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.