The angular velocity vector $\boldsymbol{\omega}$ is defined as: $$\boldsymbol{\omega}=\frac{d\boldsymbol{\phi}}{dt}.$$ For the circular motion on the $xy-$plane, $\boldsymbol{\omega}$ is perpendicular to the $xy$-plane i.e., $\boldsymbol{\omega}=\omega\hat{\textbf{z}}$ and the infinitesimal angular displacement vector $\delta{\boldsymbol{\phi}}=\delta{\phi}\hat{\boldsymbol{\phi}}$ is directed tangential (or is it?) to the circular path.

This relation $\boldsymbol{\omega}=\frac{d\boldsymbol{\phi}}{dt}$ cannot be justified (as a vector equation) if the LHS and RHS have different directions. But how can, a scalar operator ($\frac{d}{dt}$), change the direction of $\delta{\boldsymbol{\phi}}$ to coincide with that of $\boldsymbol{\omega}$?

EDIT: Consider the position vector of a particle in 2-D (in Plane polar coordinates $(r,\phi)$) moving in any arbitrary path: $$\textbf{r}=r\hat{\textbf{r}}$$ Taking the time-derivative, one obtains, $$\textbf{v}=\frac{d}{dt}\textbf{r}=\dot{r}\hat{\textbf{r}}+r\dot{\phi}\hat{\boldsymbol{\phi}}$$ We have acted $d/dt$ on $\textbf{r}$ (which was directed along $\hat{r}$) but the velocity has both $\hat{r}$ and $\hat{\phi}$ components. How is this possible when $d/dt$ is a scalar operator?

  • $\begingroup$ It is because the $\hat r$ and $\hat {\phi}$ are not fixed in time($\hat i$, $\hat j$, $\hat k$ are !). Their time evolution is intertwined. $\endgroup$ – Prasad Mani Oct 20 '16 at 14:28
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    $\begingroup$ One chan show that, under arbitrary rotation $d\textbf{r}=\textbf{r}^\prime-\textbf{r}=\textbf{r}\times d\boldsymbol{\phi}$. Specializing to circular motion in $xy$-plane one obtains $d\textbf{r}=ds\hat{\phi}=r(\hat{r}\times d\boldsymbol{\phi})$ which forces $d\boldsymbol{\phi}$ to the along $\hat{z}$. Therefore, both $d\boldsymbol{\phi}$ and $\boldsymbol{\omega}$ point along $\hat{z}$. $\endgroup$ – SRS Oct 20 '16 at 15:16
  • $\begingroup$ The angular velocity vector is not defined as $\omega = \frac{{\rm d} \phi}{{\rm d}t}$ it is defined as $$\frac{{\rm d}\vec{s}}{{\rm d}t} = \vec{\omega} \times \vec{s}$$ $\endgroup$ – ja72 Oct 20 '16 at 18:36

When you differentiate a vector quantity $\mathbf x$ with respect to time the direction of the differential is going to be the direction of the infinitesimal $\mathrm d\mathbf x$. That is the direction of the vector:

$$\mathrm d\mathbf x= \mathbf x(t+\mathrm dt) - \mathbf x(t)$$

The direction of an angular displacement isn't tangential. The rotation vector (actually a pseudovector) is obtained by multiplying the angle by a unit vector that points along the axis. So the rotation vector points in the same direction as the angular velocity. Ths means $\mathrm d\phi$ also points along this axis. So the operator $\mathrm d/\mathrm dt$ is not changing the direction of the vector.

However this is a bit of a special case because in rotation all the vectors point in the same direction. You give the example of differentiating a position vector, where the differential doesn't point in the same direction as the position vector. But that is because the direction of the position vector $\mathbf r$ does change with time. If you took some special case like the particle moving radially outwards then $\mathbf r$ and $\mathrm d\mathbf r/\mathrm dt$ would point in the same direction.

  • $\begingroup$ In response to your answer, I have edited my question include a counter-example where the scalar operator $d/dt$ does unambiguously alter the direction of the vector it operates on. But I don't understand physically how this happens. $\endgroup$ – user133658 Oct 20 '16 at 14:18
  • $\begingroup$ Last line of your answer (agreeing to my idea that $d/dt$, being a scalar operator, cannot change direction) contradicts my example as explained in EDIT. $\endgroup$ – user133658 Oct 20 '16 at 14:21
  • $\begingroup$ Dear John, I'd appreciate if you take a look at this (and if possible answer it): physics.stackexchange.com/questions/287730/… $\endgroup$ – user.3710634 Oct 21 '16 at 2:40

The first thing is that $\hat r$ and $\hat \phi$ are not fixed vector as are $\hat i, \hat j$ and $\hat k$ and they are real vectors.

To answer your question.
All the vectors drawn in the diagram are are coplanar.

enter image description here

So your angular velocity $\frac{d \phi}{dt}$ vector must point at right angles to this plane in the $\hat z$ direction.

Although difficult to draw I have found that drawing the digram makes interpretation of the Mathematics easier.

On going from $A$, position vector $\vec r$, to $B$, position vector $\vec r + d\vec r$, there is a rotation of $d \phi \;\hat z$ and a change of position of $d\vec r$.

The unit vector $\hat r$ changes with time.
$|d \vec r| = |\vec r| d \phi = d \phi$ in the direction of $\hat \phi$.

$\Rightarrow \dfrac{d \hat r}{dt} = \dfrac{d \phi}{dt} \hat \phi$

Similarly it can be shown that $d \vec \phi = - d \phi \;\hat r$.
The negative sign is there because the rotation of this unit vector is radially inwards ie in the opposite direction to the unit vector $\hat r$.
Note that you wrote $\delta{\boldsymbol{\phi}}=\delta{\phi}\hat{\boldsymbol{\phi}}$ albeit with a question mark.

$\vec v = \dfrac {d \vec r}{dt} = \dfrac {d(r \hat r)}{dt} = \dfrac{dr}{dt} \hat r + r\dfrac {d \hat r}{dt} = \dot r \hat r + r \dot \phi \hat \phi$ as you showed.


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