# Problem in deriving elements of Lorentz Group

I am following a lecture series for Classical Field Theory in which they the lecturer uses the invariance of the length of a vector during a Lorentz transformation to derive the equation $$O^{T}\eta~O=\eta.$$ He derives this equation by the following these steps: Taking a vector $v$ and taking its length $$|v|^{2}=v^{T}\eta~v$$to be constant in a transformation we would get $$|v'|^{2}=v'^{T}\eta~v'$$ where $v'=O~v$, $O$ being the Lorentz transformation matrix. Further he writes $$|v'|^{2}=(v^{T}~O^{T})\eta~(O~v)$$ and then $$O^{T}\eta~O=\eta$$ for the length of the vector to be constant. Now my question is shouldn't the metric tensor $\eta$ itself not change under coordinate transformation and the equation should be $$O^{T}\eta'~O=\eta?$$ Because he then moves forth to derive some properties of $O$ using the equation $$O^{T}\eta~O=\eta.$$

This is a common confusion and it's just due to some tricky notation.

• Every rank 2 tensor $A$ transforms as $A' = O^T A O$.
• Your book has shown that for Lorentz transformations, the metric $\eta$ satisfies $\eta = O^T \eta O$.
• Therefore, the metric transforms as $\eta' = \eta$ under Lorentz transformations. This is a very special condition that isn't true for a general rank 2 tensor.
• where does this condition of $\eta=\eta'$ come from? – Naman Agarwal Dec 12 '17 at 12:42
• @NamanAgarwal Just combine the first two equations. – knzhou Dec 12 '17 at 12:45
• I have learnt the general way in which tensors transform but have never heard of the in variance of the metric tensor...so i am asking why is it so? – Naman Agarwal Dec 12 '17 at 12:49

Let's start with 3-dimensional euclidian world. Consider 3D rotations represented by matrices R. This should be demonstrative enough to show that the length of a 3-dim. vector is invariant under rotations.

$|v|^2 =v^T v$ is the length of a 3D arbitrary vector. In the rotated system the length is $|v'|^2 =v'^T v' = (v^T R^T)(Rv) = v^T v = |v|^2$ since the rotation matrices fulfill $R^T R = id \equiv 1$ It can be also be written like $R_{ij}\delta_{jk}R_{kl}= \delta_{il}$ with $\delta$ as Kronecker symbol which now has the role of the metric, i.e. in component-free language we would write for it : $R^T\delta R=\delta$. Of course one can imagine a coordinate transformation to polar coordinates, then the corresponding transformation matrices P no longer fulfill $P^T P = id$. But of course the length of an arbitrary vector is the same in polar coordinates than in cartesian coordinates.

Now we'll consider Minkowski space. Actually the Lorentz-transformations $O$ are those types of coordinate transformations which fulfill $\eta = O^T \eta O$. The Minkowski metric is not changed by the Lorentztransformation is another phrasing of this fact. And as in euclidian 3D space we can carry out a coordinate transformation $C$ where $\eta' = C^T \eta C$. But imagine a transformation which would only change the diagonal elements in $\eta$.

Normally:

$\eta = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 &-1 & 0\\ 0 & 0& 0 & -1 \end{array} \right)$

Imagine the matrix element $\eta'_{00}=1.2$ after the coordinate transformation instead of $\eta_{00}=1$. Then the invariant line element would be $ds^2 = 1.2 c^2 dt'^2 -dx'^2 -dy'^2 -dz'^2$. In these coordinates the speed of light would be no longer $c$. Actually such kind of coordinate transformations exist, but evidently these are no longer Lorentz transformations.