# Definition of metric tensor for Minkowski space

I learned that for a 4D Minkowski space with a set of basis vectors $${e^\mu}$$ the metric tensor is defined as $$g_{\mu\nu}=e_\mu \cdot e_\nu.$$

Does this mean that if one choose different sets of basis vectors, the metric tensor for Minkowski space will change?

What are some examples of the sets of basis vectors that can describe 4D Minkowski space?

I learned that for a 4D Minkowski space with a set of basis vectors $${e^\mu}$$ the metric tensor is defined as $$g_{\mu\nu}=e_\mu \cdot e_\nu.$$

I would put this a little differently. This reads to me as a definition of one of the components of the metric tensor. You can think of a tensor as the collection of its components (a matrix), and people did used to think about it that way ca. 1940.

But a nicer and more modern way to think about it is that a tensor is a linear function. The metric is a linear function that takes two vectors as inputs and gives a real scalar as an output. It treats its inputs symmetrically, and it has the interpretation of measuring the interval, which can often be interpreted as a squared time or squared distance.

Does this mean that if one choose different sets of basis vectors, the metric tensor for Minkowski space will change?

This would be the 1940 way of describing it. People today would typically say that the metric stays the same, but its components are different when described in the new basis.

What are some examples of the sets of basis vectors that can describe 4D Minkowski space?

For example, you could have one observer's orthonormal t, x, y, z basis, or the t', x', y', z' basis of another observer in a different state of motion. You could have a basis with some null vectors in it. You can have any basis you want, as long as it's linearly independent -- this is the same as for any 4-dimensional vector space.

Does this mean that if one choose different sets of basis vectors, the metric tensor for Minkowski space will change?

If by metric tensor you mean the whole tensor ($$\bf{g}$$) then no the metric tensor will not change. But each component may change. Basis vectors transform according to $$e'_\mu = \frac{\partial x^{\nu}}{\partial x'^\mu} e_\nu$$ Hence the metric tensors components will transform as $$g'_{\mu\nu} = \frac{\partial x^{\alpha}}{\partial x'^\mu}\frac{\partial x^{\beta}}{\partial x'^\nu} g_{\alpha \beta}$$

What are some examples of the sets of basis vectors that can describe 4D Minkowski space?

Spherical coordinates and cylindrical coordinates also describe $$(3+1)$$ dimensional Minkowski space. It can be easily shown that Cartesian, spherical and cylindrical coordinates describe the same space time.

Nope. The metric is defined as such to be the same in all inertial frames (if it wasn’t then you could have a frame in which the ‘length’ $$ds = g_{\mu \tau}dx^\mu dx^\tau$$ was different. An equivalent and more axiomatic way to see this is to define the metric as a tensor, thus by definition is the same in all reference frames.

This means that there is a restriction on the basis vectors, that being they must transform like

$$e_{j’}= \frac{\partial x^i}{\partial x^{j’}} e_i$$

That is they transform as covariant tensors.

So instead of the metric changing, the basis vectors (and thus the entire apace of vectors) change under such a way as to keep the metric, and therefore the structure of the 4D space the same.