# What are the correct values for relativistic 3-momentum $p=mv$?

Is the value of $$m$$ in this formula relativistic mass or real mass? Just trying to figure out if this is the right equation for my problem.

• There is no relativistic mass. It's time to get rid of that nonsensical idea. The formula for the momentum is $p = \gamma m v$ where $m$ is the (rest) mass of the object. Oct 20 '21 at 8:27
• "There is no relativistic mass" is an unscientific statement. Whether there is relativistic mass or not depends on your definition of mass. Oct 20 '21 at 13:57
• @md2perpe : The concept of relativistic mass has been abandoned in contemporary Physics. Oct 20 '21 at 19:37
• Downvoted for use of term 'relativistic mass' Oct 21 '21 at 11:04
• Why is there a controversy on whether mass increases with speed? Oct 21 '21 at 11:41

The four-momentum has $$E/c=p^0=\gamma m_0c,\,p^i=\gamma m_0c\beta^i$$ if $$m_0\ne0$$, but $$E/c=p^0,\,p^i=E\beta^i$$ whether or not $$m_0=0$$.

• Downvoted for use of the term 'relativistic mass' . Oct 21 '21 at 11:01
• @JunSeo-He Thanks; fixed.
– J.G.
Oct 21 '21 at 11:24

p = mv

Is the value of m in this formula relativistic mass or real mass?

In your formula, $$m$$ is the "relativistic mass," not the rest mass (or "real mass" as you say).

If the rest mass is denoted as $$m_0$$ then $$p = \gamma m_0 v\;,$$ where $$v=\frac{dx}{dt}$$ and $$\gamma = 1/\sqrt{1-v^2/c^2}$$.