
#1
Oct613, 12:29 PM

P: 365

How do you take the cross product of two 4Vectors?
[tex]\vec{r} = \left( \begin{array}{ccc}c*t & x & y & z \end{array} \right)[/tex] [tex]\vec{v} = \left( \begin{array}{ccc}c & vx & vy & vz \end{array} \right)[/tex] [tex]\vec{v} \times \vec{r} = ?[/tex] [tex][/tex] 



#2
Oct613, 02:09 PM

Thanks
P: 1,353

The vector cross product does not have a definite meaning except in 3D space.
See http://en.wikipedia.org/wiki/Cross_p...terior_product What is the purpose of your proposed construction? 



#3
Oct613, 02:16 PM

C. Spirit
Sci Advisor
Thanks
P: 4,941

Actually it's also well defined nontrivially for 7 dimensions.




#4
Oct613, 02:28 PM

P: 365

Cross product of two 4Vectors
I can do a Triple Product of xyz. I just do not know what to do with the t.
The triple product is:[tex]v \times r = \bar{x}*(vy*zvz*y) + \bar{y}*(vz*xvx*z) + \bar{z}*(vx*yvy*x)[/tex] 



#5
Oct613, 02:45 PM

Thanks
P: 1,353

Still not clear what you are trying to accomplish with the 4vectors here.




#6
Oct613, 03:27 PM

Mentor
P: 4,499

Philosophaie, what your asking is how to apply an operation which is not defined for the objects you have produced. It's analogous to me asking you "how do I add two circles together?" It doesn't make any sense to ask the question.
If you give the context under which you want to ask this kind of question, we can probably identify what you actually want to do with your two vectors. 



#7
Oct613, 03:40 PM

P: 2,080

As UltrafastPED said, the vector cross product really only works in three dimensions. In four dimensions, you can form what is called the "wedge product" or "exterior product" of two four vectors, but this object will not be another fourvector. It will be a different geometric object referred to a bivector. This object can be defined with a rank2 antisymmetric tensor, and it has six components instead of just four.




#8
Oct713, 11:01 AM

P: 365

I only got 6 dimensions. Check my math:
[tex]\vec{v} \times \vec{r} = 2*(v_z*c*tc*z)*\bar{x}\bar{y}+2*(v_y*c*tc*y)*\bar{x}\bar{z}+2*(v_x*c*tc*x)*\bar{y}\bar{z}+2*\bar{t}*(\bar{x}*(v_x*yyv_y*x)+\bar{y}*(v_z*xv_x*z)+\bar{z}*(v_x*yv_y*x))[/tex] 



#9
Oct713, 11:28 AM

Mentor
P: 14,477

Before we go any further, what are you trying to do, Philosophaie? You have a penchant for asking "XY" questions. In trying to solve problem X you run into a problem with Y. So you ask about Y. The problem is that in an "XY" question, Y is typically a dead end. The ultimate problem isn't how to solve Y. The problem is how to solve X, and you haven't told us what X is.




#10
Oct713, 11:54 AM

P: 365

I am just looking for an easy way to take the cross product of two 4vectors to find the Magnetic Field of a Black Hole.




#11
Oct713, 12:06 PM

C. Spirit
Sci Advisor
Thanks
P: 4,941

Well that's definitely not how you would go about doing it. For one, the magnetic field is frame dependent. Secondly, you have to first find the covariant electromagnetic field tensor ##F_{ab}## by solving Maxwell's equations, which will be coupled to the metric tensor ##g_{ab}## describing the electrovacuum exterior to the black hole. Once you have ##F_{ab}## you must choose an observer with 4velocity ##u^a## with respect to whom you can split ##F_{ab}## into the electric and magnetic field. The formulas are ##E^a = F^{a}{}{}_{b}u^b## and ##B^a = \frac{1}{2}\epsilon^{abcd}u_b F_{cd}## where ##\epsilon^{abcd}## is the natural volume element on spacetime; ##\epsilon^{abcd}u_b## is (up to a sign) the 3dimensional levicivita symbol which you use to take cross products in the 3space orthogonal to ##u^a##.



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