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I've seen this question asked a few times on Stack Exchange, but I'm still quite confused why the following "contradiction" seems to arise.

By definition:

  1. $(\Lambda^T)^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}$
  2. $\Lambda^T \eta \Lambda = \eta$, which is $\Lambda^{\rho}{}_{\mu} \eta_{\rho \sigma} \Lambda^{\sigma}{}_{\nu} = \eta_{\mu \nu}$ in index notation.

We can further manipulate the second definition (as done in Tong's lecture notes):

$\begin{align} \Lambda^{\rho}{}_{\mu} \eta_{\rho \sigma} \Lambda^{\sigma}{}_{\nu} &= \eta_{\mu \nu} \\ \Lambda^{\rho}{}_{\mu} \Lambda_{\rho \nu} &= \eta_{\mu \nu} \\ \Lambda^{\rho}{}_{\mu} \Lambda_{\rho \nu} \eta^{\nu \sigma} &= \eta_{\mu \nu} \eta^{\nu \sigma} \\ \Lambda^{\rho}{}_{\mu} \Lambda_{\rho}{}^{\sigma} &= \delta_{\mu}^{\sigma} \\ \Lambda_{\rho}{}^{\sigma} \Lambda^{\rho}{}_{\mu} &= \delta_{\mu}^{\sigma} \end{align}$

Recalling that $({\Lambda^{-1}})^{\sigma}{}_{\rho}$ is defined through:

$$({\Lambda^{-1}})^{\sigma}{}_{\rho} \Lambda^{\rho}{}_{\mu} = \delta^{\sigma}_{\mu}$$

This then implies that $$({\Lambda^{-1}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma}\tag{A}$$ but according to definition 1, doesn't $$({\Lambda^{T}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma}~?\tag{B}$$ This seems to incorrectly imply that $$({\Lambda^{T}})^{\sigma}{}_{\rho} = ({\Lambda^{-1}})^{\sigma}{}_{\rho}\tag{C}.$$ I'm not really sure what step of my logic is incorrect.

Tong makes the following comment on the (A) result:

The result is analogous to the statement that the inverse of a rotation matrix is the transpose matrix. For general Lorentz transformations, we learn that the inverse is sort of the transpose where “sort of” means that there are minus signs from raising and lowering. The placement of indices in tells us where those minus signs go.

This comment seems to suggest that (B) is incorrect - although it just seems like mere application of definition 1.

Edit to clarify question after initial responses:

From this analysis, why is it incorrect to conclude that $$\Lambda^{-1} = \Lambda^T~?\tag{D}$$ We know this matrix equation is not true, but why is this not implied by $({\Lambda^{T}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma} = ({\Lambda^{-1}})^{\sigma}{}_{\rho}$ since the indices in $\Lambda^{T}$ and $\Lambda^{-1}$ are the same?

Further Clarification Question:

Some of the responses will reveal that in fact only the matrix equation D is incorrect because the index structure of $\Lambda$ is $\Lambda^{\mu}{}_{\nu}$, the index structure of $\Lambda^{-1}$ is ${(\Lambda^{-1})}^{\mu}{}_{\nu}$, but the index structure of $\Lambda^T$ is ${(\Lambda^{T})}_{\mu}{}^{\nu}$ (not ${(\Lambda^{T})}^{\mu}{}_{\nu}$).

However, this leaves one final question: how can we show explicitly that the matrix $\Lambda^T$ should correspond to this different index structure? Using this structure does make everything consistent again, but how does this follow from defining the matrix $\Lambda$ as corresponding to $\Lambda^{\mu}{}_{\nu}$?

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    $\begingroup$ physics.stackexchange.com/q/456640 seems to be the most relevant link. Doublefelix's answer for how $(\Lambda^T)^\sigma_{\,\,\alpha}\Lambda^\alpha _{\,\,\mu}=\delta_\mu ^{\,\,\sigma}$ should be interpreted does help, but I think it would be nice to see explicitly why there is no contradiction. $\endgroup$ – Shrey Jul 21 '20 at 17:21
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    $\begingroup$ physics.stackexchange.com/q/144371 $\endgroup$ – Shrey Jul 21 '20 at 17:29
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    $\begingroup$ The index structure of $\Lambda^T$ is $(\Lambda^T)_\mu{}^\nu = \Lambda^\nu{}_\mu$. Everything you've done will be fixed if you take this as the starting point. $\endgroup$ – Prahar Mitra Jul 21 '20 at 17:41
  • $\begingroup$ I have edited my post to clarify the question. @Prahar, by index structure do you mean that $\Lambda^T$ is always to be interpreted as $(\Lambda^T)_{\mu}{}^{\nu}$? Also, I presume $\Lambda$ is always to be interpreted as $\Lambda = \Lambda^{\alpha}{}_{\beta}$. I think this helps to see why there is no contradiction: $({\Lambda^{T}})^{\sigma}{}_{\rho} = ({\Lambda^{-1}})^{\sigma}{}_{\rho}$ but the LHS is not what we would commonly call $\Lambda^{T}$ due to the index structure point? $\endgroup$ – Shrey Jul 21 '20 at 19:17
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    $\begingroup$ That is also totally valid as long as you understand what both sides mean. The LHS is what I described in my last comment whereas the RHS is $\eta_{\nu\beta} \Lambda^\beta{}_\alpha \eta^{\alpha\mu}$ which using the defining equation for the Lorentz matrix is $(\Lambda^{-1})^\mu{}_\nu$. You can freely manipulate all the indices as much as you want. Just be sure to understand exactly what each side means when you move back to matrix notation. $\endgroup$ – Prahar Mitra Jul 21 '20 at 19:32
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$\def\th{\theta} \def\sp{\hspace{1ex}} \def\b{\beta} \def\a{\alpha} \def\m{\mu} \def\n{\nu} \def\g{\gamma} \def\d{\delta} \def\mt{\eta} \def\mti{\mt^{-1}} \def\F{\Phi} \def\Ft{\widetilde{\F}} \def\L{\Lambda} \def\Li{\L^{-1}} \def\Lt{\L^T} \def\id{\mathbb{I}}$From the invariant interval on derives \begin{align*} \L^\b_{\sp\a} \mt_{\b\g} \L^\g_{\sp\d} &= \mt_{\a\d}.\tag{1} \end{align*} Let $\F_\a^{\sp\b}=\L^\b_{\sp\a}$, so (1) is written as \begin{align*} \F_\a^{\sp\b} \mt_{\b\g} \L^\g_{\sp\d} &= \mt_{\a\d}\tag{2} \end{align*} with matrix interpretation \begin{align*} \F\mt\L=\mt.\tag{3} \end{align*} By index gymnastics (2) is massaged into the form $$\F^\a_{\sp\g} \L^\g_{\sp\d} = \d^\a_\d$$ so $\F^\a_{\sp\b} = (\Li)^\a_{\sp\b}$. Note that, critically, the first index has been raised and the last lowered. Letting $\Ft$ be the matrix determined by $\F^\a_{\sp\b}$ we have $$\Ft=\Li.$$ We are, however, interested in $\F_\a^{\sp\b}$. We find $\F_\a^{\sp\b} = \mt_{\g\a}\F^\g_{\sp\d}\mt^{\b\d} = \mt_{\g\a}(\Li)^\g_{\sp\d}\mt^{\b\d}$. This has the matrix interpretation \begin{align*} \F &= \mt\Li\mti. \tag{4} \end{align*} In fact, (4) follows immediately from (3), illustrating the usefulness of the matrix representation. The confusion boils down to one between $\F$ and $\Ft$. It is straightforward to show from (4) and the general form for $\L$ that $\F=\Lt$. (See comment below.) Thus, $$\Lt\mt\L=\mt$$ is the correct matrix representation of (1).

Let us illustrate the difference between $\F$ and $\Ft$ with a specific nontrivial example. Represent $\L^\a_{\sp\b}$ by $$\L = \left[ \begin{array}{cccc} \gamma & 0 & 0 & -\beta \gamma \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{array} \right].$$ Then $$\F = \Lt = \left[ \begin{array}{cccc} \gamma & 0 & 0 & -\beta \gamma \\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & -\sin \theta & \cos \theta & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{array} \right].$$ But $$\Ft = \Li = \left[ \begin{array}{cccc} \gamma & 0 & 0 & \beta \gamma \\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & -\sin \theta & \cos \theta & 0 \\ \beta \gamma & 0 & 0 & \gamma \\ \end{array} \right].$$ (This is a boost in the $z$ direction and rotation about the $z$-axis.)

Comment

When one studies the general solutions to (1) one finds that they are combinations of rotations and boosts. Note that for a rotation $\mt \L(\th)^{-1}\mti=\mt\L(-\th)\mti=\L(-\th)=\L(\th)^T$ and that for a boost, $\mt \L(\b)^{-1}\mt=\mt \L(-\b)\mti= \L(\b)=\L(\b)^T$.

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  • $\begingroup$ Thank you for your answer. "It is straightforward to show [...] that $\Phi=\Lambda^T$." Sorry, this part is not too clear to me. Isn't equation 1 (or 3) the defining equation for $\Lambda$? Not sure how to use that equation and (4) to show this. Maybe understanding this will make it clear why $\Lambda^T = (\Lambda^T)_{\mu}{}^{\nu}$, whereas $\Lambda = \Lambda^{\mu}{}_{\nu}$ and $\Lambda^{-1} = (\Lambda^{-1})^{\mu}{}_{\nu}$. i.e. why the matrix representation for $\Lambda^T$ corresponds to this specific index placement to be consistent with the indices in the matrix rep of $\Lambda$ $\endgroup$ – Shrey Jul 21 '20 at 23:18
  • $\begingroup$ I am glad to help! I added a short comment above to address this. $\endgroup$ – user26872 Jul 21 '20 at 23:37
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After a very helpful discussion in the comments section and reading the responses, I thought I would type up (from my perspective) what I have learnt in case it will help anyone with the same question.

$$({\Lambda^{T}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma} = ({\Lambda^{-1}})^{\sigma}{}_{\rho}$$

is in fact a correct statement, but we have to be careful when converting this back into a matrix equation.

We should interpret $\Lambda$ as $\Lambda^{\mu}{}_{\nu}$, $\Lambda^{-1}$ as $({\Lambda^{-1}})^{\mu}{}_{\nu}$, but $\Lambda^T$ should be interpreted as $(\Lambda^T)_{\mu}{}^{\nu}$.

Therefore, we cannot interpret $({\Lambda^{T}})^{\sigma}{}_{\rho}$ as $\Lambda^T$ so equation D is incorrect. Instead, using the metric, $({\Lambda^{T}})^{\sigma}{}_{\rho} = \eta^{\sigma \alpha} (\Lambda^T)_{\alpha}{}^{\beta} \eta_{\beta \rho}$. So, instead of matrix equation D, we really should have:

$$\Lambda^{-1} = \eta \Lambda^T \eta$$

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  1. Note that the conventional definition of transposed matrix $$(\Lambda^T)_{\nu}{}^{\mu} ~:=~ \Lambda^{\mu}{}_{\nu}.\tag{1'} $$ is slightly different than OP's definition (1).

  2. In plain English: When we don't apply the metric, the Lorentz matrix $\Lambda$ has conventionally NW-SE slanted indices, while the transposed matrix $\Lambda^T$ has SW-NE slanted indices.

    See also e.g. this & this related Phys.SE post.

  3. Incidentally, OP's eq. (1) is consistent with eq. (1') after appropriately raising and lowing indices with the metric.

  4. OP's eq. (D) is wrong because it doesn't comply with above convention.

  5. More details: In matrix form OP's eqs. (A)-(C) read $$ \Lambda^{-1}~=~ (\eta\Lambda\eta^{-1})^T,\tag{A} $$ $$ \eta^{-1}\Lambda^T\eta~=~ (\eta\Lambda\eta^{-1})^T,\tag{B} $$ $$ \eta^{-1}\Lambda^T\eta~=~\Lambda^{-1},\tag{C} $$ respectively, which are indeed all true. Eq. (C) follows from the definition $$ \Lambda^T\eta\Lambda~=~\eta$$ of a Lorentz matrix.

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  • $\begingroup$ Thank you for your answer. I have attempted to justify the matrix form of A and B to ensure I have understood correctly (denoting the inverse metric as the metric for brevity) - I think I've shown A but got something different for B. I will post these in separate comments since it's too long. However, from A and B in index form, why can I not then conclude that $\Lambda^{-1} = \Lambda^T$ in matrix form? $\endgroup$ – Shrey Jul 21 '20 at 18:41
  • $\begingroup$ For A: \begin{align} (\Lambda^{-1})^{\sigma}{}_{\rho} &= \eta^{\sigma \nu} (\Lambda^T)_{\nu}{}^{\mu} \eta_{\mu \rho} \\ &= \eta^{\sigma \nu} (\Lambda^T)_{\nu}{}^{\mu} \eta_{\mu \rho}\\ &= \eta^{\nu \sigma} \eta_{\rho \mu} (\Lambda)^{\mu}{}_{\nu} \\ &= \Lambda_{\rho}{}^{\sigma} \\ \end{align} $\endgroup$ – Shrey Jul 21 '20 at 18:42
  • $\begingroup$ For B: \begin{align} \Lambda_{\rho}{}^{\sigma} &= (\eta_{\rho \mu} \Lambda^{\mu}{}_{\nu} \eta^{\nu \sigma}) \\ \eta^{\sigma \nu} (\Lambda^T)_{\nu}{}^{\mu} \eta_{\mu \rho} &= (\eta_{\rho \mu} \Lambda^{\mu}{}_{\nu} \eta^{\nu \sigma}) \\ \eta \Lambda^T \eta &= \eta \Lambda \eta \end{align} However, this is a different matrix equation to the one you arrived at. $\endgroup$ – Shrey Jul 21 '20 at 18:43
  • $\begingroup$ Under the convention that $\Lambda$ has NW-SE slanted indices, are there a few lines of algebra that can show $\Lambda^T$ has SW-NE indices? Equivalently, is it possible to show why (1') should be the conventional definition, consistent with the slanted indices of $\Lambda$? I understand the transpose switches the row and column indices, keeping their vertical positions intact, but this doesn't feel like enough. $\endgroup$ – Shrey Jul 22 '20 at 11:23
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jul 22 '20 at 11:32
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I think the reason this is confusing is that tensor and matrix notations are being mixed up in ways that do not actually make sense. Also, the notation fails to observe that the Lorentz transformation takes one coordinate system to another. Usually one would need a prime on either $\mu$ or $\nu$ (one talks of primed and unprimed coordinates). The Lorentz transformation is a particular instance of a general coordinate transformation, which can be written

$$k^{\mu'}_\nu = x^{\mu'}_{,\nu} = \frac{\partial x^{\mu'} }{\partial x^{\nu}}$$

In such a case ${\mu'}$ runs over rows and ${\nu}$ runs over columns. It does not matter which index is "first" (I definitely prefer accounts like that of Dirac, General Theory of Relativity which explicitly does not put either index first in this case). The transpose would swap covariant and contravariant indices, which kind of makes a nonsense of your definition 1. Transpose is used for matrices, because the ordering of indices is important for matrix multiplication. But in general relativity this is already taken care of in index notation through Einstein's summation convention. I don't recall any of my preferred texts for gtr using transpose, but I have to confess, if an author did use it, I think I would rapidly find another author.

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The problem here is that your definition (1) is incorrect if the metric isn't the identity in your coordinates. The only correct way of raising/lowering indices is by contraction with the metric.

A simple proof that your definition can't be right: Suppose our metric is $\def\d{{\rm d}}\d s^2={\d x^2\over4}-\d v^2$, and $\Lambda_\mu{}^\sigma=\begin{bmatrix}5\over4&3\over2\\3\over8&5\over4\end{bmatrix}$.

Then your definition gives $\Lambda^\mu{}_\sigma=\begin{bmatrix}5\over4&3\over8\\3\over2&5\over4\end{bmatrix}$, while mine gives $\Lambda^\mu{}_\sigma=\begin{bmatrix}5\over4&-{3\over2}\\-{3\over8}&5\over4\end{bmatrix}$.

Then if you use your definition to evaluate $\Lambda^\rho{}_\mu\eta_{\rho\sigma}\Lambda^\sigma{}_\nu$, you get $\begin{bmatrix}-{119\over64}&-{225\over128}\\-{225\over128}&-{391\over256}\end{bmatrix}$; if you use mine, you get $\begin{bmatrix}1\over4&0\\0&-1\end{bmatrix}$.

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