Reaching the nearest star

[eburacum45]

We would detect the infra-red signature of the swarm as an apparently spherical dust cloud. Dyson swarms could be as massive or as lightweight as necessary - in some respects we have already started building one, since there are several hundred square metres of solar panels in orbit around the Earth and elsewhere in the Solar System.

[md-2000]

As I understood, one of the problems too is that a Dyson sphere, like a Ringworld, is dynamically unstable, like balancing a ball bearing on top of an upside down bowl. In Ringworld sequels, Niven described fusion jets on the rim to maintain stability. I imagine for a sphere they would be even more complex. At least for Ringworld, it would spin, creating a form of artificial gravity. I’m not clear how gravity would work for a Dyson sphere.

[Chronos]

A solid Dyson sphere has neutral stability: If it drifts off-center, there’s nothing to stop the drift, but also nothing to speed it up. It’s like a marble resting on a flat surface.

But there’s no need for it to be solid. And also no need for it to have anything resembling gravity. The point isn’t to have a lot of habitable real estate; the point is just to capture all of the star’s energy.

[tampora]

@Chronos

The total energy of a moving object is E = \gamma m c^2E=γmc2E = \gamma m c^2. Of that, m c^2mc2m c^2 is just the object’s rest energy that it’d have anyway, so we want \gamma - 1γ−1\gamma - 1.

I believe you skipped an important step. You simply state “we want \gamma - 1” without giving any reason why or where this quantity comes from. If we pick units to give the rest energy of the object 1, we come up with E = \gamma . So, were does \gamma - 1 come from?

I’m not saying you’re wrong. I’m just saying the proof has a glaring hole.

[Chronos]

It already has mc^2. It needs \gamma m c^2. So it needs to gain (\gamma-1)mc^2.

[Saffer]

@Whack-a-Mole

Alpha Centauri at 10% light speed would seem to the travelers to take 43.45 years instead of 43.67 year but, at 70% light speed, the trip would seem to take 4.45 years instead of 6.24 years; at 95% it is 1.44 years instead of 4.6 years)

I wonder what these numbers are for accelerating half way at 1g then turning around and decelerating the rest of the distance at 1g. Probably needs some sort of integral.

[Sam_Stone]

@Chronos

Only about 1% of Earthlike planets would be detectable by transit, because it depends on the orbit of the planet being oriented right relative to us. The good news, for exoplanet science, is that that fraction is very easy to calculate for any given sort of planet, so once we make a decent number of observations, we can make very good estimates of the numbers of various sorts of planets in the Universe. The bad news is that if we’re interested in specific planets, we’re 99% likely to miss the one we want.

Stellar wobble can detect planets that transits can’t, but that method is still dependant on inclination, so we still couldn’t detect planets in orbits with high inclination relative to us.

Yeah, I added stellar wobble because between that and transit we can characterize most of the star systems around us. There will be some where the olanetary alignment is such that we just can’t detect the planets without direct observaation. I guess I should add that too - if we get simething like Luvoir and a starshield, we should be able to directly image the closest Earth-like planets.

@Chronos

That’s the gold standard, of course, but the downside is that you can’t re-use the scope: You’d need a separate mission for every target you want to look at that way (well, you could probably reposition it for different objects in the same system). So that’s a step that we’d only take for the most interesting planets (whatever our criterion for “interesting” is). Still much cheaper than an interstellar mission, of course.

You can reuse it within limits, by moving the ship so that it images other stars close to the original one. Repositioning might take months or even years, though. I suspect if we had a surplus of targets, we’d pick the ones that had reasonable secondary targets that could be feasibly repositioned for.

But you’re certainly right that this isn’t a general observation method to look for planets. We’d have to have one already picked out and observed enough and interesting enough to warrant the expense of a dedicaated mission to image it.

@eburacum45

We would detect the infra-red signature of the swarm as an apparently spherical dust cloud. Dyson swarms could be as massive or as lightweight as necessary - in some respects we have already started building one, since there are several hundred square metres of solar panels in orbit around the Earth and elsewhere in the Solar System.

Yes, but just because we technically can identify them doesn’t mean that we should have by now. Take Tabby’s star for example: it had all the signatures of a Dyson swarm at first, and we didn’t even know it existed until recently, and it took us some time after to rule out a Dyson swarm. It’s only about 1/20 of the Galaxy’s width from us. It is entirely possible that the galaxy is full of Dyson swarms and other megastructures and we just haven’t found them yet.

Speaking of Tabby’s star… We still aren’t certain what’s going on there, but it’s on the list for this cycle’s JWST observations!

2757 - Understanding the origin of Boyajian’s Star occultations

[Chronos]

@Saffer

I wonder what these numbers are for accelerating half way at 1g then turning around and decelerating the rest of the distance at 1g. Probably needs some sort of integral.

For the time measured by the travelers, you can actually just use the Newtonian result.

For the time measured by the stars’ reference frame, it’s more complicated, and IIRC hyperbolic trig functions are involved.

Of course, both the Newtonian formula and the one involving the hyperbolic trigs are the result of an integral, but the easiest method for doing an integral is realizing that someone else has already done it for you.

[Steve_MB]

@Saffer

I wonder what these numbers are for accelerating half way at 1g then turning around and decelerating the rest of the distance at 1g. Probably needs some sort of integral.

The equations are given here (in the “Summing Up” section). They are worked out for an acceleration leg of the trip, but doing the calculations for an acceleration-decleration trip just means plugging in half the distance and doubling the final time.

Addendum: It turns out that 1g (9.8 m/s^2) works out to almost unity in lightyear-year units (1.03 lightyear/year^2).

[Chronos]

@Steve_MB

Addendum: It turns out that 1g (9.8 m/s^2) works out to almost unity in lightyear-year units (1.03 lightyear/year^2).

Didn’t I already mention that in this thread? Or, wait, there are two different interstellar travel threads going on right now, aren’t there? I must have posted it in the other one.