I saw some problems in special relativity that use conservation of total energy and momentum and the conclusions are spectacular. My problem is the following: how can a massless particle like neutrino have mementum not equal to 0? If their rest mass is 0 then the momentum $$p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$$ should be also 0.
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1Similar: http://physics.stackexchange.com/questions/2229/if-photons-have-no-mass-how-can-they-have-momentum – BowlOfRed Apr 24 '15 at 20:34
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Yes, I have read that. But where is the mistake in the relation I've written? – SebiSebi Apr 24 '15 at 20:36
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2The term on the right is undefined (not zero) when $m_0=0$ because $v=c$ and it becomes $\frac{0}{0}$ – BowlOfRed Apr 24 '15 at 23:19
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Neutrinos are not massless. See https://en.wikipedia.org/wiki/Neutrino#Mass – user12205 Apr 25 '15 at 00:16
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That formula for momentum is only true for massive particles.
Here's what is always true: A particle with a mass $m$ ($\geq 0$) can have an arbitrary momentum $p$ (in some direction, with magnitude $\geq 0$). The energy of such a particle is
$$ E = \sqrt{m^2c^4 + p^2c^2}$$
The velocity of a particle is equal to
$$ v = \frac{pc^2}{E} $$
When $m = 0$, $E = pc$ and so $v=c$ -- the particle must travel at the speed of light. For $m \neq 0$, you can solve this for $p$ and find that
$$ p = \gamma m v $$
where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$.
jwimberley
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Ok. The relation between E and p can be proved without using relation? $$p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$$ All the proofs I've seen so far use this. – SebiSebi Apr 24 '15 at 20:43
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To be more precise, your equation is a result rather than the principle. $E=\sqrt{m^2c^4 + p^2 c^2}$ is a result of combining energy and momentum into a four vector $p^\mu =(E/c, \vec{p})$ and then defining its invariant norm to be $mc$, which is the only dimensionally reasonable guess. This has higher logical precedence in SR than $p=m_0 v/\sqrt{1-v^2/c^2}$ – Ali Moh Apr 25 '15 at 06:16
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Is there an absolutely true relation between momentum and velocity, without involving energy? If a particle has zero speed, then it must have no momentum? – SebiSebi Apr 25 '15 at 08:04
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@SebiSebi That's just a change in syntax. Take the equation I gave you for $E$ and plug it in: $$v = \frac{pc^2}{\sqrt{m^2c^4+p^2c^2}}.$$ Now, the issue is just that this equation has to be solved for $p$ in terms of $v$, but that when $m=0$, there is no solution for $p$ in terms of $v$ (because $v=c$ is a constant). So, when $m = 0$, $p$ is entirely unrelated to $v$. There certainly is a momentum $p$ (which is a more fundamental quantity than velocity); it's just that for massless particles, speed is not an indication of momentum. – jwimberley Apr 25 '15 at 13:26