Relative velocity between two points on a line of sight

Velocity components along the line of sight   [l] Consider one particular line of sight which will be described by equation 4.8. This is depicted in a 2-D version in figure 4.10 (ie. the z component is not shown). The figure shows just two rings with the line of sight (labelled `los') and starting at position $(r_{mid},0,0)$ (ie. the centre of the innermost ring). Two of the velocity components are shown (namely ${\bf V_r}$ and ${\bf V_\theta}$) for the intersection with the outer edge of the second ring and the line of sight vector, ${\bf X}$. It is necessary to calculate the components of each velocity vector that lies along the line of sight at each intersection with a disk or a cylinder (as well as at the start of the line of sight). In order to do this the angle between the line of sight and each component of velocity must be known. This is best calculated using the definition of the dot product for vectors:

$${\bf V}.{\bf X}=\left\vert V\right\vert \left\vert X \right\vert \cos \theta$$ (4.12)

where

$${\bf X}=\left(\cos \theta_l,\sin \theta_l, z_l \right)$$

Note that the length of ${\bf X}$ is unimportant for these purposes (ie. the value of $b$ in equation 4.8) as is the position (ie. ${\bf x_1}$ in equation 4.8) as it is only the angle between this line and ${\bf V}$ that is of interest.

The vector ${\bf V}$ (the total velocity vector) takes on different forms depending on the location along the line of sight for   which the angle is being calculated. At the origin of the line of sight it has components:

 

$${\bf V_r(x,y,z)} = (1,0,0)$$

$${\bf V_\theta(x,y,z)} = (0,1,0)$$ (4.13)

$${\bf V_z(x,y,z)} = (0,0,1)$$

Note that these are the cylindrical co-ordinate system components, but their components are given in cartesians (the dot product needs orthogonal axes to work). For a general position along the line of sight the situation is more complex (these are the components drawn in figure 4.10 - again the components are cartesians):

 

$${\bf V_r(x,y,z)} = (r_{mid}+\frac{b}{\sqrt{1+z_{los}^2}}\cos \theta_l, \frac{b}{\sqrt{1+z_{los}^2}} \sin \theta_l,0)$$

$${\bf V_\theta(x,y,z)} = (-\frac{b}{\sqrt{1+z_{los}^2}} \sin \theta_l, r_{mid}+\frac{b}{\sqrt{1+z_{los}^2}} \cos \theta_l,0)$$ (4.14)

$${\bf V_z(x,y,z)} = (0,0,1)$$

where $r_{mid}$ is the starting position for the line of sight and $\theta_l$ is the angle of the line of sight (as previously defined). Now by picking the required value of $b$ the angle between the line of sight and the velocity components at any point on the line of sight can be calculated. Using equations 4.13, 4.14 & 4.12 the solutions are:

$$\cos \theta_{r_1} = \frac{\cos \theta_l}{\sqrt{1+z_l^2}}$$

$$\cos \theta_{\theta_1} = \frac{\sin \theta_l}{\sqrt{1+z_l^2}}$$ (4.15)

$$\cos \theta_{z_1} = \frac{z_l}{\sqrt{1+z_l^2}}$$

where the 1 subscript signifies the angle at the start of the line of sight. For the general case:

$$\cos \theta_{r_2} = \frac{r_{mid} \cos \theta_l+b}{\sqrt{r_{mid}^2+2br_{mid}\cos \theta_l +b^2} \hspace*{2mm} \sqrt{1+z_l^2}}$$

$$\cos \theta_{\theta_2} = \frac{r_{mid} \sin \theta_l} {\sqrt{r_{mid}^2+2br_{mid}\cos \theta_l +b^2}\hspace*{2mm} \sqrt{1+z_l^2}}$$ (4.16)

$$\cos \theta_{z_2} = \frac{z_l}{\sqrt{1+z_l^2}}$$

where the 2 subscript represents the angle at the desired end position on the line of sight.

So the velocity component along the line of sight is given by

 

$$V_1=V_{r_1} \cos \theta_{r_1} + V_{\theta_1} \cos \theta_{\theta_1} + V_{z_1} \cos \theta_{z_1}$$

$$V_2=V_{r_2} \cos \theta_{r_2} + V_{\theta_2} \cos \theta_{\theta_2} + V_{z_2} \cos \theta_{z_2}$$ (4.17)

The difference between these two values is the relative velocity between the start position of the line of sight and the intersection of interest. This now means that each segment on the line of sight has its velocity relative to the starting point of the line of sight defined at both ends. In the main program these values will be averaged or interpolated to yield a relative velocity for the entire segment.