Problem 2.3b5

A First Course in String Theory

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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}}

So

\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

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One defines (as a matter of notation),

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

and may in this notation write

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

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

{\displaystyle {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,

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

— Wikipedia on Lorentz transformation

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So

\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}}

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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}}

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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}}

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

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

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

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\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}}

Therefore,

\displaystyle{ \begin{aligned}   \frac{\partial}{\partial (x')^\mu} &= [L^{-1}]^\nu_{~\mu} \frac{\partial}{\partial x^\nu} \\   \end{aligned}}

It is the same as the Lorentz transform for covariant vectors:

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

— Me@2020-11-23 04:27:13 PM

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2020.11.24 Tuesday (c) All rights reserved by ACHK