 ## with comments and two Appendiceson the correct Lorentz transform of the basis vectors

### by G.W. Bruhn

This is given by Carroll on page 43 of his notes. The
complete vector field must be invariant:

V = Vμμ = Vμ'μ'                 (1)

under the general coordinate transformation. In the special case of
the Lorentz transform the complete vector field is:

V = Vμeμ = Vμ'eμ'                 (2)

Jackson considers the vector:

Vμ = xμ = (ct, x,y, z)                 (3)

Under the Lorentz transform:

xμ' = (γ(ct−βz), x, y, γ(z−βct)) .                 (4)

The unit vector is:

eμ = (1, −1, −1, −1) .                 (5)

and is changed to eμ'. We have:

xo eo + x3 e3 = xo' eo' + x3' e3'                 (6)

from eq. (2), so one solution is

eo' = 1/γ(ct−βz),         e3' = 1/γ(z−βct),                 (7)

For the B Cyclic theorem we are considering the unit vector, so

Vμ = (1, 1, 1, 1) .                 (8)

so

V = Vμeμ = −2 .                 (9)

This is an invariant scalar.         Q.E.D.

Due to eq. (1) V is a vector (of dimension 4) while due to eq. (9) V is a scalar (−2). Since 4-dimensional vectors are different from numbers, the above proof must be wrong.

### Where is the error?

The error is contained in eq.(5), which in eq. (9) is falsely interpreted as

eo = 1, e1 = −1, e2 = −1, e3 = −1 .                 (5')

The right hand side of eq. (5) is no vector, only a quadruple of coordinates without naming the basis which the quadruple is related to. E.g. the equation

e = 1 ∂o + (−1) ∂1 + (−1) ∂2 + (−1) ∂3                 (5")

or, by using the basis h = ∂o, i = ∂1, j = ∂2, k = ∂3, the equation

e = 1 h + (−1) i + (−1) j + (−1) k = hijk                 (5"')

would make sense. However, the denotation "unit vector" would require a vector of length 1 in some sense.

### Conclusion: Evans' note under consideration is wrong beyond repair.

For a correct treatment of the problem the reader is kindly requested to have a look at the section Lorentz transform of a z-boost .

## Appendix 1: Derivation of the correct transform of the basis vectors

The same result can be attained on the basis of the correct eqs. (1-4): The eqs. (2),(3) and (4) yield the identity

(i)         ct eo + x e1 + y e2 + z e3 = γ(ct−βz) eo' + x e1' + y e2' + γ(z−βct) e3'

valid for arbitrary choices of the quadrupel (ct,x,y,z).

The choices (ct,x,y,z) = (0,1,0,0) and (ct,x,y,z) = (0,0,1,0) give

(ii)        e1 = e1'         and         e2 = e2'

respectively. Thus the above identy reduces to

(iii)         ct eo + z e3 = γ(ct−βz) eo' + γ(z−βct) e3'

valid for arbitrary choices of (ct,z). Then the special choices (ct,z) = (1,0) and (ct,z) = (0,1) yield

(iv)         eo = γ (eo' − β e3')         and         e3 = γ (e3' − β eo')

## Appendix 2: Following Carroll . . .

In his "Correct Method" quoted above Evans refers with his eq. (1) to p.43 of S.M. Carroll's Lecture Notes on General Relativity.

The correctly copied equation from Carroll's text is

Vμμ = Vμ'μ' = Vμ' ∂xμ/∂xμ'μ                 (2.12)

Carroll's immediately previous equation gives the transformation rule of the basis vectors:

∂μ' = ∂xμ/∂xμ'μ                                         (2.11)

to be written out here in some more detail:

From (3) and (4) we may conclude (ct',x',y',z') = (γ(ct−βz), x, y, γ(z−βct)), i.e.

ct' = γ(ct−βz),     x' = x,     y' = y,     z' = γ(z−βct).

For the evaluation of the coefficients ∂xμ/∂xμ' we need the inversion of the above system, i.e.

ct = γ(ct'+βz'),     x = x',     y = y',     z = γ(z'+βct').

(left to be checked by the reader). From this (with xo=ct, x1=x, x2=y, x3=z, xo'=ct', . . .) we obtain the coefficients

∂xo/∂xo' = γ,    ∂xo/∂x1' = 0,    ∂xo/∂x2' = 0,    ∂xo/∂x3' = γβ,
∂x1/∂xo' = 0,    ∂x1/∂x1' = 1,    ∂x1/∂x2' = 0,    ∂x1/∂x3' = 0,
∂x2/∂xo' = 0,    ∂x2/∂x1' = 0,    ∂x2/∂x2' = 1,    ∂x2/∂x3' = 0,
∂x3/∂xo' = γβ,    ∂x3/∂x1' = 0,    ∂x3/∂x2' = 0,    ∂x3/∂x3' = γβ,

thus from Carroll's eq. (2.11)

∂o' = ∂xμ/∂xo'μ = γ (∂o + β∂3)
∂1' = ∂xμ/∂x1'μ = ∂1
∂2' = ∂xμ/∂x2'μ = ∂2
∂3' = ∂xμ/∂xo'μ = γ (β∂o + ∂3)

in total agreement with our above results by other methods (see e.g. the inverse resolution of the eqs. (ii) and (iv) in Appendix 1.

### Conclusion: If Dr Evans would have read (and understood) "his Carroll" more accurately, then probably his readers would not have been bothered by his numerous nonsensical "rebuttals".

Final remarks:

All the calculations correcting Evans' "rebuttals" might appear somewhat difficult and sophisticated to unexperienced readers. Nevertheless, what has been commented here on Evans' methods is completely textbooks' folklore.