How to find the displacement vector in Spherical coordinate

[Aswin Jagadeesh A]

Is there a way of subtracting two vectors in spherical coordinate system without first having to convert them to Cartesian or other forms?

Since I have already searched and found the difference between Two Vectors in Spherical Coordinates as,

$$|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'[\cos(\theta-\theta')+\sin\theta\sin\theta'(cos(\phi-\phi')-1))])^{\frac{1}{2}}$$

which I believe the radius of displaced vector. I still didn't get any way to find the theta (angle from positive z axis)and psi(angle from positive x axis).

For more information please refer this link. math.stackexchange.com/a/1365667/613522 This helped me with the cylindrical coordinate case. I Just want a similar solution in Spherical Coordinate system.
Please help.

 

[BvU]

Well, see the comment by david K in your link

I do not think I want to attempt this in spherical coordinates or in any higher dimension

If you insist, take a few simple cases (i.e pare down to 2D polar first) and then generalize.

 

[Aswin Jagadeesh A]

I have already tried that and failed to get the result.
I believe there will be a separate formula for calculating the angles $\theta$ and $\phi$ like the way $\beta$ had found as mentioned in the link.
I'm seeking for the same or other better options.
Thanks for your reply though.

 

[jasonRF]

The standard way to compute $|\mathbf{r}-\mathbf{r}^\prime|$ is to write each position vectors as
$$ \mathbf{r} = \hat{\mathbf{x}}\, x + \hat{\mathbf{y}}\, y + \hat{\mathbf{z}}\, z = \hat{\mathbf{x}}\, r\, \cos\phi\, \sin\theta + \hat{\mathbf{y}}\, r\, \sin\phi\, \sin\theta + \hat{\mathbf{z}}\, r\, \cos\theta$$.
Then the algebra is easy.

Also, in the future you should mark threads at this level as "intermediate", not "advanced".Hope that helped.

jason

 

[Aswin Jagadeesh A]

The standard way to compute $|\mathbf{r}-\mathbf{r}^\prime|$ is to write each position vectors as
$$ \mathbf{r} = \hat{\mathbf{x}}\, x + \hat{\mathbf{y}}\, y + \hat{\mathbf{z}}\, z = \hat{\mathbf{x}}\, r\, \cos\phi\, \sin\theta + \hat{\mathbf{y}}\, r\, \sin\phi\, \sin\theta + \hat{\mathbf{z}}\, r\, \cos\theta$$.
Then the algebra is easy.

Also, in the future you should mark threads at this level as "intermediate", not "advanced".Hope that helped.

jason

Thank you for the info, I'll keep the prefix thing in mind.
But the expression you have mentioned is not helping. I need to calculate the difference in $\theta, \phi$ without converting it into Cartesian form.
Since the problem which I'm dealing here is kind of complex,
lets say there are two points, one initial and the other final. I have the $\r,\theta,\phi$ values of each. Now without changing it into any other form is it possible to find their difference like $\beta$ found in the link mentioned, I believe they used pure trigonometry.

 

[jasonRF]

You don't need to turn it into Cartesian form. Following through with what I wrote yields,
$$|\mathbf{r}-\mathbf{r}^\prime|^2 = \left( r\, \cos\phi\, \sin\theta - r^\prime\, \cos\phi^\prime\, \sin\theta^\prime \right)^2 + \left( r\, \sin\phi\, \sin\theta - r^\prime\, \sin\phi^\prime\, \sin\theta^\prime \right)^2 + \left(r\, \cos\theta - r^\prime \, \cos\theta^\prime\right)^2$$
Now just expand and use trig identities. The differences between angles will show up if you use the right trig identities to simplify this expression. You just need to remember that $\cos a \cos b + \sin a \sin b = \cos(a-b)$.

jason

 

[Aswin Jagadeesh A]

You don't need to turn it into Cartesian form. Following through with what I wrote yields,
$$|\mathbf{r}-\mathbf{r}^\prime|^2 = \left( r\, \cos\phi\, \sin\theta - r^\prime\, \cos\phi^\prime\, \sin\theta^\prime \right)^2 + \left( r\, \sin\phi\, \sin\theta - r^\prime\, \sin\phi^\prime\, \sin\theta^\prime \right)^2 + \left(r\, \cos\theta - r^\prime \, \cos\theta^\prime\right)^2$$
Now just expand and use trig identities. The differences between angles will show up if you use the right trig identities to simplify this expression. You just need to remember that $\cos a \cos b + \sin a \sin b = \cos(a-b)$.

jason

Thank You so much for the contribution. But still one doubt left. Through this we can only calculate the difference in $r$ right, not $\theta,\phi$? So how to find those?

 

[jasonRF]

What exactly are you trying to do? I thought i understood but clearly I do not. Please give us a clear question so we can help (and sending us a link to a long stackexchange thread doesn't count as a question).

Jason

 

[Aswin Jagadeesh A]

What exactly are you trying to do? I thought i understood but clearly I do not. Please give us a clear question so we can help (and sending us a link to a long stackexchange thread doesn't count as a question).

Jason

This is my case,
I have two points in $r,\theta,\phi$ form. I need to find out their difference. And in the end I need to represent this difference in $x,y,z$ , (which I believe I can do it by spherical to Cartesian conversion.). The ling to stackexchange which I provided help me to find the same in cylindrical coordinate, the $r, \beta$ mentioned in the same helped me to find the answer. Now here I'm looking for similar equations.
I hope I explained my requirement clearly.

 

[jasonRF]

I have two points in $r,\theta,\phi$ form. I need to find out their difference. And in the end I need to represent this difference in $x,y,z$ , (which I believe I can do it by spherical to Cartesian conversion.).

If you need the answer in Cartesian coordinates why don't you just convert first, then take the difference?

 

[Aswin Jagadeesh A]

If you need the answer in Cartesian coordinates why don't you just convert first, then take the difference?

with all due respect, I already tried all those and failed to get the exact values. That is why I'm asking for a solution here.
For more info see the link provided below,
https://www.physicsforums.com/threa...-cylindrical-coordinates.729099/#post-4609839
(thread number 12)
This also helped me in case of cylindrical coordinates. Is there any similar equation for spherical coordinates.?

 

[jasonRF]

Wow - I wish you had started this thread the way you started that cylindrical coordinates thread (with an actual explicit question!). Anyway, the way I would do it woult be based on the equation I gave in post #4, yielding $$
\mathbf{r}_1 - \mathbf{r}_2 = \hat{\mathbf{x}}\,\left( r_1\, \cos\phi_1\, \sin\theta_1 - r_2\, \cos\phi_2\, \sin\theta_2\right)
+ \hat{\mathbf{y}}\, \left( r_1\, \sin\phi_1\, \sin\theta_1 - r_2\, \sin\phi_2\, \sin\theta_2\right) + \hat{\mathbf{z}}\, \left( r_1\, \cos\theta_1\, - r_2\, \cos\theta_2\right) $$
If you wanted, you could then express the Cartesian unit vectors in terms of spherical coordinate unit vectors, but those of course depend on location. For example,
$$\hat{\mathbf{r}}(\theta,\phi) = \hat{\mathbf{x}}\, \cos\phi\, \sin\theta + \hat{\mathbf{y}}\, \sin\phi\, \sin\theta + \hat{\mathbf{z}}\, \cos\theta $$
So you need to either pick a location (most logically $(\theta_1,\phi_1)$ or $(\theta_2,\phi_2)$) or leave the unit vectors as functions of arbitrary $(\theta,\phi)$.

Of course, my approach does use Cartesian unit vectors at the outset, which may not be what you want. Presumably you could also use Chestermiller's approach (post #2 from the cylindrical coordinates thread you linked in post #11 of this thread) and start with $\mathbf{r}_1 = \hat{\mathbf{r}}_1\, r_1$ and $\mathbf{r}_2 = \hat{\mathbf{r}}_2\, r_2$, but now you need to either write $\hat{\mathbf{r}}_1$ (which is just $\hat{\mathbf{r}}(\theta_1,\phi_1)$ in my notation above) in terms of $(\hat{\mathbf{r}}_2,\hat{\mathbf{\theta}}_2,\hat{\mathbf{\phi}}_2)$ or write $\hat{\mathbf{r}}_2$ in terms of $(\hat{\mathbf{r}}_1,\hat{\mathbf{\theta}}_1,\hat{\mathbf{\phi}}_1)$.