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[Starver]

Well, the "mostly empty" portion only works out if you're not considering infinite distance. If you draw a straight line, you're bound to hit something; it just might be... far, is all.

Only if the big rip or big freeze aren't true. Since space is expanding and that rate is also expanding, if the thing that you would hit is far enough away that by the time you get there space will be expanding faster than the speed of light you would never hit anything at all, since for every bit forward the blast moved it would have two bits of space created between it and it's eventual impact.

This is somewhat the reversal of the "why is the night sky dark?" question.  Find an area of sky that isn't obviously a star and zoom in and there are (further, fainter) stars there.  Zoom in on a gap between those stars, and there'll be more stars.

In a universe without an 'edge' (i.e. infinite or wrap-around), there'd be something at every possibly discrete point you could choose, and thus (theoretically) it should all be light.  Even where there'd be a dust cloud obscuring anywhere, the dust is as consistently lit by the infinitely summed 'star shells' around it, so would also be shining out (albeit exhibiting a 'texture')... assuming one did not live on Krikkit.

Of course, that ignores the prior expansion of the universe (which also lets us argue that the CBR counts, etc) and the fact that an infinite series can still add up to a low/practically-zero value (the inverse square law and a low density of stars could mean that even such a virtual continuum of distant stars upon the celestial background wouldn't collectively dump enough light energy in our direction to even approximate the intensity represented by the direct view of a more nearby star...  the ultimate example being the Sun).

But it was a useful argument to consider, in developing various theories about the universe (steady state/continuous creation/etc).

[TheDarkStar]

Some quick math (PPE: I made a mistake somwhere - if anyone has any better knowledge, can they look over my work?):

The only things that really matter here are the stars and the distance. Adding more stars makes things brighter because more light is produced. Increasing the distance causes the light the disperse following an inverse square law. I used these things to get an estimate of how much energy should reach the Earth overall.

Call the amount of energy produced by a star E and the distance from us D. The amount that reaches an observer, if every part of it is in the observer's direction, is going to be E/D² time a constant (for this I'll assume 1). Google tells me that the average distance between stars is 4150 ly (so D = 4150 ly), so on average, stars in a given radius = 4pi(D)³/3. Each one, on average, adds the average luminosity of a star - some research tells me that 4x10²⁶ J of energy are released by the Sun every second. We'll assume every star is like the Sun for simplicity and because the Sun is probably close to average. Combining the prior facts requires some calculus:

One shell of stars is (Volume of shell, radius x) - (Volume of shell, radius x - dx) and the energy reaching the observer is E/x²*stars in the shell. The rate of change is the limit of

((⁴/₃*pi(x)³ - (⁴/₃*pi(x - dx)³)) * (E/x²) / dx as dx approaches 0

which gets us

⁴/₃*p * E/x² *((x³ - (x + dx)³)/dx) as dx approaches 0.

This is a constant times the derivative of x³, which is 3x². We end up with 4*pi*E after cancelling, which is a constant roughly equal to 5x10^{27 J}/_s. This seems incorrect - it indicates an infinite amount of energy hitting the earth assuming an infinite universe and an Earth-vaporizing amount within only a few hundreds of thousands of ly. Did I mess up or do I get this result from assuming that all of the energy from all the stars starts off going towards the observer?

[nogoodnames]

Well, without looking at the math there's the fact that the universe is expanding uniformly which means that in an infinite universe, stars extremely far away from Earth will be moving away at faster than the speed of light. Also, space, even intergalactic space, still has some matter strewn throughout it which would absorb some of the energy. It's not much, but when you're talking about infinite distances eventually all the energy will be absorbed.

[RedKing]

Although, given that it takes the fastest probe we've ever built nine and a half years to get to the end of the driveway, so to speak, a lot of that high-level cosmology isn't practically important at this stage. It's enough to know that space is really, really big and really, really old.

Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space.

[Il Palazzo]

This seems incorrect - it indicates an infinite amount of energy hitting the earth assuming an infinite universe and an Earth-vaporizing amount within only a few hundreds of thousands of ly. Did I mess up or do I get this result from assuming that all of the energy from all the stars starts off going towards the observer?

The conclusion is correct. Your maths were somewhat hard to follow. I couldn't get what you did with star density there, or why approximate 4pi to 1 for luminosity but leave it for shells etc. But that might be just me being rusty.

Spoiler: I'd go like this (click to show/hide)

For energy flux coming from a shell you need:
(shell volume)*(avg. star density)*(avg. luminosity)/(inverse square law = area of a sphere)
i.e.
(4piR^2dr)*q*L/(4piR^2)
=qLdr
Which you then integrate over r from 0 to infinity to get total energy from all shells.

Under the assumptions of the Olbers' paradox you get infinite energy. The assumptions are an eternal static universe, and point source stars.

The energy being infinite is due to the point-like stars, which gives them infinite surface brightness.

If you take into account that stars have a finite, non-zero extent, then you end up with a sky as bright as the temperature of the surface of the average star you used in calculations. This simply means that some stars obscure farther ones.

The actual distance when the Earth is 'fried' depends on the stellar density parameter and avg. luminosity you adopt.

In any case, that the sky is not as hot as the Sun is a good indicator that at least one of the two remaining assumptions of the paradox is invalid, which is exactly what it intends to show.

[Reelya]

This is a constant times the derivative of x³, which is 3x². We end up with 4*pi*E after cancelling, which is a constant roughly equal to 5x10^{27 J}/_s. This seems incorrect - it indicates an infinite amount of energy hitting the earth assuming an infinite universe and an Earth-vaporizing amount within only a few hundreds of thousands of ly. Did I mess up or do I get this result from assuming that all of the energy from all the stars starts off going towards the observer?

The shell argument is actually a classic argument against an infinite universe. If you take each shell (of constant thickness) of stars centered on Earth and assume an even spread of stars, then each shell can be shown to have a uniform amount of energy aimed at Earth. Assuming infinite shells, you get the constant times infinite amount of light which must hit us.

So the infinite energy part is correct, but maybe you over-estimated the percentage of light from each star that is aimed at us specifically. You assumed the entire output is in our direction I think.

[Starver]

The easier and simpler mental calculation might be to go with the idea of the inverse square law for a source at a given distance and the (non-inverse) square law for "how many sources are at a given distance", and (barring any other subtleties you want to introduce) realise that the squares cancel out to produce what is effectively a constant.

That is: For any given fraction of brightness of a set of 'standard' stars, there's an exactly corresponding multiple of such stars shining with that brightness across any particular solid angle, including the whole 4π steradians of the sphere around you, floating in space, or ~2π steradians of the 'above horizon' hemisphere, standing on a planet's surface¹.

(So then the sum of shells is a basic Σk (where k is any given shell's derivative luminosity/power/whatever, whether that be based on δx or dx) for the range of shells 0→lim, and if lim=∞ then any non-zero k results in the sum reaching infinity, also.)

It pretty much proves the problem with the initial naive assumption (as per Olber/etc, that is), as has already been said.


¹ Which, illuminated by the same light, potentially can shining upwards with at least as much light as the sky above, but of course will be less.  Assuming it doesn't burst into flames.

[Arx]

Kepler's picked up an exoplanet that, while 60% larger than Earth, is in the liquid water Goldilocks zone.

[Cryxis, Prince of Doom]

I can't remember if this is the thread where I can ask scifi related questions about space.
I'm assuming it is and continuing to my main question.

Would an ion (pulse?) cannon work in space?
Basically a cannon that fires charged particles at a target to destroy electronics

[MetalSlimeHunt]

Ion engines shoot charged particles into space, but they're very slow and definitely no good as a weapon. Unparalleled efficiency though!

[Culise]

The issue with a charged particle gun is maintaining coherence over long distances - same-charge particles repulse each other.  You'll also have to deal with the fact that if you're shooting, say, positively-charged particles, you're building up just as powerful a negative charge in your own ship, which will need to be resolved (as with ion drives, by the bye, which include a neutralizer to fire off electrons and counteract the loss of protons).  These two problems can partially resolve each other by firing two beams in a helical structure - the net charge leaving your ship is neutral (no charge build-up) and the electromagnetic attraction between the two beams, along with the initial velocity imparted by the spiral, keeps the two beams from losing coherence for a while, but that introduces problems of its own, and still is unlikely to give you ranges on the level of other available weaponry. 

If combat is being dealt with at significant distances greater than LEO (that is, your sci-fi 'verse has effectively settled the Earth-Moon system or beyond), you'll also get to deal with interactions with ambient magnetic fields - solar wind, fluctuations in planetary magnetic fields if you're fighting near a planet...basically, "windage."  These won't affect weapons that are lighter but uncharged (lasers) or heavier (missiles, kinetics), but they will affect particle beams with a significant electric charge (that is, not firing neutrons). 

Finally, I don't believe it's likely to be practical to use such a weapon to break electronics specifically, I don't think - anything shielded enough to operate in an environment that includes stuff like solar flares or other similar radiation spikes is likely going to be shielded enough to handle particle beams.  Something powerful enough to kill electronics should be powerful enough to kill crew, actually.  Your charged particle beam impacting a metal ship is going to become an x-ray generator par excellence through the notion of bremsstrahlung.  That said, anyone worth their salt is going to include some sort of non-metallic ablative - high-density ceramic, say - to block precisely this effect from turning their ship into the Fleet's newest CT scanner. 

That's all just me hypothesizing out of my rear, though. 

[wierd]

Assuming your space ship is epically huge...... (like several miles across huge)

You could have something similar to the LHC beam generator on board, but using electrostatic charges to accellerate neutrons instead of protons.  You will need a much larger track, because you wont have the same confinement that you have with protons, because neutrons dont have a charge of their own, and thus would be much harder to confine. (You get around this limited ability to control vector by making only very small changes to the vector, over a much longer track.)  A strongly charged track will still pull on neutrons even if they are neutrally charged, because they are "more positive" or "More negative" than the rails used in the track, and will thus still be atrracted-- just less so.

The resulting beam will be a highly energetic beam of neutral particles traveling at a significant fraction of C.  It would be pretty bad ass as a primary cannon.  But again-- is basically a souped up version of the LHC on board your ship.  Not something small, to say the least. For the power requirements, you would probably get more bang for the buck with high powered lasers.

Props if you use whole helium atoms instead of straight neutrons. Those can be turned into a charged neucleus, accellerated in the accellerator track, then colimated with an electron beam going in the same direction at the end of the track, and thus still have a coherent beam of particles-- with a smaller and more energy efficient track.

[PTTG??]

Or, instead of one city-sized spaceship, you have one thousand tiny robot spaceships that love city-sized spaceships and want to get as close as possible to them as fast as possible.

[Cryxis, Prince of Doom]

@weird- wouldn't that just be an alpha radiation cannon at that point?

[wierd]

Yes, but with sufficient energy that fusion detonations would occur at the target. Typically, helium nuclei accelerated in a particle accelerator are not called alpha particles.

These bad boys would have so much energy behind them from their tour in the accelerator, that they would blast holes through most substances quite effortlessly.

*Note, it is NOT the radiation of the beam that is lethal to the crew of the enemy ship. It is the secondary radiation produced from the cascading fusion and fission reactions the beam causes on the hull that is lethal. (Biproducts include gamma rays.)