Explicitly, my question is: Suppose we have $f(x-x\_p(t))$ in our unprimed frame. We now consider a primed frame moving along the x-axis, where $x' = \gamma(x-\beta ct)$ and $ct' =\gamma(ct-\beta x).$ How do we write $f(x-x\_p(t))$ using the primed coordinates? Is this a well-phrased question? Here $x$ refers to the distance on an x-axis.

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My question originates in the context of showing that the usual definition of a point charge $q$'s 4-current $J^\mu$ in the rest frame of the particle will net us $\int d^3x J^0=cq$ in every Lorentz frame; that is, $\int d^3x J^0=cq$ is a Lorentz scalar. I am very convinced of this fact. However, in giving a quick proof of this fact for a specific case, I have bumped up against how to find the Lorentz transformation of a 3-delta function when the argument of the delta function depends on coordinate time. I will give the context of my question below:

Suppose we have a particle lying still on the x-axis of an unprimed frame. Then our usual definition of the 4-current is 
$J^\mu = (cq\delta(x)\delta(y)\delta(x-x\_p), 0,0,0)$. Clearly $\int d^3x J^0 = cq$.

On boosting to a frame moving with speed $c\beta'$ on the x-axis, we will have, via Lorentz transformations, $J'^\mu = cq\delta(y)\delta(z)\big(\gamma'\delta(x-x\_p), -\gamma' \beta'\delta(x-x\_p),0,0\big)$. 

*Using* $x-x\_p = \gamma(x'-x\_p'(t'))$, $y'=y$, $z'=z$, *we have* 

$J'^\mu = cq\delta(y')\delta(z')\big(\gamma'\delta(\gamma'(x'-x'\_p(t'))), -\gamma' \beta'\delta(\gamma'(x'-x'\_p(t'))),0,0\big)$. 

Using $\delta(ax) = \frac{\delta(x)}{|a|}$, we have 

$J'^\mu = cq\delta(y')\delta(z')\big(\delta(x'-x'\_p(t')), -\beta'\delta(x'-x'\_p(t')),0,0\big)$

We then see immediately that $\int d^3x' J'^0=cq$, which is exactly as I would anticipate. 

However, now let's take one more boost in the x' direction to a double primed frame, which will be at velocity $c\beta''$ relative to the primed frame. *Here is where my naive transformations in the style $x-x\_p = \gamma'(x'-x\_p'(t'))$ appear to break down.*

We get 

$J'^\mu = cq\delta(y')\delta(z')(\gamma''\delta(x'-x'\_p(t'))(1+\beta''\beta'), -(\beta'+\beta'')\gamma''\delta(x'-x'\_p(t')),0,0)$

Immediately, it is likely we *need* $\gamma''\delta(x'-x'\_p(t')) = \frac{1}{1+\beta''\beta'},$ as this will get us the correct conserved charge and will also ensure the correct relativistic velocity addition in the spatial components.

Thus, naively saying $x'-x'\_p(t') = \gamma'' (x''-x''\_p(t''))$ doesn't cut the bill.

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Thus, in general, when we have a spatial function of coordinate time, how do we write that function in the coordinates of some frame moving relative to our original one? I'm most interested in the case above, where we're considering some function of the spatial function of coordinate time, like $f(x-x\_p(t))$.

Please let me know if you want any clarifications.