Question about completely antisymmetric unit pseudotensor In a book, it says for transforming a completely antisymmetric unit pseudotensor from non-curvilinear system $x'$ to  curvilinear system $x$, one can use
$$E^{iklm}=\frac{\partial x^i}{\partial x^{'p}}\frac{\partial x^k}{\partial x^{'r}}\frac{\partial x^l}{\partial x^{'s}}\frac{\partial x^m}{\partial x^{'t}}e^{prst}$$
I understand this part. 


*

*But then it says it is also:
$$E^{iklm}=Je^{iklm}$$


Why is that? 


*In the above formula $J$ is defined as
$$J=\frac{\partial (x^0,x^1,x^2,x^3)}{\partial (x^{'0},x^{'1},x^{'2},x^{'3})}$$


What is the meaning of $J$ in this definition? Is it a 4x4 matrix?
 A: Concerning subquestion 2: $$J~=~\det J^{i}{}_{j}$$ is the determinant of the Jacobian matrix with entries $$J^{i}{}_{j}~=~\frac{\partial x^i}{\partial x^{\prime j}}.$$ ($J$ is also known as the Jacobian determinant, cf. deleted answer by Michael Seifert.) So a better notation for OP's last formula would be  
$$J~=~\det\frac{\partial (x^0,x^1,x^2,x^3)}{\partial (x^{'0},x^{'1},x^{'2},x^{'3})}.$$
Subquestion 1 follows from properties of the determinant and the chain rule.
A: $$
E^{iklm}=\frac{\partial x^i}{\partial x^{'p}}\frac{\partial x^k}{\partial x^{'r}}\frac{\partial x^l}{\partial x^{'s}}\frac{\partial x^m}{\partial x^{'t}}e^{prst}\;, \\
=\delta_{[a}^i \delta_b^k \delta_c^l \delta_{d]}^m \frac{\partial x^a}{\partial x^{'p}}\frac{\partial x^b}{\partial x^{'r}}\frac{\partial x^c}{\partial x^{'s}}\frac{\partial x^d}{\partial x^{'t}}e^{prst} \;,\\
=\frac 1 {4!} e^{iklm} e_{abcd}\frac{\partial x^a}{\partial x^{'p}}\frac{\partial x^b}{\partial x^{'r}}\frac{\partial x^c}{\partial x^{'s}}\frac{\partial x^d}{\partial x^{'t}}e^{prst} \;,\\
=\Big(\frac 1 {4!}e_{abcd}e^{prst}\frac{\partial x^a}{\partial x^{'p}}\frac{\partial x^b}{\partial x^{'r}}\frac{\partial x^c}{\partial x^{'s}}\frac{\partial x^d}{\partial x^{'t}}\Big)e^{iklm}\;,\\
=J\;e^{iklm}
$$
