Forest UCM NLM Ch1 CoordSys

Cartesian, Spherical, and Cylindrical coordinate systems are commonly used to describe three-dimensional space.

Cartesian

Vector Notation convention:

Position:

$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k} = (x,y,z) = \sum_1^3 r_i \hat{e}_i$

Velocity and Acceleration vector in cartesian coordinates

$\vec{v}$ = $\frac{d \vec{r}}{dt}$ = $\frac{d x}{dt}\hat{i} + x\frac{d \hat{i}}{dt} + \cdots$

cartesian unit vectors do not change with time (unit vectors for other coordinate system types do)

$\frac{d \hat{i}}{dt} =0 =\frac{d \hat{j}}{dt} =\frac{d \hat{k}}{dt}$

$\vec{v}$ = $\frac{d \vec{r}}{dt}$ = $\frac{d x}{dt}\hat{i} + \frac{d y}{dt}\hat{j} + \frac{d z}{dt}\hat{k}$

Similarly Acceleration is given by

$\vec{a}$ = $\frac{d \vec{v}}{dt}$ = $\frac{d^2 x}{dt^2}\hat{i} + \frac{d^2 y}{dt^2}\hat{j} + \frac{d^2 z}{dt^2}\hat{k}$

Polar

Vector Notation convention:

Position:

Because $\hat{r}$ points in a unique direction given by $\hat{r} = \frac{\vec{r}}{|r|}$ we can write the position vector as

$\vec{r} = r \hat{r}$

$\vec{r} \ne r \hat{r} +\phi \hat{\phi}$: $\phi$ does not have the units of length

The unit vectors ($\hat{r}$ and $\hat{\phi}$ ) are changing in time. You could express the position vector in terms of the cartesian unit vectors in order to avoid this

$\vec{r} = r \cos(\phi) \hat{i} + r \sin(\phi)\hat{j}$

The dependence of position with $\phi$ can be seen if you look at how the position changes with time.

Velocity in Polar Coordinates

Consider the motion of a particle in a circle. At time $t_1$ the particle is at $\vec{r}(t_1)$ and at time $t_2$ the particle is at $\vec{r}(t_2)$

If we take the limit $t_2 \rightarrow t_1$ ( or $\Delta t \rightarrow 0$) then we can write the velocity of this particle traveling in a circle as

$\hat{r} (t_2)-\hat{r}(t_1) \equiv \Delta \hat{r} = \Delta \phi \hat{\phi}$

or

$\frac{ d \hat{r}}{dt} = \frac{d \phi}{dt} \hat{\phi}$

Thus for circular motion at a constraint radius we get the familiar expression

$\vec{v} = \lim_{\Delta t \rightarrow 0} \frac{\vec{r}(t_2)-\vec{r}(t_1)}{\Delta t}= \lim_{\Delta t \rightarrow 0} \frac{r\left( \hat{r}(t_2) - \hat{r}(t_1)\right)}{\Delta t} = r \frac{\Delta \phi}{\Delta t} \hat{\phi} = r \omega \hat{\phi}$

$\vec{v} = r \frac{d \phi}{dt} \hat{\phi}$

If the particle is not constrained to circular motion ( i.e.: $r$ can change with time) then the velocity vector in polar coordinates is

$\vec{v}$ = $\frac{d r}{dt}\hat{r} + r\frac{d \phi}{dt} \hat{\phi}$

or in more compact form

$\vec{v}=\vec{\dot{r}} = \dot{r} \hat{r} + r \dot{\phi} \hat{\phi}= v_r \hat{r} + v_{\phi} \hat{\phi}$

linear velocity $\equiv v_r$ Angular velocity $\equiv v_{\phi}$

Finding the derivative directly

Cast the unit vector in terms of cartesian coordinates and take the derivative.

$\hat r = \cos \phi \hat{i} + \sin \phi \hat{j}$

$\hat \dot{r} = \frac{d \hat{r}}{d \phi} \frac{d \phi}{d t} =\left( \sin \phi \hat{i} - \cos \phi \hat{j} \right ) \dot{\phi}$

$= \left ( \hat{\phi} \right ) \dot{\phi}$

Acceleration in Polar Coordinates

Taking the derivative of velocity with time gives the acceleration

$\vec{a} = \frac{d \vec{v}}{dt} =\vec{\ddot{r}}$

$= \frac{ d \left (\dot{r} \hat{r} + r \dot{\phi} \hat{\phi}= v_r \hat{r} + v_{\phi} \hat{\phi}\right)}{dt}$

$= \left ( \frac{ d \dot{r}}{dt} \hat{r} + \dot{r} \frac{ d\hat{r}}{dt} \right) + \left ( \frac{d r}{dt} \dot{\phi} \hat{\phi} +r \frac{d \dot{\phi}}{dt} \hat{\phi} +r \dot{\phi} \frac{d \hat{\phi}}{dt} \right )$

$= \left ( \ddot{r} \hat{r} + \dot{r} \dot{\phi}\hat{\phi} \right) + \left ( \dot{r} \dot{\phi} \hat{\phi} +r \ddot{\phi} \hat{\phi} +r \dot{\phi} \frac{d \hat{\phi}}{dt} \right )$

We need to find the derivative of the unit vector $\hat{\phi}$ with time.

Consider the position change below in terms of only the unit vector $\hat{\phi}$

Using the same arguments used to calculate the rate of change in $\hat{r}$:

If we take the limit $t_2 \rightarrow t_1$ ( or $\Delta t \rightarrow 0$) then we can write the velocity of this particle traveling in a circle as

$\hat{\phi} (t_2)-\hat{\phi}(t_1) \equiv \Delta \hat{\phi} = \Delta \phi (- \hat{r})$

or

$\frac{ d \hat{\phi}}{dt} = -\frac{d \phi}{dt} \hat{r}$

$\frac{d \hat{\phi}}{dt}= -\dot{\phi} \hat{r}$

Finding the derivative of $\hat{\phi}$ directly

Cast the unit vector in terms of cartesian coordinates and take the derivative.

$\hat{\phi} = \sin \phi \hat{i} - \cos \phi \hat{j}$

$\hat \dot{\phi} = \frac{d \hat{\phi}}{d \phi} \frac{d \phi}{d t} =\left( -\cos \phi \hat{i} - \sin \phi \hat{j} \right ) \dot{\phi}$

$= \left (- \hat{r} \right ) \dot{\phi}$

Substuting the above into our calculation for acceleration:

$\vec{a} = \left ( \ddot{r} \hat{r} + \dot{r} \dot{\phi}\hat{\phi} \right) + \left ( \dot{r} \dot{\phi} \hat{\phi} +r \ddot{\phi} \hat{\phi} +r \dot{\phi} \frac{d \hat{\phi}}{dt} \right )$

$= \left ( \ddot{r} \hat{r} + \dot{r} \dot{\phi}\hat{\phi} \right) + \left ( \dot{r} \dot{\phi} \hat{\phi} +r \ddot{\phi} \hat{\phi} +r \dot{\phi} \left( -\dot{\phi} \hat{r}\right) \right )$

$= \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} + \left ( 2\dot{r} \dot{\phi} +r \ddot{\phi} \right ) \hat{\phi}$

For the case of circular motion at constant $r (\dot{r} = 0)$

$\vec{a} = -r\dot{\phi}^2 \hat{r} + r \ddot{\phi} \hat{\phi}$

radial (centripetal, center seeking) acceleration $\equiv -r\dot{\phi}^2 \hat{r} = -r \omega^2 \hat{r}$

angular (tangential) acceleration $\equiv r \ddot{\phi} \hat{\phi} = r \alpha \hat{\phi}$

If $\dot{r} \ne 0$

Then there are two additional terms

$\ddot{r} \hat {r}$ = radial acceleration

$2\dot{r} \dot{\phi} \hat {\phi}$ = Coriolis acceleration (to be described later)

Cylindrical

Cylindrical coordinates are polar coordinates with a third dimension usually labeled $(z)$

change picture so angle is [math]\phi[/math] not [math]\theta[/math]

We just need to add $z\hat{k}$ to all the vectors (remember $\hat{\dot{k}} = 0$)

$\vec{r} = r \hat{r} + z \hat{k}$

$\vec{v}=\vec{\dot{r}} = \dot{r} \hat{r} + r \dot{\phi} \hat{\phi} + \dot{z} \hat{k}$

$\vec{a} = \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} + \left ( 2\dot{r} \dot{\phi} +r \ddot{\phi} \right ) \hat{\phi} + \ddot{z} \hat{k}$

Spherical

Position:

$\vec{r} = r \hat{r}$

Velocity vector in Spherical coordinates

$\vec{v}$ = $\frac{d \vec{r}}{dt}$ = $\frac{d r}{dt}\hat{r} + r\frac{d \hat{r}}{dt}$

another way to determine the unit vector derivative is to cast them in terms of cartesian coordinate.

$\hat{r} = \frac{\vec{r}}{|\vec{r}|} =\cos \phi \sin \theta \hat{i} + \sin \phi \sin \theta \hat{j} + \cos \theta \hat{k}$

$\hat{\phi} = \frac{\hat {z} \times \hat {r} }{\sin \theta}=- \sin \phi \hat{i} + \cos \phi \hat{k}$

$\hat{\theta} = \hat{\phi} \times \hat {r} =-\cos \phi \cos \theta \hat{i} - \sin \phi \cos \theta \hat{j} + \sin \theta \hat{k}$

The derivative of the above unit vectors are

$\dot{\hat{r}} = \dot{\theta} \hat{\theta} + \dot{\phi} \sin \theta \hat {\phi}$

$\dot{\hat{\phi}} = -\dot{\phi} \left ( sin \theta \hat{r} + \cos \theta \hat {\theta}$

$\dot{\hat{\theta}} = -\dot{\theta} \hat{r} + \dot{\phi} \cos \theta \hat {\phi}$

Spherical coordinates simply have an additional degree of freedom labeled the $\theta$ direction which is usually the deflection angles as projected along the Z-axis of a cartesian coordinate system. When describing the process of a scattering particel you would orient the Z-axis along the direction of the particles momentum vector. The polar angle is denoted as $\theta$ and the $\phi$ angle is the azimuthal angle.

$=\dot{r}\hat{r} + r\dot{\theta}\sin \theta\hat{\theta} + r \dot{\phi} \hat{\phi}$

cartesian unit vectors do not change with time (unit vectors for other coordinate system types do)

$\frac{d \hat{i}}{dt} =0 =\frac{d \hat{j}}{dt} =\frac{d \hat{k}}{dt}$

$\vec{v}$ = $\frac{d \vec{r}}{dt}$ = $\frac{d x}{dt}\hat{i} + \frac{d y}{dt}\hat{j} + \frac{d z}{dt}\hat{k}$

Acceleration vector in Spherical coordinates

$\vec{a}$ = $\frac{d \vec{v}}{dt}$ = $\frac{d^2 x}{dt^2}\hat{i} + \frac{d^2 y}{dt^2}\hat{j} + \frac{d^2 z}{dt^2}\hat{k}$

$\ddot{r} -r \dot{\theta}^2 \cos^2{\phi} -r \dot{\phi}^2) \frac{d^2 x}{dt^2}\hat{i} + \frac{d^2 y}{dt^2}\hat{j} + \frac{d^2 z}{dt^2}\hat{k}$

Forest_UCM_NLM#Space