Can Newtonian kinetic energy and Einsteinian rest-mass energy be expressed by the binomial expansion of the Lorentz factor? And how valid is using Einstein’s E = mc2 to calculate the nuclear fission energy?

Talal Al-Ameen, Maythem Mahmud, Imad Muhi
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Abstract

The binomial (Taylor) expansion of the Lorentz factor has been reconsidered here in an attempt to find out whether the Newtonian kinetic energy and the Einsteinian rest-mass energy are implicitly embedded in the mathematical structure of the binomial expansion of the Lorentz factor (as Einstein postulated in his Special Theory of Relativity). Advocates of Standard Special Relativity show that it is possible to obtain these two kinds of energy by multiplying both sides of the expansion of the Lorentz factor by the moving object’s rest mass m 0 and the square of the speed of light c 2. This study shows that the apparent reconciliation between classical and relativistic physics made possible by employing the binomial expansion of the Lorentz factor γ = [ 1 ( v 2 c 2 ) ] 1 2 , where v is the moving object’s velocity, is challengeable. Also it is unclear how Einstein’s famous rest-mass energy equation E = m 0 c 2 from Ekinetic net = mrelativistic c 2m 0 c 2 can be used to calculate the amount of nuclear fission energy released?
牛顿动能和爱因斯坦静止质量能可以用洛伦兹系数的二项式展开来表示吗?使用爱因斯坦的 E = mc2 计算核裂变能量的有效性如何?
这里重新考虑了洛伦兹系数的二项式(泰勒)展开,试图找出牛顿动能和爱因斯坦静止质量能是否隐含在洛伦兹系数二项式展开的数学结构中(如爱因斯坦在他的狭义相对论中所假设的那样)。标准狭义相对论的倡导者指出,将洛伦兹系数展开的两边分别乘以运动物体的静止质量 m 0 和光速 c 2 的平方,就可以得到这两种能量。这项研究表明,利用洛伦兹系数 γ = [ 1 - ( v 2 c 2 ) 的二项式展开,可以在表面上调和经典物理学和相对论物理学。 ] - 1 2,其中 v 是运动物体的速度,这一点值得商榷。此外,爱因斯坦著名的静止质量能量方程 E = m 0 c 2(从动能网 = mrelativistic c 2 - m 0 c 2 得出)如何用于计算核裂变释放的能量?
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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