Because that's what it needs to be for the units to work out.
  No, really.
  The [math]c^4[/math] in question is the one on the right hand side of the Einstein Field Equations, which reads:
  [math]\displaystyle R_{\mu \nu} -\frac{1}{2} g_{\mu \nu} R + g_{\mu \nu} \Lambda = \frac{8\pi G}{c^4} T_{\mu \nu} \tag*{}[/math]
... 
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