# Proof derivative of a vector following precession motion

I do not get some points of this proof about the time derivative of a unit vector $\hat{u}$ (costant magnitude) which is following a precession motion. The picture is the following. I want to prove that $$\frac{d\hat{u}}{dt}=\vec{\Omega}\wedge \hat{u}.$$

I'm ok with almost all the proof except the following points.

Consider the red vector $d\hat{u}$. Since $\hat{u}$ has constant magnitude, $d\hat{u}$ must be orthogonal to $\hat{u}$. That's ok.

But why must $d\hat{u}$ be orthogonal to $\vec{\Omega}$ too (i.e. be tangential to a circle orthogonal to $\vec{\Omega})$?

Secondly how can I geometrically prove that the angle $d\theta$ (the one in purple) that separates $\hat{u}(t)$ and $\hat{u}(t+dt)$ is the same that separates the projection of these two vectors on the circle orthogonal to $\vec{\Omega}$ (the projections are $\hat{u}(t) Sin(\alpha)$ and $\hat{u}(t+dt) Sin(\alpha)$)?

• $\mathrm d\hat u$ is orthogonal to $\boldsymbol \Omega$ because $\mathrm d\hat u=(\boldsymbol\Omega\times\hat u)\mathrm dt\perp \boldsymbol\Omega$. – AccidentalFourierTransform Apr 10 '16 at 16:27
• Note that $\wedge$ is the wedge operator that yields a bivector, use $\times$ for cross product for the result to be a vector. – ja72 Apr 10 '16 at 16:28
• In your diagram the ${\rm d}\theta$ on the top is not equal to the ${\rm d}\theta$ on the bottom. The triangles are not similar. – ja72 Apr 10 '16 at 16:31
• @ja72 Thanks for the reply, I got it but then does this imply that the angular velocity of the precession and the angular velocity of the circular motion are different? – Sørën Apr 10 '16 at 21:18
• No it does not. – ja72 Apr 11 '16 at 2:27

## 2 Answers

I think sections 4.1.2 and 4.1.3 of this lecture on dynamics explains it by looking at each component separately. Since $\frac{{\rm d}}{{\rm d}t} \sin \theta = \dot{\theta} \cos\theta$ and $\frac{{\rm d}}{{\rm d}t} \cos \theta = -\dot{\theta} \sin\theta$ the components of ${\rm d}u$ are perpendicular to $u$.

Our first step is to choose cartesian axes, even though our aim is eventually to abandon them. We now assume, for convenience, that our given set of axes are such that $\hat{k} = (0, 0, 1)$, with the origin at a point on the axis of rotation a distance h from the plane of motion. In these axes, the position vector of the particle is given by $$r = (a \cos θ, a \sin θ, h)$$

We can differentiate this to obtain the velocity of the particle: $$\dot{r} = (−a \dot{θ} \sin θ, a \dot{θ} \cos θ, 0) ≡ ω × r$$ where $ω = \dot{θ} \hat{k}$

But why must $d\hat{u}$ be orthogonal to $\vec{\Omega}$ too (i.e. be tangential to a circle orthogonal to $\vec{\Omega})$?

To get such a precession there must be a clockwise torque in the plane of the screen acting on the system which means that the torque vector must be pointing into the screen.
That torque produces a change in the angular momentum into the screen and in the diagram the direction of the change in angular momentum is the direction of $d\hat u$.
Given that $\vec \Omega$ is in the plane of the screen then $d\hat u$ and $\vec \Omega$ are orthogonal.

I want to prove that $\dfrac{d\hat{u}}{dt}=\vec{\Omega}\times \hat{u}$

$\vec \Omega \times \hat u$ has a magnitude of $\Omega u \sin \alpha$ and is in the direction of $d\hat u$ (into the screen).

For the sector in the plane orthogonal to $\vec \Omega$

$du = d\theta \; u \sin \alpha\Rightarrow \dfrac {du}{dt} = \dfrac {d\theta}{dt} \; u \sin \alpha= \Omega \; u \sin \alpha$

$\Rightarrow \dfrac{d\hat{u}}{dt}=\vec{\Omega}\times \hat{u}$

As was pointed out the two angles $d\theta$ in your diagram are not the same.

• Thanks, but if $\Omega$ is the angular velocity of the precession ($d\theta$ is the one below), can I still write $\frac{d\theta}{dt}=\Omega$ (where this time $d\theta$ is the one on the circle), considered that the two $d\theta$ are different? – Sørën Apr 10 '16 at 21:21
• I think that $\vec \Omega$ relates to the change of angle in a plane which is orthogonal to $\vec \Omega$? – Farcher Apr 10 '16 at 21:26