5 To check this expression simply differentiate with respect to time and keep the leading order term in 1 / ( mt ). This agrees with ˙ E = g ˙ φ = j $^{x}$. Note that the static part of the electric field plays no role. In fact, well away from the capacitor plates, the static electric field dies off expo- nentially fast and can be ignored. So we will define the decaying part of the electric field as E$_{d}$ ≡ E − E$_{static}$ and refer to this as the electric field. The expression for the electric field shows that it is 90 ◦ out of phase with the current but the amplitude has the same 1 $_{/}$√$_{t}$ decay as the current. The rms value of the electric field is This leads to Ohm’s law where σ$_{E}$ is the electrical conductivity of the vacuum This result is independent of Q and L . Let us now consider the energy in the capacitor. At late times, the fields approach the static solution whose energy can be computed using (20) This shows that the final energy is smaller if the coupling g = m is stronger. Or stronger coupling implies more complete radiation of the capacitor energy. We can identify a typical time scale for energy loss from the capacitor by considering the ratio of the decaying part of the energy in the electric field within the capacitor at time t to the initial energy ( Q $^{2}$L/ 2). The ratio is where Hence larger couplings imply longer decay times i.e. slower decay. The capacitor is more effectively discharged when the coupling constant is large but it takes a longer time for the discharge to happen. In the zero coupling limit, the rate of pair production is rapid, but the original electric field E remains relatively undissipated. In the second setup we do not wish to introduce ex- ternal charges. Instead the capacitor is charged with the III. SETUP II: EXTERNAL PLATES FIG. 2: Schematic view of setup II. The dashed line is the double well potential V ( x + L ) + V ( $^{x}$− L ). The thick black line is the initial φ configuration, tanh( x + L $^{)}$− tanh( $^{x}$− L ), and the thick grey line is a bound state solution in the double well. Note that the figure is meant to be schematic and the horizontal axis is not the zero of the potential. same fermionic field, ψ , or its bosonized version, φ . How- ever we still need to have some capacitor “plates” that we can charge. These plates have to be external to the system. To implement this scheme, we add a double well potential to the action in (2) where the form of V ( x ) is chosen so that we can find a non-dissipative solution for the scalar field in a single well. A sketch of the setup is shown in Fig. 2. 2 It is convenient to choose where M is some mass scale that sets the depth and width of the well. With this choice the solution to the single well problem where we have chosen the initial condition ˙ φ ( t = 0 , x ) = 0. The solution describes a positive charge in the well at t = 0, which then oscillates – due to pair creation in the electric field – but does not dissipate. The charge in the well at any given time can be found from Eq. (6) to be 2 m cos( mt ). Now consider a capacitor with two plates, one at x = L and the other at x = − L . Since we do not have any ex- ternal charges, we take the constant background electric E$_{d}$$_{,}$$_{rms}$ = mQL $_{2}$√$_{πmt}$ (39) j x rms = σ$_{E}$E$_{d}$$_{,}$$_{rms}$ (40) σ$_{E}$ = g = e √$_{π}$ (41) ∫ L/ 2 − L/ 2 T $^{00}$dx = Q 2 4 m ( 1 − e − 2 $^{Lm}$) . (42) E ( t ) (0) ≈ m $^{2}$Q$^{2}$L$^{3}$/ (8 πmt ) Q $^{2}$L/ 2 ≡ τ t (43) τ = gL 2 4 π (44) S V ≡ − 1 2 ∫ d $^{2}$x [ V ( x + L ) + V ( x − L )] φ 2 (45) V ( x ) = − 2 M $^{2}$sech$^{2}$( Mx ) (46) ( ∂ 2 + m 2 + V ( x )) φ = 0 (47) φ ( t, x ) = cos( mt ) tanh( Mx ) , (48) 2 See [4] for a setup in a similar spirit, but in the context of QCD$_{2}$. GLYPH<Minus> L GLYPH<Plus> L is given by the λφ 4 kink