- #1

TroyElliott

- 59

- 3

From page 22 of P&S we want to show that ##\delta^{3}(\vec{p}-\vec{q})## is not Lorentz invariant. Boosting in the 3-direction gives ##p_{3}' = \gamma(p_{3}+\beta E)## and ##E' = \gamma(E+\beta p_{3})##. Using the delta function identity ##\delta(f(x)-f(x_{0})) = \frac{1}{|f'(x_{0})|}\delta(x-x_{0}),## we get $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'}) \frac{dp'_{3}}{dp_{3}}.$$

This is where I am confused, how is this step arrived at? From what I can see we have

$$\delta^{3}(\vec{p'}-\vec{q'}) = \delta(p_{1}-q_{1})\delta(p_{2}-q_{2})\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q})).$$

We can further write $$\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q}))) = \delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma.$$

From here I don't see how to transform ##\delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma## into the form ##\delta(p_{3}-q_{3})\frac{dp_{3}}{dp'_{3}}.##

Any ideas on how to go about this? Thanks!

This is where I am confused, how is this step arrived at? From what I can see we have

$$\delta^{3}(\vec{p'}-\vec{q'}) = \delta(p_{1}-q_{1})\delta(p_{2}-q_{2})\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q})).$$

We can further write $$\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q}))) = \delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma.$$

From here I don't see how to transform ##\delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma## into the form ##\delta(p_{3}-q_{3})\frac{dp_{3}}{dp'_{3}}.##

Any ideas on how to go about this? Thanks!

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