Help with proof involving the Minkowski metric tensor.

Question

$\newcommand{\LF}[2]{\Lambda^#1_{\hspace{.2cm} #2}} \newcommand{\LL}[2]{\Lambda^{\hspace{.2cm} #2}_#1} \newcommand{\af}{\alpha} \newcommand{\be}{\beta}$

I'm trying to prove that the Minkowski metric tensor $\eta_{\nu\mu}$ has the same components in every refernce frame, that is $\eta_{\nu\mu} = \eta'_{\nu\mu}.$ I tried to prove it using the fact that $\eta'_{\nu\mu} = \LL{\mu}{\af} \LL{\nu}{\be} \eta_{\af\be},$ and that $d s^2 $ is invariant, that is: the fact that it is a contravariant second rank tensor and it is related to a relativistic invariant. Any help/insight would be deeply appreciated. I'm familiar with the (+,-,-,-) convention ;).

Answer 1

With abuse of terminology a coordinate transformation as a Lorentz transformation

consider $T: \mathcal{M} \rightarrow \mathcal{M}$ an orthogonal transformation of $\mathcal{M}$ and $\left\{e_{1}, e_{2}, e_{3}, e_{4}\right\}$ an orthonormal basis for $\mathcal{M} .$ As orthogonal transformations carry any orthonormal basis to another orthonormal basis, $\hat{e}_{1}=T e_{1}$, $\hat{e}_{2}=T e_{2}$, $\hat{e}_{3}=T e_{3}$ and $\hat{e}_{4}=T e_{4}$ also form an orthonormal basis for $\mathcal{M}$. For each $e_{u}, u=1,2,3,4,$ we may write $\hat{e}_{a}$

$$ e_{u}=\Lambda_{u}^{\! 1} \hat{e}_{1}+\Lambda_{u}^{\! 2} \hat{e}_{2}+\Lambda_{u}^{\! 3} \hat{e}_{3}+\Lambda_{u}^{\! 4} \hat{e}_{4}=\Lambda_{u}^{\! a} \hat{e}_{a}, \quad u=1,2,3,4 $$

where the $\Lambda_{u}^{\! a}$ are constants. By orthogonality conditions $g\left(e_{c}, e_{d}\right)=\eta_{c d}$ where $c, d=1,2,3,4$ (please, plug in and verify yourself) we may write

$$ \Lambda_{c}^{\! 1} \Lambda_{d}^{\! 1}+\Lambda_{c}^{\! 2} \Lambda_{d}^{\! 2}+\Lambda_{c}^{\! 3} \ \Lambda_{d}^{\! 3}-\Lambda_{c}^{\! 4} \Lambda_{d}^{\! 4}=\eta_{c d} $$

with Einstein summation convention,

$$ \Lambda_{c}^{\! a} \Lambda_{d}^{\! b} \eta_{a b}=\eta_{c d}, \quad c, d=1,2,3,4 $$

In special relativity, the space is considered to be a flat manifold. By definition of a flat manifold, the metric tensor g is constant. In this sense there is nothing to prove as the constancy of the metric is due to the definition.

Answer 2

It can be shown by some tensorial manipulations.

Consider a map; $\Lambda_{a}^{\mu}dx^{a}=dx^{'\mu}$

Now, $ds^2$ being a Lorentz invariant object, is written as:

$$ds^2 = \eta_{ab}dx^{a}dx^{b} = \eta_{\mu\nu}^{'}dx^{'\mu}dx^{'\nu}$$

With $\eta_{ab}$ and $\eta_{\mu\nu}^{'}$ being the Minkowski Metric of corresponding Co-ordinate system.

now,

$$\eta_{\mu\nu}^{'}dx^{'\mu}dx^{'\nu} = \eta_{\mu\nu}^{'}\Lambda_{a}^{\mu}\Lambda_{b}^{\nu}dx^{a}dx^{b} =ds^2= \eta_{ab}dx^{a}dx^{b}$$

Which gives us: $$\eta_{\mu\nu}^{'}\Lambda_{a}^{\mu}\Lambda_{b}^{\nu} = \eta_{ab}$$

Now, it is to be noted that: $\eta_{\mu\nu}^{'}\Lambda_{a}^{\mu}\Lambda_{b}^{\nu} = \eta_{ab}^{'}$

$\therefore$ It is evident that: $$\eta_{ab}^{'} = \eta_{ab}$$.

So, in this way, the in-variance of Components of Minkowski Metric, under arbitrary co-ordinate transformations, is established.