Suppose there exist particles with zero rest mass ($m_0=0$) which are not photons. Let them have a non-zero finite energy $E$. Do they also, just like photons, travel through space with the speed of light?

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    $\begingroup$ Yes, en.wikipedia.org/wiki/Massless_particle $\endgroup$ Apr 10, 2020 at 7:43
  • $\begingroup$ It's unfortunate that that Wikipedia article uses relativistic mass. (It also has some ungrammatical English and poor typography). $\endgroup$
    – PM 2Ring
    Apr 10, 2020 at 7:59

1 Answer 1


Yes, it can be derived from the energy-momentum-relation of SRT. $$ E^2 - (pc)^2 = (m_0c^2)^2 $$ which for $m_0 = 0$ becomes $$ E^2 - (pc)^2 = 0$$ $$\implies \frac E p = c \quad(*)$$ Since $$ E = \gamma m_0c^2$$ $$ p = \gamma m_0v$$ $$\implies \frac E p =\frac{c^2}{v} \quad(**)$$ $(*) = (**)$ yields $$ v = c$$

  • $\begingroup$ That derivation is a little shaky because it involves the indeterminate form $0/0$ in the 2nd last step. $\endgroup$
    – PM 2Ring
    Apr 10, 2020 at 7:56
  • $\begingroup$ Thank you for the explanation $\endgroup$
    – Sanket J H
    Apr 10, 2020 at 8:09
  • $\begingroup$ No, that step is fine. $E$ and $p$ are nonzero. You can take the ratio for finite $m_0,\gamma$ and then the limit as $m_0 \to 0, \gamma \to \infty $ $\endgroup$ Apr 10, 2020 at 8:38
  • $\begingroup$ @RogerJBarlow Sure, it's fine if you take limits. That's why I said it's shaky, rather than saying it's definitely wrong. $\endgroup$
    – PM 2Ring
    Apr 10, 2020 at 8:43

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