Dispersion Relation: Photons vs Bosons

[rayman123]

Homework Statement

I wonder if dispersion relation for photons is $$\omega= ck$$ what it will be for bosons?

Homework Equations

The Attempt at a Solution

 

[cepheid]

Photons ARE bosons. Bosons are a general class of particle, so your question doesn't have a unique answer.

 

[rayman123]

okej, I thought photons were fermions...

 

[fzero]

A nonrelativistic particle has a dispersion relation which is

$$\omega = \frac{\hbar k^2}{2m}.$$

As $$k\rightarrow m$$ the dispersion relation approaches the relativistic $$\omega = k c$$.

 

[paranormal]

The dispersion relation for particles, specifically photons and bosons, depends on their respective properties and interactions. For photons, the dispersion relation is given by the equation \omega= ck, where c is the speed of light and k is the wave vector. This relationship is a result of the fact that photons are massless particles and travel at the speed of light in vacuum.

On the other hand, the dispersion relation for bosons can vary depending on the type of boson. For example, the dispersion relation for phonons, which are bosonic quasiparticles responsible for sound propagation, is given by \omega= c_s k, where c_s is the speed of sound in the medium. This is because phonons have a non-zero mass and therefore do not travel at the speed of light.

For other types of bosons, such as W and Z bosons, the dispersion relation is more complex and is described by the Standard Model of particle physics. In general, the dispersion relation for bosons can be written as \omega= \sqrt{c^2k^2 + m^2c^4}, where m is the mass of the boson.

In summary, the dispersion relation for photons and bosons can differ due to their different properties and interactions. While photons have a simple dispersion relation due to their masslessness, the dispersion relation for bosons can vary depending on the type of boson and its associated interactions.