula for $\mathbf{F}_{\text{centrifugal}}$ is valid for time-varying $\mathbf{\omega}$ as well. In that case the potential is also time-dependent. – Tob Ernack Aug 22 at 15:27
• For the azimuthal force (by which I think you mean the force due to change in $\mathbf{\omega}$ in the tangential direction, also called Euler force), you have $\mathbf{F}_{\text{Euler}} = m\dot{\mathbf{\omega}}\times\mathbf{r} = -\frac{1}{2}m\dot{\mathbf{\omega}}\times\nabla|\mathbf{r}|^2 = -\nabla\times\left(\frac{1}{2}|\mathbf{r}|^2\dot{\mathbf{\omega}}\right)$ so it is not a conservative force. Still you can view $-\frac{1}{2}|\mathbf{r}|^2\dot{\mathbf{\omega}}$ as a kind of vector potential. – Tob Ernack Aug 22 at 15:37

## 2 Answers

In elementary Newtonian mechanics only the centrifugal force, and for a uniform rotation, admits potential energy in addition to the fictitious force due to uniform acceleration provided some conditions are valid. If forces are time dependent, a notion of potential ( not potential energy ) can be introduced as well. The Coriolis force is much more problematic to treat.

If however passing to the Lagrangian formulation of classical mechanics for a system of points (also subjected to ideal constraints), all inertial forces can be described in terms of a so-called generalized potential.

That is a term subtracted to the kinetical-energy pert of the Lagrangian which linearly depends on velocities and may also depend on time in addition to coordinates.

Formally, this part of the Lagrangian is similar to the non-kineric part of the Lagrangian of charged particles immersed in a generic given electromagnetic field. The Coriolis force corresponds here to the part of Lorentz force ascribed to the magnetic field only.

1. Generalized/velocity-dependent potentials $$U(q,\dot{q},t)$$ are discussed in many textbooks, see e.g. Refs. 1 and 2. Let us just mention here that by definition the functional derivative $$\frac{\delta}{\delta q^i}\int\! dt~ U(q,\dot{q},t) ~=~-Q_i\tag{1}$$ of the time-integrated potential $$\int\! dt~ U(q,\dot{q},t)$$ should be minus the generalized force $$Q_i$$.

2. The generalized/velocity-dependent potential for an arbitrary fictitious force, which is a combination of

is $$\frac{U}{m} ~=~ \vec{A}\cdot \vec{r} - \vec{v}\cdot (\vec{\Omega} \times \vec{r})-\frac{1}{2} (\vec{\Omega} \times \vec{r})^2, \tag{39.6}$$ cf. e.g. Ref. 2.

References:

1. H. Goldstein, Classical Mechanics, Chapter 1.

2. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1 (1976); $$\S$$39.