content_image/443504.jpg content_image/443508.jpg content_image/443509.jpg content_image/443503.jpg content_image/443506.jpg content_image/443510.jpg content_image/443507.jpg content_image/443505.jpg content_image/443512.jpg content_image/443513.jpg content_image/443514.jpg content_image/443515.jpg content_image/443511.jpg content_image/443516.jpg
Note that we have set the constant of integration to zero so that the electric field at spatial infinity vanishes. 

The charge within some interval ( a, b ) is given by Gauss’ law 

Inserting (9) into the scalar field equation, (5), gives 

<img src='content_image/443504.jpg'>

where the effective mass of the scalar field is given by the coupling constant, 

Hence our problem reduces to solving the Klein-Gordon equation for a scalar field of mass m , sourced by the “electric field”, E$_{1}$ = F$_{01}$ = Q , within the capacitor. The initial condition at t = 0 is given by the requirement that no fermions be present, or in terms of the bosonic variables, 

Before we evolve the equations, however, it is inter- esting to find static solutions into which the system can evolve asymptotically. 

In the asymptotic future, t → ∞ , we expect the φ solution to be simply the static solution to (12). Since the Klein-Gordon equation (12) is linear, we may first solve it with ¯ F = ( Q/ 2)(Θ( x ) − Θ( − x )) = ( Q/ 2)sgn( x ) i.e. due to a single point charge at the origin. The static solution to the present problem would then follow using appropriate linear superposition. For now, 

Using the integral representation of the step-function and that of the retarded Green function, G$_{r}$ ( x − y ), 

<img src='content_image/443508.jpg'>

provides us with the integral representation 

<img src='content_image/443509.jpg'>

<img src='content_image/443503.jpg'>

<img src='content_image/443506.jpg'>

## A. Static Solution 

<img src='content_image/443510.jpg'>

<img src='content_image/443507.jpg'>

<img src='content_image/443505.jpg'>

The integral in (18) may be evaluated by perform- ing a partial fraction decomposition of the denomina- tor k ( k 2 + m $^{2}$) and converting the resulting three sub- integrals into appropriate contour integrals, which may then be computed straightforwardly. The answer is 

The static solution to (12) is therefore 

<img src='content_image/443512.jpg'>

The solution to (12) we are seeking must satisfy the initial conditions in (14). To obtain this dynamical solution, we add a homogeneous solution, φ$_{h}$ , obeying ( ∂ 2 + m $^{2}$) φ$_{h}$ = 0, to the static solution φ$_{s}$ such that the initial conditions are satisfied. Again, it helps to first solve the problem with a single charge. Then we have to solve (15) with the initial conditions corresponding to (14). From the conditions in (14), we observe that the integral representation of the solution is 

<img src='content_image/443513.jpg'>

The solution to (12) with the same initial conditions is thus 

<img src='content_image/443514.jpg'>

From this, we can extract the flux of energy passing through a given spatial point x and integrate over all time to get the total energy radiated. It is 

<img src='content_image/443515.jpg'>

We first use 

<img src='content_image/443511.jpg'>

## B. Dynamical Solution 

<img src='content_image/443516.jpg'>

