What is the magnitude of the relativistic wave-vector?

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SUMMARY

The discussion centers on the relativistic wave-vector defined as \( k^\mu = (k^0, k^1, k^2, k^3) = \left(\frac{\omega_k}{c}, \vec{k}\right) \). The magnitude of the spatial part, \( |\vec{k}| \), is derived from the dispersion relation, where \( k_0 = \frac{\omega}{c} \). It is established that \( |\vec{k}|^2 = k_0^2 - m^2 \), confirming that \( |\vec{k}|^2 \) equals \( k_0^2 \) only when the mass \( m \) is zero.

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Abigale
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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
 
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Yes it's just given by the dispersion relation.
 
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k_0=\omega/c.
{|\bf k|}^2=k^2_0-m^2, and equals k^2_0 only if m=0.
 
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