Determinant of a Jacobian of a space-time coordinate transformation In Lewis Ryder's Quantum field theory's Classical field theory part, a certain variation of field is mentioned $$ \phi(x^\mu)\longrightarrow\phi(x^{\prime\mu})= \phi(x^\mu)+\delta\phi(x^\mu)$$
$$x^\mu\longrightarrow x^{\prime \mu}=x^\mu +\delta x^\mu $$
And then,
$$\frac{\partial x^{\prime \mu}}{\partial{x^\nu}}=\delta^\mu_\nu + \partial_\nu\delta x^\mu$$ is calculated. But then they had calculated the Jacobian directly by writing
$$ J\left(\frac{x^\prime}{x}\right)=det\left(\frac{\partial x^{\prime \mu}}{\partial{x^\nu}}\right)=1+\partial_\mu (\delta x^\mu) $$
I dont understand this result. Can someone help me to see how this determinant is being calculated?
 A: Try to write the Jacobian in a matrix form with the $\mu,\nu$ indices labelling the matrix elements. You have already given the formula for each element.
$\begin{pmatrix} 1+\partial_{0}\delta x^{0} & \partial_{1}\delta x^{0} & \dots \\ \partial_{0}\delta x^{1} & 1 + \partial_{1}\delta x^{1} & \dots\\ .& .& \dots \\ .& .& \dots \\ \end{pmatrix}$
If you now take determinant, then upto first order in $\delta x$ we have just the product of diagonal elements. This gives us
$J = 1 + \partial_{\mu}\delta x^{\mu}$
A: $$ \begin{pmatrix}
\partial_0 \delta x^0 & \partial_1 \delta x^0 & . . .\\
\partial_0 \delta x^1 & \partial_1 \delta x^1 & . . .\\
\partial_0 \delta x^2 & \partial_1 \delta x^2 & . . .\\
. & . & . . .\\
. & . & . . .
\end{pmatrix} $$
I have written the matrix in this form, only excluding the $\delta ^\mu_\nu$ term because I know that matrix would give me a determinant of 1. Now the second part, which is the above one needs it's determinant to be calculated. Some of the typical terms in the expansion of the determinant are $\rightarrow$
$ 1. \frac{\partial}{\partial x^0} \delta x^0 \frac{\partial}{\partial x} \delta x \frac{\partial}{\partial y} \delta y \frac{\partial}{\partial z} \delta z  \\
 2. -\frac{\partial}{\partial x^0} \delta x^0 \frac{\partial}{\partial x} \delta x \frac{\partial}{\partial y} \delta z \frac{\partial}{\partial z} \delta y \\
 3.-\frac{\partial}{\partial x^0} \delta x^0 \frac{\partial}{\partial x} \delta y \frac{\partial}{\partial y} \delta x \frac{\partial}{\partial z} \delta z \\
 4. \frac{\partial}{\partial x^0} \delta x^0 \frac{\partial}{\partial x} \delta z \frac{\partial}{\partial y} \delta x \frac{\partial}{\partial z} \delta y$
and etc.
where we see that, as the determinant is calculated it will be as a sum of 4! terms, first 4 of which can be written in the above way. Number 1 is a diagonal term and the others are non diagonal. SO here is my question, why the non diagonals would be 0? I mean what is in those non-diagonal terms that is making it 0?
And as the result should be in the form of $\partial_\mu \delta x^\mu$. which, when expanded in terms of the indices looks like $\partial_0 \delta x^0+\partial_1 \delta x+\partial_2 \delta y+\partial_3 \delta z$.
so even if the non-diagonals are 0, then also how this form is reached? since the diagonal terms are in forms mentioned in Number 1?
