- #1

asdf1

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Why is the narrorer the wave group, the greater the range of wavelengths involved?

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- Thread starter asdf1
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- #1

asdf1

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Why is the narrorer the wave group, the greater the range of wavelengths involved?

- #2

Galileo

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Have you tried to look at the expression for a wave-packet and considered its Fourier transform?

- #3

jtbell

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Set up the waves so that they're all in phase at one point (say, the origin). If their wavelengths are nearly the same, they are still nearly in phase and interfere mostly constructively at a large distance from the origin. On the other hand, if they have a wide range of wavelengths, the destructive interference is significant at a short distance from the origin.

- #4

asdf1

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@@a

still a little confused~

- #5

asdf1

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i haven't studied the Fourier transformation yet, so i don't know how to use it~

but i believe that most complicated things can be explained in simple ways~

- #6

jtbell

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asdf1 said:so you mean that the smaller the amount of waves, the less they interfere with each other, so the bigger the range of wavelengths?

No, I'm talking about the

[tex]\psi(x,t) = \int_{\infty}^{-\infty} {A(k) e^{i(kx - \omega t)} dk} [/tex]

[itex]A(k)[/itex] gives the amplitude of the wave with wavelength [itex]2 \pi / k[/itex]. Although we normally write the integral using in infinite range (limits) in [itex]k[/itex], what counts is the range where [itex]k[/itex] is significantly different from zero.

You might say that a large range in [itex]k[/itex] has a larger number of waves in it (i.e. a larger number of values of [itex]k[/itex]), but that's not really correct because any continuous range has an infinite number of values in it!

- #7

asdf1

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@@a

sorry, but I'm still a little confused...

sorry, but I'm still a little confused...

- #8

jtbell

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Well, I guess I'm still a little confused about what you're confused about...

- #9

reilly

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However try building a wave packet graphically -- take several waves with different frequencies, and add them, by hand or whatever, and graph the result. Add some more -- big freqs and small ones. As you do this, you will see the interference patterns that ultimately produce a delta function -- make sure that all waves are of the form exp(kx), where k is the wave number, so that the exp functions are all = one at x=0.

Regards,

Reilly Atkinson

- #10

quasar987

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- #11

asdf1

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thanks! :)

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