Problem 2.3b3

…

Now we lower the indices, by expressing the upper-index coordinates (contravariant components) by lower-index coordinates (covariant components), in order to find the Lorentz transformation for the covariant components:

$\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ \eta^{\rho \mu} (x')_\rho &= L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \sum_\rho \eta^{\rho \mu} (x')_\rho &= \sum_\sigma \sum_\nu L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \end{aligned}}$

After raising the indices, we lower the indices again:

$\displaystyle{ \begin{aligned} \eta_{\alpha \mu} \eta^{\rho \mu} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \eta_{\alpha \mu} \eta^{\mu \rho} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \delta_{\alpha \rho} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ (x')_\alpha &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \end{aligned}}$

Prove that $\displaystyle{\eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} = \left[L^{-1}\right]^\sigma_{~\alpha}}$ .

By index renaming, $\displaystyle{ \begin{aligned} (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} x_\nu \\ \end{aligned}}$ , the question becomes

Prove that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} = \left[L^{-1}\right]^\nu_{~\mu}}$ .

Denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ . Then the question is simplified to

Prove that $\displaystyle{ L^{~\nu}_{\mu} = \left[L^{-1}\right]^\nu_{~\mu}}$ .

$\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \left( \sum_{\nu = 0}^4 L^\mu_{~\nu} x^\nu \right) \left( \sum_{\beta = 0}^4 {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &(x')^0 (x')_0 + (x')^1 (x')_1 + (x')^2 (x')_2 + (x')^3 (x')_3 \\ &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right) \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right) \\ \end{aligned}}$

The right hand side has 64 terms.

Since the spacetime interval is Lorentz-invariant, $\displaystyle{ (x')^\mu (x')_\mu = x^\mu x_\mu }$ . So the left hand side can be replaced by $\displaystyle{ x^\mu x_\mu }$ .

$\displaystyle{ \begin{aligned} &x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 \\ &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right) \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right) \\ \end{aligned}}$

Note that the 4 terms on the left side also appear on the right hand side.

$\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\ \end{aligned}}$

Since this equation is true for any coordinates, it is an identity. By comparing coefficients, we have:

1. For any terms with $\displaystyle{\nu \ne \beta}$ , such as $\displaystyle{\nu = 0}$ and $\displaystyle{\beta=1}$ ,

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} x^0 x_1 &\equiv 0 \\ \left( \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} \right) x^0 x_1 &\equiv 0\\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} &= 0 \\ \end{aligned}}$

2. For any terms with $\displaystyle{\nu = \beta}$ .

$\displaystyle{ \begin{aligned} x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = \nu} L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\ x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \left( L^\mu_{~0} L^{~0}_{\mu} x^0 x_0 + L^\mu_{~1} L^{~1}_{\mu} x^1 x_1 + L^\mu_{~2} L^{~2}_{\mu} x^2 x_2 + L^\mu_{~3} L^{~3}_{\mu} x^3 x_3 \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~0}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~1} L^{~1}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~2} L^{~2}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~3} L^{~3}_{\mu} &= 1 \\ \end{aligned}}$

Denoting $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ is misleading, because that presupposes that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ is directly related to the matrix $\displaystyle{L}$ .

To avoid this bug, instead, we denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{M ^\nu_{~\mu}}$ . So

$\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ \end{aligned}}$

…
$\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}$

— Me@2020-09-12 09:33:00 PM

Physics Town

continues

Problem 2.3b3

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