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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Komar mass ： ウィキペディア英語版 Komar mass The Komar mass (named after Arthur Komar〔A. Komar'', Positive-Definite Energy Density and Global Consequences for General Relativity'', Physical Review, vol. 129, Issue 4, pp. 1873–1876 (1963)〕) of a system is one of several formal concepts of mass that are used in general relativity. The Komar mass can be defined in any stationary spacetime, which is a spacetime in which all the metric can be written so that they are independent of time. Alternatively, a stationary spacetime can be defined as a spacetime which possesses a timelike Killing vector field. The following discussion is an expanded and simplified version of the motivational treatment in (Wald, 1984, pg 288). == Motivation == Consider the Schwarzschild metric. Using the Schwarzschild basis, a frame field for the Schwarzschild metric, one can find that the radial acceleration required to hold a test mass stationary at a Schwarzschild coordinate of r is: $a^\backslash hat\; =\; \backslash frac\}\}$ Because the metric is static, there is a well-defined meaning to "holding a particle stationary". Interpreting this acceleration as being due to a "gravitational force", we can then compute the integral of normal acceleration multiplied by area to get a "Gauss law" integral of: :$\backslash frac\}\}$ While this approaches a constant as r approaches infinity, it is not a constant independent of r. We are therefore motivated to introduce a correction factor to make the above integral independent of the radius r of the enclosing shell. For the Schwarzschild metric, this correction factor is just $\backslash sqrt\; \backslash ,\; dt^2\; +\; \backslash mathrm(dx\; \backslash ,\; dy\; \backslash ,\; dz)\; \backslash ,$ where gtt and the quadratic form are functions only of the spatial coordinates x,y,z and are not functions of time. In spite of our choices of variable names, it should not be assumed that our coordinate system is Cartesian. The fact that none of the metric coefficients are functions of time make the metric stationary: the additional fact that there are no "cross terms" involving both time and space components (such as dx dt) make it static. Because of the simplifying assumption that some of the metric coefficients are zero, some of our results in this motivational treatment will not be as general as they could be. In flat space-time, the proper acceleration required to hold station is du/d tau, where u is the 4-velocity of our hovering particle and tau is the proper time. In curved space-time, we must take the covariant derivative. Thus we compute the acceleration vector as: :$a^b\; =\; \backslash nabla\_u\; u^b\; =\; u^c\; \backslash nabla\_c\; u^b$ :$a\_b\; =\; u^c\; \backslash nabla\_c\; u\_b$ where ub is a unit time-like vector such that ub ub = -1. The component of the acceleration vector normal to the surface is :$a\_\}\; c^2\; \backslash right)\; /\; \backslash left(\; 2\; g\_\; \backslash sqrt\}\}$ as expected - we have simply re-derived the previous results presented in a frame-field in a coordinate basis. We define $a\backslash mathrm\; =\; \backslash sqrt\_b\; =\; m/r^2\backslash ,$. We can, if we desire, derive the accelerations ab and the adjusted "acceleration at infinity" ainfb from a scalar potential Z, though there is not necessarily any particular advantage in doing so. (Wald 1984, pg 158, problem 4) $a\_b\; =\; \backslash nabla\_b\; Z\_1\; \backslash qquad\; Z\_1\; =\; \backslash ln$ $a\backslash mathrm\_b\; =\; \backslash nabla\_b\; Z\_2\; \backslash qquad\; Z\_2\; =\; \backslash sqrt\; \backslash int\_A\; N^b\; a\backslash mathrm\_b\; dA$ To make this demonstration, we need to express this surface integral as a volume integral. In flat space-time, we would use Stokes theorem and integrate $-\backslash nabla\; \backslash cdot\; a\backslash mathrm$ over the volume. In curved space-time, this approach needs to be modified slightly. Using the formulas for electromagnetism in curved space-time as a guide, we write instead. $F\_\; =\; a\backslash ,\backslash mathrm\_a\; \backslash ,u\_b\; -\; a\backslash ,\backslash mathrm\_b\; \backslash ,u\_a\; \backslash ,$ where F plays a role similar to the "Faraday tensor", in that $a\backslash mathrm\_a\; =\; F\_\; u^b\backslash ,$ We can then find the value of "gravitational charge", i.e. mass, by evaluating $\backslash nabla^a\; F\_\; u^b$ and integrating it over the volume of our sphere. An alternate approach would be to use differential forms, but the approach above is computationally more convenient as well as not requiring the reader to understand differential forms. A lengthly, but straightforward (with computer algebra) calculation from our assumed line element shows us that $-u^b\; \backslash nabla^a\; F\_\; =\; \backslash sqrt\; u^a\; u^b\; =\; \backslash sqrt\; u^a\; u^b$ Thus we can write $m\; =\; \backslash frac\; \backslash int\_V\; R\_\; u^a\; u^b$ In any vacuum region of space-time, all components of the Ricci tensor must be zero. This demonstrates that enclosing any amount of vacuum will not change our volume integral. It also means that our volume integral will be constant for any enclosing surface, as long as we enclose all of the gravitating mass inside our surface. Because Stokes theorem guarantees that our surface integral is equal to the above volume integral, our surface integral will also be independent of the enclosing surface as long as the surface encloses all of the gravitating mass. By using Einstein's Field Equations :$G^u\; R\; I^u\}\; \backslash left(\; 2\; T\_\; -\; T\; g\_\; \backslash right)\; u^a\; u^b\; dV$ :where V is the volume being integrated over :Tab is the Stress–energy tensor :ua is a unit time-like vector such that ua ua = -1 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Komar mass」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.