4 Momentum and 4 velocity relationship

[bayners123]

$$P = \left( \begin{array}{c} E/c \\ \bar{p} \end{array}\right)$$

and

$$U = \left( \begin{array}{c} \gamma c \\ \gamma \bar{v} \end{array}\right)$$

right? But I frequently see in textbooks that $P = m_0 U$. Surely,
$$m_0 U = \left( \begin{array}{c} \gamma m_0 c \\ \gamma m_0 \bar{v} \end{array}\right) = \left( \begin{array}{c} E/c \\ \gamma \bar{p} \end{array}\right) \neq \left( \begin{array}{c} E/c \\ \bar{p} \end{array}\right)$$

So how does this work?
Yours confusedly

 

[PAllen]

$$P = \left( \begin{array}{c} E/c \\ \bar{p} \end{array}\right)$$

and

$$U = \left( \begin{array}{c} \gamma c \\ \gamma \bar{v} \end{array}\right)$$

right? But I frequently see in textbooks that $P = m_0 U$. Surely,
$$m_0 U = \left( \begin{array}{c} \gamma m_0 c \\ \gamma m_0 \bar{v} \end{array}\right) = \left( \begin{array}{c} E/c \\ \gamma \bar{p} \end{array}\right) \neq \left( \begin{array}{c} E/c \\ \bar{p} \end{array}\right)$$

So how does this work?
Yours confusedly

The p in (E/c,p) is three momentum but it is still relativistic three momentum not Newtonian 3-momentum. Thus p=γmv, and your confusion is resolved.

 

[bayners123]

The p in (E/c,p) is three momentum but it is still relativistic three momentum not Newtonian 3-momentum. Thus p=γmv, and your confusion is resolved.

Ah, thanks