Cross product of two 4-Vectors


by Philosophaie
Tags: 4vectors, cross, product
Philosophaie
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#1
Oct6-13, 12:29 PM
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How do you take the cross product of two 4-Vectors?

[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]
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UltrafastPED
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#2
Oct6-13, 02:09 PM
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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?
WannabeNewton
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#3
Oct6-13, 02:16 PM
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Actually it's also well defined non-trivially for 7 dimensions.

Philosophaie
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#4
Oct6-13, 02:28 PM
P: 365

Cross product of two 4-Vectors


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*z-vz*y) + \bar{y}*(vz*x-vx*z) + \bar{z}*(vx*y-vy*x)[/tex]
UltrafastPED
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#5
Oct6-13, 02:45 PM
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Still not clear what you are trying to accomplish with the 4-vectors here.
Office_Shredder
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#6
Oct6-13, 03:27 PM
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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.
phyzguy
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#7
Oct6-13, 03:40 PM
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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 four-vector. It will be a different geometric object referred to a bi-vector. This object can be defined with a rank-2 anti-symmetric tensor, and it has six components instead of just four.
Philosophaie
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#8
Oct7-13, 11:01 AM
P: 365
I only got 6 dimensions. Check my math:
[tex]\vec{v} \times \vec{r} = 2*(v_z*c*t-c*z)*\bar{x}\bar{y}+2*(v_y*c*t-c*y)*\bar{x}\bar{z}+2*(v_x*c*t-c*x)*\bar{y}\bar{z}+2*\bar{t}*(\bar{x}*(v_x*y-y-v_y*x)+\bar{y}*(v_z*x-v_x*z)+\bar{z}*(v_x*y-v_y*x))[/tex]
D H
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#9
Oct7-13, 11:28 AM
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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.
Philosophaie
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#10
Oct7-13, 11:54 AM
P: 365
I am just looking for an easy way to take the cross product of two 4-vectors to find the Magnetic Field of a Black Hole.
WannabeNewton
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#11
Oct7-13, 12:06 PM
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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 4-velocity ##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 space-time; ##\epsilon^{abcd}u_b## is (up to a sign) the 3-dimensional levi-civita symbol which you use to take cross products in the 3-space orthogonal to ##u^a##.


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