Timeline for Negative frequency contributions for very short pulses?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
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| Nov 5, 2014 at 16:02 | history | edited | Floris | CC BY-SA 3.0 |
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| Oct 30, 2014 at 11:31 | comment | added | Floris | That's how I understand it, yes. | |
| Oct 30, 2014 at 10:22 | comment | added | thyme | Oh, I see. The fourier transform of my wave just shows the frequency contributions of a decomposition of my wave into plane waves. And negative frequencies then just mean plane waves propagating in the opposite direction... right? | |
| Oct 29, 2014 at 18:40 | comment | added | Floris | There is nothing unphysical about negative frequency - you can write $\sin(\omega t)$ or $\sin(-\omega t)$ and will see immediately that one is just the inverse of the other - if you are looking at just one point, it's got 180° phase shift, or if you look at a wave, it's traveling in the opposite direction. Does that clear it up for you? | |
| Oct 29, 2014 at 18:14 | comment | added | thyme | Ok, maybe my problem with negative frequencies is solved, when you can explain their meaning. I thought any negative frequency contribution is unphysical... or am I wrong? | |
| Oct 29, 2014 at 17:12 | comment | added | Floris | Maybe I just don't understand what your problem is with negative frequencies. My point was that when the pulse width is much shorter than the period of the wave, then the latter doesn't matter any more and you just have the transformation of a pure Gaussian (which looks "more and more" like a delta function) and whether that is centered at f=0 or f=$10^{15}$ doesn't matter when its width is so much greater than the offset. | |
| Oct 29, 2014 at 16:13 | comment | added | thyme | Hm , sorry I don't understand that. What do you mean with an "an essentially flat response"? How can I see that the negative frequency contributions do not come in? When I fourier transform a delta I get negative frequency distributions, too. And what do you mean with "just becomes a delta function"? If I want to distinguish attosecond pulses from femtosecond pulses I can not just treat both as a delta function, since they have different pulse widths. I think they always have a shape, even if they are very short... | |
| Oct 29, 2014 at 15:59 | history | answered | Floris | CC BY-SA 3.0 |