# A Normal velocity to the surface in Spherical Coordinate System

1. May 17, 2016

### Wisam

Let's say we have r=R( theta, phi, t) on the surface of the particle and need to find the normal vector in Spherical Coordinate system. We know that, the unit vector =grad(r-R( theta, phi, t)) / |grad((r-R( theta, phi, t))|
where grad is Spherical gradient operator in term of e_r, e_\theta, e_\phi.

Can you please help me to calculate the normal velocity to the surface in Spherical Coordinate system.

2. May 18, 2016

### andrewkirk

I'm afraid your question is too hard to understand as it is written. If you write it again more clearly, making clear in particular
- what the surface is to which you refer
- what the path is, for which you want to calculate a velocity

3. May 19, 2016

### George Jones

Staff Emeritus
If you have a unit normal to the surface, then:

1) take the dot product of this unit normal vector and the particle's velocity vector;

2) multiply the result of 1) by the unit normal vector.

Together, 1) and 2) give the part of the particle's velocity that is normal to the surface.

4. May 23, 2016

### Wisam

We know the normal for spherical particle ( for sphere we know how the normal) but I need to find the normal for non-spherical shape.
If we say the radius r=R(t)+epsilon R(theta,t)
then how can I find the normal for that form.?
I think the normal will be

n=n0+epsilon n1 (theta,t)
where n0 is the vector for (for spherical shape)??
Is that right ?

5. May 23, 2016

### Wisam

We know the normal for spherical particle ( for sphere we know how the normal) but I need to find the normal for non-spherical shape.
If we say the radius r=R(t)+epsilon R(theta,t)
then how can I find the normal for that form.?
I think the normal will be

n=n0+epsilon n1 (theta,t)
where n0 is the vector for (for spherical shape)??
Is that right ?
thank you