{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Lec5 - proportional to the(second spatial derivative of the...

This preview shows pages 1–2. Sign up to view the full content.

proportional to the (second) spatial derivative of the wave function (recall the wave equa- tion!), we may write the condition on the acceleration as ∂ψ / x | x =0 - = ∂ψ / x | x =0 + . (The second spatial derivative being continuous guarantees that the first derivative is continuous and di ff erentiable.) We choose to represent the waves in complex representation. That is, a harmonic wave traveling to the right would be given by e i ( ω t - kx ) and wave traveling to the left would be e i ( ω t + kx ) . What is the meaning of a complex valued wave? Recall that we may superpose two solutions of the wave equation, and the result would still be a solution, because of the superposition principle and the linearity of the wave equation. Therefore, the complex valued wave function is a superposition of cosine and sine waves: e i ( ω t kx ) = cos( ω t kx ) + i sin( ω t kx ) . Because of the superposition principle, we may be interested only in the cosine wave, say, work with the full complex wave, but take only the real part for the cosine wave. This way, we may benefit form all the niceties of complex analysis, yet retain the “realness” of the physical field. We now write the incident wave as ψ I = A I e i ( ω t - k 1 x ) , the reflected wave as ψ R = A R e i ( ω t + k 1 x ) , and the transmitted wave as ψ T =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Lec5 - proportional to the(second spatial derivative of the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online