Physics Asked on May 23, 2021
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 $p$, $p^2 = pcdot p$)
How do you derive this relation?
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 $$ p^0=mcosh s p^1=msinh s $$ then $ dp^1= mcosh s ,ds$ and $$ frac{dp^1}{p^0}=ds. $$ Under a Lorentz boost with rapidity $s_1$ we have $sto s + s_1$ and $d(s+s_1)=ds$ as $s_1$ is a constant.
Correct answer by mike stone on May 23, 2021
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