B Notation for a "scalar absolute field"?

AI Thread Summary
The discussion centers on a proposed notation for describing a scalar absolute field derived from a vector field in three-dimensional space. The notation suggests integrating the absolute values of vector components without cancellation, leading to a scalar result that represents the field's magnitude. An example illustrates how to calculate the resultant force from two vectors, emphasizing the importance of summing absolute values before normalizing. The conversation also touches on the application of this method to gravitational and other force fields, advocating for a straightforward approach without the need for new notation. The proposed method effectively simplifies the integration of scalar magnitudes for various forces.
timjdoom
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I have a mathematical expression which I'm unsure of the formal representation or name of. The best way I can think to describe it is as a "scalar absolute field".
The notation I think best describes it is

## F = \lVert\int^{space}_s|\vec{V}|ds\rVert ##

So you have a vector field V in a 3d space. For each point you integrate over all of space (similar to a gravitational or electromagnetic field) *but* vectors in opposite directions do not cancel, they add to the magnitude (i.e. you integrate the absolute of the vector). Then finally you want the scalar value, so you normalise the vector.

For context the python (numpy) code to express it is given say a 4-dimensional array `vec` (a 3D vector in 3D space) might be

Python:
xyz = np.array([np.sum(dimension) for dimension in np.abs(vec)])
field = np.sum(xyz**2, axis=0)**0.5

Example for gravitation (or charge) in a 2D space
- You have a mass ##M_a## at position a, exerting a force ##\vec{F_a} = -3\hat{x} +3\hat{y}## on point p.
- Next you have a force ##M_b## at position b, exerting a force ##\vec{F_b} = +2\hat{x} +2\hat{y}## on point p.
- The vector force at point p is therefore ##\vec{F_p} = \vec{F_a}+\vec{F_b} = -1\hat{x} +5\hat{y}##
- Meaning the absolute of the vector is ##|\vec{F_p}| = 1\hat{x} + 5\hat{y}##, and normalising would give you the scalar of ##\lVert\vec{F_p}\rVert = \sqrt{1^2+5^2} = \sqrt{26}##
- What I need is ##|\vec{F_a}|+|\vec{F_b}| = (|-3|+|2|)\hat{x}+ (|3|+|2|)\hat{y} = 5\hat{x}+5\hat{y}##, which gives you the scalar ##\lVert|\vec{F_p}|\rVert = \sqrt{5^2+5^2} = \sqrt{50}##
 
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