# 试析网上常见的一个E = Mc2推导过程中的错误An Analysis for the Error of a Common Derivation Process of E = Mc2 on the Internet

• 全文下载: PDF(668KB)    PP.277-281   DOI: 10.12677/MP.2019.96027
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A derivation process of E = Mc2, which is widely spread on the internet, is checked in detail by the engineering calculation software MATLAB. The results show that it is impossible to deduce E = Mc2 by using existing mathematical tools correctly. This paper introduces the method of deducing E = Mc2 from physical and mathematical point of view based on dimensional analysis, which is independent of relativity.

1. 引言

E = Mc2是一个著名的公式(其中，E为能量，M为质量，c为光速)，在(狭义)相对论中称为质量–能量关系式，或者称为质能方程，它是相对论的基础之一。一个源自国外出版物的基于微积分推导E = Mc2的计算过程在网上流传较广。作者用MATLAB对该计算过程逐步进行了详细分析和验算，结果表明在正确使用现有数学工具的前提下，该过程不可能推导出E = Mc2。可以基于量纲分析方法，从物理和数学角度推导E = Mc2，该方法与相对论无关。

2. 用MATLAB验算网上常见的一个推导E = Mc2过程

Figure 1. A common derivation process of E = Mc2 on the internet

MATLAB是一款功能强大的工程计算软件，是科学技术领域应用和影响最广泛的三个计算机数学语言之一。利用MATLAB的符号运算工具箱，可以直接求解一般的微积分计算问题。当然MATLAB不是万能的，对于某些特殊的微积分问题，MATLAB同样无能为力。

$y=f\left(x\right)$ 的微分定义为 [2]

$\text{d}y={f}^{\prime }\left(x\right)\text{d}x$ (1)

$y=y\left(v\right)=\frac{{m}_{0}v}{\sqrt{1-{\left(v/c\right)}^{2}}}$ (2)

>> syms m0 c v;

>> simple(diff((m0*v)/sqrt(1-(v/c)^2),v))

simple( )为MATLAB的化简命令。化简后的最终结果为

ans =m0/(-(-c^2+v^2)/c^2)^(3/2)

${y}^{\prime }\left(v\right)=\frac{{m}_{0}}{{\left[1-{\left(v/c\right)}^{2}\right]}^{3/2}}$ (3)

$\text{d}y=\text{d}\left(\frac{{m}_{0}v}{\sqrt{1-{\left(v/c\right)}^{2}}}\right)={y}^{\prime }\left(v\right)\text{d}v=\frac{{m}_{0}}{{\left[1-{\left(v/c\right)}^{2}\right]}^{3/2}}\text{d}v$ (4)

$\underset{0}{\overset{v}{\int }}v\text{d}\left(\frac{{m}_{0}v}{\sqrt{1-{\left(v/c\right)}^{2}}}\right)={m}_{0}\underset{0}{\overset{v}{\int }}\frac{v\text{d}v}{{\left[1-{\left(v/c\right)}^{2}\right]}^{3/2}}$ (5)

>> syms v c m0

>> m0*int(v/(1-(v/c)^2)^(3/2),v,0,v)

??? Error using ==> sym/maple

Error, (in limit) invalid limiting point

$\underset{a}{\overset{b}{\int }}f\left(x\right)\text{d}x=\left(b-a\right)f\left(\frac{a+b}{2}\right)+\frac{1}{24}{\left(b-a\right)}^{3}{f}^{″}\left(\xi \right)$ (6)

$\begin{array}{l}F\left(x\right)=\underset{a}{\overset{x}{\int }}f\left(x\right)\text{d}x=\underset{a}{\overset{\frac{a+b}{2}}{\int }}f\left(x\right)\text{d}x+f\left(\frac{a+b}{2}\right)\left(x-\frac{a+b}{2}\right)\\ \text{}+\frac{1}{2!}{f}^{\prime }\left(\frac{a+b}{2}\right){\left(x-\frac{a+b}{2}\right)}^{2}+\frac{1}{3!}{f}^{″}\left({\xi }_{1}\right){\left(x-\frac{a+b}{2}\right)}^{3}\end{array}$ (7)

$f\left(v\right)=\frac{v}{{\left[1-{\left(v/c\right)}^{2}\right]}^{3/2}}$ (8)

>> diff(v/(1-(v/c)^2)^(3/2),v)

ans =1/(1-v^2/c^2)^(3/2)+3*v^2/(1-v^2/c^2)^(5/2)/c^2

${f}^{\prime }\left(v\right)=\frac{1}{{\left[1-{\left(v/c\right)}^{2}\right]}^{3/2}}+\frac{3}{{\left[1-{\left(v/c\right)}^{2}\right]}^{5/2}}\cdot {\left(\frac{v}{c}\right)}^{2}$ (9)

>> diff(v/(1-(v/c)^2)^(3/2),v,2)

ans = 9/(1-v^2/c^2)^(5/2)*v/c^2+15*v^3/(1-v^2/c^2)^(7/2)/c^4

${f}^{″}\left(v\right)=\frac{9v}{{c}^{2}{\left[1-{\left(v/c\right)}^{2}\right]}^{5/2}}+\frac{15{v}^{3}}{{c}^{4}{\left[1-{\left(v/c\right)}^{2}\right]}^{7/2}}$ (10)

>> syms v c m0

>> simple(m0*int(v/(1-(v/c)^2)^(3/2),v,0,w))

ans =m0*c^2/(1-1/c^2*w^2)^(1/2)-m0*c^2

$E=\cdots ={m}_{0}\underset{0}{\overset{w}{\int }}\frac{v\text{d}v}{{\left[1-{\left(v/c\right)}^{2}\right]}^{3/2}}={m}_{0}{c}^{2}\left(\frac{1}{{\left[1-{\left(w/c\right)}^{2}\right]}^{1/2}}-1\right)$ (11)

(12)

3. 从物理和数学角度推导E = Mc2的方法

$能量\equiv 质量×{\left(速度\right)}^{2}$ (13)

4. 结束语

 [1] 质能方程E=Mc2的推导过程是怎样的？http://www.360doc.com/content/19/0101/15/45506057_805832643.shtml [2] M. R. Spiegel著. 高等数学的理论和习题[M]. 谢国瑞, 蒋司勋, 宣月华, 等, 译. 上海: 上海科学技术出版社, 1978. [3] 朱有清, 贺才兴. 高等数学复习十五讲(上) [M]. 上海: 上海交通大学出版社, 1985. [4] 王忆锋. 基于量纲分析从物理和数学角度推导光速原理[J]. 现代物理, 2019, 9(4): 183-190. [5] 王忆锋. 光速原理及其推论[J]. 现代物理, 2019, 9(5): 227-245.