# Maxwell's equations, verification

Equations f Maxwell: $$\begin{cases} \nabla \cdot \bf E=0\\ \nabla \cdot \bf H = 0\\ \nabla \wedge \bf E =-μ_o \frac{\partial \bf H}{\partial t}\\ \nabla \wedge \bf H =ε_ο \frac{\partial \bf E}{\partial t}\\ \end{cases}$$

I want to show that if $\bf u$ verifies the equations of waves: $$\begin{cases} \frac{\partial^2 \bf E}{\partial t^2}\ =\ (C_0)^2Δ\bf E\\ \frac{\partial^2 \bf H}{\partial t^2}\ =\ (C_0)^2Δ\bf H\\ \end{cases}$$

where $$(C_0)=\frac{1}{\sqrt{μ_οε_ο}}$$

then $$\begin{cases} \bf E=a\nabla\wedge(\nabla\wedge\bf u )\\ \bf H=aε_ο\nabla\wedge\left(\frac{\partial\bf u}{\partial t}\right) \end{cases}$$

and

$$\begin{cases} \bf E= -βμ_ο\nabla\wedge\left(\frac{\partial \bf u}{\partial t}\right)\\ \bf H= β\nabla\wedge(\nabla\wedge \bf u)\\ \end{cases}$$

are two solutions of the Maxwell equations, no matter the value of the constants $a, β$

• I'm assuming then that those $\bf{a}$'s should be $\alpha$'s? Jan 3, 2018 at 19:28
• Doesn't really make a difference i think, but yes they're all $α$ Jan 3, 2018 at 19:31
• Oh for some reason I read alpha and beta, I suppose it doesn't matter. Jan 3, 2018 at 19:36
• If you replace $\mathbf E, \mathbf H$ with $\mathbf u$ in the first set of equations, you get $$\nabla \wedge \bf u =-\mu_0 \frac{\partial \bf u}{\partial t}\\ \nabla \wedge \bf u =\varepsilon_0 \frac{\partial \bf u}{\partial t}$$which doesn't make sense unless you are using a measurement system where $-\mu_0 = \varepsilon_0$. Which would be highly unusual with the sign change. Jan 4, 2018 at 1:02
• You also get $-\mu_0 \dfrac{\partial{\mathbf u}}{\partial t} = \varepsilon_0 \dfrac{\partial{\mathbf u}}{\partial t}$ which implies $\dfrac{\partial{\mathbf u}}{\partial t} = 0$. Assuming $\mu_0, \varepsilon_0 > 0$. Jan 4, 2018 at 3:40

We recall the general identities from vector calculus

$\nabla \times (\nabla \times \mathbf v) = \nabla (\nabla \cdot \mathbf v) - \nabla^2 \mathbf v, \tag 1$

$\nabla \cdot (\nabla \times \mathbf v) = 0, \tag 2$

and we are given that $\mathbf u$ satisfies

$\mathbf u_{tt} = \dfrac{\partial^2 \mathbf u}{\partial t^2} = C_0^2 \nabla^2 \mathbf u. \tag 3$

We first use (1) and (2) to show that

$\mathbf E = \mathbf a \nabla \times (\nabla \times \mathbf u), \tag 3$

$\mathbf H = \mathbf a \mathbf \varepsilon_0 \nabla \times \mathbf u_t \tag 4$

satisfy the Maxwell system. It is evident via (2) that

$\nabla \cdot \mathbf E = 0 \tag 5$

and

$\nabla \cdot \mathbf H = 0; \tag 6$

also,

$\nabla \times \mathbf H = \mathbf a \varepsilon_0 \nabla \times (\nabla \times \mathbf u_t) = \mathbf a \varepsilon_0 (\nabla \times (\nabla \times \mathbf u))_t = \varepsilon_0 \mathbf E_t; \tag 7$

furthermore,

$\nabla \times \mathbf E = \mathbf a \nabla \times (\nabla \times (\nabla \times \mathbf u)) = \mathbf a \nabla(\nabla \cdot (\nabla \times \mathbf u))- \mathbf a \nabla^2(\nabla \times \mathbf u)$ $= -\mathbf a \nabla^2(\nabla \times \mathbf u) = -\mathbf a \nabla \times \nabla^2 \mathbf u = -\mathbf a \nabla \times C_0^{-2}u_{tt}= -\mathbf a \nabla \times \mu_0 \varepsilon_0 \mathbf u_{tt}$$= -\mathbf a \mu_0 \varepsilon_0 \nabla \times \mathbf u_{tt} = -\mathbf a \mu_0 \varepsilon_0 (\nabla \times \mathbf u_t)_t = -\mu_0 \mathbf H_t. \tag 8 Next, we use (1) and (2) again to show that \mathbf E = -\beta \mu_0 \nabla \times u_t, \tag 9 \mathbf H = \beta \nabla \times (\nabla \times u), \tag{10} are aslo a Maxwellian pair of fields: as before, (2) yields (5) and (6); from (9), \nabla \times \mathbf E = -\beta \mu_0 \nabla \times (\nabla \times u_t) = -\beta \mu_0 (\nabla \times (\nabla \times u))_t = -\mu_0 \mathbf H_t; \tag{11} from (10), \nabla \times \mathbf H = \beta \nabla \times (\nabla \times (\nabla \times u)) = \beta \nabla(\nabla \cdot (\nabla \times \mathbf u)) - \beta \nabla^2(\nabla \times \mathbf u) = -\beta \nabla \times \nabla^2 \mathbf u = -\beta \nabla \times C_0^{-2} u_{tt} = -\beta \mu_0 \varepsilon_0 \nabla \times u_{tt} = -\beta \mu_0 \varepsilon_0 (\nabla \times u_t)_t = \varepsilon_0 \mathbf E_t. \tag{12} There are two general equations that will help you here. One is the differential version of the rule that W \cdot W \times V = 0 (the cross-product of two vectors is perpendicular to both of them). The differential version is$$\nabla \cdot \nabla \times V = 0$$The other is the "BAC-CAB" rule: A\times (B \times C) = B(A\cdot C) - C(A \cdot B). The differential version is:$$\nabla \times \nabla \times V = \nabla(\nabla \cdot V) - \nabla \cdot \nabla V = \nabla(\nabla \cdot V) - \Delta V$$• Using what you shwed i managed to show in both cases that$\nabla\cdot \bf E$and$\nabla \cdot \bf H$are equal to zero but i am still having troube with the second one,$\bf H=aε_ο\nabla\wedge\left(\frac{\partial\bf u}{\partial t}\right)\$. Could you be a bit more explicit? Jan 4, 2018 at 4:45