Transverse Waves

A plane wave solution to the electromagnetic wave equation for the E ⃗ E ⃗ field is E ⃗ ( r ⃗ , t ) = E 0 ⃗ e i ( k ⃗ ⋅ r ⃗ − ω t ) E ⃗ ( r ⃗ , t ) = E 0 ⃗ e i ( k ⃗ ⋅ r ⃗ − ω t ) In vacuum with no currents present we know that: ∇ ⃗ ⋅ E ⃗ = 0 ∇ ⃗ ⋅ E ⃗ = 0 . Recall that earlier we showed ∇ ⃗ ⋅ E ⃗ = i k ⃗ ⋅ E 0 ⃗ e i ( k ⃗ ⋅ r ⃗ − ω t ) ∇ ⃗ ⋅ E ⃗ = i k ⃗ ⋅ E 0 ⃗ e i ( k ⃗ ⋅ r ⃗ − ω t ) So ∇ ⃗ ⋅ E ⃗ = i k ⃗ ⋅ E 0 ⃗ e i ( k ⃗ ⋅ r ⃗ − ω t ) = 0 ∇ ⃗ ⋅ E ⃗ = i k ⃗ ⋅ E 0 ⃗ e i ( k ⃗ ⋅ r ⃗ − ω t ) = 0 implies that the E ⃗ E ⃗ associated with our plane wave is perpendicular to its direction of motion.

Likewise ∇ ⃗ ⋅ B ⃗ = 0 ∇ ⃗ ⋅ B ⃗ = 0 implies that the B ⃗ B ⃗ field is also perpendicular to the direction of motion Lets pick a specific simple case: E ⃗ =  ̂ E y ( x , t ) E ⃗ =  ̂ E y ( x , t ) Then Faraday's law ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t tells us that (since ∂ E y ∂ z = 0 ∂ E y ∂ z = 0 ) ∂ E y ∂ x k ̂ = − ∂ B z ∂ t k ̂ ∂ E y ∂ x k ̂ = − ∂ B z ∂ t k ̂ That is the B ⃗ B ⃗ field is at Right angles to the E ⃗ E ⃗ field.Also B z = − ∫ ∂ E y ∂ x ⅆ t = − ∫ ∂ ∂ x E 0 e i ( k x − ω t ) ⅆ t = − ∫ ∂ ∂ x E 0 e i k x e − i ω t ⅆ t = − ∂ ∂ x E 0 e i k x ∫ e − i ω t ⅆ t = − i k E 0 e i k x ∫ e − i ω t ⅆ t = − i k E 0 e i k x e − i ω t − i ω = 1 c E 0 e i k x e − i ω t = 1 c E y B z = − ∫ ∂ E y ∂ x ⅆ t = − ∫ ∂ ∂ x E 0 e i ( k x − ω t ) ⅆ t = − ∫ ∂ ∂ x E 0 e i k x e − i ω t ⅆ t = − ∂ ∂ x E 0 e i k x ∫ e − i ω t ⅆ t = − i k E 0 e i k x ∫ e − i ω t ⅆ t = − i k E 0 e i k x e − i ω t − i ω = 1 c E 0 e i k x e − i ω t = 1 c E y I leave as an exercise showing k ω = 1 c k ω = 1 c

A movie demonstrating a plane wave can be seen at

http://www.cs.brown.edu/stc/outrea/greenhouse/nursery/physics/gfx/emwave.mov

An applet can be viewed at

http://www.phy.ntnu.edu.tw/java/emWave/emWave.html