# I How do we divide one vector field by another?

1. Apr 29, 2016

Let's say we have two vector fields, described by 6 functions: Ax, Ay, Az and Bx, By, Bz.

We want to divide field A by field B.

Do we take Ax/Bx , Ay/By and Az/Bz individually?

But in this case we might end up with Three different scalar fields.

What's the proper way to do this?

2. Apr 29, 2016

### andrewkirk

There is no standard definition of division of vector fields. But in a given context, for a particular purpose, there may be a use in defining something that shares some properties with division as defined for other algebraic objects.

The answer to the question will come from reflecting on why you want to do this.

3. Apr 30, 2016

Thanks. I've figured it out the context of the division.

4. Apr 30, 2016

### mathwonk

you can normally only divide two vectors if they have the same direction, so on a smooth curve you can divide two tangent vector fields, since the tangent space is everywhere one dimensional. i.e. when one vector is a number times the other, then their quotient is that number.

5. Apr 30, 2016

### zinq

Of course, two vector fields on a 1-dimensional manifold can be divided, as long as the denominator vector field is nowhere zero. Because, at each point they belong to the tangent space at that point, which is the real numbers.

For example, on the circle S1 parametrized by t, 0 ≤ t ≤ 2π, we could have the vector fields

V(t) = e d/dt = e1 d/dt

and

W(t) = esin(t)2 d/dt.

In this case, we have

f(t) = V(t) / W(t) = e1 - sin(t)2,

= ecos(t)2

which is just a scalar field, or in other words just a real-valued function.

This is just saying that

f(t) W(t) = V(t).

(The same thing is also possible if the space on which the two vector fields are defined is a Riemann surface M. Since then at each point of M the vector of each vector field lies in the complex numbers.)

Last edited: Apr 30, 2016