# Fields of a very short laser pulse, pulse a fraction of wavelength.

1. ### Spinnor

Scientists can produce laser pulses of order one wavelength in "length" see,

http://www.bbc.co.uk/news/science-environment-19489384

Do Maxwell's equations admit solutions of such short pulses?

Can one approximate such a solution by multiplying a plane electromagnetic wave by the appropriate space decaying exponentials?

Thanks for any help or suggestions!

2. ### waveandmatter

16
From a purely classical EM theory viewpoint, you only need a temporally equally short 'pulse' of current at some point and the only thing that Maxwell's equations do is prescribe how the resulting em-field propagates. But they do not forbid you to create any kind of however short temporal profile if you manage to do so experimentally: Which is probably the difficult part.
Also, after propagation in a dispersive medium, the pulse will broaden (If this is not compensated by nonlinear effects).

A short pulse will be very broad in the frequency domain so i would not think that it makes sense to approximate them with a plane wave.

3. ### Spinnor

Try E_x = cos(-t)cos(z)exp(-x^2)exp(-y^2)exp([z-t]^2)
B_y = cos(-t)cos(z)exp(-x^2)exp(-y^2)exp([z-t]^2)

in Maxwell's equations and see what you get?

4. ### Spinnor

You don't have to as both the divergence of both E_x and B_y above are not equal to zero as they should be for a electromagnetic wave far from charges. ---> Both B and E need a component in the z direction?

5. ### Spinnor

See J.D. Jackson, 2nd ed. page 333, problem 7.20 for a solution closer to what you want.

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6. ### Spinnor

Take Jackson's approximate fields for a circularly polarized wave of finite beam width, set ω and k equal to one, and multiply each by exp([z-t]^2) and see how well these new fields satisfy Maxwell's equations. Is this something the software Mathematica would handle?

Edit, that should be exp(-[z-t]^2) above.

Last edited: Dec 30, 2012
7. ### Spinnor

Would a better attempt be to take Jackson's approximate solution above and multiply by an unknown function of (z-t), f(z-t), (where f(z-t) decays rapidly from the line z-t =0) and apply Maxwell's equations, one might then find differential equations the unknown function f(z-t) must satisfy?

Thanks for any help!

8. ### pumila

113
This is perhaps more fundamental than Maxwell's equations, being really to do with wave theory, which applies to electromagnetic, sound and all other propagating waves equally. All that happens as the pulse is shortened is that the bandwidth increases. This can be seen by taking the Fourier transform of the pulsed wave of different lengths. An infinitely long pulse of monochromatic radiation has a zero bandwidth. An infinitely short one has an infinite bandwidth which in turn means it cannot be considered monochromatic. Chopping up monochromatic light into short pulses creates a spread of frequencies centred around the original monochromatic frequency.

9. ### einstein1921

73

why Chop up monochromatic light into short pulses creates a spread of frequencies centred around the original monochromatic frequency? what effect cause the spread?thank you!

10. ### pumila

113
An infinitely long waveform is pure and unadulterated. If however, you want a short pulse of only (say) three wavelengths, you are essentially multiplying that pure waveform with another gating signal in the time domain, in this case a square pulse three of the wavelength in length; this pulse allows those three wavelengths through and blocks the rest. That gating signal has its own frequency spectrum.

That means we end up with two separate signals multiplied together in time, each with its own frequency spectrum. Multiplication in time, when converted to the frequency domain, becomes convolution (or 'spreading') in frequency, where the frequency spectrums of the two source signals spread each other out to make a more complex spectrum for the composite signal.

(note that although convolution is generally thought of as "spreading", it can do the opposite under certain circumstances, but that does not apply in this case).