Spherical coordinates, vector field and dot product

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Homework Help Overview

The discussion revolves around demonstrating that two vector fields, A and B, expressed in spherical coordinates, are everywhere parallel. Participants are exploring the implications of the dot product and the conditions for parallelism in vector fields.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the dot product and the angle between the vector fields, questioning how to determine if the fields are parallel based on the dot product result. There is confusion regarding the conditions for parallelism versus perpendicularity.

Discussion Status

The conversation is ongoing, with participants providing insights and corrections regarding the properties of the dot product. Some participants are reconsidering their understanding of the angle between the vector fields and the implications of their calculations.

Contextual Notes

There is mention of potential confusion regarding the definitions of angles in spherical coordinates and how they relate to the vector fields in question. Participants are also reflecting on the impact of sleep on their cognitive performance while tackling the problem.

Rombus
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Homework Statement



Show that the vector fields A = ar(sin2θ)/r2+2aθ(sinθ)/r2 and B = rcosθar+raθ are everywhere parallel to each other.

Homework Equations


\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos(0)

The Attempt at a Solution



So, if the dot product equals 1. They should be parallel correct?

A={sin(2θ)/(r2),2(sin(θ)/r2),0}
B={rcos(θ),r,0}

if this is the dot product how do I determine the angle between the vectors?
(2 Sin(θ))/r + (Cos(θ) Sin(2 θ))/r

Do i need to transform to rectangular coordinates?
 
Last edited:
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Rombus said:
|A||B|=ABcosθ

You'll want to double check this equation :wink:
So, if the dot product equals 1. They should be parallel correct?

Careful, \mathbf{i}+2\mathbf{j} and \mathbf{i} are not parallel, but their dot product is 1. Likewise, \mathbf{i}+\mathbf{j} and 2\mathbf{i}+2\mathbf{j} are parallel but their dot product is not equal to 1.

If 2 vector fields are parallel, what can you say about the angle between them at every point? What does the dot product formula then tell you?
 
Hello, thanks for the reply. I blame lack of sleep on my dot product equation mishap. :zzz:

So, the angle between the vector fields is 90 degrees and the dot product would be 0 correct?
 
Rombus said:
Hello, thanks for the reply. I blame lack of sleep on my dot product equation mishap.

So, the angle between the vector fields is 90 degrees and the dot product would be 0 correct?

:zzz: a 15 minute nap can sometimes do a world of good for one's studies :wink:

If the angle between two vector fields is 90 degrees, then they are perpendicular, not parallel:wink:
 
of course! Okay, so the angle is zero or 180. So upon finding the dot product how would I determine the angle between these two fields from the result of this dot product? (2 Sin(θ))/r + (Cos(θ) Sin(2 θ))/r

Do I just plug in zero for theta?
 
Rombus said:
of course! Okay, so the angle is zero or 180.

Wouldn't 180 degrees mean the vector fields were anti-parallel?:wink:

So upon finding the dot product how would I determine the angle between these two fields from the result of this dot product? (2 Sin(θ))/r + (Cos(θ) Sin(2 θ))/r

Do I just plug in zero for theta?

No, the θ in the equations for your 2 vector fields is either the polar angle (the angle between the position vector and the polar axis) in spherical coordinates, or the azimuthal angle (the angle between the projection of the position vector onto the xy-plane, and the x-axis), depending on which naming convention you are using for spherical coordinates.

That θ is, in general, not the same as the angle between the two vector fields.

If the angle between the two vector fields is 0, then the (correct) dot product equation tells you \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos(0)
 

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