Atom optics is typically done in the last years with gratings of light. These gratings successfully implement mirrors and beam-splitters. Two counter-propagating laser beams cross one another, and in the crossing region there appear fringes - see figure.
For the case when the beams have the same wavelength it is no problem to calculate the grating constant, it is equal to 1/(2k cosθ), where θ is half the angle between the directions of propagation of the two beams.
However, recently appeared in the literature "moving gratings". That means, the two laser beams differ slightly in wavelength. Such an arrangement is said to produce the effect of a moving grating.
How to calculate the grating constant of this moving lattice, and with which velocity does it move with respect to the static emitting lasers?
Dear Sofia! You placed the good question on the agenda. I can now give You answer in just zero approximation. Interaction of two coherent waves with different frequencies is characterised instead interference by beats like on your picture. In the case of the same propagation beat velocity in media without dispersion equals light velocity and beat period can be determined by usual way. Grating as response of media on optical field beat will be smashed on the similar light its movement. in any media. Will counter propagation of beams change group velocity of beat I do not know now..
With best wishes,
Eugene Tikhonov
Remi,
Thanks for your kindness, but I don't understand. What means "sum x envelope"?
Also, I don't need the speed of the envelope but the speed of the fringes of the moving lattice. The running lattice has dark and bright fringes along the direction z, and this patters of fringes runs with a certain velocity. In your calculus I don't see the space-variable at all.
Kind regards!
Dear @Remi,
you are very kind. I think that your simple calculus is very close to the answer. What still seems to me incomplete in the calculus, is that a pattern of fringes, if static, should have the expression
(1) cos(kx) cos(ωt),
i.e. all the minima (maxima) are reached at the same time. Of course, in a running lattice the minima (maxima) should move with a given velocity, but still they should appear reached at the same time.
So, I'll try to see if I can get this effect starting from your formulas. Let me refine them a bit. I'll denote
(2) k̅ = (k1 + k2)/2 and δk = (k1 - k2)/2,
ω̅ = (ω1 + ω2)/2 and δω = (ω1 - ω2)/2.
So, we have
(4) cos(k2x - ω2t) + cos(k1x - ω1t) = 2cos(k̅ x - ω̅ t) cos(δkx - δωt)
If δk
Dear @Remi,
you are very lovely. Thank you very much for the drawings. To your attention, the writing is so small that I can't read it, can you replace the picture by a bigger one?
Let me restrict a bit my problem, for being more specific. In the experiment that I read, and on which I am building a certain proof, what the authors say is that they take two counter-propagating laser beams. If the frequencies of the two beams were equal we would have obtained a standing lattice of minima and maxima. But there is a very small difference in frequency, and thus, they claim that they obtain a moving grating, i.e. the minima and maxima move with a certain velocity.
Well, I had to bother the experimenters with all sort of questions already, and this is why I ask certain things over the RG for (hopefully) economizing questions to them.
Now, why do you think that the formula cos(k̅ x - ω̅ t) cos(δkx - δωt) is a standing wave fixed in space? The first cosine is an ordinary running wave, not running lattice but running wave with velocity ω̅ /k̅ = c. It would have been a fixed standing wave if the term ω̅ t under the cosine weren't present.
About the second cosine the data are as follows: k̅ ~ 2π×104cm-1, δω = 2π×105Hz therefore δk = 2×10-5cm-1; x varies in a tiny range between 0 and 100μm because x sweeps the region of intersection between the two laser beams - see figure, and t varies between 0 and 3×10-4sec. So, you can see that δkx < 2×10-7, while δωt varies in the range [0, 60π]. What I understand is that the running grating sweeps all the time the crossing region (as if it enters continuously from one extremity of this region end vanishes at the other extremity).
Again with thanks for your kind help,
Sofia
Dear @Remi,
you are lovely as always. Don't hurry for me, take your time. I am glad that my problem succeeds to be interesting for you.
Kindest regards and wishes!
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