# Non-symmetric Lorentz Matrix

I was working out a relatively simple problem, where one has three inertial systems $S_1$, $S_2$ and $S_3$. $S_2$ moves with a velocity $v$ relative to $S_1$ along it's $x$-axis, while $S_3$ moves with a velocity $v'$ along $S_2$s $y$-axis.

So I constructed the Lorentz transformation by multiplying the transformation from 1 to 2 with the transformation from 2 to 3, and I obtain the transformation. Now, this is all nice and well, and it is also where the question for which I'm doing this stops. However, it made me think: the resulting Lorentz matrix is not symmetric!

Somehow I always thought that they always were (simply because I hadn't encountered one that was not), which I suppose is naive. Is there any information inherent to the fact that the transformation is not symmetric? Does this in some way mean that there is a rotation happening? This is what seemed most plausible to me, as if you boost 'harder' in one direction than in the other, your axis is essentially rotating. Or am I going about this the wrong way?

• By not symmetric you mean the observation that for two Lorentz-trafos in general $\Lambda_1\Lambda_2\neq\Lambda_2\Lambda_1$ ? – Nephente Feb 23 '14 at 18:11
• Ah no, I should have been more clear. I mean that the Lorentz transformation from system 1 to 3, which is a 4x4 matrix, is not symmetric in it's diagonal. – user129412 Feb 23 '14 at 18:13
• Do you mean $\Lambda_{ij} \neq \Lambda_{ji}$? – Hunter Feb 23 '14 at 19:06
• Yes, indeed, that is what I mean. – user129412 Feb 23 '14 at 20:12
• "Simply because I hadn't encountered one that was not." A rotation is a lorentz transformation which is not symmetric. – Brian Moths Feb 23 '14 at 21:50

You may know already that "symmetry" is not always important when it comes to non-Euclidean spaces. For instance, in quantum mechanics, a symmetric operator is seldom important, but one that is equal to its Hermitian conjugate--one that is Hermitian or "self-adjoint"--is incredibly important, for those operators have real eigenvalues and thus correspond to observables.

While Lorentz boost matrices may appear symmetric, they do not obey the basic propery of being self-adjoint: for Minkowski space (and in general spaces), the adjoint $A^\dagger$ of a matrix $A$ is the matrix that obeys

$$A_{\alpha \beta} a^\alpha b^\beta = (A^\dagger)_{\alpha \beta} b^\alpha a^\beta$$

Lorentz transformation matrices do not have $A = A^\dagger$, but they do have $A^{-1} = A^\dagger$--they are "orthogonal" in the same sense that orthogonal matrices in Euclidean space obey this basic property of their inverses being equal to their adjoints.

Long story short: you should pay little attention to the transpose once you get to special relativity. It's only in Euclidean space that the transpose operation is equivalent to the adjoint, but the adjoint is more interesting and more useful in general.

Edit: you may look at the equation and realize that all it implies is that $(A^\dagger)_{\alpha \beta} = A_{\beta \alpha}$, so it looks like a transpose. There is a critical difference, though, in the placement of indices compared to a usual matrix transpose.

A usual matrix corresponds to a placement of indices like ${A^\alpha}_\beta$. For instance, if $A$ is a Lorentz boost in 1+1 spacetime, then its components are as follows:

\begin{align*} {A^t}_t &= \gamma \\ {A^t}_x &= \gamma \beta \\ {A^x}_t &= \gamma \beta \\ {A^x}_x &= \gamma \end{align*}

You should see, then, that ${A^\alpha}_\beta b^\beta$ is then, for any vector $b^\beta$:

$${A^t}_\beta b^\beta = \gamma (b^t + \beta b^x), \quad {A^x}_\beta b^\beta = \gamma (b^x + \beta b^t)$$

So this is a Lorentz transformation.

But if we want to consider an inner product of a vector and a boosted vector, as is necessary to consider in the definition of the adjoint, we need to lower an index somewhere. This uses the metric.

If we choose a $(-, +)$ signature, then $\eta_{tt} = -1$ and $\eta_{xx} = +1$, and we get

\begin{align*} A_{tt} &= \eta_{t t} {A^t}_t = -\gamma \\ A_{tx} &= \eta_{tt} {A^t}_x = -\gamma \beta \\ A_{xt} &= \eta_{xx} {A^x}_t = \gamma \beta \\ A_{xx} &= \eta_{xx} {A^x}_x = \gamma \\ \end{align*}

So any apparent symmetry you thought was there, in looking at the matrix ${A^\alpha}_\beta$, is gone. The non-Euclidean metric spoils it.

"Simply because I hadn't encountered one that was not." A rotation is a lorentz transformation which is not symmetric.

Indeed the transpose of a rotation matrix is its inverse, and only trivial rotations or rotations through half a turn are involutary (self inverse). To see this in detail, write a rotation matrix as $\exp(H)$, where $H$ is skew-symmetric and real, then $\exp(H)^\dagger = \exp(H)^T$. More explicitly, partition the transformation matrix into $2\times 2$ blocks and consider a rotation about the $x$-axis:

$$\Lambda = \left(\begin{array}{cc}I&0\\0&R\end{array}\right)$$

where $I$ is the identity and $R$ is a $2\times2$ rotation matrix; the latter's general form is, of course,:

$$\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)$$

which is clearly not symmetrix.

Yes, your intuition is correct: two different boosts do contain one rotation, and precisely two boosts along two orthogonal axes contain one rotation around the third orthogonal axis --- the most direct way to see that is by considering that the commutator of two different boosts is one rotation, and more completely the Lorentz algebra of rotations $R_{a}$ and boosts $B_{a}$ is given by the commutation relationships $$[R_{a},R_{b}]=\varepsilon_{abc}R^{c}$$ $$[B_{a},B_{b}]=-\varepsilon_{abc}R^{c}$$ $$[R_{a},B_{b}]=\varepsilon_{abc}B^{c}$$ where $\varepsilon_{abc}$ is the Levi-Civita symbol and indices are moved by employing the Minkowski matrix.

So to summarize, yes, you were right.