# A Invariance of metric tensor

1. Jan 11, 2017

### Fermiat

I'm stuck on an apparently obvious statement in special relativity, so I hope you can help me. Can I define Lorenz transformations as transformations that don't change the spacetime interval in M4 and from this deduct that the metric tensor is invariant under LT? I've always read that the invariance of the metric tensor under LT was assumed, but I've never seen this way of proceeding

2. Jan 11, 2017

### Staff: Mentor

This would not be a deduction, it would be a tautology. "Not changing the spacetime interval" is the same thing as "leaving the metric tensor invariant".

3. Jan 14, 2017

### vanhees71

Tensors don't change under the transformations they are tensors for. That's a definition. What changes in general are the components of tensors with respect to basis transformations.

Denoting the Minkowski product of two four vectors with $\boldsymbol{x} \cdot \boldsymbol{y}$ the metric (or better pseudometric!) tensor's components with respect to an arbitrary basis $\boldsymbol{b}_{\mu}$ are given by
$$g_{\mu \nu} = \boldsymbol{b}_{\mu} \cdot \boldsymbol{b}_{\nu}.$$
A Lorentz transformation by definition is a linear transformation which leaves the Minkowski products between any two vectors invariant. So defining a new basis via a Lorentz transformation $\boldsymbol{b}_{\mu}'=\Lambda \boldsymbol{b}_{\mu}$ implies
$$g_{\mu \nu}'= \boldsymbol{b}_{\mu}' \cdot \boldsymbol{b}_{\nu}' = (\Lambda \boldsymbol{b}_{\mu}) \cdot (\Lambda \boldsymbol{b}_{\nu}) = \boldsymbol{b}_{\mu} \cdot \boldsymbol{b}_{\nu}=g_{\mu \nu},$$
i.e., if you change a basis by using Lorentz transformations, the components of the pseudometric don't change.

This is particularly true for pseudoorthonormal bases, for which
$$g_{\mu \nu}=\eta_{\mu \nu} =\mathrm{diag}(1,-1,-1,-1).$$

4. Jan 14, 2017

### Fermiat

Thank you for the replies, so I was just confused by the fact I was considering a tautology.