The Traveling Electromagnetic Wave, Quantitatively — Question 3

Section 33-3 · Physics pages 5416–5417

Question 3 A Gaussian electric pulse is an exact one-direction vacuum field profile Let ξ = x −ct and Ey = F(ξ) = E0e−ξ2/a2 with a positive. Seek a +x vacuum wave without static field offsets. Use the one-dimensional curl equations ∂xEy = −∂tBz and −∂xBz = c−2∂tEy. ξ Ey Schematic; labels, not drawing lengths, specify the data. (a) Find Bz and verify both curl equations for arbitrary differentiable F. (b) Find the full width at half maximum of the electric-field amplitude in space and in a fixed-point time trace.
Show solutionHide solution
Question 3 - Solution A Gaussian electric pulse is an exact one-direction vacuum field profile ξ Ey (a) Method and calculation Bz = F(ξ)/c ; ∂xEy = F ′, −∂tBz = −(−c)F ′/c = F ′. −∂xBz = −F ′/c = c−2(−cF ′) = c−2∂tEy. (b) Interpretation and further calculation F = E0/2 ⇒ξ2/a2 = ln 2 ⇒∆x = 2a √ ln 2 . ∆t = ∆x/c = 2a √ ln 2/c . Physical check These widths describe field amplitude; the intensity half-maximum width is smaller by a factor of square root of two.

Question 3

A Gaussian electric pulse is an exact one-direction vacuum field profile

Let ξ=x−ct\xi=x-ct and Ey=F(ξ)=E0e−ξ2/a2E_y=F(\xi)=E_0e^{-\xi^2/a^2} with a positive. Seek a +x vacuum wave without static field offsets. Use the one-dimensional curl equations ∂xEy=−∂tBz\partial_xE_y=-\partial_tB_z and −∂xBz=c−2∂tEy-\partial_xB_z=c^{-2}\partial_tE_y.

Diagram shown in the original SVG worksheet.

  1. Find Bz and verify both curl equations for arbitrary differentiable F.

  2. Find the full width at half maximum of the electric-field amplitude in space and in a fixed-point time trace.

Question 3 - Solution

A Gaussian electric pulse is an exact one-direction vacuum field profile

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Bz=F(ξ)/c;∂xEy=F′,−∂tBz=−(−c)F′/c=F′.\boxed{B_z=F(\xi)/c};\quad\partial_xE_y=F^{\prime},\quad-\partial_tB_z=-(-c)F^{\prime}/c=F^{\prime}.−∂xBz=−F′/c=c−2(−cF′)=c−2∂tEy.-\partial_xB_z=-F^{\prime}/c=c^{-2}(-cF^{\prime})=c^{-2}\partial_tE_y.

(b) Interpretation and further calculation

F=E0/2⇒ξ2/a2=ln⁡2⇒Δx=2aln⁡2.F=E_0/2\Rightarrow\xi^2/a^2=\ln2\Rightarrow\boxed{\Delta x=2a\sqrt{\ln2}}.Δt=Δx/c=2aln⁡2/c.\boxed{\Delta t=\Delta x/c=2a\sqrt{\ln2}/c}.

Physical check

These widths describe field amplitude; the intensity half-maximum width is smaller by a factor of square root of two.