Post History
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 $p=\gamma m_0 v$ $$=(m_0c^2)^2+(\g...
Answer
#6: Post edited
- I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
- $$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$$
- So energy-(relativistic)mass relation is true for all observer. But here mass is relativistic in mass-energy relation.
#4: Post edited
- I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
- $$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$$
- So energy-(relativistic)mass relation is true for all observer. But here mass is relativistic in mass-energy relation.
#2: Post edited
- I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,
- $$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 $p=\gamma m_0 v$
- $$=(m_0c^2)^2+(\gamma m_0 v c)^2$$
- $$=m_0^2c^4(1+\gamma \beta^2)$$
- $$=m_0^2c^4(1+\frac{1}{1-\beta^2} \cdot \beta^2)$$
- $$E=m_0c^2\gamma$$
- $$=mc^2$$
- 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
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.