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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.

Qmechanic
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Mathwiz
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    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. – Prahar Oct 20 '21 at 08:27
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    "There is no relativistic mass" is an unscientific statement. Whether there is relativistic mass or not depends on your definition of mass. – md2perpe Oct 20 '21 at 13:57
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    @md2perpe : The concept of relativistic mass has been abandoned in contemporary Physics. – Frobenius Oct 20 '21 at 19:37
  • @Frobenius. Not using the concept of relativistic mass is quite different from saying that there is no relativistic mass. – md2perpe Oct 20 '21 at 19:56
  • I would say that using it is terrible teaching practice and irresponsible. It is misleading; read this site to see why! People answering questions here should know better. – m4r35n357 Oct 21 '21 at 10:45
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    Downvoted for use of term 'relativistic mass' – Jun Seo-He Oct 21 '21 at 11:04
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    Why is there a controversy on whether mass increases with speed? – PM 2Ring Oct 21 '21 at 11:41
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    @PM2Ring there is no controversy whatsoever; mass does not increase with speed! If you disagree, give a physical mechanism for this to happen. – m4r35n357 Oct 21 '21 at 12:33
  • Gravitational mass doesn't change between different inertial frames. At high velocities inertial mass stops being equal to the gravitational mass – Jun Seo-He Oct 21 '21 at 12:38
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    @m4r35n357 I most certainly don't disagree! I linked that post because the top answer there (by Ben Crowell) is excellent. – PM 2Ring Oct 21 '21 at 12:39
  • @PM2Ring Of course I recognize your nick, please pardon my directness ;) But it seems there is so much residual "controversy" here about all this that I needed to make the point about the title. There are not two sides to this argument. – m4r35n357 Oct 21 '21 at 12:43
  • The title of that question is a little unfortunate, but Ben forcefully insists that there is no controversy... unless you're stuck in some kind of time warp. In which case, you probably think that "modern music" means The Beatles. :) – PM 2Ring Oct 21 '21 at 12:46
  • @PM2Ring I like Tsamparlis' approach to this! – m4r35n357 Oct 21 '21 at 13:02

2 Answers2

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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 $m_0=0$.

J.G.
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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 $\gamma = 1/\sqrt{1-v^2/c^2}$.

hft
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