I Minimum number of cycles in a short pulse laser

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In a short pulse laser setup, it is not feasible to emit a pulse with a duration shorter than its period, as this would violate Maxwell's laws, which require the integral of the electric field over a full pulse to equal zero. While half-cycle pulses exist, they are asymmetric and still adhere to this principle. The discussion emphasizes that a pulse length must be an integer multiple of one wavelength to maintain the integrity of electromagnetic fields. Mechanical chopping of a pulse does not allow for instantaneous cessation of light, complicating the generation of non-integer cycle pulses. Ultimately, the physics of wave propagation dictates that only full cycles can be accurately represented in electromagnetic theory.
Uchida
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Hello to all,
In a short pulse laser emission setup, can a pulse be emmited with beam length shorter than one wavelenght? (can a pulse have a duration shorter than its period?)

Lets say a laser emmiter shoots a quarter cycle pulse, what would happen to this short beam?

(lets supose the existence of a very fast chopper, and a very agile laser emmiter on the setup)EDIT:

I've found another thread within this subject (couldn't find it before):
https://www.physicsforums.com/threads/how-short-can-a-pulse-be.251777/
 
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The integral of the electric field of the full pulse must be zero, otherwise it would violate Maxwell's laws (the Fourier component of frequency 0 must be zero if no charges are present). In that sense, a single cycle is the shortest one can get.

Note that there are things called half-cycle pulses, which are in fact highly asymmetric, with a strong positive half-wave followed by a long and weaker negative tail. But even then, the total field integrates to zero.
 
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DrClaude said:
The integral of the electric field of the full pulse must be zero, otherwise it would violate Maxwell's laws (the Fourier component of frequency 0 must be zero if no charges are present). In that sense, a single cycle is the shortest one can get.

Note that there are things called half-cycle pulses, which are in fact highly asymmetric, with a strong positive half-wave followed by a long and weaker negative tail. But even then, the total field integrates to zero.

So, a pulse length must be integers multiples of one wavelenght?
 
What would happen if a mechanical chopper choped a linearly polarized laser pulse with length of 1.125 cycle?

sine wave.png

(consider propagation from right to left)This would be feaseble with a mechanichal chopper. And it gives a non zero area EM fields.
What would happen to the wave form far from the source?
 
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Uchida said:
This would be feaseble with a mechanichal chopper. And it gives a non zero area EM fields.
No, this is not feasible. You have to include all the physics of the blade moving across the beam front. Shutting off the light instantly is simply not possible.
 
DrClaude said:
The integral of the electric field of the full pulse must be zero, otherwise it would violate Maxwell's laws (the Fourier component of frequency 0 must be zero if no charges are present). In that sense, a single cycle is the shortest one can get.

How do I square the above fact with the image below where it looks like we have a propagating E&M pulse of just one half pulse? I apologize if this is obvious.

Thank you.
1577796119999.png
 
Spinnor said:
How do I square the above fact with the image below where it looks like we have a propagating E&M pulse of just one half pulse?
That picture was intended only to show qualitatively how accelerating an electric charge produces radiation. It’s not precisely and quantitatively accurate so you cannot rely on it for anything else.

When we work through all the physics of the accelerated charge, we find that the pulse in that picture is a superposition of full cycles of waves of much shorter wavelength.
 
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