- #1
timjdoom
- 6
- 3
- TL;DR
- 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
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}##
## 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)**0.5Example 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}##