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Could I, purely in quantity of material only, build a solid bridge to a star?

Ignore anything like strength, relative motions of systems, that is all taken care of, using lalalaicanthearyouium - this is not remotely based in any actual science.

All I want to know is whether there is enough material for a decently powerful race to build a solid bridge spanning actual light years from things they have lying around nearby, i.e. a solar system, a couple of stars - but not a whole cluster or galaxy.

The bridge should be along the lines of the classic film trope - "it looks like metal, but it's not any material we've seen before, Bob!" - that sort of solid.

Although using iron held together with lalalaicanthearyouium would also be fine if there is enough iron lying around. "My God, Bob, it's just iron ... but how can that be?!?"

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Well... if your structure is thin enough, it sounds feasible. BTW i would use unobtanium, if i were you, or handwavium. Both are much more convenient. – Burki Jan 26 at 12:43
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The tricky part isn't getting enough material to build the bridge; it's maintaining it when any two given stars are moving relative to each other at a significant speed. That's not a design requirement ordinary bridges have to deal with! – Mason Wheeler Jan 26 at 15:29
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+1 for lalalaicanthearyouium – The Anathema Jan 26 at 17:09
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Vanity project? – Whelkaholism Jan 27 at 14:31
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Neither the Earth, nor any star have fixed positions relative to each other, nor even fixed distances. In space, everything is in motion. – RBarryYoung Jan 27 at 18:52
up vote 57 down vote accepted

That is a surprisingly reasonable project!

Let's dismantle the Earth, and use it as building material to Alpha Centauri. The Earth is $5.974 \times 10^{24}~\text{kg}$, and Alpha Centauri is roughly $4.13 \times 10^{16}~\text{m}$ away, so that is almost 150,000 tons of building material per metre. That is enough for any bridge. Even if you stretch it all the way to the galactic centre, you still have about 20 tons of materials per meter.

Given a cost of about £14,000 per metre for a motorway, a bridge to the star will cost you 600 billion billion ($6×10^{20}$) £. (assuming the same cost...)

Happy building!

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This is going down in my book of cases where, against all odds, the stellar distance do not make something impossible! That also suggests that the tensile strength of the bridge might be enough to prevent it from collapsing back into a planetoid! – Cort Ammon Jan 26 at 19:36
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The closest star to Earth is Sun. You should be able to keep the planet in place. – Jan Dvorak Jan 26 at 19:42
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For comparison, this is larger than the UK GDP multiplied by the population of the UK ($2×10^{12}*6×10^{7} = 1.2×10^{20}$) – March Ho Jan 26 at 23:46
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@MichaelT: In the English language, a billion ceased to be a million million since the 70s (before then it was a million million in the British Empire but a thousand million in the US). It is a thousand million. Granted, some languages may have numbers that sounds like "billion" that means a million million but that's no longer true in modern English. – slebetman Jan 27 at 11:37
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@CortAmmon: Did you really mean tensile strength, rather than compressive strength? – LarsH Jan 29 at 12:12

Yeah, ok, you've got your billion-billion tons of lalalaicanthearyouium, but:

How long would it take?

I don't have very good data on this, but let's assume that the 1973 expansion to the Chesapeake Bay Bridge is a reasonable model (powers of ten are more important here, you'll see that even if I'm off by a full factor of 10 it won't really change much). The bridge is 4.3 miles long and took 4 years to construct.

6920 meters in 1461 days, or about 4.75 meters per day (really 4.73, but 4.75 is a nice round number). And that's assuming that a bridge is constructed from one bank, across the river, to the other and not bottom-up: remember, we're building with lalalaicanthearyouium which is both weightless and has an infinite tensile strength (not to mention plentiful, cheap, and as easy to work with as steel, if not more so).

The nearest star, the sun, is 149,604,618,000 meters away. A little math and... 86.2 million years later you've got a bridge to the sun!

Job well done.

You're going to need some serious industry to shorten that duration. You'd need thousands of sites building bridge segments and flying them into space in a near-continuous stream and that'd still only take you down to a tens of thousands of years. The good news is that because lalalaicanthearyouium is weightless, you simply need to drive down the constructed portion of the bridge, to the end, maneuver the new section into place, bolt it down, then drive back down the other side (traffic flowing as a loop up and over the rising vertical spire of the bridge).

TL;DR, I'm pretty sure you won't be taking a night train to Rigel any time soon.

Edit:

So I thought about this a bit more, and at the x1000 speed mentioned via multiple construction sites and combining the results, we get the following numbers:

  • It still takes 86,200-ish years to complete
  • Construction progresses at 4750 meters per day, or just under 200 meters per hour.
  • This is also 0.2 km/h or about 4% walking speed.
  • This means that if we scale up another 25 fold, our construction speed equates to a comfortable walking speed.
  • Under the idea that The Anathema (see comments below) proposes of building from the bottom, someone could walk along the structure once it is half-finished and reach the far end at the same time it reaches the destination (walking the first half with the second half constructed behind them). The trip would take 1,724 years.
    • Better pack a lunch.
  • This still only gets us to the Sun. Reaching Alpha Centauri (nearest extra-solar star) increases the distance and build time 276,173.784 times.
  • If we want to increase our build speed by the same amount, then our construction speed reaches 0.127% the speed of light...assuming that the speed of sound within lalalaicanthearyouium is at least this value.
    • Bad news, this is greater than the speed of sound in steel by two powers of 10 (0.00127c ~= 1119 Mach, Steel: 17.78 Mach) and even greater than that of diamond (34.98 Mach) and beryllium (37.58).
    • Which would require our lalalaicanthearyouium to not only be weightless, but unimaginably dense/rigid [citation needed: cannot find a relationship between a material's properties and its speed of sound].
      • Which lead me to just learn that a neutron star can't exceed 3.2 solar masses in size or its density results in a material-speed-of-sound that exceeds c. Turns out that that's pretty darn accurate.
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But since it's weightless, it should be reasonable to build it from the origin, and simply have the new pieces push the rest of it closer to the star. Then you wouldn't have to fly all the way to the end. I wonder if that would work, considering that push occurs in an object at the speed of sound. What do you think? – The Anathema Jan 26 at 17:01
    
@TheAnathema The effect on the calculation doesn't really effect anything. You're still limited by how quickly you can move the pieces into place. – Draco18s Jan 26 at 17:42
    
Night Train to Rigel isn't a good example--while that book doesn't explain how the system works later books do to some degree. The trains aren't giga-engineering, the handwave is elsewhere. – Loren Pechtel Jan 26 at 19:00
    
@Draco18s By building them at the end, you're increasing some variables. Resources to fly the pieces to the end. Each ship would need to travel the increasing distance to the end of the bridge, and that costs resources - you need to build the ships that are great in number enough to have the continuous stream that you're looking for, and that might influence how much material you have available to convert to lalalaicanthearyouium. It might be more efficient to build it from the start and push it out to the sun. Or, even better, build it on both ends. This is all just word vomit, though. : ) – The Anathema Jan 26 at 19:08
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@TheAnathema I just did some math based on construction speeds relative to walking speeds and discovered really neat numbers. :) – Draco18s Jan 26 at 22:05

It's not completely ridiculous

For simplicity's sake, I assumed you're making a bridge that's one lightyear long, has a cross-section of one square metre and plugged that into Wolfram Alpha. We're looking at roughly 9.461 quadrillion m3 ($ 9.461 * 10^{15} $ m3) of material which corresponds to a solid sphere with a radius of 131.2 km.

If we consider that the Death Star II had a diameter of 160km, I'd say what you're setting out to accomplish is not entirely without precedent.

You could for instance grab 15 Eunomia, turn it entirely into lalalaicanthearyouium and have enought material to build yourself a whole lightyear of interstellar bridge.

Won't it collapse under its own gravity?

Let's see. Imagine a bunch infinite amount of 1 metre wide lalalaicanthearyouium spheres lined up next to each other. We start by calculating the gravitational force the second sphere exerts on the first using Newton's law of gravity:

\begin{align} F_1 & = G\frac{m_1m_2}{r^2} \\ & = G\frac{m^2}{1^2} \\ \end{align}

For the third sphere's force on the first one we get:

$$ F_1 = Gm^2 $$

What we're getting here is an infinite sum, so let's write that out and see what it gets us.

\begin{align} F_t & = \sum_{r=1}^\infty G\frac{m^2}{r^2} \\ & = Gm^2\sum_{r=1}^\infty \frac{1}{r^2} \\ & = Gm^2\frac{\pi^2 }{6} \end{align}

This is great news, The gravitational pull of an infinite series of spheres lined up next to each other on the first sphere is equal to $\frac{\pi^2 }{6} \approx 1.645$ times the gravitational pull between the first two. Of course, a bridge is not actually a series of spheres, but this should not be a problem since this approximation work better when the two objects are further apart. at a few kilometres, there will hardly be a difference.

You can improve on this calculation by making the distance between the segments (except for the first one) variable, and scaling the mass with the distance between, so they still accurately represent slices of the bridge. And then take the limit of the distance going to 0:

$$ F_t = \lim_\limits{d \to 0} Gdm^2 \sum_r^\infty \frac{1}{(\frac{1}{2}+rd)^2} $$

Calculating this limit is left as an exercise to the reader.

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One light year long? That doesn't even get you out of the solar system by some measures (e.g. outer limits of the Oort Cloud). – Brian Jan 26 at 16:20
    
@overactor: Build five of these and you can reach Alpha Centauri – Wouter Lievens Jan 27 at 7:07
    
The bridge requires the material of a solid sphere with r=131km; the Death Star is relatively hollow and with a radius of 80km. – chepner Jan 27 at 19:28
    
@Brian maybe it's not a story about OUR solar system :) – DrCopyPaste Jan 28 at 9:46
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@overactor This is great news indeed! I shall start building this bridge immediately! – Whelkaholism Jan 29 at 11:38

Other people looked carefully at the amount of materials required and found that it would not be unreasonable to make a "bridge to the stars."

However, there's more to this than just building the bridge. You also need some sort of vehicle to traverse the bridge.

Yes

Bridge Crawlers

In most ways the devices used to move stuff along the bridge to the stars would act like elevators (just like for a space elevator).

I assume that this would be the case for your bridge to the stars.

Eliminates Propellant Needs

The most important benefit of using elevator cars to crawl on your bridge to the stars is that it removes the need to bring along propellants - since the bridge becomes your propellant. You will still need to bring the other stuff you require to survive in space (life support, power, etc.).

Might eliminate the need to generate power

Depending upon how you make your bridge, you may be able to transmit electrical power through the lalalaicanthearyouium structure of the bridge. Lalalaicanthearyouium had better be a superconductor though or transmission losses will kill you, lol.

Conclusion

Basically, not only are the material requirements reasonable, building one would actually make the travel easier.

We just need to pretend stellar motion doesn't occur :)

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – James Jan 27 at 18:52
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If you knew you weren't going to use this post to answer the question, why did you post it as an answer? – Frostfyre Jan 27 at 19:36
    
You should rephrase your answer. You can expand more on the need for power and energy since that does indeed count as a resource for building. You could have done so, but instead you prefaced your answer with "Not an answer." which is peculiar given the Stack Exchange guidelines. – The Anathema Jan 27 at 19:51
    
I originally wrote it as a comment - but it seemed more significant and relevant to the discussion than a comment. I'm thinking about how to redo it as a relevant answer. – Jim2B Jan 28 at 4:11

What would prevent your superstructure from rolling up to a giant hank of lalalaicanthearyouium (handwavium) under its own mass?


I mean it has to be flexible to link two solar systems (which are of course in motion related to each other). I mean if it has the mass of 100kg/m, then it weights 4.0680272*10¹⁸ kg which is all gravitated to its center of mass. I would assume if it is strong enough to not brake apart, it would roll up in a spiral then entangle itself.

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But if you add a reasonable percentage of anti-higgs to your unobtanium-handwavium-composite you still get a decent bridge but without this nasty gravity thing messing up a beautiful structure. :-) – Burki Jan 26 at 14:54
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@Burki Yes, some fine-tuned anti-higgs would help building the bridge - you wouldn't even need unobtanium-handwavium-composite , you could use i.e. PVC since it would not collapse uder it's own mass and you could balance the amount of an anti-higgs to keep the whole structure apart. – mg30rg Jan 26 at 14:59
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true... but unobtanium-handwavium-composite looks so much cooler! – Burki Jan 26 at 15:16
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lalalaicanthearyouium has some suprisingly strong anti-Higgs properties :) Thanks for this answer, but I did state to ignore strength / motion etc. If there had been a i-laugh-in-the-face-of-science tag I'd have used it :) – Whelkaholism Jan 26 at 15:17
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It's so spread out that the forces will be weak. Tiny, even. The bridge isn't a spherical shell, so the simplification that it's gravitation equals the gravitation of a point mass at the center doesn't apply. It's not even close to spherical, so it doesn't work even as an approximation. The gravity of the stars at either end acting on the the bridge will put it under tension. Since gravity increases closer to the star, every part of the bridge will be pulled away from every other part (tidal effect). I haven't done the math either, but star gravity > bridge self-gravity for sure. – Peter Cordes Jan 27 at 8:00

Say your bridge has 1 square metre cross section and a density of 1 metric ton per cubic metre. Then one lightyear of bridge would be approx $10^{16}$ tons. The Earth is about $10^{22}$ tons. If Earth was to contain a percent unobtainium and if it was dismantled to get material, it should work.

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Unobtainium isn't much use for this project, what it needs is lalalaicanthearyouium! – Separatrix Jan 26 at 14:53
    
Water is about one metric ton per cubic metre; which raises an important question. Does unobtainium float? – sh1 Jan 27 at 20:06
    
@user16295 If you use a compound of lalalaicanthearyouium and handwavium most of the problems will simply vanish into thin air. – mg30rg Jan 28 at 9:02

Several answers have addressed the numbers--you can build it.

What nobody seems to have touched is what good is it? Bridges have an implicit assumption of gravity being normal to the bridge surface. What gravity you will encounter is parallel to the bridge, not normal to it. You have a strip of lalalaicanthearyouium to the stars that can only be climbed, not walked upon.

If you're going to do something I would think it would have to be a ladder, not a bridge.

(I'm ignoring the problem of a lack of fixed endpoints. In reality that's going to make it useless anyway.)

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Space elevators work by attaching cars that crawl up or down the surface of the strands of the elevators. I assume that this would be the case here. The practical impact is that it removes the need to bring along propellants since the bridge becomes your propellant. You still need the other stuff (life support, power, etc.). – Jim2B Jan 26 at 19:32
    
You might not even need to bring power, as it's possible to transmit electrical power through the lalalaicanthearyouium. It'd better be a superconductor though or transmission losses will kill you, lol. – Jim2B Jan 26 at 19:33
    
@Jim2B It makes a reasonable support for an elevator to the stars. I was specifically objecting to "bridge", though. – Loren Pechtel Jan 26 at 20:14
    
I wasn't really critiquing your answer. I just thought I'd provide that little bit of info. I ended up creating a separate "answer" and putting that info into it. However, it really isn't an answer to the question, just additional information to think about. – Jim2B Jan 26 at 20:30
    
I think you may be underestimating the qualities of lalalaicanthearyouium somewhat. – Whelkaholism Jan 27 at 10:23

Really, this is just an upscaled space elevator.

Most others have already pointed out that it may be possible from a materials point of view, but I don't think anyone's pointed out what other issues you might have.

Let's look at an Earth-Moon bridge for a more local example:

One problem we have is that the Moon orbits at a different velocity to the Earth's surface - so anchoring it at this end could be a problem. The Moon itself is tidally locked, with the same face always pointing at Earth, but there is still a small wobble during its orbit, so anchoring might be easier there. Oh, and the Moon is receding (slowly) and the Earth's rate of spin is slowing, too (one is a consquence of the other).

Then, we have this giant ribbon of material undergoing gravitational attraction towards the Earth, gravitational attraction towards the Moon, and so it will be under massive tension. Not too mention, the Sun and other planets will have some gravitational influence.


Let's get back to the Sun-star bridge:

As as been mentioned - all stars are in relative motion to each other - but locally, this isn't all that fast.

One problem is that to prevent the bridge from just collapsing into the Sun, its will need to be in orbit around the Sun. I'm pretty sure the maths on this is going to get very complicated for an elongated body (compared to the relative small spherical bodies we're familiar with).

This in itself will setup tension along the bridge (some of the bridge is being pulled toward the Sun, some of the bridge is accelerating away from the Sun as it orbits).

Then, eventually, as the bridge is extruded out towards the other star, we're going to have to account for additional tension of the bridge being pulled in towards that star.

And then there's the gravitational effects of all the other local stars.


I think what all this adds up to is a mathematical nightmare of orbital mechanics...

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Shouldn't build "from StarA to StarB". Rather build from the mid_point out towards each star, keeping the rate balanced against the two stars' gravities. Bridge shouldn't exactly "orbit" either star. Definitely a "mathematical nightmare", though. – user2338816 Jan 28 at 8:14
    
You'd have to shift a lot of material to the mid-point... starting at one star or another, you can "extrude" the bridge out. Even from the mid-point, you're going to end up with two parts of the bridge under different gravitational influences - maybe orbit wouldn't be technically correct. – HorusKol Jan 28 at 21:20
    
Yep, a lot. OTOH, it's much easier to avoid or vastly minimize the "orbit" problem. The varying gravities may be countered by controlling the rates that mass is added to either end. – user2338816 Jan 29 at 12:07

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