1
$\begingroup$

By definition a matrix representing a Lorentz transformation is orthogonal, so that its inverse is equal to its transpose.

Consider a pure boost in the t-x plane; $$\Lambda_x=\begin{pmatrix} \cosh(\gamma) && \sinh(\gamma) && 0 && 0\\ \sinh(\gamma) && \cosh(\gamma) && 0 && 0\\ 0 && 0 && 1 && 0\\ 0&&0&&0&&1 \end{pmatrix}.$$ $\Lambda_x$ has inverse $$\Lambda_x^{-1}=\begin{pmatrix} \cosh(\gamma) && -\sinh(\gamma) && 0&&0\\ -\sinh(\gamma) && \cosh(\gamma) && 0&&0\\ 0 && 0 && 1 &&0\\ 0&&0&&0&&1 \end{pmatrix}$$ but tranpose $$\Lambda_x^T=\begin{pmatrix} \cosh(\gamma) && \sinh(\gamma) && 0&&0\\ \sinh(\gamma) && \cosh(\gamma) && 0&&0\\ 0 && 0 && 1&&0\\ 0&&0&&0&&1 \end{pmatrix}.$$ These are not equal. Where have I gone wrong?

$\endgroup$
  • $\begingroup$ Why should the inverse of a boost be equal to the original boost? $\endgroup$ – SRS Jul 28 '17 at 9:56
  • $\begingroup$ I believe the OP's asking why the inverse is not equal to the transpose? $\endgroup$ – Philip Cherian Jul 28 '17 at 9:58
  • $\begingroup$ I thought that part of the definition of a Lorentz transformation is that its inverse is equal to its transpose (The O in SO(3,1)). Isn't this the reason why we can perform manipulations such as $$(\Lambda^{-1})^a_{~~b}=\Lambda_b^{~~a}?$$ $\endgroup$ – SigmaAlpha Jul 28 '17 at 9:58
  • 2
    $\begingroup$ I am curious to know the reasons for any downvotes. $\endgroup$ – SigmaAlpha Jul 28 '17 at 10:04
  • $\begingroup$ For what it's worth, I think the question is something many people, at least those learning the subject themselves, will face (I know I did) and I don't think it should be down-voted. $\endgroup$ – Philip Cherian Jul 28 '17 at 11:16
3
$\begingroup$

The matrix representing a Lorentz boost is orthogonal with respect to the Minkowski metric $\eta = \mathrm{diag}(-1,1,1,1)$ (or reversed signs), which means $$ \Lambda \eta \Lambda^T = \eta \text{ or } \Lambda^{-1} = \eta \Lambda^T\eta.$$

$\endgroup$
4
$\begingroup$

This is quite a common problem, and I feel it's usually made a bit clearer by using indices.

The Lorentz Group is $SO(1,3)$, not $SO(4)$, since (using the $(-+++)$ convention) the line element has an extra negative sign:

$$\text{d}s^2 = -c^2\text{d}t^2 + \text{d}x^2 + \text{d}y^2 + \text{d}z^2 $$

which is why Lorentz boosts are different from pure rotations in 4-dimensional space.

Such a boost $\Lambda$ takes $$x^\mu \longrightarrow x'^\mu = \Lambda^\mu_{\;\sigma} x^\sigma$$

Requiring that the element $\text{d}s^2$ be invariant, we have that

$$x^\mu x_\mu = x'^\alpha x'_\alpha$$ i.e.

$$x^\mu x^\nu \eta_{\mu\nu} = \left(\Lambda^\alpha_{\;\mu} x^\mu\right) \, \left(\Lambda^\beta_{\;\nu} x^\nu\right) \,\eta_{\alpha\beta} $$

Or in other words,

$$\eta_{\mu\nu} = \Lambda^\alpha_{\;\mu}\Lambda^\beta_{\;\nu}\eta_{\alpha\beta}$$

In terms of matrix multiplication, this is simply

$$\eta = \Lambda^\text{T} \, \eta \,\, \Lambda$$

which means, as you've shown yourself, $\Lambda^\text{T}\neq \Lambda^{-1}$, but rather

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

$\endgroup$
1
$\begingroup$

There is a little more that can be added to the previous answers.

  • To see that your $\Lambda_x$ is not orthogonal, remember that orthogonal matrices preserve the square of the length $\vec a\cdot \vec a$ of a vector $\vec a$. Using your $\Lambda_x$ and $\vec a=(1,0)$ one find, with $\vec a'=\Lambda_x \vec a$, that $\vec a'\cdot\vec a'=\cosh(2\gamma)\ne 1$, showing clearly that $\Lambda_x$ is NOT orthogonal.
  • The representation you have for $\Lambda_x$ is not only not orthogonal it is not unitary either. If $\Lambda_x$ were unitary, then $\Lambda_x^{-1}=\Lambda_x^\dagger=\Lambda_x^T$ since $\Lambda_x$ is real. The non-unitarity of $\Lambda_x$ is rooted in the fact that the Lorentz group is non-compact, and that there are no non-trivial finite dimensional representations of non-compact groups that are unitary.
$\endgroup$

protected by Qmechanic Jul 28 '17 at 12:58

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.