# Post History

##
**#6: Post edited**

- 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**

##
**#4: Post edited**

- 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**

##
**#2: Post edited**

- 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 $p=\gamma m_0 v$
- $$=(m_0c^2)^2+(\gamma m_0 v c)^2$$
- $$=m_0^2c^4(1+\gamma \beta^2)
**$$** - $$=m_0^2c^4(1+\frac{1}{1-\beta^2}
- $$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