- #1
Hill
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- Homework Statement
- Show that $$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0)=\frac 1 {2 \omega_k}$$ where ##\theta(x)## is the unit step function and ##\omega_k \equiv \sqrt {\vec k^2 +m^2}##.
- Relevant Equations
- ##k^2={k^0}^2 - \vec k ^2##
##\omega _k ^2 = \vec k^2 +m^2##
##k^2 - m^2 = {k^0}^2 - \omega_k^2##
##dk^0= \frac {d{k^0}^2} {2k^0}##
##\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0) = \int_{-\infty}^{\infty} \frac {d{k^0}^2} {2k^0} \delta ({k^0}^2 - \omega_k^2) \theta (k^0) = \frac 1 {2 \omega_k} \theta (\omega_k) = \frac 1 {2 \omega_k}##
Wouldn't the result be the same without the unit step function?
##k^2 - m^2 = {k^0}^2 - \omega_k^2##
##dk^0= \frac {d{k^0}^2} {2k^0}##
##\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0) = \int_{-\infty}^{\infty} \frac {d{k^0}^2} {2k^0} \delta ({k^0}^2 - \omega_k^2) \theta (k^0) = \frac 1 {2 \omega_k} \theta (\omega_k) = \frac 1 {2 \omega_k}##
Wouldn't the result be the same without the unit step function?