Showing the Lorentz Invariance of Integral measure $dp^4$ using Levi-Civita Symbols

Question

It's probably being a bit pedantic.

I understand that since the transformation of $p^\mu$ is $p^\mu \rightarrow \Lambda^\mu_\nu p^\nu$. The integral measure transforms as the determinant of the Jacobian, which in this case is $\Lambda$ itself by virtue of it being a linear transformation. Leading to $d^4p \rightarrow \det(\Lambda) d^4p$.

I just wanted to work it out directly as: $$p^\mu \rightarrow \Lambda^\mu_\nu p^\nu$$ $$dp^\mu \rightarrow \Lambda^\mu_\nu dp^\nu$$ $$dp^0dp^1dp^2dp^3 \rightarrow \Lambda^0_\alpha \Lambda^1_\beta \Lambda^2_\gamma \Lambda^3_\delta dp^\alpha dp^\beta dp^\gamma dp^\delta $$ How do we show that $$\Lambda^0_\alpha \Lambda^1_\beta \Lambda^2_\gamma \Lambda^3_\delta dp^\alpha dp^\beta dp^\gamma dp^\delta = \epsilon^{\alpha \beta \gamma \delta} \Lambda^0_\alpha \Lambda^1_\beta \Lambda^2_\gamma \Lambda^3_\delta dp^0 dp^1 dp^2 dp^3 = \det(\Lambda) dp^0 dp^1 dp^2 dp^3$$

Answer 1

$d^4p$ is actually something called a differential form, and has a different notation $dp_0 \wedge dp_1 \wedge dp_2 \wedge dp_3 $. Differential forms have the property that $ dx \wedge dy = - dy \wedge dx$. With this, you get the epsilon for free.

The alternative is to just recognize that if you change coordinates, you need to include the Jacobian for the coordinate change. In this case, that's exactly $\det(\Lambda)$. Actually, that's exactly why we demand that differential forms are antisymmetric: it makes any coordinate change affect them with a Jacobian.