/sci/ - Science & Math

[Anonymous No.3150585]

How powerful would my laser have to be before I could successfully communicate with Alpha Centauri?

[Anonymous No.3150591]

>>3150585
Why use a laser when you could use radio?

[Anonymous No.3150599]

>>3150591
Radio is less likely to be successfully detected, among other reasons.

[Anonymous No.3150600]

>>3150599
bullshit

[Anonymous No.3150603]

You would need to have a really powerful lazer so you could shoot faster light out of it, and you could do femtosecond pulses with gradual ramp-up in power such that when the light reaches alpha centurai all of the light that's been emitted over the course of 4.6 billion light-years (because alpha century is 4.6 light years away, it takes 4.6 billion years for laser to get to it) is coalessed into the span of a femtosecond pulse at the surface of alpha century and the star receives all of that extra energy and becomes voltron.

[Anonymous No.3150608]

>>3150603
Goddammit, I'm looking for a serious answer here.

(Also, producing high powered radio signals is illegal in the united states! You learn something new every day, children!)

[Anonymous No.3150611]

Inverse square law, wakka wakka

[Anonymous No.3150614]

>>3150608
You'd have to use a more powerful laser because faster light has more energy, otherwise you'd have to wait 4.6 billion years between sending and receiving (because of casualty inversion, you can send something and have it grandfather paradox then send a response before it gets there) so powerful laser is required but other than that i dunno what you want, I mean power/speed doesn't reach an asymptote so the more powerful the better I think

[Anonymous No.3150621]

>>3150614
For fuck's sake. Shut the fuck up.

[Krakengineer !!5XY+x7grkpt No.3150635]

The lasers used for that moon mirror bounce experiment are a gigawatt according to http://www.spacetoday.org/SolSys/Moons/TheMoon/Retroreflectors.html

With all that they still only get a few photons back.

Alpha Centauri is 4.365 light years away (4.12952065 × 10^13 m). The distance to the moon and back is 768 806 km.


So its going to take about a 53 713 429 gigawatts to get a few photons to Alpha Centauri.

[Krakengineer !!5XY+x7grkpt No.3150642]

>>3150635
I meant to write 4.12952065 × 10^13 km

[Anonymous No.3150664]

>>3150635

You would think though that most of the light would be lost by the reflection process, what is there to obstruct it in empty space?

[Anonymous No.3150691]

>>3150635
how much is that in jigawatts?

[Anonymous No.3150724]

>>3150691
53 713 429

or 53.7 Petawatts

[Krakengineer !!5XY+x7grkpt No.3150753]

>>3150664
Even a mirror you can get at a dollar store reflects about 95% of the light you shine at it, so probably not, but I'm just guessing at this.

Another way to look at this problem is: There's already a great deal of light coming at Alpha Centauri from our sun, so a laser needs to be really fucking bright to be noticed at all.

[Anonymous No.3150784]

>>3150753
Not really, if you use a wavelength not emitted by the sun you could still communicate wouldn't you?

[Anonymous No.3151516]

I haven't done physics in quite a while but doesn't shorter wavelengths mean faster dissipation? So lasers would be much much more inefficient compared to radio signals. Also you could probably direct the radio signals too, so that it won't be detected by too many undesirable people.

[The Doctor No.3151567]

>>3150585
If you must strike the first blow, make sure it is the last.