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歡迎光臨散文網(wǎng) 會員登陸 & 注冊

場變分的注解

2023-06-27 05:56 作者:小小物理學(xué)家 0人讀過 | 我要投稿

我們證明: $%5Cdelta(%5Cpartial_a%5Cphi)%3D%5Cpartial_a(%5Cdelta%5Cphi).$

證: $%5Cdelta(%5Cpartial_a%5Cphi)%3D%5Clim_%7B%5Clambda%20%5Cto0%7D%20%5Cfrac%7B%5Cpartial_a%5Cphi(%5Clambda)-%5Cpartial_a%5Cphi(0)%7D%7B%5Clambda%7D%3D%5Cfrac%7Bd%7D%7Bd%5Clambda%7D(%5Cpartial_a%5Cphi)%7C_%7B%5Clambda%3D0%7D$

$%5Cqquad%5Cqquad%3D%5Cpartial_a(%5Cfrac%7Bd%7D%7Bd%5Clambda%7D%7C_%7B%5Clambda%3D0%7D%5Cphi)%3D%5Cpartial_a(%5Cdelta%5Cphi).$
$%5Cpartial_b%5Cphi$ 對 $%5Cpartial_a%5Cphi$ 的求導(dǎo): $%5Cfrac%7B%5Cpartial(%5Cpartial_b%5Cphi)%7D%7B%5Cpartial(%5Cpartial_a%5Cphi)%7D%3D%5Cfrac%7B%5Cpartial(%5Cdelta%5Ea_%7B%5C%3Bb%7D%5Cpartial_a%5Cphi)%7D%7B%5Cpartial(%5Cpartial_a%5Cphi)%7D%3D%5Cdelta%5Ea_%7B%5C%3Bb%7D%5Cfrac%7B%5Cpartial(%5Cpartial_a%5Cphi)%7D%7B%5Cpartial(%5Cpartial_a%5Cphi)%7D%3D%5Cdelta%5Ea_%7B%5C%3Bb%7D.$
求度規(guī)行列式 $g$ 的變分:

由克氏符的計算公式: $%5CGamma%5E%7B%5Csigma%7D_%7B%5C%3B%5Cmu%5Cnu%7D%3D%5Cfrac12g%5E%7B%5Csigma%5Crho%7D(g_%7B%5Crho%5Cmu%2C%5Cnu%7D%2Bg_%7B%5Cnu%5Crho%2C%5Cmu%7D-g_%7B%5Cmu%5Cnu%2C%5Crho%7D)$ 可得:

$%5CGamma%5E%5Cmu_%7B%5C%3B%5Cmu%5Csigma%7D%3D%5Cfrac12g%5E%7B%5Cmu%5Clambda%7D(g_%7B%5Csigma%5Clambda%2C%5Cmu%7D%2Bg_%7B%5Cmu%5Clambda%2C%5Csigma%7D-g_%7B%5Cmu%5Csigma%2C%5Clambda%7D)$

$%5Cqquad%5C%3B%3D%5Cfrac12g%5E%7B%5Cmu%5Clambda%7Dg_%7B%5Cmu%5Clambda%2C%5Csigma%7D%2Bg%5E%7B%5Cmu%5Clambda%7D%5B%5Cfrac12(g_%7B%5Csigma%5Clambda%2C%5Cmu%7D-g_%7B%5Csigma%5Cmu%2C%5Clambda%7D)%5D$

$%5Cqquad%5C%3B%3D%5Cfrac12g%5E%7B%5Cmu%5Clambda%7Dg_%7B%5Cmu%5Clambda%2C%5Csigma%7D%2Bg%5E%7B%5Cmu%5Clambda%7Dg_%7B%5Csigma%5B%5Clambda%2C%5Cmu%5D%7D$

$%5Cqquad%5C%3B%3D%5Cfrac12g%5E%7B%5Cmu%5Clambda%7Dg_%7B%5Cmu%5Clambda%2C%5Csigma%7D%2Bg%5E%7B%5B%5Cmu%5Clambda%5D%7Dg_%7B%5Csigma%5Clambda%2C%5Cmu%7D%3D%5Cfrac12g%5E%7B%5Cmu%5Clambda%7Dg_%7B%5Cmu%5Clambda%2C%5Csigma%7D%3B$

將上式改為: $%5CGamma%5E%5Cmu_%7B%5C%3B%5Cmu%5Csigma%7D%3D%5Cfrac12g%5E%7B%5Cmu%5Clambda%7D%5Cfrac%7B%5Cpartial%20g_%7B%5Cmu%5Clambda%7D%7D%7B%5Cpartial%20x%5E%5Csigma%7D%3B$

而另一方面，由線性代數(shù)知識可知,度規(guī)行列式 $g$ 可展為: $g%3Dg_%7B%5Cmu%5Clambda%7DA%5E%7B%5Cmu%5Clambda%7D$ ,其中 $A%5E%7B%5Cmu%5Clambda%7D$ 是分量 $g_%7B%5Cmu%5Clambda%7D$ 的代數(shù)余子式,從而便有: $%5Cfrac%7B%5Cpartial%20g%7D%7B%5Cpartial%20g_%7B%5Cmu%5Clambda%7D%7D%3DA%5E%7B%5Cmu%5Clambda%7D%3B$ 按照逆矩陣的求法有 $g%5E%7B%5Cmu%5Clambda%7D%3D%5Cfrac%7BA%5E%7B%5Cmu%5Clambda%7D%7D%7Bg%7D%2C$ 所以

$%5Cfrac%7B%5Cpartial%20g%7D%7B%5Cpartial%20g_%7B%5Cmu%5Clambda%7D%7D%3DA%5E%7B%5Cmu%5Clambda%7D%3Dgg%5E%7B%5Cmu%5Clambda%7D%2C$ 所以對度規(guī)行列式的變分便是:

$%5Cdelta%20g%3D%5Cfrac%7Bdg(%5Calpha)%7D%7Bd%5Calpha%7D%7C_%7B%5Calpha%3D0%7D%3D%5Cfrac%7B%5Cpartial%20g%7D%7B%5Cpartial%20g_%7B%5Cmu%5Cnu%7D%7D%5Ccdot%5Cfrac%7B%5Cpartial%20g_%7B%5Cmu%5Cnu%7D%7D%7B%5Cpartial%20%5Calpha%7D%7C_%7B%5Calpha%3D0%7D$

$%5Cquad%3Dgg%5E%7B%5Cmu%5Cnu%7D%5Cdelta%20g_%7B%5Cmu%5Cnu%7D%3Dgg%5E%7Bab%7D%5Cdelta%20g_%7Bab%7D.$ ?????????????????????

標簽：

場變分的注解的評論 (共條)

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