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user36790
user36790

The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have, \begin{equation} {\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4 \end{equation} Now for extremely relativistic case , we can use the condition that $E>>m_0c^2$$E\gg m_0c^2$, thus this yields $p=\frac{E}{c}$.

The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have, \begin{equation} {\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4 \end{equation} Now for extremely relativistic case , we can use the condition that $E>>m_0c^2$, thus this yields $p=\frac{E}{c}$.

The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have, \begin{equation} {\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4 \end{equation} Now for extremely relativistic case , we can use the condition that $E\gg m_0c^2$, thus this yields $p=\frac{E}{c}$.

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vbj
vbj
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The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have, \begin{equation} {\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4 \end{equation} Now for extremely relativistic case , we can use the condition that $E>>m_0c^2$, thus this yields $p=\frac{E}{c}$.