I calculated the energy and momentum of electromagnetic field generated by a moving particle with constant velocity $v\hat{z}$ using the general solution of Maxwell's equation. A particle of charge $q$ is moving in $z$-direction with constant velocity $v$ and passes through the origin at $t=0$. My results are as follows: $$ \varphi(x, y, z, t) = \frac{q}{4\pi\varepsilon_0} \cdot \frac{\gamma}{\sqrt{x^2 + y^2 + \gamma^2 (z - vt)^2}} \\ \mathbf{A}(x, y, z, t) = \varphi \cdot \frac{v \hat{z}}{c^2}\\ \boldsymbol{\Pi} = \frac{q^2 v}{16\pi^2 \varepsilon_0} \cdot \frac{\gamma^2}{r^6} \left( -x(z - vt), -y(z - vt), x^2 + y^2 \right)\\ u = \frac{q^2}{32\pi^2 \varepsilon_0} \cdot \frac{(2\gamma^2 - 1)(x^2 + y^2) + \gamma^2 (z - vt)^2}{r^6} $$ where $\varphi, \mathbf{A}, \mathbf{\Pi}, u$ are scalar and vector potential, poynting vector and energy density, respectively, with $r = \sqrt{x^2 + y^2 + \gamma^2 (z - vt)^2}$.
I tried to calculate the momentum and energy of the whole electromagnetic field only to find it diverging at the origin. So, I calculated the same thing, but this time, by a spherical shell charge. Imagine a spherical shell with radius $a$, homogeneously charged with total charge $q$ moving in the same manner as the point charge. The object is spherical in its reference frame. And I thought the electromagnetic field would be the same outside the shell because those at the frame of the object are same(static point charge) and the transformation is the same. And there is no field inside the shell. So, I calculated them at all space out of the shell. My calculations are as follows:
With $\xi = \gamma(z - vt)$, the equation of the shell surface at the frame of $(x, y, z, t)$ is $x^2+y^2+\xi^2=a^2$. Because of the symmetry, what we have is: $$ \mathbf{P} = \frac{q^2}{16\pi^2 \varepsilon_0} \cdot \frac{\mathbf{v} \hat{z}}{c^2} \int \frac{\gamma^2 (x^2 + y^2)}{r^6} \, d\tau \\ U = \frac{q^2}{32\pi^2 \varepsilon_0} \int \frac{(2\gamma^2 - 1)(x^2 + y^2) + \gamma^2 (z - vt)^2}{r^6} \, d\tau $$ The integration is calculated in all space where $x^2+y^2+\xi^2>a^2$. With $x = \rho \cos\theta, y = \rho \sin\theta$, the integration is calculated: $$ \frac{\gamma}{2\pi} \int \frac{x^2 + y^2}{r^6} \, d\tau = \left( \int_{-\infty}^{-a} + \int_{a}^{\infty} \right) \left( \int_{0}^{\infty} \frac{\rho^3}{(\rho^2 + \xi^2)^3} \, d\rho \right) d\xi + \int_{-a}^{a} \int_{\sqrt{a^2 - \xi^2}}^{\infty} \frac{\rho^3}{(\rho^2 + \xi^2)^3} \, d\rho d\xi\\ = \left( \int_{-\infty}^{-a} + \int_{a}^{\infty} \right) \left( \frac{1}{4\xi^2} \right) d\xi + \int_{-a}^{a} \frac{2a^2 - \xi^2}{4a^4} \, d\xi = \frac{4}{3a}\\ \frac{\gamma}{2\pi} \int \frac{\gamma^2 (z - vt)^2}{r^6} \, dV = \left( \int_{-\infty}^{-a} + \int_{a}^{\infty} \right) \left( \int_{0}^{\infty} \frac{\rho \xi^2}{(\rho^2 + \xi^2)^3} \, d\rho \right) d\xi + \int_{-a}^{a} \int_{\sqrt{a^2 - \xi^2}}^{\infty} \frac{\rho \xi^2}{(\rho^2 + \xi^2)^3} \, d\rho d\xi\\ = \left( \int_{-\infty}^{-a} + \int_{a}^{\infty} \right) \left( \frac{1}{4\xi^2} \right) d\xi + \int_{-a}^{a} \frac{\xi^2}{4a^4} \, d\xi = \frac{2}{3a} $$ So, the answer is $$ \mathbf{P} = \frac{q^2}{8\pi\varepsilon_0} \cdot \frac{\mathbf{v} \hat{z}}{c^2} \left( \frac{4\gamma}{3a} \right)\\ U = \frac{q^2}{8\pi\varepsilon_0} \left( \frac{4\gamma - 1/\gamma}{3a} \right) $$
The momentum is as expected like the 4/3-problem in E&M. But I expected the energy to be either $\frac{q^2}{8\pi\varepsilon_0} \left( \frac{\gamma}{a} \right)$ or $\frac{q^2}{8\pi\varepsilon_0} \left( \frac{4\gamma}{3a} \right)$. I want to know what is the right answer and what would have went wrong in my calculations. And I want to know why the momentum and energy are like that if $(U, \vec{P})$ is a 4-vector, where $U'$ should be $\gamma U_0$ since $P_0=0$..
