What is the purpose of these mathematical operations?

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Homework Help Overview

The discussion revolves around the mathematical operations of the del operator, divergence, and Laplacian, focusing on their definitions and applications in vector and scalar fields.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions and applications of the del operator, divergence, and Laplacian, questioning their correct usage and the nature of the outputs when applied to scalars and vectors.

Discussion Status

Some participants express uncertainty about the definitions and applications of these operators, while others challenge the original poster's interpretations and suggest a need for clearer definitions and proper notation.

Contextual Notes

There is a mention of the necessity to apply these differential operators to vector or scalar fields, indicating a potential misunderstanding in their application by the original poster.

Bucky
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I've been reading up on these three recently, and wondered if anyone could confirm what I think they do. I'm not 100% I understand these.

del (\bigtriangleup), when applied to a scalar, creates a vector with that scalar as each of the XYZ values. eg

\bigtriangleup . x = (x,x,x)
\bigtriangleup . 3 = (3,3,3)

divergence is applied to a vector, and sums the components of the vector into a scalar. eg

\bigtriangleup . (x,y,z) = x+y+z
\bigtriangleup . (1,2,3) = 1+2+3 = 6


finally, laplacian. This is the one I'm not as sure about. It's applied to a scalar I think?

\bigtriangleup ^2 = \bigtriangleup(\bigtriangleup)
\bigtriangleup ^2 . x = \bigtriangleup(\bigtriangleup . x)
= \bigtriangleup((x,x,x))
= 3x

That doesn't seem right (I think I'm meant to end up with a vector). Can laplacian be broken up like that or does it have a special rule?
 
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Bucky said:
I've been reading up on these three recently, and wondered if anyone could confirm what I think they do. I'm not 100% I understand these.

del (\bigtriangleup), when applied to a scalar, creates a vector with that scalar as each of the XYZ values. eg

\bigtriangleup . x = (x,x,x)
\bigtriangleup . 3 = (3,3,3)

divergence is applied to a vector, and sums the components of the vector into a scalar. eg

\bigtriangleup . (x,y,z) = x+y+z
\bigtriangleup . (1,2,3) = 1+2+3 = 6


finally, laplacian. This is the one I'm not as sure about. It's applied to a scalar I think?

\bigtriangleup ^2 = \bigtriangleup(\bigtriangleup)
\bigtriangleup ^2 . x = \bigtriangleup(\bigtriangleup . x)
= \bigtriangleup((x,x,x))
= 3x

That doesn't seem right (I think I'm meant to end up with a vector). Can laplacian be broken up like that or does it have a special rule?

No.

Those are differential operators so they must applied to vector fields or scalar fields

The divergence applied to a vector field gives a number .
The gradient applied to a scala field gives a vector.
The laplacian may be applied to a scalar field or to a vector field producing something of the same nature as what it was applied to.

You can't apply those things to a single vector.
 
ok, is method right? Regardless of the size of the field, are you still doing "that" to each element?
 
Bucky,

Nothing in your original post is correct. You need to use the definition of the del operator, together with the definition of a dot product. To write a dot product of del with some scalar is nonsense.

Also, I've never seen anyone write the del operator the way that you have done it. It is always written \nabla. The Laplacian on the other hand can be written either as \nabla^2 or \bigtriangleup.
 

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