# How the Lorentz transformation affects the metric tensor?

After performing a Lorentz transformation, the orthogonal coordinates will become askew, as in the following figure:

and in such coordinate system, according to this Wikipedia article, the metric will have off-diagonal non-zero elements: $$g_{ij}=\mathbf{e}_i.\mathbf{e}_j$$

which is not that of a flat spacetime: $$g = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$

What's the problem?

• A Lorentz transformation is just a change of coordinates, and although it does happen to leave the components of the metric invariant, it's not really a problem if a change of coordinates doesn't do that. All that really matters is the signature of the metric (the signs of its eigenvalues), and that's guaranteed not to change (assuming the transformation is nonsingular) because of Sylvester's law of inertia. For example, there is nothing wrong with a transformation in which you double all the coordinates, so that $g=\operatorname{diag}(1/4,-1/4,-1/4,-1/4)$. – user4552 Sep 16 '13 at 0:27
• @user4552 It indeed matters. (1) It does not "happen to“ leave the components invariant", it is destined to leave it invariant, as the answers say. (2) If it indeed doesn't leave the components invariant (which is destined to never happen) then there won't be the concept "metric tensor" discussed in physics at all, all the reason "metric tesor" is discussed in physics is that it leave the components invairant under the specific tranformation that happens in SR- the Lorentz transformation. – jw_ Jan 17 at 10:15

You need to be very careful. The $\mathbf{e}_i$ are vectors, so they have a Lorentz index: $\mathbf{e}_i^\mu.$ When you write $$\mathbf{e}_i \cdot \mathbf{e}_j$$ you actually mean $$g_{\mu \nu} \mathbf{e}_i^\mu \mathbf{e}_j^\nu$$ where $g_{\mu \nu}$ is the flat Minkowski metric (not the Euclidean metric). Once you know this, it's straightforward to check that the flat metric is preserved under Lorentz transformations.

Edit: in the theory of relativity, physicists refer to these objects as a frame or a vierbein. Might help you look for other resources if you want to.

• Not really. When you write $g_{\mu \nu} \mathbf{e}_i^\mu \mathbf{e}_j^\nu$ it's a shorthand for $\mathbf{e}_i \cdot \mathbf{e}_j$, and that (the inner product structure, in this case a Minkowski indeterminate one) is the fundamental object. – Emilio Pisanty Sep 16 '13 at 16:22
• Yes, but that's not what OP is asking. He/she wants to know why get an inconsistent result when you calculate the inner product $\mathbf{e}_i \cdot \mathbf{e}_j$ naively (in the Euclidean sense). It's a more fundamental view but you need to assign the "right" value to this inner product. – Vibert Sep 17 '13 at 11:00
• Then the problem is that you need to rethink what you mean by inner product and orthogonality. The component version is a secondary construct. – Emilio Pisanty Sep 17 '13 at 11:04

Vibert is, of course, completely correct. I'm gong to propose a slightly more geometric version of what he says. The minkowski metric tensor is given by:

$$ds^{2} = g_{ab}dx^{a}dx^{b} = -dt^{2} + dx^{2} + dy^{2} + dz^{2}$$

Now, knowing that $v = \tanh\phi$, it is easy enough to show that the Lorentz transformations are given by:

\begin{align} t' &= t \cosh \phi - x \sinh\phi\\ x' &= x \cosh \phi - t \sinh\phi \end{align}

Taking the differential, solving for $dt$ and $dx$ and substituting into the line element above, we find:

$$ds'^{2} = -\left(\cosh^{2}\phi - \sinh^{2}\phi\right)dt'^{2} + \left(\cosh^{2}\phi - \sinh^{2}\phi\right)dx'^{2} + dy^{2} + dz^{2}$$

Since we know the basic trigonometric identity $\cosh^{2} - \sinh^{2} = 1$, we find that

$$ds'^{2} = g_{ab}'dx'^{a}dx'^{b} = -dt'^{2} + dx'^{2} + dy^{2} + dz^{2}$$

and it is clear that the components of the metric tensor are the same.

EDIT: to see how this becomes the ordinary lorentz transform:

\begin{align} \tanh\phi &= v\\ \frac{\sinh\phi}{\cosh\phi} & = v\\ \sinh^{2}\phi &= v^{2} \cosh^{2}\phi\\ \cosh^{2}\phi -1 &= v^{2}\cosh^{2}\phi\\ \cosh^{2}\phi\left(1 - v^{2}\right) &= 1\\ \cosh\phi &= \frac{1}{\sqrt{1-v^{2}}}\\ \sinh\phi &= \sqrt{cosh^{2}\phi -1}\\ &= \sqrt{\frac{1}{1-v^{2}} - 1}\\ &= \sqrt{\frac{1 - (1-v^{2}) }{1-v^{2}}}\\ &= \frac{v}{\sqrt{1-v^{2}}} \end{align}

You can figure out the rest

The transformation law for a covariant vector $e$ (like a basis vector), is :

$$e_i = \frac{\partial x^\mu}{\partial x^i} e_\mu \tag{1}$$

For a 2-covariant tensor like the metrics $g$, the transformation law is :

$$g_{ij} = \frac{\partial x^\mu}{\partial x^i} \frac{\partial x^\nu}{\partial x^j} g_{\mu\nu} \tag{2}$$

This simply means that $g_{ij}$ transforms as $e_i e_j$

The relation between $e_i^\mu$ and the partial derivatives is simply :

$$e_i^\mu = \frac{\partial x^\mu}{\partial x^i}\tag{3}$$ So, you may write :

$$e_i = e_i^\mu e_\mu, \qquad g_{ij} = e_i^\mu e_j^\nu g_{\mu\nu}\tag{4}$$

So, $e_i^\mu$ is the component of the new basis vector $e_i$, in the old basis vector $e_\mu$