Spherical coordinates, vector field and dot product

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SUMMARY

The discussion centers on the relationship between two vector fields, A and B, defined in spherical coordinates. The vector fields are A = ar(sin2θ)/r² + 2aθ(sinθ)/r² and B = rcosθar + raθ. Participants clarify that for the fields to be parallel, the angle between them must be 0 or 180 degrees, resulting in a dot product of either 1 or -1, respectively. Misunderstandings regarding the dot product and angles are addressed, emphasizing that a dot product of 1 does not necessarily indicate parallelism.

PREREQUISITES
  • Understanding of spherical coordinates and their notation
  • Familiarity with vector fields and their representations
  • Knowledge of the dot product and its geometric interpretation
  • Basic trigonometry, particularly sine and cosine functions
NEXT STEPS
  • Study the properties of vector fields in spherical coordinates
  • Learn about the geometric interpretation of the dot product
  • Explore the conditions for parallel and anti-parallel vectors
  • Investigate transformations between spherical and rectangular coordinates
USEFUL FOR

Students and educators in physics or mathematics, particularly those focusing on vector calculus and coordinate systems. This discussion is beneficial for anyone seeking to understand the relationship between vector fields in spherical coordinates.

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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