Problem interpreting the divergence result

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Homework Help Overview

The discussion revolves around interpreting the divergence of a vector field, specifically the function V defined as V=x^2 \boldsymbol{\hat {x}}+3xz^2\boldsymbol{\hat {y}}-2xz\boldsymbol{\hat {z}}. The original poster expresses confusion regarding the result of zero divergence, questioning its implications and the expected behavior of the vector field.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why the divergence results in zero and questions whether this implies certain properties about the vector field at different points. Some participants explore the analogy of fluid dynamics to clarify the meaning of divergence, while others question the implications of divergence being zero at a point versus over a volume.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of divergence and its relationship to fluid dynamics. There is an exploration of different interpretations of divergence, particularly regarding point-wise versus volume considerations, but no consensus has been reached.

Contextual Notes

Participants are navigating assumptions about the behavior of the vector field and its implications in the context of divergence, including the relationship between vector magnitude and direction at different points.

Karol
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Homework Statement


I take the divergence of the function:
V=x^2 \boldsymbol{\hat {x}}+3xz^2\boldsymbol{\hat {y}}-2xz\boldsymbol{\hat {z}}
And get zero. the answer doesn't make sense, since i expect to get a zero divergence only for a function that looks like the one in the drawing attached.

The Attempt at a Solution


\nabla \cdot V=2x+0-2x=0
I test to see whether the function V behaves like in the drawing.
The function V at an arbitrary point, (1,1,1) is:
V_(1,1,1)=1\boldsymbol{\hat {x}}+3\boldsymbol{\hat {y}}-2\boldsymbol{\hat {z}}
And, on another arbitrary point, let's say (2,2,2):
V_(2,2,2)=4\boldsymbol{\hat {x}}+6\boldsymbol{\hat {y}}-6\boldsymbol{\hat {z}}
The vectors are different, not the same like i expected.
 

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Hi Karol! :smile:

Think of a vector field V as showing the velocity of a fluid at each point.

If the fluid has constant density ("incompressible"), then divV = 0.

divV = 0 is another way of saying that, for a fixed region of space, the flow in equals the flow out.

Alternatively, if we look at a fixed mass of fluid, divV = 0 says that that fixed mass may change shape, but it will always have the same volume. :wink:
 
Does Div V=0 also mean that the vector V at point B must have the same length as the vector at point A, but may change it's direction, like in the new drawing attached?
But the function V doesn't even support this state since the magnitude of the two vectors i have calculated, at points (1,1,1) and (2,2,2) isn't the same!
I see no regularity in V, but, of course, i cannot know from inspecting only 2 points
 

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Hi Karol! :smile:

(just got up :zzz:)
Karol said:
Does Div V=0 also mean that the vector V at point B must have the same length as the vector at point A …

No.

Imagine V is the velocity of water flowing through a pipe that's wider near the end …

the water will slow down, but it won't change density …

|V| is less, but divV = 0 …

A disc of water will get wider but thinner … same volume, different shape :wink:

If you drew arrows to show the flow, you would have to draw shorter arrows, but more of them.​
 
Thanks, Tim, but the divergence is in a point, and you are talking about volumes of water
 
Karol said:
Thanks, Tim, but the divergence is in a point, and you are talking about volumes of water

Agreed, but div is a derivative, so …

i] I'm allowed to talk about an infinitesimal volume

ii] I'm allowed to integrate over a finite volume :wink:

(also, you are talking about divV = 0 everywhere: so divV = 0 over a volume)
 

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