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Notations conventions: $p$ stands for the momentum (so $d^3p$ is the differential element according to which we integrate, for the $3$ space coordinates). A Lorentz transformation is denoted by $\Lambda$. I denote $\vec p = (p^0,p^1,p^2,p^3)$ the 4-vector energy-momentum while I denote $p$ the spacial momentum.

I was reading a course about QFT, and inside an integral $\int u(\vec p)d^3p$, after the use of substitution $p \leftrightarrow \Lambda p$, they use this relation:

$$ \dfrac{d^3(\Lambda p)}{(\Lambda p)^0} = \dfrac{d^3p}{p^0} $$

It seems to be pretty obvious for the author, so I might have missed something.

I know that $p^0 = \sqrt{m^2 + p^2}$ and $(\Lambda p)^0 = \sqrt{m^2 + (\Lambda p)^2}$ (where I use the notation convention for a 3-vector $p$, $p^2 = p\cdot p$) How do you derive this relation?

Qmechanic
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MiKiDe
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1 Answers1

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The measure $d^3p$ is unaffected by rotations, neither is $p^0$.

For boosts can restrict to 1+1 dimensions as the transverse momentum components are not affected by a boost.

Now parameterise the mass-shell by the rapidity $s$, so that $$ p^0=m\cosh s\\ p^1=m\sinh s $$ then $ dp^1= m\cosh s \,ds$ and $$ \frac{dp^1}{p^0}=ds. $$ Under a Lorentz boost with rapidity $s_1$ we have $s\to s + s_1$ and $d(s+s_1)=ds$ as $s_1$ is a constant.

mike stone
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