I found a way to visualise 4D vectors! - Very simple.

Just take a 4D vectoruand draw it in 3D as if the 4'th component doesn't matter,

(i.e. ignore the 4'th component). Then draw its "ghost" or imageu' in the same axis

system with just the 4'th and 1'st component "interchanged" via the cross product

as follows:

for two 4D vectorsv,wlet the result of vxw be u, then

u' = [1R4](vxw)_1[1] - (vxw)_2[2] - (vxw)_3[3] + [4R1](vxw)_4[4]

where the [1R4] means index 1 replaces index 4 in the operand in ( ) after taking

the m'th component for each _m, and [n] is the n'th unit vector.

This formula comes from:

u' =

[1] [2] [3] [4]

v_4 v_2 v_3 v_1

w_4 w_2 w_3 w_1

\textibf {u'} =

\left|

\begin {array}

{r r r r}

i&j&k&l\\

v_4&v_2&v_3&v_1\\

w_4&w_2&w_3&w_1\\

\end {array}\right|

and use my definition for the 4D cross product (posted here in December 2007).

There is a way to get somevandwfrom u (will be posted soon).

One can also exchange 4'th and 3'rd, or 4'th and 2'nd but since we give the axises

consecutive labels the usual convention should be to exchange the 4'th and 1'st.

One can now re-look at the light cone of 3D + 1D of relativity for the case where

the particle (or everything) is not restricted to move in a 2D plane.

For larger dimensions (n) one would have one restricted and n-3 ghost vectors.

Edited 25072008.

Thanks Jane.