Natural Science Forum / Physics / Research / August 2007

I think I've learned something by the result.
I'll do a bit of contraction to (1) to get rid of
the dummies,

A_u B^v -*A_u *B^v = (1/2)lambda_u^v AB ,   (1 contracted).

The Kronecker leaves only 4 terms, one is,

A_1 B^1 -*A_1 *B^1 = (1/2) AB ,

and the rest 2,3,4 are the same, (I'm using Pauli's
"Theory of Relativity" pg. 34 , he uses 1,2,3,4).
A summation over 4D will give,

A_u B^u -*A_u *B^u = 2AB .

The index "u" is dummied so I write that as,

AB - A*B* = 2AB,

where

A*B* = - AB is a solution and the result.

Here's my understanding.
In the usual 3D vector analysis we write a vector
product as, AxB = -BxA.

But in 4D we use

A*B* = - AB = - BA = B*A*,

where the order of multiplying is suspended,
and instead, a "dual invariance" surfaces,
notated by the "*".
Rebuttal appreciated, i.e. did I get that wrong?

Welcome, thanks for the question.
Ken S. Tucker
PS: An unsolved problem is a wasted solution.
____________________________

I assume that *A_ab was intended here.

I think you have a typo in equation (1).  For example, let

 g_00 = -1
 g_11 = +1
 g_22 = +1
 g_33 = +1,

 A_12 = +1
 A_21 = -1,

and

 B_23 = +1
 B_32 = -1,

with all other components zero.  Then the nonzero components of the dual
tensors I get (using eta_0123 = +1) are

 *A_03 = +1
 *A_30 = -1

and

 *B_01 = +1
 *B_10 = -1.

Then considering equation (1) with u = 1 and v = 3, I get

 A_1a B^3a = -1,

 *A_1a *B^3a = 0,

and

 (1/2) lambda_1^3 A_ab B^ab = 0,

which gives me -1 for the left side and 0 for the right.

I can, however, give you a proof of

 A^va B_ua - *A_ua *B^va = (1/2) lambda_u^v A_ab B^ab,

assuming that, or something equivalent, was the intended equation.

Starting with the product of the duals, we have

 *A_ba *B^qa

  = (1/2) (-g)^.5 eta_bacd A^cd (1/2) (-g)^.5 eta^qars B_rs.

  = (1/4) (-g)^.5 eta_abcd (-g)^.5 eta^aqrs A^cd B_rs.

We can use an identity about the Levi-Civita tensor here, namely

 - (-g)^.5 eta_abcd (-g)^.5 eta^aqrs

 = + lambda_b^q lambda_c^r lambda_d^s
   + lambda_b^r lambda_c^s lambda_d^q
   + lambda_b^s lambda_c^q lambda_d^r
   - lambda_b^q lambda_c^s lambda_d^r
   - lambda_b^r lambda_c^q lambda_d^s
   - lambda_b^s lambda_c^r lambda_d^q,

where in the sum of product of the Kronecker deltas, every even permutation
of (qrs) is listed with a positive sign, and every odd permuation with a
negative sign.

In case you don't have a proof of that handy, we'll prove this first before
applying it to the original question.

We have

 eta^abcd

  =   g^ap g^bq g^cr g^ds eta_pqrs

  = - g^ap g^bq g^cr g^ds eta_qprs

  = - eta^bacd,

showing eta^abcd is antisymmetric in the first two indices.  Similarly, we
can show eta^abcd is antisymmetric for all other pairs of indices.  Since
eta^abcd is totally antisymmetric, it is proportional to eta_abcd.  We only
have to work out one component:

 eta^0123

  = g^0p g^1q g^2r g^3s eta_pqrs

  = det(g^uv)

  = det(g_uv)^-1

Now let's look at

 - (-g)^.5 eta_abcd (-g)^.5 eta^aqrs.

We have

 - (-g)^.5 eta_abcd (-g)^.5 eta^aqrs

  = - (-g)^.5 eta_abcd (-g)^.5 g^-1 eta_aqrs

  = eta_abcd eta_aqrs.

If (qrs) contains any repeated indices, then the expression above is zero.
Otherwise, the only term in the implied sum over (a) is the one where (a)
is the index not in (qrs).  In order for the expression to be nonzero, none
of (bcd) can be (a), making (bcd) a permutation of (qrs).  If (bcd) is an
even permutation of (qrs), eta_abcd and eta_aqrs will have the same sign,
and the expression will be +1.  Similarly, if (bcd) is an odd permutation
of (qrs), it will be -1.

Now consider

 + lambda_b^q lambda_c^r lambda_d^s
 + lambda_b^r lambda_c^s lambda_d^q
 + lambda_b^s lambda_c^q lambda_d^r
 - lambda_b^q lambda_c^s lambda_d^r
 - lambda_b^r lambda_c^q lambda_d^s
 - lambda_b^s lambda_c^r lambda_d^q.

It's totally antisymmetric in (qrs), so it is zero if (qrs) contains any
repeated indices.  If the indices (qrs) are all distinct, then it evaluates
to +1 if (bcd) is any of the even permutations of (qrs), it evaluates to -1
if (bcd) is one of the odd permutations, and it is zero otherwise.

Plugging the identity into the expression above, we get

 *A_ba *B^qa

  = (-1/4) (   lambda_b^q A^cd B_cd + A^cq B_bc + A^qd B_db
             - lambda_b^q A^cd B_dc - A^qd B_bd - A^cq B_cb ).

By the antisymmetry of B, this simplifies to

 (-1/2) (lambda_b^q A^cd B_cd + A^cq B_bc - A^qd B_bd),

and using the antisymmetry of A, we can combine the last two terms,
obtaining

 - (1/2) lambda_b^q A^cd B_cd + A^qd B_bd.

Thus we have

 A^qd B_bd - *A_ba *B^qa = (1/2) lambda_b^q A^cd B_cd,

or, going back to the original indices and manipulating the last term
slightly,

 A^va B_ua - *A_ua *B^va = (1/2) lambda_u^v A_ab B^ab.