## APPENDIX 10: REBUTTAL OF G. BRUHN'S COMMENTS ON THE LORENTZ COVARIANCE OF THE B CYCLIC THEOREM.

### Gerhard W. Bruhn, Darmstadt University of Technology

July 19, 2008
(Quotations from Evans' original text in

It is well known that special relativity demands that any equation of physics be covariant under the Lorentz transformation. It has been shown elsewhere {8} that the B Cyclic Theorem is correctly Lorentz invariant. This is self-evident from the fact that Theorem is, within Lorentz invariant phase factors, the frame of reference itself, and it is well known that a frame of reference is Lorentz covariant. Bruhn {14} incorrectly asserts that a frame of reference must be Lorentz invariant. . . .

This is a distortion of facts. And Evans does not specify the spot of that alleged flaw. − However, let's see below how Evans attempts to prove his Theorem which asserts that if the cyclic relations (10.1) are valid in a inertial frame S then the primed version of (10.1) is valid also in another inertial frame S' moving at constant velocity v (|v|<c) relative to the un-primed frame. Therewith the transformation rules of the electromagnetic field for that Lorentz boost must be taken into account. Then to prove or disprove the Theorem is merely a matter of patience and skill [2].

However, an impatient reader needs not to check all the following tedious calculations in detail. Dr. Evans has provided a very surprising conclusion just before eq.(10.22) which make all further efforts superfluous.

. . .

The B Cyclic Theorem is {8}:

B(1)×B(2) = i B(o) B(3)                                 (10.1)
et cyclicum

where the plane wave magnetic flux densities are:

B(1) = B(2)* = B(o)/2½ (iij) eiΦ
B(3) = B(3)* = B(o) k.                                 (10.2)

Here, the electromagnetic phase factor is

exp(iΦ) = exp(i(ωt−κz))                                 (10.3)

where ω is the angular frequency at instant t, κ the wave number at position z. it is well known that the phase factor is Lorentz invariant. The frame of reference being used is the complex circular basis {8}

e(1)×e(2) = i e(3)*                                 (10.4)

whose unit vectors are related to the Cartesian unit vectors by

e(1) = e(2)* = 1/2½ (iij) eiΦ
e(3) = e(3)* = k.                                 (10.5)

where * denotes ''complex conjugate''.

The general rule {13} (ref. {13} is S.M. Carroll's book Spacetime and Geometry, eqs.(2.17-19). The same can be found in Carroll's Lecture Notes [5], eqs.(2.11-13)) for the covariant transformation of a basis vector is well known, and is:

ê(μ') = (∂xμ/∂xμ') ê(μ)                                 (10.6)

∂μ' = ∂xμ/∂xμ'μ                                 (Carroll [5] (2.11))

where

xμ = (ct, x, y, z)                                     (10.7)

and where the prime denotes a different frame of reference moving arbitrarily with respect to the original (unprimed) frame of reference. The rule is found in innumerable textbooks and shows that basis elements such as unit vectors are changed under the transformation. In other words they are COVARIANT. The Lorentz transform is a special case:

ê(a') = Λa'a(x) ê(a)                                     (10.8)

The complete vector field {13}:

V = Vμ ê(μ)                                                 (10.9)

is invariant under the transformation:

V = Vμ ê(μ) = Vμ' ê(μ')                                 (10.10)

Vμμ = Vμ'μ                                     (Carroll [5] (2.12))

Therefore the rule for general coordinate transformation {13} is:

V(μ') = (∂xμ'/∂xμ) V(μ)                                 (10.11)

Vμ' = ∂xμ'/∂xμ Vμ                                 (Carroll [5] (2.13))

The Lorentz boost considers a frame of reference moving at velocity in an axis with respect to another frame. If this is the Z axis then:

V = Vz k = Vz' k'                                 (10.12)

from eq. (10.10) In general the Lorentz transformation produces the result

i × j = k
↓                                                         (10.13)
i' × j' = k'

in the Cartesian basis . . .

That's a mere claim, no proof! What is required here for the application of the eqs. (10.6) and (10.11) are the transformation rules of coordinates, i.e. xμ = xμ(xμ') for the eq.(10.6) and xμ' = xμ'(xμ) for the eq.(10.11). These well known eqs. using x instead of z can be found at Wikipedia. Hence we have the coordinate transformation to be used here:

t' = γ (ct−βz)/c,         z' = γ (z−vt)         (to be supplied by x' = x and y' = y)

where β = v/c and γ = (1−β²)−½. Hence for

x0 = ct, x1 = x, x2 = y, x3 = z         and         x0' = ct', x1' = x', x2' = y', x3' = z',

we obtain the transformation xμ' = xμ'(xμ)

x0' = γ (x0− β x3),         x1' = x1,         x2' = x2,         x3' = γ (x3− β x0),         (L)

and the inverse transformation xμ = xμ(xμ') is

x0 = γ (x0'+ β x3'),         x1 = x1',         x2 = x2',         x3 = γ (x3'+ β x0'),         (L')

Now we can check by calculation Evans' eq. (10.12) where he asserts k | | k'. This is evidently wrong as can be seen by the well known Lorentz transformation graph e.g. at p.7 of Carroll's Lecture notes [5]:

### Evans has not realized that the vectors k and k' are 4-dimensional.

Due to eq.(10.6) we obtain by using (L')

k' = ∂3' = ∂x0/∂x3'0 + ∂x1/∂x3'1 + ∂x2/∂x3'2 + ∂x3/∂x3'3
=    γβ ∂0     +       0       +         0         +     γ ∂3
=   γ (∂3+β ∂0) =   γ (kh)

where h = ∂0 is the (fourth axial) unit vector in the t axis that does not appear anywhere in Evans' considerations.

By a similar calculation we can see that the axial unit vectors perpendicular to the Lorentz boost remain unchanged, i' = i and j' = j.

Thus, with respect to Evans' claim (10.13) for a Lorentz boost L in z-direction we have the contradicting result

i = i'         and         j = j',         but         NOT         k | | k'

Evans' eq. (10.12) is wrong.

### While the basis vectors i and j remain invariant under a Lorentz boost in z-direction the basis vector k does not. The vectors k and k' are not even parallel.

The same holds for Evans' next assertion

. . . or:

e(1) × e(2) = i e(3)*
↓                                                         (10.14)
e(1)' × e(2)' = i e(3)*'

in the complex circular basis. These results follow directly from Eq.(10.8)

This is wishful thinking only: The vectors e(1) and e(2) remain unchanged as being linear combinations of the invariant vectors i and j, but not so the vector e(3) = k which is NOT equal to k' = e(3)', not even e(3)' | | e(3).

In the B Cyclic Theorem (10.1) the following definitions are used:

B(1) = B(2)* = B(o) eiΦ e(1)
(10.15)
B(3) = B(3)* = B(o) k.

It is well known that the electromagnetic phase is Lorentz invariant, because is a scalar, so:

eiΦ = eiΦ'.                                                 (10.16)

Therefore we need only consider the Lorentz covariance of B(o) and the frame itself (the unit vectors). If the boost takes place in z the complete vector field to be considered is:

B(3)B(3)                                                 (10.17)

Probably a prime is missing at the right hand side.

This means that:

B(o) k = B(o)' k'
(10.18)
B(o) i = B(o) i , B(o) j = B(o) j .

The first line (10.18)1 does not hold since as we have seen the unit vectors k and k' are not parallel. The second line is trivial, perhaps there are missing primes. Whatsoever, the eqs. (10.18) do not describe the transformation conditions valid for the magnetic flux that were derived in the eqs.(5-9) in [2]. Especially the following equ.(10.19) is wrong: Due to [2], eq.(7) the quantity B(o) must be invariant: B(o)' = B(o). And just this discrepancy will lead Evans to a very curious conclusion below.

It has been shown that in this case:

B(o)' = (1−v/c/1+v/c)½ B(o)                                                 (10.19)

(Ref.{8}, eq.(3.111)), where v is the velocity or the primed frame with respect to the unprimed frame. The Cartesian frame after Lorentz transformation is:

i×j = k'                                                                 (10.20)

This is not true: we have

i×j = kk' = i'×'j'                                                 (10.20')

where × and ×' are different cross products defined in the different subspaces of spacetime spanned by {i,j,k} and {i',j',k'} respectively.

and the complex circular frame after Lorentz transformation is:

e(1) × e(2) = i e(3)*'                                                 (10.21)

Wrong again! We have

e(1) × e(2) = i e(3)*i e(3)*' = e(1)' ×' e(2)'                 (10.21')

The frames are distorted, indicating the length contraction, but the overall form of the cyclic relations (10.20) and (10.21) is the same as the original cyclic relations (10.1) and (10.4). This is what is meant by COVARIANCE.

Furthermore, as indicated in ref.(8), the velocity v is zero, because the fields are already propagating at c, and cannot propagate any faster. Thus:

B(o)' = B(o)                                                 (10.22)

The above is elementary and well known. Why should such an elementary error by Bruhn be published.

The remark before eq.(10.22) tops all! After having discussed in full the effect of a Lorentz boost on Evans' Cyclical Theorem he tells us that all was a joke only: The velocity v of the primed frame of reference relative to the unprimed frame must be ZERO to yield the validity of his Covariance Theorem for his Cyclical basis:

### Evans' ''Covariance Theorem'' is valid only for Lorentz boosts of velocity v=0.

Which is the reason for that surprising turn? Probably Evans became aware of the well known fact that the size of the longitudinal magnetic field component remains unchanged under a Lorentz boost. B(o)' = B(o). Then eq.(10.19) would contradict unless we assume v=0.

However, Evans' argumentation for v=0 is rather curious: ''. . . the velocity v is zero, because the fields are already propagating at c, and cannot propagate any faster.''

Did he never take notice of the relativistic addition theorem of velocities in case that one of the summands is c?

w(u,v) = u+v/1+uv/c²         Þ         w(c,v) = c+v/1+cv/c² = c .

#### Summary

Evans' calculation of the transformation behavior fails due to wrong handling of the transformation laws, especially of eq.(10.12), which yields wrong transformation laws (10.18)1 and eq.(10.19). At the end the wrong eq.(10.19) leads to the absurd conclusion v=0.

### References

[2] G.W. Bruhn, On the Lorentz Behavior of M.W. Evans' O(3)-Symmetry Law,

[3] G.W. Bruhn, No Lorentz Property of MW Evans' O(3)-Symmetry Law,
Physica Scripta 74 2006, pp.1-2

[4] G.W. Bruhn, On the Non-Lorentz-Invariance of M.W. Evans’ O(3)-Symmetry Law,
Foundations of Physics, 38, 1, pp.3-6

[5] S.M. Carroll, Lecture Notes on General Relativity,
http://xxx.lanl.gov/pdf/gr-qc/9712019