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How can one prove that $\mathbf{\Lambda^T \eta \Lambda} = \mathbf{\eta}$ in special relativity, where $\mathbf{\Lambda}$ is the Lorentz transformation and $\eta$ is the Lorentz metric? Also, how does this correspond to the intuitive definition of the invariance of the Lorentz metric that is $s(\mathbf{a}, \mathbf{b}) = s(\mathbf{\Lambda a}, \mathbf{\Lambda b})$?

If we could stay clear of anything explicitly related to tensors or Einstein notation here, that would be great, since they haven't been introduced at this stage in the course that I'm taking.

Qmechanic
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Noldorin
  • 7,547

2 Answers2

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Let $\mathbf{a}$ and $\mathbf{b}$ be some arbitrary 4-vectors. Begin with the invariance of the product $s(\mathbf{a}, \mathbf{b})$ with respect to Lorentz transformation $\mathbf{\Lambda}$: $$s(\mathbf{a}, \mathbf{b}) = s(\mathbf{\Lambda a}, \mathbf{\Lambda b})$$

Using the definition $s(\mathbf{a}, \mathbf{b})=\mathbf{a}^T \eta \mathbf{b}$ both for the left and the right side we get: $$\begin{align} \mathbf{a}^T \eta \mathbf{b} &= (\mathbf{\Lambda}\mathbf{a})^T \eta \mathbf{\Lambda}\mathbf{b} \\ &= \mathbf{a}^T \mathbf{\Lambda}^T \eta \mathbf{\Lambda}\mathbf{b} \end{align}$$

Since the above is true for every $\mathbf{a}$ and $\mathbf{b}$ we can immediately conclude: $$\eta = \mathbf{\Lambda}^T \eta \mathbf{\Lambda}$$

Thomas Fritsch
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$x^u x_u = n_{u\beta} x^{\beta} x^u = n_{u \beta} \Lambda^u _\gamma x^\gamma \Lambda^\beta _\omega x^\omega = n_{u \beta} \Lambda^\beta _\omega \Lambda^u _\gamma x^\gamma x^\omega$

$[m] \equiv$ matrix of m.

By the definition of matrix multiplication,

$n_{u \beta} \Lambda^\beta _\omega = n \Lambda =[n \Lambda]^u _\omega$

$\to$

$\Lambda^u _\gamma [n \Lambda]^u _\omega=[\Lambda^T]^{\gamma} _u [n \Lambda]^u _\omega = [\Lambda^T n \Lambda]^\gamma _{\omega}$

and

$x^T [n]^u _b x = \text{a number} =\overline{x}^T [\Lambda^T n \Lambda]^\gamma _{\omega} \overline{x}$

$\to$

$ [n]^u _b = [\Lambda^T n \Lambda]^\gamma _{\omega}$

because these matrices define the same dot product in their respective bases.

$n = \Lambda^T n \Lambda$.