Putting phase factor in amplitude in Lorentz oscillator

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SUMMARY

The discussion centers on the mathematical equivalence of the Lorentz oscillator's representation using complex amplitudes. The transformation from the sinusoidal form X(t) = X_0sin(-ωt+α) to the exponential form X(t) = X_0 exp(-iωt) is validated through the application of Euler's formula, e^{-iωt} = cos(ωt) + i sin(-ωt). The use of complex numbers simplifies calculations, allowing the phase factor to be incorporated into the amplitude, represented as X_0 = |X_0|e^{iα}. Ultimately, extracting the imaginary part yields the original sinusoidal function.

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Dreak
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Hi there,

In my course solid state physics, there is a part about the Lorentz oscillator. At a certain part, this is written:

"X(t) = X_0sin(-ωt+α)

This changes into:

X(t) = X_0 exp(-iωt)

by choosing X_0 as a complex number and putting the phase factor into the complex amplitude."


But I just don't see how you can do/prove this mathematically?
 
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Those two things aren't mathematically equivalent. They're using the Euler formula, e^{-i\omega t} = \cos(\omega t) + i \sin(-\omega t) and making the assumption that at the end of the day, you'll take the imaginary part of X(t) to get the actual value for the Lorentz oscillator. This is done because it is often more convenient to work with exponentials than trig functions.

To answer your second question, if X_0 = |X_0|e^{i\alpha} is complex, then you have
X(t) = |X_0|e^{i\alpha} e^{-i \omega t}
X(t) = |X_0| e^{-i \omega t + \alpha}
and if you take the imaginary part then you have
X(t) = |X_0| \sin(-\omega t + \alpha)
 
Thanks for the clear answer! :)
 

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