Problem 2.3b5

A First Course in String Theory

2.3 Lorentz transformations, derivatives, and quantum operators.

(b) Show that the objects $\displaystyle{\frac{\partial}{\partial x^\mu}}$ transform under Lorentz transformations in the same way as the $\displaystyle{a_\mu}$ considered in (a) do. Thus, partial derivatives with respect to conventional upper-index coordinates $\displaystyle{x^\mu}$ behave as a four-vector with lower indices – as reflected by writing it as $\displaystyle{\partial_\mu}$ .

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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

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

…

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

Using the Kronecker Delta and Einstein summation notation, we have

$\displaystyle{ \begin{aligned} L^\mu_{~\nu} M^{\beta}_{~\mu} &= M^{\beta}_{~\mu} L^\mu_{~\nu} \\ &= \delta^{\beta}_{~\nu} \\ \end{aligned}}$

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

$\displaystyle{ \begin{aligned} M^{\beta}_{~\mu} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}$

In other words,

$\displaystyle{ \begin{aligned} \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}$

— Me@2020-11-23 04:27:13 PM

One defines (as a matter of notation),

${\Lambda _{\nu }}^{\mu }\equiv {\left(\Lambda ^{-1}\right)^{\mu }}_{\nu },$

and may in this notation write

${A'}_{\nu }={\Lambda _{\nu }}^{\mu }A_{\mu }.$

Now for a subtlety. The implied summation on the right hand side of

${A'}_{\nu }={\Lambda _{\nu }}^{\mu }A_{\mu }={\left(\Lambda ^{-1}\right)^{\mu }}_{\nu }A_{\mu }$

is running over a row index of the matrix representing $\displaystyle{\Lambda^{-1}}$ . Thus, in terms of matrices, this transformation should be thought of as the inverse transpose of $\displaystyle{\Lambda}$ acting on the column vector $\displaystyle{A_\mu}$ . That is, in pure matrix notation,

$A'=\left(\Lambda ^{-1}\right)^{\mathrm {T} }A.$

— Wikipedia on Lorentz transformation

$\displaystyle{ \begin{aligned} M^{\beta}_{~\mu} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}$

In other words,

$\displaystyle{ \begin{aligned} \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}$

Denote $\displaystyle{[L^{-1}]^{\beta}_{~\mu}}$ as

$\displaystyle{ \begin{aligned} N^{~\beta}_{\mu} \\ \end{aligned}}$

In other words,

$\displaystyle{ \begin{aligned} N^{~\beta}_{\mu} &= M^{\beta}_{~\mu} \\ [N^T] &= [M] \\ \end{aligned}}$

The Lorentz transformation:

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

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

$\displaystyle{ \begin{aligned} x^\mu &= [L^{-1}]^\mu_{~\nu} (x')^\nu \\ (x')_\mu &= M^{\nu}_{~\mu} x_\nu \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} x^\mu &= [L^{-1}]^\mu_{~\nu} (x')^\nu \\ (x')_\mu &= [L^{-1}]^{\nu}_{~\mu} x_\nu \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} &= \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} \\ &= \frac{\partial x^0}{\partial (x')^\mu} \frac{\partial}{\partial x^0} + \frac{\partial x^1}{\partial (x')^\mu} \frac{\partial}{\partial x^1} + \frac{\partial x^2}{\partial (x')^\mu} \frac{\partial}{\partial x^2} + \frac{\partial x^3}{\partial (x')^\mu} \frac{\partial}{\partial x^3} \\ \end{aligned}}$

Now we consider $\displaystyle{f}$ as a function of $\displaystyle{x^{\mu}}$ ‘s:

$\displaystyle{f(x^0, x^1, x^2, x^3)}$

Since $\displaystyle{x^{\mu}}$ ‘s and $\displaystyle{(x')^{\mu}}$ ‘s are related by Lorentz transform, $\displaystyle{f}$ is also a function of $\displaystyle{(x')^{\mu}}$ ‘s, although indirectly.

$\displaystyle{f(x^0((x')^0, (x')^1, (x')^2, (x')^3), x^1((x')^0, ...), x^2((x')^0, ...), x^3((x')^0, ...))}$

For notational simplicity, we write $\displaystyle{f}$ as

$\displaystyle{f(x^\alpha((x')^\beta))}$

Since $\displaystyle{f}$ is a function of $\displaystyle{(x')^{\mu}}$ ‘s, we can differentiate it with respect to $\displaystyle{(x')^{\mu}}$ ‘s.

$\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} f(x^\alpha((x')^\beta))) &= \sum_{\nu = 0}^4 \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} f(x^\alpha) \\ \end{aligned}}$

Since

$\displaystyle{ \begin{aligned} x^\nu &= [L^{-1}]^\nu_{~\beta} (x')^\beta \\ \end{aligned}}$ ,

$\displaystyle{ \begin{aligned} \frac{\partial f}{\partial (x')^\mu} &= \sum_{\nu = 0}^4 \frac{\partial}{\partial (x')^\mu} \left[ \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} (x')^\beta \right] \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} \frac{\partial (x')^\beta}{\partial (x')^\mu} \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} \delta^\beta_\mu \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 [L^{-1}]^\nu_{~\mu} \frac{\partial f}{\partial x^\nu} \\ &= [L^{-1}]^\nu_{~\mu} \frac{\partial f}{\partial x^\nu} \\ \end{aligned}}$