3 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 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 Inserting (9) into the scalar field equation, (5), gives The static solution to (12) is therefore 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. 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 The solution to (12) with the same initial conditions is thus 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 ), 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 provides us with the integral representation We first use q = F$_{01}$ ( t, x = b ) − F$_{01}$ ( t, x = a ) = g [ φ ( t, x = b ) φ ( t, x = a )] (11) ( ∂ 2 + m $^{2}$) φ = − m ¯ F = − mQ (Θ( x + L/ 2) − Θ( x − L/ 2)) (12) m = g (13) φ ( t = 0 , x ) = 0 = ˙ φ ( t = 0 , x ) (14) Θ( x ) = ∫ dk 2 πi e ikx k − i 0 $_{+}$, (16) G$_{r}$ ( x − y ) = ∫ d $^{2}$k (2 π ) 2 e ik · ( x − y ) − ( k$_{0}$ − i 0 $^{+}$)2 + ( k$_{1}$ ) 2 + m 2 (17) φ ( x ) = − m Q 2 ∫ d $^{2}$y G$_{r}$ ( x − y )(Θ( y ) − Θ( − y )) = − mQ dk 2 π sin( kx ) k ( k 2 + m $^{2}$) (18) ( ∂ 2 + m $^{2}$) φ = − m Q 2 sgn( x ) (15) φ$_{0}$ ( x ) = − Q 2 m sgn( x ) ( 1 − e − m | x | ) (19) φ$_{s}$ ( x ) = − φ$_{0}$ ( x − L/ 2) + φ$_{0}$ ( x + L/ 2) = − Q m × { e m | x $^{|}$sinh( mL/ 2) , | x | > L/ 2 1 − e − mL/ $^{2}$cosh( mx ) , x < L/ 2 (20) ¯ φ ( t, x ) = φ$_{s}$ ( x ) + φ$_{h}$ ( t, x ) = − mQ ∫ dk 2 π sin( kx ) k ( k 2 + m $^{2}$) × ( 1 − cos ( t √ k 2 + m 2 )) (21) φ ( t, x ) = − ¯ φ ( t, x − L/ 2) + ¯ φ ( t, x + L/ 2) = − 2 mQ ∫ dk 2 π cos( kx ) sin( kL/ 2) k ( k 2 + m $^{2}$) × ( 1 − cos ( t √ k 2 + m 2 )) (22) F ( x ) = ∫ ∞ 0 dt T 01 = − ∫ ∞ 0 dt ∂$_{0}$φ∂$_{1}$φ = − (2 mQ ) 2 ∫ dk 2 π dp 2 π sin( kL/ 2) cos( kx ) k ( k 2 + m $^{2}$) × sin( pL/ 2) sin( px ) k 2 − p 2 (23) ∫ + ∞ −∞ dk 2 π exp( ikx ) k 2 − a 2 = − 1 2 sin( a | x | ) a , (24) A. Static Solution B. Dynamical Solution