# Pure Lorentz boost; transpose $\neq$ inverse?

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?

• Why should the inverse of a boost be equal to the original boost? – SRS Jul 28 '17 at 9:56
• I believe the OP's asking why the inverse is not equal to the transpose? – Philip Cherian Jul 28 '17 at 9:58
• 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 $$(\Lambda^{-1})^a_{~~b}=\Lambda_b^{~~a}?$$ – SigmaAlpha Jul 28 '17 at 9:58
• I am curious to know the reasons for any downvotes. – SigmaAlpha Jul 28 '17 at 10:04
• 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. – Philip Cherian Jul 28 '17 at 11:16

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

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

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

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.

## protected by Qmechanic♦Jul 28 '17 at 12:58

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