Laplacian of a vector function

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Homework Help Overview

The problem involves evaluating the Laplacian of a vector function defined in spherical polar coordinates. The vector function A(r) is specified as having only a radial component.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the components of the vector field and the application of the Laplacian operator in spherical coordinates. There are questions about the missing components of the vector function and how to properly express the Laplacian of a vector field.

Discussion Status

Some participants have provided clarifications regarding the components of the vector field, while others are exploring the correct formulation of the Laplacian. There is an acknowledgment that the Laplacian of a vector field results in a vector field, and attempts are being made to ensure all components are accounted for.

Contextual Notes

There is uncertainty regarding the application of specific equations and the definitions of the components in the context of spherical coordinates. Participants are navigating through the implications of having a vector field with only one non-zero component.

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Problem: The vector function A(r) is defined in spherical polar coordinates by A = (1/r) er

Evaluate ∇2A in spherical polar coordinates

Relevant equation: I'm assuming I have to use the equation 1671 on this website

But I haven't got a clue as to how I would apply it since, for example, I don't know what Aθ is. Any hint on how to get started would help.
 
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You're vector field has only a r component, other components are zero!
[itex] \vec A=A_r \hat e_r +A_\theta \hat e_\theta+A_\varphi \hat e_\varphi[/itex]
 
Shyan said:
You're vector field has only a r component, other components are zero!
[itex] \vec A=A_r \hat e_r +A_\theta \hat e_\theta+A_\varphi \hat e_\varphi[/itex]

So it would just be ∇2A = ∇2Ar - 2(Ar/r2)?
 
No. The Laplacian of a vector field, is a vector field. What you wrote, is only the r component of that vector field.
 
Shyan said:
No. The Laplacian of a vector field, is a vector field. What you wrote, is only the r component of that vector field.

Right, so will it be ∇2 A = (∇2Ar - 2(A2/r2))er? Or am I missing the other components?
 
Shyan said:
That's correct.

EDIT: Oops, I meant the answer I got is (-2/r3)er
 
Last edited:

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