For the 1D case on this page we calculated the full table for the outer product for null basis, but here the full table is a bit big so I have put it at the bottom of this page, here we just have the entries for cross multiplying two vectors:
a^b |
b.n0 | b.n∞ | b.n1 | b.n2 |
a.n0 | 0 |
n0∞ -1/2 | n01 | n02 |
a.n∞ | -n0∞+ 1/2 | 0 | n∞1 | n∞2 |
a.n1 | -n01 | -n∞1 | 0 | n12 |
a.n2 | -n02 | -n∞2 | -n12 | 0 |
Each term in this table is calculated from:
n0= (e1 + e2)/2
n∞= (e1 - e2)/2
n1= e3
n2= e4
So we just multiply out each term by converting to 'e' basis, doing inner product, then converting back to 'n' basis.
The outer product of two null vectors
Lets take the point 'p' (x1,y1) in Euclidean space, this gives,
p=(-1,x1²+y1²,x1,y1)
and we want to take the outer product with q:
q=(-1,x2²+y2²,x2,y2)
So, multiplying out the terms using the above table, the outer product is the multivector:
scalar = (x1²+y1² -x2²- y2²)/2
n0∞ = x2²+y2² - x1²-y1²
n01 = x1 - x2
n02 = y1 - y2
n∞1 = x1²x2 - x2²x1 = x1x2(x1 - x2)
n∞y = y1²y2 - y2²y1 = y1y2(y1 - y2)
n12 = x1 y2 - x2 y1
Meet
In the above example if x1=x2 and y1=y2 then:
scalar = 0
n0∞ = 0
n01 = 0
n02 = 0
n∞1 = 0
n∞y = 0
n12 = 0
So this gives us a way to test if x1 and x2 represent the same point since, if they do then,
p^q=0