Q&A

# Post History

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posted 10mo ago by deleted user  ·  edited 8mo ago by deleted user

#6: Post edited by deleted user · 2022-02-20T02:21:17Z (8 months ago)
• I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
• $$E^2=(m_0c^2)^2+(pc)^2$$
• Here momentum is relativistic hence $p=\gamma m_0 v$
• $$=(m_0c^2)^2+(\gamma m_0 v c)^2$$
• $$=m_0^2c^4(1+\gamma \beta^2)$$
• $$=m_0^2c^4(1+\frac{1}{1-\beta^2} \cdot \beta^2)$$
• $$=m_0^2c^4(\frac{1}{1-\beta^2})$$
• $$E=m_0c^2\gamma$$
• $$=mc^2$$
• So energy-(relativistic)mass relation is true for all observer. But here mass is relativistic in mass-energy relation.
• I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
• $$E^2=(m_0c^2)^2+(pc)^2$$
• Here momentum is relativistic hence $p=\gamma m_0 v$
• $$=(m_0c^2)^2+(\gamma m_0 v c)^2$$
• $$=m_0^2c^4(1+\gamma^2 \beta^2)$$
• $$=m_0^2c^4(1+\frac{1}{1-\beta^2} \cdot \beta^2)$$
• $$=m_0^2c^4(\frac{1}{1-\beta^2})$$
• $$E=m_0c^2\gamma$$
• $$=mc^2$$
• So energy-(relativistic)mass relation is true for all observer. But here mass is relativistic in mass-energy relation.
#5: Post undeleted by deleted user · 2021-12-10T08:23:09Z (10 months ago)
#4: Post edited by deleted user · 2021-12-10T08:23:02Z (10 months ago)
• I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
• $$E^2=(m_0c^2)^2+(pc)^2$$
• Here momentum is relativistic hence $p=\gamma m_0 v$
• $$=(m_0c^2)^2+(\gamma m_0 v c)^2$$
• $$=m_0^2c^4(1+\gamma \beta^2)$$
• $$=m_0^2c^4(1+\frac{1}{1-\beta^2} \cdot \beta^2)$$
• $$E=m_0c^2\gamma$$
• $$=mc^2$$
• So energy-mass relation is true for all observer. But here mass is relativistic in mass-energy relation. And mass is rest for the main equation.
• I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
• $$E^2=(m_0c^2)^2+(pc)^2$$
• Here momentum is relativistic hence $p=\gamma m_0 v$
• $$=(m_0c^2)^2+(\gamma m_0 v c)^2$$
• $$=m_0^2c^4(1+\gamma \beta^2)$$
• $$=m_0^2c^4(1+\frac{1}{1-\beta^2} \cdot \beta^2)$$
• $$=m_0^2c^4(\frac{1}{1-\beta^2})$$
• $$E=m_0c^2\gamma$$
• $$=mc^2$$
• So energy-(relativistic)mass relation is true for all observer. But here mass is relativistic in mass-energy relation.
#3: Post deleted by deleted user · 2021-12-10T08:18:31Z (10 months ago)
#2: Post edited by deleted user · 2021-12-10T08:16:16Z (10 months ago)
• I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
• $$E^2=(m_0c^2)^2+(pc)^2$$
• Here momentum is relativistic hence $p=\gamma m_0 v$
• $$=(m_0c^2)^2+(\gamma m_0 v c)^2$$
• $$=m_0^2c^4(1+\gamma \beta^2) •$$=m_0^2c^4(1+\frac{1}{1-\beta^2}\cdot \beta^2)$$•$$E=m_0c^2\gamma$$•$$=mc^2$$• So energy-mass relation is true for all observer. But here mass is relativistic in mass-energy relation. And mass isrest for the main equation. • I will use m_0 for rest mass in the answer and m for relativistic mass. The main equation is, •$$E^2=(m_0c^2)^2+(pc)^2$$• Here momentum is relativistic hence p=\gamma m_0 v •$$=(m_0c^2)^2+(\gamma m_0 v c)^2$$•$$=m_0^2c^4(1+\gamma \beta^2)$$•$$=m_0^2c^4(1+\frac{1}{1-\beta^2} \cdot \beta^2)$$•$$E=m_0c^2\gamma$$•$$=mc^2$$• So energy-mass relation is true for all observer. But here mass is relativistic in mass-energy relation. And mass is rest for the main equation. #1: Initial revision by deleted user · 2021-12-10T08:13:22Z (10 months ago) I will use m_0 for rest mass in the answer and m for relativistic mass. The main equation is,$$E^2=(m_0c^2)^2+(pc)^2$$Here momentum is relativistic hence p=\gamma m_0 v$$=(m_0c^2)^2+(\gamma m_0 v c)^2=m_0^2c^4(1+\gamma \beta^2)
$$=m_0^2c^4(1+\frac{1}{1-\beta^2}\cdot \beta^2)$$
$$E=m_0c^2\gamma$$
$$=mc^2$$

So energy-mass relation is true for all observer. But here mass is relativistic in mass-energy relation. And mass isrest for the main equation.