Coordinate conversion formula in Jackson’s Electrodynamics Firstly, I have very little familiarity with Einstein notation, but would definitely like to improve my skills. So, excuse any difficulties I may have understanding. I am reading Jackson's 'Classical Electrodynamics' (3rd Edition) and have gotten to Chapter 11, where Jackson begins talking about special relativity and introduces the idea of four vectors as shown below:

My questions/confusions are:

*

*Is the {$x^\alpha$, $x'^\alpha$} notation still relating to a frame at rest (unprimed) and frame moving (primed).


*I am confused about the structure of equation 11.60. I am assuming $x'^\alpha$ on the left-hand side is a four vector, but then what is the $x'^\alpha$ on the right-hand side? The right-hand side is two four vectors, $x'^\alpha$ and $(x^0, x^1, x^2, x^3)$, multiplied together (inner product)? The structure is very confusing to me.


*After equation 11.61, Jackson states, "...the derivative is computed from 11.60". Seeing that I don't even understand fully what 11.60 is trying to say, I am having a difficult time understanding how the derivative is derived from 11.60. Also, I am trying to discern how the derivative is defined in this formalism: The change in $x'^\alpha$ per change in $x^\beta$?
Any help/clarity on these points would be greatly appreciated!
 A: You’ve been bitten by an abuse of notation common among physicists: we often say “four-vector $u^\alpha$” when what we should have said is “four-vector with components $u^0$, $u^1$, $u^2$, $u^3$ [in the coordinate system we are implicitly using]”. This makes for some very confusing moments when $u^\alpha$ actually needs to refer to the $\alpha$th component of the vector $u$ in a non-covariant way for some reason. Because of this, in the following $u\equiv (u^\alpha)\equiv (u^\alpha)_{\alpha=0}^3$ is the vector with the components $u^0,\ldots,u^3$, and $u^\alpha$ is its $\alpha$th component.
Now the point of (11.60) is to write down an arbitrary (not necessarily linear) mapping of quadruples $x\equiv (x^\alpha)$ to quadruples $x'\equiv (x'^\alpha)$. (For now, I am deliberately not calling them “vectors”.) A functional relationship between the two looks like
$$x' = f(x)$$
with a quadruple-valued function $f$; or explicitly for all components
$$\left\{\begin{align}
x'^0 &= f^0(x^0, x^1, x^2, x^3),\\
x'^1 &= f^1(x^0, x^1, x^2, x^3),\\
x'^2 &= f^2(x^0, x^1, x^2, x^3),\\
x'^3 &= f^3(x^0, x^1, x^2, x^3);
\end{align}\right.$$
or finally, keeping the components but compressing similar equations,
$$x'^\alpha = f^\alpha(x^0, x^1, x^2, x^3)\quad(\alpha = 0,\ldots, 3),$$
where the qualification at the end means “repeat the preceding line four times substituting these values for $\alpha$”. The Jacobian matrix (multidimensional derivative) of this transformation will then be written as
$$\frac{\partial f(x)}{\partial x}\equiv\left(\frac{\partial f^\alpha(x^0,x^1,x^2,x^2)}{\partial x^\beta}\right)_{\alpha,\beta=0}^3\,.$$
At this point, another abuse of notation comes into view: we write $x'$ instead of $f$, so $x'$ is simultaneously the new quadruple and the transformation function. The very same equations are now written as (11.60),
$$x'^\alpha = x'^\alpha(x^0,x^1,x^2,x^3)\quad(\alpha = 0,\ldots, 3),$$
and the Jacobian matrix is (as, indeed, “computed from (11.60)”)
$$\frac{\partial x'}{\partial x}\equiv\left(\frac{\partial x'^\alpha}{\partial x^\beta}\right)_{\alpha,\beta=0}^3\,.$$

I am actually quite surprised by what Jackson is doing here, because non-linear changes of coordinates and, consequently, vectors transforming differently from coordinate quadruples is something you have most probably never seen before. This is a technique for curvilinear coordinates and/or curved spacetime that usually only comes into play in GR. Lorentz transformations are linear, so you probably won’t actually need to worry about that once chapter 11 is over, but that’s why I avoided referring to $(x^\alpha)$ as a vector: in curvilinear coordinates, it isn’t. For example, on a plane adding the $(x,y)$ pairs and the $(r,\phi)$ pairs representing the same points yields two different results, while vector addition should make sense independent of coordinates. Meanwhile, things like velocities and momenta are always vectors. Once it’s proven that Lorentz transformations are linear, you won’t need to worry about it, so ignore it for now; if you are interested, try reading the first few sections of Misner, Thorne, and Wheeler’s classic, “Gravitation”.
A: in 11.60 it is not saying multiplied together it is saying "function of" like y=f(x).
A: 
I am confused about the structure of equation 11.60. I am assuming x′α on the left-hand side is a four vector, but then what is the x′α on the right-hand side? The right-hand side is two four vectors, x′α and (x0,x1,x2,x3), multiplied together (inner product)? The structure is very confusing to me.

Remember that you can use any coordinate system you like, it's just some are easier to use than others and you may need to transform between them.
11.60 is saying:
Start with a set of 4 coordinates $x^{′α}$, there are 4 because $\alpha$ goes through 0, 1, 2, 3.
Each of these 4 components can be transformed into a different coordinate system  using a function (which he keeps undefined for now) and this function uses  all the new coordinates $x^{\alpha}$ as  variables in the undefined function.
An example would be   $x^{′0}$ = 5 $x^0$ + 8 $x^1$ + 2 $x^2$ + 9 $x^3$, and then this would be done again for $x^{′1}$ and $x^{′2}$ and $x^{′3}$
I am using 5, 8, 2 and 9 just as placeholders for how the actual undefined function will act on the r.h.s variables.
It's a while since I had to do this stuff, so if anybody want to sort me out or edit the post, feel free.  His notation is terrible,  the primes are very easy  to miss.
If you find the primes confusing, use some other notation that you are happy with in your own notes.
