Could SETI Signals be Detected?
Most recent answer: 04/25/2014
- Eric Van Loocke (age 72)
Belgie
Eric- Your electromagnetic calculations look correct. The key to SETI searches for evidence of extraterrestrial life, however, is not to rely on finding a signal large compared to the various types of background radiation. It's just to find some component of the radiation that has peculiar signal-like features. These would primarily consist of a narrow-band frequency spectrum together with some signal-like modulation of the amplitude or frequency. WIthin a narrow enough frequency band, it's much easier to beat the background.
Mike W.
p.s. Here's a Google translation of the question.
Is the following reasoning correct? If not . Where is the mistake ?
Possible buzzing Galaxy intellectual life ?
Seti is already more than half a century searching for a signal from the intellectual ruimte.Tevergeefs.Ik think you know why ?
Never can one make contact with E. T. with electromagnetic waves .
One can not keep parallel to a parallel circular beam of radiation ( parabolic antenna ) by the phenomenon of diffraction.
I use the book Serway : Physics for scientists and engineers with modern physics third edition ECR 1075 figure below and we work here with small angles sin ( theta ) is approximately equal to tan ( theta ) and I get the formula R = 2D.lambda / r . R is the diameter circle D at the end of the distance r and lambda is the wavelength of the beginning diameter circle .
The radio waves have a wavelength of 1 mm to several meters and the receiver must be as sensitive wavelength longer so I assume wavelength 0.001 m
Since the dichtsbijstaande star about 4 light years ( galaxy about 100,000 light years) I take D 1 light year and I take a parabolic channel with diameter 100 m.Ik take the central maximum and the other I . Neglect
After calculation, I get R = 1.89E11 m.Als I start with a certain energy , then that at least approximately 2.80E -19 as small . ( In first approximation the energy I homogeneously distributed over the final circle suppose )
So I have to start with a huge ( practically not retrieve ) energy to rise above the noise present everywhere to come out.
So we feel lonely even though this probably is not so .
I 'm interested if my great formula is correct .
I also calculated the following : I am leaving with a circular laser beam of wavelength 500 nm in diameter and 1 dm , I get on the moon a circle with a diameter of plus or minus 3.5 km .
Sincerely ,
Eric
(published on 04/25/2014)