simple harmonic oscillation

d2ydt2=ω2y

  • where:
    • ω=KM=2πf: angular frequency
    • y: displacement
    • M: mass
    • K: spring constant
    • t: time variable

wave equation

2ydz21v22ydt2=0

  • where
    • v=Tρ wave velocity
    • T: tension in string,
      • influenced by mass and acceleration due to field potential
    • ρ=ML: density of string
    • z: spatial variable
    • t: time variable
  • this is for a wave in 1D over time

solution

  • the solutions to this equations are of the form
    • f(zvt): wave moving to right
    • g(z+vt): wave moving left

generalizing wave equation to 3D

  • applying gradient operator to 1D wave equation gives
    • 2ϕ1c22ϕt2=0
    • where:
      • 2=2x2+2y2+2z2
  • 3D wave equation with unit vectors x^,y^,z^ becomes
    • 2=
    • 2=(x^x+y^y+z^z)(x^x+y^y+z^z)

plane wave solutions

  • a (monochromatic) plane wave is given by
    • exp[i(krωt)]

    • where

      • r=xx^+yy^+zz^

      • k=kxx^+kyy^+kzz^

        • this is the wave vector
  • a plane wave is a solution to the 3D wave equation when
    • k=ωt

gradient operator for space partials of plane wave

  • first partial:
    • exp[i(krωt)]=ikexp[i(krωt)]
  • second partial:
    • 2exp[i(krωt)]=k2exp[i(krωt)]
    • where: k2=kx2+ky2+kz2

time partials of plane wave

  • 2t2exp[i(krωt)]=ω2exp[i(kt)ωt]

verify solution plane wave in 3D wave

  • 3D wave:
    • 2ϕ1c22ϕt2=0
  • plugging in space and time partials of plane wave:

    • 2exp[i(krωt)]1c22exp[i(krωt)]t2=0

    • (k2+ω2c2)exp[i(krωt)]=0

    • (k2+k2)exp[i(krωt)]=0

    • 0exp[i(krωt)]=0

conclusion

  • exp[i(krωt)] is a solution for any vector direction k provided k=ωc
    • where: k2=kx2+ky2+kz2

wave interference

  • consider a plane wave with wave-vector k1
    • which is a solution to the 3D wave equation

K-1

  • consider a second plane k2
    • which is a also a 3D wave equation solution

K-2

  • now, the linear combination of these two waves is a solution
    • the two waves k1 and k2 show interference

K1 + K2