Graduate Is the Metric Tensor Invariant under Lorenz Transformations in M4?

[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

 

[PeterDonis]

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?

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".

 

[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
$$g_{\mu \nu}=\eta_{\mu \nu} =\mathrm{diag}(1,-1,-1,-1).$$

 

[Fermiat]

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

 

[pedro_deoliveira]

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".

I see what you mean here, but just for logical consistency: a (deducted) theorem T have the same truth value as the premise P used to prove the theorem. So the statement (P ^ T) is always a tautology. Tecnically, even if it was the case that there was a deduction involved, it would still be a tautology.

 

[PeterDonis]

Tecnically, even if it was the case that there was a deduction involved, it would still be a tautology.

"Tautology" doesn't mean "has the same truth value". It means something stronger: it means "doesn't even need to be deduced because the two statements have exactly the same meaning and refer to exactly the same concept".