Special relativity - four current and dirac's delta

[luxux]

Homework Statement

Hello,
in my special relativity course we are making the (electrical charge) current a four-current. I have serious problems understanding the following:
http://imageshack.com/a/img540/2046/8IYpxu.png
In particular, on line 2, in the last equation, I don't understand why it is implicit (as the professor said) that there is an integral (formally - we are working with distributions) (over the 3d space) that, with a test function Φ(x) (also implicit), selects one spatial position (which one?!). That integral should be implicit in both the last term and in the one preceding.
I have no problems with the calculations, but to me the fact that there is an implicit integration makes no sense:
How can i have on line 1 j⁰(x) that, after the calculations, is equal to cρ(x), that should be integrated, thus giving cρ(z)?
j⁰(x)=...=cρ(x)=c∫ρ(x-z(x⁰))δ⁽³⁾(x-z(x⁰))Φ(x)d⁽³⁾x=cρ(z(x⁰)). j should be general (depends on coordinates x) and it is equal to ρ in point z?!

I asked to my fellow students and the professor both, but without success.

Could anyone explain me what is the meaning of that formal integral, why is it implicit and what would the physical meaning (if any) of the test function (implicit) be?

I think i have a good understanding of dirac's delta, at least I could make all the calculations, but when it comes to applying it to physics, I miss something.

x are the coordinates of the IRF, z are the coordinates of the charged particle, s is a parameter that describes the worldline of the particle, c is the speed of light and e is the charge of the particle, ρ is the charge density.
x⁰, z⁰ are the temporal coordinates. j⁰ is the temporal component of the four-current.

Homework Equations

I guess the same problem applies here, when we define the charge density as:
http://imageshack.com/a/img540/9453/mydsBa.png
Also here there should be the formal integration.
If I think about this particle in space time, since it always existed and will always exist, I imagine it as a 2d plane (which represents the 3d space), described by the coordinates x, and on the vertical axis the value of the charge density. It's a very sharp peak, indeed an extremely sharp one. It is centered on the point of coordinates z (so x=z) and looks like a sharp peak (as we imagine a dirac's delta); in the usual limit, it is a line perpendicular to the plane.
I don't understand why I need the integral.
ρ is a function of x. It means that I'm not selecting any point of space-time. Indeed, the only place where ρ would be different from zero, it would be where the particle is, that is z(x⁰). this for the instant x⁰. If i put the integration over x, thus selecting z(x⁰), I get the value of ρ in z(x⁰).Sorry for my bad english and the confusion: I am really confused.

EDIT: is it just saying that, a priori, j could depend on any point of spacetime and after the calculations we discover that, by integrating, it only depends on the position z of the particle? So it also works that if I have xA and xB they will both have j equal to zero unless they are part of the wordline of the particle.

 

[_coyote_]

The Attempt at a SolutionI'm not sure if this is the right way to answer this question, but I think the implicit integration in these equations is meant to represent the fact that the current density depends on the position of the particle. It's saying that j0(x) is equal to cρ(x), where ρ(x) is the charge density of the particle at the position x. In other words, the integral is meant to represent the fact that the current density is only non-zero at the position of the particle z(x0). So, if you have two points xA and xB, they will both have j equal to zero unless they are part of the worldline of the particle. The test function Φ(x) is used to select one spatial position, which in this case is the position of the particle z(x0). This is why the integral is implicit, because it is selecting the position where the charge density is non-zero. Hope this helps!