on the correct Lorentz transform of the basis vectors

This is given by Carroll on page 43 of his notes. The

complete vector field must be invariant:

V = V^{μ}∂_{μ} = V^{μ'}∂_{μ'}
(1)

under the __general__ coordinate transformation. In the special case of

the Lorentz transform the complete vector field is:

V = V^{μ}e_{μ} = V^{μ'}e_{μ}'
(2)

Jackson considers the vector:

V^{μ} = x^{μ}
= (ct, x,y, z)
(3)

Under the Lorentz transform:

x^{μ}' =
(γ(ct−βz), x, y, γ(z−βct)) .
(4)

The unit vector is:

e_{μ} =
(1, −1, −1, −1) .
(5)

and is __changed__ to e_{μ}'. We have:

x^{o} e_{o} + x^{3} e_{3} =
x^{o'} e_{o}' + x^{3'} e_{3}'
(6)

from eq. (2), so one solution is

e_{o}' = ^{1}/_{γ(ct−βz)},
e_{3}' = ^{1}/_{γ(z−βct)},
(7)

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

V^{μ} =
(1, 1, 1, 1) .
(8)

so

V = V^{μ}e_{μ} = −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**.

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

e_{o} = 1, e_{1} = −1, e_{2} = −1, e_{3} = −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** =
**h** − **i** − **j** − **k**
(5"')

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

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 e_{o} + x e_{1} + y e_{2} + z e_{3} =
γ(ct−βz) e_{o}' + x e_{1}' + y e_{2}' + γ(z−βct) e_{3}'

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)
e_{1} = e_{1}' and
e_{2} = e_{2}'

respectively. Thus the above identy reduces to

(iii)
ct e_{o} + z e_{3} =
γ(ct−βz) e_{o}' + γ(z−βct) e_{3}'

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

(iv)
e_{o} =
γ (e_{o}' − β e_{3}')
and
e_{3} =
γ (e_{3}' − β e_{o}')

in agreement with the Lorentz transformation rules of the basis vectors
in section
"Lorentz transform of a z-boost K → K' " of
my note
Commentary on Evans' Key Derivation 6 .

The correctly copied equation from Carroll's text is

V^{μ}∂_{μ}
= V^{μ'}∂_{μ'}
= V^{μ'} ^{∂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 x^{o}=ct, x^{1}=x, x^{2}=y, x^{3}=z,
x^{o}'=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.

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.**