I Contravariant first index, covariant on second, Vice versa?

[vanhees71]

Okay, but there is a sense in which the ambiguity is an artifact of the use of the same name for a tensor, F in this case, after being contracted with g. F^{\mu \nu}, F^\nu_{\ \ \mu}, F_\mu^{\ \ \nu} are really three different tensors.

Usually in spaces with a fundamental form (aka "metric" or "pseudo-metric") all these are different components for the same tensor since vectors and one-forms are identified via the canonical mapping provided by the fundamental form, but that's indeed a matter of semantics. You can as well take the standpoint that your examples are all different objects (i.e., just ignoring the identification of vectors and covectors via the fundamental form), but that's unusual for physicists.

 

[pervect]

This is not the correct transformation rule for dual vectors. It would imply
$$
V_{\mu'}W^{\mu'} = \Lambda^\mu_{\phantom\mu\mu'} \Lambda^{\mu'}_{\phantom\mu\nu} V_\mu W^\nu,
$$
but generally $\Lambda^\mu_{\phantom\mu\mu'} \Lambda^{\mu'}_{\phantom\mu\nu} \neq \delta^\mu_\nu$ (it is correct only for rotations). Instead, you can get the correct transformation rule by first raising the index, applying the transformation rule for contravariant indices, and then lower the index, i.e.,
$$
V_{\mu'} = \eta_{\mu'\nu'}V^{\nu'} = \eta_{\mu'\nu'} \Lambda^{\nu'}_{\phantom\nu\nu} V^{\nu} = \eta_{\mu'\nu'} \Lambda^{\nu'}_{\phantom\nu\nu} \eta^{\nu\rho} V_{\rho}.
$$
We also note that
$$
\Lambda^{\mu'}_{\phantom\mu\gamma}\eta_{\mu'\nu'} \Lambda^{\nu'}_{\phantom\nu\nu} \eta^{\nu\rho} = \eta_{\gamma\nu}\eta^{\nu\rho} = \delta^\rho_\gamma,
$$
just by the defining property of the Lorentz transformation preserving the metric tensor, and thus recovering the appropriate inner product.

For convenience we can apply the lowering and raising of the indices using the metric to write $\eta_{\mu'\nu'} \Lambda^{\nu'}_{\phantom\nu\nu} \eta^{\nu\rho} = \Lambda_{\mu'}^{\phantom\mu\rho}$.

This doesn't seem to match my text. At the moment I'm not sure what's wrong.

See MTW gravitation pg 203, the section entitled "Change of Basis" in box 8.4

Specifically:

$$\sigma_{a'} = \sigma_{\beta} \, L^\beta{}{}_{a'}$$

 

[Orodruin]

This doesn't seem to match my text. At the moment I'm not sure what's wrong.

See MTW gravitation pg 203, the section entitled "Change of Basis" in box 8.4

Specifically:

$$\sigma_{a'} = \sigma_{\beta} \, L^\beta{}{}_{a'}$$

Note that $L^\beta_{\phantom \beta\alpha'}$ are not the same transformation coefficients as $L^{\beta'}_{\phantom\beta\alpha}$. The primes on the indices are important. In particular, note the comment below the equations.

Edit: In particular, for Lorentz transformations, note that you would have $x'^{\mu'} = \Lambda^{\mu'}_{\phantom\mu\mu} x^\mu$ but $x'_{\mu'} = \Lambda^\mu_{\phantom\mu\mu'} x_\mu$. Here,
$$
\Lambda^{\mu'}_{\phantom\mu\mu} = \frac{\partial x'^{\mu'}}{\partial x^\mu}
\quad\mbox{and}\quad
\Lambda^\mu_{\phantom\mu\mu'} = \frac{\partial x^{\mu}}{\partial x'^{\mu'}}.
$$
It holds that $\Lambda^{\mu'}_{\phantom\mu\mu}\Lambda^\nu_{\phantom\mu\mu'} = \delta^\nu_\mu$.

 

[vanhees71]

This doesn't seem to match my text. At the moment I'm not sure what's wrong.

See MTW gravitation pg 203, the section entitled "Change of Basis" in box 8.4

Specifically:

$$\sigma_{a'} = \sigma_{\beta} \, L^\beta{}{}_{a'}$$

I don't know, what your symbols mean, but concerning the question of the Lorentz transformation, it's not that difficult. You MUST (!) be careful with the horizontal order of the indices. Lorentz transformations are in general not represented by symmetric matrices (rotation-free boosts are, by the way, but it's nevertheless good to stick to the notational rools of the Ricci calculus).

The defining property of the Lorentz-transformation matrix is
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma},$$
which implies
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} \eta^{\rho \alpha}=\delta_\sigma^{\alpha}.$$
Applying the "raising-and-lowering rules" for tensor indices also to the Lorentz-transformation matrix (which is not denoting tensor components!), you get
$${\Lambda_{\nu}}^{\alpha} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha} \; \Rightarrow {(\Lambda^{-1})^{\alpha}}_{\nu} = {\Lambda_{\nu}}^{\alpha}.$$
From these few examples you see how important the horizontal order of the indices (no matter whether upper or lower ones) really is.

It's also easy to see, how the components of tensors transform. It's sufficient to consider only vectors (i.e., first-rank tensors). The generalization to higher ranks is trivial: For each index just use the rules for the corresponding index of the vector components.

By definition the Lorentz transformation of contravariant vector components is
$$A^{\prime \mu}={\Lambda^{\mu}}_{\rho} A^{\rho}.$$
This implies, since $\eta_{\mu \nu}'=\eta_{\mu \nu}$ (under Lorentz transformations!),
$$A_{\nu}'=\eta_{\nu \mu} A^{\prime \mu}= \eta_{\nu \mu} {\Lambda^{\mu}}_{\rho} A^{\rho} = \eta_{\nu \mu} {\Lambda^{\mu}}_{\rho} \eta^{\sigma \rho} A_{\sigma} = {\Lambda_{\nu}}^{\sigma} A_{\sigma}={(\Lambda^{-1})^{\sigma}}_{\nu} A_{\sigma},$$
i.e., the covariant components transform contragrediently to the contravariant ones, as it must be.

As @Orodruin already stressed tensors are of course invariant objects. Indeed by definition the basis vectors transform as
$$\boldsymbol{e}_{\mu}'={\Lambda^{\rho}}_{\mu} \boldsymbol{e}_{\rho}$$
and thus
$$\boldsymbol{A}=A^{\prime \mu} \boldsymbol{e}_{mu}'=A^{\prime \mu} {\Lambda^{\rho}}_{\mu} \boldsymbol{e}_{\rho} = A^{\rho} \boldsymbol{e}_{\rho}.$$

 

[pervect]

I don't know, what your symbols mean, but concerning the question of the Lorentz transformation, it's not that difficult.

$L^{a}{}_{b}$ is a general change of basis, as per the following quote:

The discussion of Lorentz transforms in equations (2.3() to (2.43) is applicable to general changes of basis if one replaces $||\Lambda^{\alpha'}{}_{\beta}||$ by an arbitrary but nonsingular matrix $||L^{\alpha'}{}_{\beta}||$.

I'm not quite sure what the || denote in this context, it's not explained in the section I looked at. Presumably, if I hunted back far enough, I'd find an explanation, but I don't have one at this point. It looks like an absolute value, but that doesn't make any sense. It could have something to do with L not being a tensor, I suppose. Hopefully it's not relevant to the issue in hand, but in case it is I'll mention it.

My current understanding is that I am indeed quoting my text correctly in my statements as how to transform one-forms, and my text is the source I feel most confident about at this point. I don't know how in general to handle situations where my textbook conventions (northwest-southeast indices on the transformation matrices) are not followed, up until this point I though everyone did it that way. I do note that you (@vanhees71) state that we raise and lower indices on the transformation matrices using the same methods we use for tensors, even though they are not tensors. Thus I would take $L^{a}{}_b$ as defined as the transformation matrix as my text states, and interpret other variants as being created from $L^{a}{}_b$ by raising and lowering indices using tensor rules.

The text results are explicitly intended to be applied to general transformations, not just the Lorentz transformations. My understanding is that @Orodruin suggests that there is a problem with what I quoted when the transformation is not a Lorentz transform, but when I check my textbook (my memory isn't what it used to be), it seems to say that I did it correctly.

The text also tells us how to transform from a primed basis back to an unprimed basis, something I didn't quote earlier, which I'll do now in case it is of interest: $\sigma_{\beta} = \sigma_{\alpha'}\,L^{\alpha'}{}_{\beta}$.

It's basically not different from the rule for transforming from the unprimed basis to the primed basis.

Doing two transformations, as a logical consequence I get:

$$\sigma_{\alpha''} = \sigma_\gamma L^\gamma{}{}_{\beta'} L^{\beta'}{}_{\alpha''}$$

If the two transforms are inverses, $\sigma_{\alpha''} = \sigma_\gamma$ so we must have that $L^\gamma{}{}_{\beta'} L^{\beta'}{}_{\alpha''}$ is $\delta^{\gamma}{}_{\alpha''}$.

My suspicion is that @odoruin is using some different conventions. If my textbook is accurate, and it's conventions are followed $L^\gamma{}{}_{\beta'} L^{\beta'}{}_{\alpha} = \delta^{\gamma}{}_{\alpha}$, where we note that the doubly-primed coordinate system must be the same as the unprimed coordinate system for the case when we transform a one-form to a different basis, and then back again.

 

[Orodruin]

My understanding is that @Orodruin suggests that there is a problem with what I quoted

No. What you quoted is fine. What you are not paying proper attention to is the difference between MTW’s $L^{\alpha’}_{\phantom\alpha\beta}$ and their $L^{\alpha}_{\phantom\alpha\beta’}$, which are different sets of transformation coefficients. The primes are important in this notation, they tell you what transformation coefficients are intended. You cannot hope to properly recover the transformation rules unambiguously if you drop the primes on the indices unless you introduce different notation for the transformation coefficients.

 

[Orodruin]

For usage, we might transform a vector using the first transformation matrix, i.e $x^{\mu} = \Lambda^{\mu}{}_{\nu} \, x^{\nu}$ and covectors using the second, $x_{\mu} = \Lambda^{\nu}{}_{\mu} x_{\nu}$. Which would be my guess as to what you wanted to do.

Just to be a bit clearer. You are using
$$
x_\mu = \Lambda^\nu_{\phantom\nu\mu} x_\nu,
$$
which is unclear for many reasons. First of all, it is not true taken at face value unless $\Lambda^\nu_{\phantom\nu\mu} = \delta^\nu_\mu$. You need to prime one of the sets of coordinates. Second, if you do this, it is not clear how you intend to handle the indices. If you just leave the indices as they are without priming, you would obtain
$$
x'_\mu = \Lambda^\nu_{\phantom\nu\mu} x_\nu.
$$
There is now nothing that tells me that the $\Lambda^\nu_{\phantom\nu\mu}$ is in any way different from the $\Lambda^\nu_{\phantom\nu\mu}$ that appeared in the contravariant transformation rule. This is the reason for priming the indices belonging to the primed coordinate system. You would end up with
$$
x'_{\mu'} = \Lambda^\nu_{\phantom\nu\mu'} x_\nu
$$
as well as
$$
x'^{\mu'} = \Lambda^{\mu'}_{\phantom\mu\nu} x^\nu,
$$
where the transformation coefficients are clearly distinct.