1 Lecture 25

2 FYS3120 – Classical mechanics and electrodynamicsRecap In terms of e.m. potential the free (no source) plane wave solution of Maxwell’s equations is where the wave (number) vector k fulfils The electric and magnetic fields are (in Coulomb gauge) given as and related through the unit vector in k-direction 𝐴 𝑟 ,𝑡 = 𝐴 0 𝑒 𝑖 𝑘 𝑟 −ω𝑡 𝑘 ⋅ 𝐴 0 = 0 Lav totalenergi første tilgjengelig eksperimentelt i våre dager. 𝐸 = 𝑖ω 𝐴 , 𝐵 = 𝑖 𝑘 × 𝐴 𝐸 = −𝑐 𝑛 × 𝐵 , 𝐵 = 1 𝑐 𝑛 × 𝐸 / Are Raklev / FYS3120 – Classical mechanics and electrodynamics

3 FYS3120 – Classical mechanics and electrodynamicsRecap The energy current density S (Poynting’s vector) and the energy density u is The momentum g density is proportional to S The energy-momentum tensor for electromagnetic fields is defined as This contains T00 = u and T0i = Si/c. 𝑆 = 1 μ 0 𝐸 × 𝐵 , 𝑢 = ϵ 0 𝐸 μ 0 𝐵 2 𝑔 = 𝑆 𝑐 2 Lav totalenergi første tilgjengelig eksperimentelt i våre dager. 𝑇 μν = 1 μ 0 − 𝐹 μρ 𝐹 ρ ν 𝑔 μν 𝐹 ρσ 𝐹 ρσ / Are Raklev / FYS3120 – Classical mechanics and electrodynamics

4 FYS3120 – Classical mechanics and electrodynamicsToday Potential and fields from static sources. Electrostatics (static electric charge) [today] Magnetostatics (constant current) [next week] General solution for electrostatics Easy to write down, difficult to calculate Approximate solution at large distance Multipole expansion / Are Raklev / FYS3120 – Classical mechanics and electrodynamics

5 FYS3120 – Classical mechanics and electrodynamicsSummary The electrostatic solution for the potential is At large distances from the charges this can be approximated in the multipole expansion with the monopole and dipole contributions where p is the dipole moment ϕ 𝑟 = 1 4π ϵ ρ 𝑟 𝑟 − 𝑟 ′ 𝑑 3 𝑟 ′ ρ 𝑟 = ρ 0 𝑟 + ρ 1 𝑟 + ρ 2 𝑟 +... Lav totalenergi første tilgjengelig eksperimentelt i våre dager. ρ 0 𝑟 = 𝑄 4π ϵ 0 𝑟 , ρ 1 𝑟 = 𝑟 ⋅ 𝑝 4π ϵ 0 𝑟 3 𝑝 = ρ 𝑟 𝑟 𝑑 3 𝑟 ′ / Are Raklev / FYS3120 – Classical mechanics and electrodynamics