The metric tensor of inertial frames in S.R is given by $$g_{\alpha \beta}=diag(1,-1,-1,-1)$$ It's inverse $$(g_{\alpha \beta})^{-1}=g_{\alpha \beta}$$

I was wondering what this means geometrically. I know in General Relativity (I haven't delved deeper into this), the metric tensor has non-zero off-diagonal components, therefore, I wouldn't expect the metric tensor to be the inverse of itself but why is this true in Special Relativity where we only deal with flat spacetime?

I'm trying to think of it right now that the properties and trajectories in flat spacetime look the same forward and backward (like multiplying it by its inverse) so it forces the inverse to be the same as the original but I'm not sure (this could be baloney but I don't have math to support to disprove it right now).

Since the metric tensor basically encodes the properties of spacetime, what does this particular result mean physically?

Note: I've just started learning classical field theory and use Landau & Lifshitz Classical Field theory (Vol.4)

  • $\begingroup$ remember that you're talking about matrices containing the components of the metric tensor $\endgroup$
    – basics
    Oct 4, 2022 at 18:50

3 Answers 3


I do not have enough reputation yet to comment (hence an answer): If you write the metric in cylindrical or spherical coordinates you'll see that the covariant metric is not the same as the contravariant metric. Regardless of the coordinate system chosen and the specific form, under all coordinate systems the curvature will be zero, which is what flat spacetime is about.


OP's title condition is not a covariant/geometric/physical condition: it depends on choice of coordinates. E.g. if it holds for coordinates $x^{\mu}$, it will not hold for coordinates $x^{\prime \mu}=2x^{\mu}$.

See also Is it foolish to distinguish between covariant and contravariant vectors? and links therein.


There isn't an "inverse" for a tensor, because a tensor (of order 2) takes two vectors and produces one scalar:

$$g: \mathbb{R}^{1,3} \times \mathbb{R}^{1,3} \rightarrow \mathbb{R}$$

or if you like, a dimensional collapse from eight dimensions to one! As a result, the "inverse", were there to be one, would be highly one-to-many (i.e. not a function): it would represent the entire class of vectors with a given inner product. Such a highly non-functional correspondence is also, perhaps unsurprisingly, not a tensor.

What you've done in your post, instead, is to take the inverse of the linear transformation represented by interpreting $g$ as a matrix. There is no unique or canonical way to convert a tensor (acting on two vectors) to a linear map (acting on one vector and outputting one vector). In particular, the action of this linear map "should" look like

$$g_{\mu \nu} v^\nu$$

which, in terms of the standard (and well-defined) tensor action involving two vectors, i.e.

$$g_{\mu \nu} u^{\mu} v^{\nu}$$

can only be obtained by taking for $\mathbf{u}$ that $\mathbf{u} = \mathbf{e}_\mu$ for $\mu = 0, 1, 2, 3$ in succession, i.e. the basis vectors for a particular basis, and likewise generating vector components also in the same said basis. Hence this single-vector action, and thus the understanding of $g$ as a unilinear map, is entirely basis-dependent.

Hence it says nothing about the dynamics taking place on that space-time manifold, and more about a peculiar property of your particular choice of coordinate system.

(You could also say that this is because the matrix representation of $g$ is basis-dependent, but that is not really a particularly convincing explanation because without care, one would be led into thinking that the "true" tensor action $g_{\mu \nu} u^{\mu} v^{\nu}$ is also basis-dependent, when that is definitely not the case!)


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