Difference between revisions of "Forest UCM NLM Ch1 CoordSys"

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another way to determine the unit vector derivative is to cast them in terms of cartesian coordinate.
 
another way to determine the unit vector derivative is to cast them in terms of cartesian coordinate.
  
:<math>\hat{r} = \frac{\vec{r}{r} =\cos \phi \sin \theta \hat{i} + \sin \phi \sin \theta  \hat{j} + \cos \theta \hat{k}</math>
+
:<math>\hat{r} = \frac{\vec{r}}{|\vec{r}|} =\cos \phi \sin \theta \hat{i} + \sin \phi \sin \theta  \hat{j} + \cos \theta \hat{k}</math>
  
 
:<math>\hat{\phi} = \frac{\hat {z} \times \hat {r} }{\sin \theta}=- \sin \phi \hat{i} + \cos \phi \hat{k}</math>
 
:<math>\hat{\phi} = \frac{\hat {z} \times \hat {r} }{\sin \theta}=- \sin \phi \hat{i} + \cos \phi \hat{k}</math>

Revision as of 20:44, 7 July 2014

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

Cartesian

TF UCM CartCoordSys.png


Vector Notation convention:

Position:

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

Velocity and Acceleration vector in cartesian coordinates

[math]\vec{v}[/math] = [math]\frac{d \vec{r}}{dt}[/math] = [math]\frac{d x}{dt}\hat{i} + x\frac{d \hat{i}}{dt} + \cdots[/math]


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


[math]\frac{d \hat{i}}{dt} =0 =\frac{d \hat{j}}{dt} =\frac{d \hat{k}}{dt}[/math]
[math]\vec{v}[/math] = [math]\frac{d \vec{r}}{dt}[/math] = [math]\frac{d x}{dt}\hat{i} + \frac{d y}{dt}\hat{j} + \frac{d z}{dt}\hat{k} [/math]


Similarly Acceleration is given by


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

Polar

TF UCM PolarCoordSys.png Vector Notation convention:

Position:

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

[math]\vec{r} = r \hat{r}[/math]
[math]\vec{r} \ne r \hat{r} +\phi \hat{\phi} [/math]: [math]\phi[/math] does not have the units of length


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

[math]\vec{r} = r \cos(\phi) \hat{i} + r \sin(\phi)\hat{j}[/math]

The dependence of position with [math]\phi[/math] 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 [math]t_1[/math] the particle is at [math]\vec{r}(t_1)[/math] and at time [math]t_2[/math] the particle is at [math]\vec{r}(t_2)[/math]


TF UCM PolarVectDiff.png


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

[math]\hat{r} (t_2)-\hat{r}(t_1) \equiv \Delta \hat{r} = \Delta \phi \hat{\phi}[/math]
or
[math]\frac{ d \hat{r}}{dt} = \frac{d \phi}{dt} \hat{\phi}[/math]

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

[math]\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}[/math]
[math]\vec{v} = r \frac{d \phi}{dt} \hat{\phi}[/math]


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



[math]\vec{v}[/math] = [math]\frac{d r}{dt}\hat{r} + r\frac{d \phi}{dt} \hat{\phi}[/math]
or in more compact form
[math]\vec{v}=\vec{\dot{r}} = \dot{r} \hat{r} + r \dot{\phi} \hat{\phi}= v_r \hat{r} + v_{\phi} \hat{\phi}[/math]


linear velocity [math]\equiv v_r [/math] Angular velocity [math]\equiv v_{\phi} [/math]


Finding the derivative directly

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

[math]\hat r = \cos \phi \hat{i} + \sin \phi \hat{j}[/math]
[math]\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}[/math]
[math]= \left ( \hat{\phi} \right ) \dot{\phi} [/math]

Acceleration in Polar Coordinates

Taking the derivative of velocity with time gives the acceleration


[math]\vec{a} = \frac{d \vec{v}}{dt} =\vec{\ddot{r}} [/math]
[math]= \frac{ d \left (\dot{r} \hat{r} + r \dot{\phi} \hat{\phi}= v_r \hat{r} + v_{\phi} \hat{\phi}\right)}{dt}[/math]
[math]= \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 )[/math]
[math]= \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 )[/math]


We need to find the derivative of the unit vector [math]\hat{\phi}[/math] with time.

Consider the position change below in terms of only the unit vector [math]\hat{\phi}[/math]


TF UCM PolarPhiUnitVectDiff.png


Using the same arguments used to calculate the rate of change in [math]\hat{r}[/math]:

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

[math]\hat{\phi} (t_2)-\hat{\phi}(t_1) \equiv \Delta \hat{\phi} = \Delta \phi (- \hat{r})[/math]
or
[math]\frac{ d \hat{\phi}}{dt} = -\frac{d \phi}{dt} \hat{r}[/math]
[math]\frac{d \hat{\phi}}{dt}= -\dot{\phi} \hat{r}[/math]


Finding the derivative of [math]\hat{\phi}[/math] directly

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

[math] \hat{\phi} = \sin \phi \hat{i} - \cos \phi \hat{j} [/math]
[math]\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}[/math]
[math]= \left (- \hat{r} \right ) \dot{\phi} [/math]


Substuting the above into our calculation for acceleration:


[math]\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 )[/math]
[math]= \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 )[/math]
[math]= \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} + \left ( 2\dot{r} \dot{\phi} +r \ddot{\phi} \right ) \hat{\phi} [/math]

For the case of circular motion at constant [math] r (\dot{r} = 0)[/math]

[math]\vec{a} = -r\dot{\phi}^2 \hat{r} + r \ddot{\phi} \hat{\phi} [/math]

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


angular (tangential) acceleration [math]\equiv r \ddot{\phi} \hat{\phi} = r \alpha \hat{\phi}[/math]

If [math]\dot{r} \ne 0[/math]

Then there are two additional terms

[math]\ddot{r} \hat {r}[/math] = radial acceleration
[math]2\dot{r} \dot{\phi} \hat {\phi}[/math] = Coriolis acceleration (to be described later)

Cylindrical

Cylindrical coordinates are polar coordinates with a third dimension usually labeled [math](z)[/math]

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

TF UCM CylCoordSys.png


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

[math]\vec{r} = r \hat{r} + z \hat{k}[/math]
[math]\vec{v}=\vec{\dot{r}} = \dot{r} \hat{r} + r \dot{\phi} \hat{\phi} + \dot{z} \hat{k}[/math]
[math]\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}[/math]

Spherical

TF UCM SphericalCoordSys.png


Position:

[math]\vec{r} = r \hat{r} [/math]

Velocity vector in Spherical coordinates

[math]\vec{v}[/math] = [math]\frac{d \vec{r}}{dt}[/math] = [math]\frac{d r}{dt}\hat{r} + r\frac{d \hat{r}}{dt} [/math]


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

[math]\hat{r} = \frac{\vec{r}}{|\vec{r}|} =\cos \phi \sin \theta \hat{i} + \sin \phi \sin \theta \hat{j} + \cos \theta \hat{k}[/math]
[math]\hat{\phi} = \frac{\hat {z} \times \hat {r} }{\sin \theta}=- \sin \phi \hat{i} + \cos \phi \hat{k}[/math]
[math]\hat{\theta} = \hat{\phi} \times \hat {r} =-\cos \phi \sin \theta \hat{i} - \sin \phi \sin \theta \hat{j} + \cos \theta \hat{k}[/math]

As shown above in polar coordinates

[math]\hat \dot{r} = \dot{\phi} \hat{\phi} [/math]
[math]\hat \dot{\phi} - \dot{\phi} \hat{r} [/math]


Spherical coordinates simply have an additional degree of freedom labeled the [math]\theta[/math] 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 [math]\theta[/math] and the [math]\phi[/math] angle is the azimuthal angle.

[math] =\dot{r}\hat{r} + r\dot{\theta}\sin \theta\hat{\theta} + r \dot{\phi} \hat{\phi}[/math]


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


[math]\frac{d \hat{i}}{dt} =0 =\frac{d \hat{j}}{dt} =\frac{d \hat{k}}{dt}[/math]
[math]\vec{v}[/math] = [math]\frac{d \vec{r}}{dt}[/math] = [math]\frac{d x}{dt}\hat{i} + \frac{d y}{dt}\hat{j} + \frac{d z}{dt}\hat{k} [/math]

Acceleration vector in Spherical coordinates

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


Forest_UCM_NLM#Space