A particle with mass m = 0 (a neutrino, possibly) carries momentum. How can this be in view of Eq. 6-1, in which we see that the momentum is directly proportional to the mass?

Question

Deepak Patra · Answer

The momentum of a particle (p) is equal to the mass of the particle (m) times the velocity of particle (v).
So, p = mv …… (1)
But according to de Broglie hypothesis, the momentum of a particle (p) is equal to the Planck’s constant (h) divided by wavelength of the particle (λ).
So, p = h/ λ …… (2)
Multiplying a constant c (velocity of light) in numerator and denominator of equation (1), we get,
p = h/ λ
= hc/ λc
= hf/c (Since, c/λ = f (frequency))
= E/c (Since, E (energy) = hf) …… (3)
From equation (3), quantum mechanically the momentum of a neutrino having zero mass (m = 0) does not depend upon the mass of the particle, it depends on the energy of the particle.
Thus a particle having zero mass still has a momentum.

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A particle with mass m = 0 (a neutrino, possibly) carries momentum. How can this be in view of Eq. 6-1, in which we see that the momentum is directly proportional to the mass?

A particle with mass m = 0 (a neutrino, possibly) carries momentum. How can this be in view of Eq. 6-1, in which we see that the momentum is directly proportional to the mass?

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