The momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as lamda=h/p and p=(mv)/(sqrt(1-v^2/c^2))

kalkulusk2

kalkulusk2

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2022-08-29

The momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as
λ = h p
and
p = m v 1 v 2 / c 2
when v c, p , so is it say that the wavelength is ZERO?

Answer & Explanation

tkaljegs

tkaljegs

Beginner2022-08-30Added 9 answers

Lorentz contraction! The measured de Broglie wavelength in the direction of propagation vanishes because that's what special relativity says happens. The wavelength has to go as h / p as you wrote, so why does it surprise you that when p gets large the wavelength gets small?
etapi9a

etapi9a

Beginner2022-08-31Added 12 answers

Yes. The energy-momentum equation E 2 = p 2 c 2 + m 2 c 4 says that a massive object's mass (relativistic mass), momentum and energy approaches as a particle is accelerated towards c. There's nothing obvious about the fact that it requires infinite energy to accelerate it to c. This strictly means that you can't accelerate the object to speed-of-light.
Substituting Planck's wave equation h ν in E = p c for photons, we get
λ = h c p c
Or in case of particles with rest energy where we can substitute for total energy,
λ = h E 2 ( m 0 c 2 ) 2 = h p 2 2 m E 0
This equation says that matter waves are observable only for matter (massive) particles which always travel at v < c. In other words, there is no matter wave for such objects.

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