0
$\begingroup$

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.

$\endgroup$
13
  • 7
    $\begingroup$ 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. $\endgroup$
    – Prahar
    Oct 20 at 8:27
  • 1
    $\begingroup$ "There is no relativistic mass" is an unscientific statement. Whether there is relativistic mass or not depends on your definition of mass. $\endgroup$
    – md2perpe
    Oct 20 at 13:57
  • 3
    $\begingroup$ @md2perpe : The concept of relativistic mass has been abandoned in contemporary Physics. $\endgroup$
    – Frobenius
    Oct 20 at 19:37
  • 1
    $\begingroup$ Downvoted for use of term 'relativistic mass' $\endgroup$
    – Jun Seo-He
    Oct 21 at 11:04
  • 1
    $\begingroup$ Why is there a controversy on whether mass increases with speed? $\endgroup$
    – PM 2Ring
    Oct 21 at 11:41
3
$\begingroup$

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 $m_0=0$.

$\endgroup$
2
  • 1
    $\begingroup$ Downvoted for use of the term 'relativistic mass' . $\endgroup$
    – Jun Seo-He
    Oct 21 at 11:01
  • 1
    $\begingroup$ @JunSeo-He Thanks; fixed. $\endgroup$
    – J.G.
    Oct 21 at 11:24
1
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.