Physics Question

[MisterPants]

Why does the wavelength of a wave packet correspond to particle momentum?

 

[TerryMathews]

AFAIK, and I'm no expert here, but particles tend to move unrestricted in three dimensions.

Wavelength is a measure of the movement of the particles, much like an osiliscope displays change in voltage. Higher wavelength = more movement.

 

[sao123]

In order for an object to have momentum, doesnt it have to have substantial mass?
I though wave packets were massless particles.

 

[f95toli]

Originally posted by: sao123
In order for an object to have momentum, doesnt it have to have substantial mass? I though wave packets were massless particles.

No, the momentum of a photon (which is a massless particle) is h/lambda (lambda=the wavelength).

 

[imgod2u]

I've heard many vague explanations. I will attempt to offer one that I'm familiar with (although I'm far from an expert). It has to do with virtual particles. Please google for an explaination of what virtual particles are. Basically, in every point in space, virtual particles are popping up everwhere. These particles, though very brief in existence (as they break into a particle and anti-particle, and then quickly collide to annihilate each other) have a certain size and in order for other particles to exist in any point in space, they have to "fit" the gaps between these virtual particles. When we say a wave packet of an EM wave fits directly with a certain particle, we mean it gives it just enough energy to knock it to another "slot" between these virtual particles. Think of it like a Chinese Checkers board. You have to give a marble a certain amount of energy before it rolls onto another hole in the board.
This is only my interpretation. Its adherence to modern theory is not guaranteed.

 

[Weyoun]

for some looking for a rather elementary proof:

E = hf
f = c/lamda
E = hc/lamda

P = mv, where v = velocity of packet, c
= mc

but E = mc^2 (this assumption requires that the particle energy hf = relativistic energy for mass mc^2 - I'm not entirely sure of the full implications for this, perhaps photons can transfer energy entirely between potential mass and kinetic and only have a finite mass once KE = 0. Don't use this, I just thought of it then )

therefore, mc^2 = hc/lamda
mc = h/lamda
P = h/lamda