What is the magnitude of the relativistic wave-vector?

[Abigale]

Hey guys,

I regard a relativistic vector:
$$

k^\mu =(k^0,k^1,k^2,k^3,)=(\frac{\omega_k}{c}, \vec{k} )

$$

What is $|\vec{k}|$ of this vector?
Is it the same as $k^0$?

THX

 

[WannabeNewton]

Yes it's just given by the dispersion relation.

 

[Meir Achuz]

$$k_0=\omega/c$$.
$${|\bf k|}^2=k^2_0-m^2,$$ and equals $k^2_0$ only if m=0.