What does it mean that the gradient is perpendicular/paralell to a vector?

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Discussion Overview

The discussion revolves around the physical interpretation of the gradient operator in relation to velocity fields, specifically solenoidal and irrotational fields. Participants explore the implications of the gradient being perpendicular or parallel to a vector, while addressing issues related to terminology and mathematical representation.

Discussion Character

  • Technical explanation, Debate/contested, Conceptual clarification

Main Points Raised

  • Some participants assert that for a solenoidal velocity field, the divergence condition implies that the gradient operator is perpendicular to the velocity vector.
  • Others argue that for an irrotational velocity field, the curl condition suggests that the gradient operator is parallel to the velocity vector.
  • A participant questions the physical meaning of having the gradient operator parallel or perpendicular to a vector, seeking clarification on this concept.
  • There is a challenge regarding the terminology used, with one participant emphasizing that the gradient operator (del) should not be described as a vector in the same sense as a physical vector in ℝ3.
  • Another participant expresses confusion due to conflicting information presented in a lecture, indicating a discrepancy between their understanding and that of their professor.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the gradient operator in relation to velocity fields. There are competing views on the terminology and the physical implications of the gradient being perpendicular or parallel to a vector.

Contextual Notes

Some participants highlight potential misunderstandings regarding the mathematical definitions and implications of the gradient operator, particularly in the context of solenoidal and irrotational fields. There are unresolved questions about the physical interpretation of these relationships.

crocomut
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For a solenoidal velocity field [ tex ] \nabla \cdot \mathbf{u} [ /tex ] which means that [ tex ] \nabla [/tex ] is perpendicular to [ tex ] \mathbf{u} [ /tex ].

Similarly, for an irrotational velocity field [ tex ] \nabla \times \mathbf{u} [ /tex ] which means that [ tex ] \nabla [/tex ] is parallel to [ tex ] \mathbf{u} [ /tex ].

So what exactly does it mean physically to have a gradient (of nothing) parallel/perpendicular to a vector?

PS - what's up with latex not working?
 
Last edited:
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crocomut said:
For a solenoidal velocity field \nabla \cdot \mathbf{u} which means that \nabla is perpendicular to \mathbf{u}.

Similarly, for an irrotational velocity field \nabla \times \mathbf{u} which means that \nabla is parallel to \mathbf{u}.

So what exactly does it mean physically to have a gradient (of nothing) parallel/perpendicular to a vector?

PS - what's up with latex not working?

Don't put spaces in square brackets for tags. I fixed this issue in the quote above.

What you've posted doesn't make much sense. There seems to be many problems with it. For a solenoidal velocity field \mathbf{u}, \nabla \cdot \mathbf{u} is the divergence of u.

\nabla is known as http://mathworld.wolfram.com/Del.html" , and not "a gradient". It may be thought of like a vector of differential operators.
 
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Sorry, let me correct and ask again:For a solenoidal velocity field \nabla \cdot \mathbf{u} = 0 which means that \nabla is perpendicular to \mathbf{u}.

Similarly, for an irrotational velocity field \nabla \times \mathbf{u} = \mathbf{0} which means that \nabla is parallel to \mathbf{u}.

So what exactly does it mean physically to have \nabla parallel/perpendicular to the velocity vector?
 
Del may be thought of as a vector of operators. Claiming that del is perpendicular to a vector in ℝ3 makes no sense, like assigning a real number value to the plus sign.
 
Your answer is exactly what I was thinking but, as you can see from http://i.imgur.com/VmbKS.jpg"in my hydrodynamics lecture, it is not what my professor claims. Hence the confusion.
 
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