Why is the divergence like that?

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Discussion Overview

The discussion revolves around the definition of divergence in vector calculus, specifically questioning why it is defined as the sum of the partial derivatives of the components of a vector field rather than the partial derivatives of the vector field itself. The conversation explores theoretical implications and potential alternative definitions.

Discussion Character

  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why divergence is defined as the sum of the partial derivatives of the components of a vector field E instead of the partial derivatives of E itself.
  • Another participant argues that if divergence were defined as the sum of the partial derivatives of E, it would yield a vector quantity rather than a scalar, which would not effectively describe sources or sinks of a field.
  • A different participant suggests that the operation of summing the partial derivatives of E itself may not be well-defined since the del operator is a vector operator and cannot act on another vector.
  • One participant proposes an alternative definition of divergence as the sum of the partial derivatives of the magnitude of E, questioning its potential meaning.
  • Another participant explains that divergence measures the difference between the amount of fluid exiting and entering an infinitesimal volume in a velocity field, emphasizing the relevance of the direction of the components.
  • A later reply indicates a participant's understanding of the discussion, acknowledging the points made.

Areas of Agreement / Disagreement

Participants express differing views on the definition and implications of divergence, with no consensus reached on the alternative definitions proposed.

Contextual Notes

The discussion includes assumptions about the nature of vector operations and the implications of defining divergence differently, which remain unresolved.

LucasGB
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Why is the concept of divergence defined to be the sum of the partial derivatives of the x, y and z components of a vector field E with respect to x, y and z, instead of being defined as the sum of the partial derivatives of E itself with respect to x, y and z? What would this operation I just defined measure, and why is it not defined nor employed?
 
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Divergence is defined as such because were it the sum of the partial derivatives of E itself, that would yield a vector quantity as opposed to a scalar, and the direction portion would not be of use in describing the source or sink of a field at the point, not to mention it would not describe the same quantity. As far as what your operation measures, I do not know.
 
To add to that, I don't even think that would be a well defined operation. Del is a vector operator so it can't act on another vector.
 
FallenRGH said:
Divergence is defined as such because were it the sum of the partial derivatives of E itself, that would yield a vector quantity as opposed to a scalar, and the direction portion would not be of use in describing the source or sink of a field at the point, not to mention it would not describe the same quantity. As far as what your operation measures, I do not know.

But what if we defined divergence to be the sum of the partial derivatives of the magnitude of E?
 
LucasGB said:
But what if we defined divergence to be the sum of the partial derivatives of the magnitude of E?

Well, take for example a velocity field of a fluid. What divergence measures is the amount of fluid that is exiting the infinitesimal volume vs. the amount that enters. In order to do this, we take the partial derivative of each component in it's respective direction(for example we take the partial of the i component, the unit vector in the x direction, with respect to x). This demonstrates the amount of fluid that flows through the volume in the direction of the field.

The operation you are describing really is a summation of the components of the gradient of the magnitude, which does not, at least at first glance to me, have any meaning.
 
I think I get it now. Thanks!
 

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