We will solve the two equations for the potentials, and put them together to generate a solution for E and B. The two equations are ∇ 2 φ − μ 0 ε 0 ∂ 2 φ ∂ t 2 = − ρ ε 0 {\displaystyle \nabla ^{2}\phi -\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\phi }{\partial t^{2}}}=-{\frac {\rho }{\epsilon _{0}}}} ∇ 2 A − μ 0 ε 0 ∂ 2 A ∂ t 2 = − μ 0 J {\displaystyle \nabla ^{2}\mathbf {A} -\mu _{0}\epsilon _{0}{\frac {\partial ^{2}\mathbf {A} }{\partial t^{2}}}=-\mu _{0}J} Imagine an infinite sheet. At time 0, a current is turned on, and thus, a magnetic field should be produced. What is the field at a very far point immediately after time 0? It is still 0, because the "news" that the current is turned on hasn't reached there yet. Electromagnetic information travel at the speed of light. This isn't surprising since light is an electromagnetic phenomena. We have already seen that electromagnetic waves travel at the speed of light; it would only make sense if all electromagnetic fields propagate at that speed. Thus, let us introduce the quantity t r e t = t − r / c {\displaystyle t_{ret}=t-r/c} called the retarded time. Suppose we have a fixed point P, and that it is currently time t. At another point Q, electromagnetic waves are sent. The waves arriving at P are not the ones generated at that instant, but the ones generated at t r e t {\displaystyle t_{ret}} . This is because light takes P Q / c {\displaystyle PQ/c} time to travel to P, so we must subtract it from t. It turns out that the most general solution to the potential equations is given by: φ ( r , t ) = 1 4 π ε 0 ∫ ρ ( r − r ′ , t r e t ) | | r − r ′ | | d V {\displaystyle \phi (\mathbf {r} ,t)={\frac {1}{4\pi \epsilon _{0}}}\int {\frac {\rho (\mathbf {r} -\mathbf {r'} ,t_{ret})}{||\mathbf {r} -\mathbf {r'} ||}}dV} and A ( r , t ) = μ 0 4 π ∫ J ( r − r ′ , t r e t ) | | r − r ′ | | d V {\displaystyle \mathbf {A} (\mathbf {r} ,t)={\frac {\mu _{0}}{4\pi }}\int {\frac {\mathbf {J} (\mathbf {r} -\mathbf {r'} ,t_{ret})}{||\mathbf {r} -\mathbf {r'} ||}}dV} with t r e t = t − | | r − r ′ | | c {\displaystyle t_{ret}=t-{\frac {||\mathbf {r} -\mathbf {r'} ||}{c}}} Note that the potentials are given at the actual time t, while the time appearing in the integral is the retarded time. Also, the integrals are over r'. Third, note that t r e t {\displaystyle t_{ret}} is a function of both r and r'; keep this in mind when doing integrals and derivatives with retarded time in it. Last of all, note that the following equation is not correct: E ( r , t ) = 1 4 π ε 0 ∫ ρ ( r − r ′ , t r e t ) | | r − r ′ | | 2 d V {\displaystyle \mathbf {E} (\mathbf {r} ,t)={\frac {1}{4\pi \epsilon _{0}}}\int {\frac {\rho (\mathbf {r} -\mathbf {r'} ,t_{ret})}{||\mathbf {r} -\mathbf {r'} ||^{2}}}dV} as might be expected from analogy. Thus, our formulas for potentials is not a trivial equation; they have to be checked against Maxwell's equations. On the other hand, since we know that ∇ × A = B {\displaystyle \nabla \times \mathbf {A} =\mathbf {B} } and − ∇ φ = − ∂ A ∂ t {\displaystyle -\nabla \phi =-{\frac {\partial \mathbf {A} }{\partial t}}} , we can "easily" calculate the fields from the potentials.
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